Solutions to Problems in Merzbacher Quantum Mechanics 3rd Ed-reid-p59

May 22, 2018 | Author: SVFA | Category: Waves, Matrix (Mathematics), Quantum Mechanics, Condensed Matter Physics, Chemistry


Comments



Description

Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid November 20, 1999 Chapter 2 Problem 2.1 A one-dimensional initial wave packet with a mean wave number k x and a Gaussian amplitude is given by Ψ(x, 0) = C exp ¸ − x 2 4(∆x) 2 + ik x x . Calculate the corresponding k x distribution and Ψ(x, t), assuming free particle motion. Plot |Ψ(x, t)| 2 as a function of x for several values of t, choosing ∆x small enough to show that the wave packet spreads in time, while it advances according to the classical laws. Apply the results to calculate the effect of spreading in some typical microscopic and macroscopic experiments. The first step is to compute the Fourier transform of Ψ(x, 0) to find the distribution of the wave packet in momentum space: Φ(k) = (2π) −1/2 ∞ −∞ Ψ(x, 0)e −ikx dx = (2π) −1/2 C ∞ −∞ exp ¸ − x 2 4(∆x) 2 + i(k 0 −k)x dx (1) 1 (I have dropped the x subscripts, and I write k 0 instead of k). To proceed we need to complete the square in the exponent: − x 2 4(∆x) 2 + i(k 0 −k)x = − ¸ x 2 4(∆x) 2 −i(k 0 −k)x −(k 0 −k) 2 (∆x) 2 + (k 0 −k) 2 (∆x) 2 = − ¸ x 2(∆x) −i(k 0 −k)∆x 2 −(k 0 −k) 2 (∆x) 2 = − 1 4(∆x) 2 x −2i(k 0 −k)(∆x) 2 2 −(k 0 −k) 2 (∆x) 2 (2) Now we plug (??) into (??) to find: Φ(k) = (2π) −1/2 C exp[−(k 0 −k) 2 (∆x) 2 ] ∞ −∞ exp − 1 4(∆x) 2 [x −2i(k 0 −k)(∆x) 2 ] 2 dx In the integral we can make the shift x → x − 2i(k 0 − k)(∆x) 2 and use the standard formula ∞ −∞ exp(ax 2 )dx = (π/a) 1/2 . The result is Φ(k) = √ 2C∆xexp −(k 0 −k) 2 (∆x) 2 To put this into direct correspondence with the form of the wave packet in configuration space, we can write Φ(k) = √ 2C∆xexp ¸ − (k 0 −k) 2 4(∆k) 2 where ∆k = 1/(2∆x). This is the minimum possible k width attainable for a wave packet with x width ∆x, which is why the Gaussian wave packet is sometimes referred to as a minimum uncertainty wave packet. The next step is to compute Ψ(x, t) for t > 0. Since we are talking about a free particle, we know that the momentum eigenfunctions are also energy eigenfunctions, which makes their time evolution particularly simple to write down. In the above work we have expressed the initial wave packet Ψ(x, 0) as a linear combination of momentum eigenfunctions, i.e. as a sum of terms exp(ikx), with the kth term weighted in the sum by the factor Φ(k). The wave packet at a later time t > 0 will be given by the same linear combination, but now with the kth term multiplied by a phase factor exp[−iω(k)t] describing its time evolution. In symbols we have Ψ(x, t) = ∞ −∞ Φ(k)e i[kx−ω(k)t] dk. For a free particle the frequency and wave number are connected through ω(k) = ¯ hk 2 2m . 2 Using our earlier expression for Φ(k), we find Ψ(x, t) = √ 2C∆x ∞ −∞ exp ¸ −(k 0 −k) 2 (∆x) 2 + ikx −i ¯ hk 2 2m t dk. (3) Again we complete the square in the exponent: −(k 0 −k) 2 (∆x) 2 + ikx −i ¯ hk 2 2m t = − [(∆x) 2 + i¯ h 2m t]k 2 −[2k 0 (∆x) 2 + ix]k + k 2 0 (∆x) 2 = − ¸ α 2 k 2 −βk + γ ¸ = −α 2 k 2 − β α 2 k + β 2 4α 4 − β 2 4α 4 −γ = −α 2 [k − β 2α 2 ] 2 + β 2 4α 2 −γ where we have defined some shorthand: α 2 = (∆x) 2 + i¯ h 2m t β = 2k 0 (∆x) 2 + ix γ = k 2 0 (∆x) 2 . Using this in (??) we find: Ψ(x, t) = √ 2C∆xexp ¸ β 2 4α 2 −γ ∞ −∞ exp ¸ −α 2 (k − β 2α 2 ) 2 dk. The integral evaluates to π 1/2 /α. We have Ψ(x, t) = √ 2πC∆x 1 α exp ¸ β 2 4α 2 −γ = √ 2πC ¸ (∆x) 2 (∆x) 2 + i¯ h 2m t ¸ 1/2 e −k 2 0 (∆x) 2 exp ¸ (ix + 2k 0 (∆x) 2 ) 2 4[(∆x) 2 + i¯ h 2m t] ¸ This is pretty ugly, but it does display the relevant features. The important point is that term i¯ ht/2m adds to the initial uncertainty (∆x) 2 , so that the wave packet spreads out with time. In the figure, I’ve plotted this function for a few values of t, with the follow- ing parameters: m=940 Mev (corresponding to a proton or neutron), ∆x=3 ˚ A, k 0 =0.8 ˚ A −1 . This value of k 0 corresponds, for a neutron, to a velocity of about 5 · 10 4 m/s; and note that, sure enough, the center of the wave packet travels about 5 nm in 100 fs. The time scale of the spread of this wave packet is ≈ 100 fs. 3 0 50 100 150 200 250 300 350 -1e-08 -5e-09 0 5e-09 1e-08 W a v e f u n c t i o n ( a r b i t r a r y u n i t s ) Distance (meters) Gaussian wave packet example t=0 s t=30 fs t=70 fs t=100 fs 4 Problem 2.2 Express the spreading Gaussian wave function Ψ(x, t) obtained in Problem 1 in the form Ψ(x, t) = exp[iS(x, t)/¯ h]. Identify the function S(x, t) and show that it satisfies the quantum mechanical Hamilton-Jacobi equation. In the last problem we found Ψ(x, t) = √ 2πC ∆x α exp ¸ β 2 4α 2 −γ = exp ¸ ln √ 2πC ∆x α + β 2 4α 2 −γ (1) where we’ve again used the shorthand we defined earlier: α 2 = (∆x) 2 + i¯ h 2m t β = 2k 0 (∆x) 2 + ix γ = k 2 0 (∆x) 2 . From (??) we can identify (neglecting an unimportant additive constant): S(x, t) = ¯ h i ¸ −lnα + β 2 4α 2 . Things are actually easier if we define δ = α 2 . Then δ = (∆x) 2 + i¯ h 2m t S(x, t) = ¯ h i ¸ − 1 2 ln δ + β 2 4δ Now computing partial derivatives: ∂S ∂t = ¯ h i ¸ − 1 2δ − β 2 4δ 2 ∂δ ∂t = − ¯ h 2 2m ¸ 1 2δ + β 2 4δ 2 (2) ∂S ∂x = ¯ h i β 2δ ∂β ∂x = ¯ hβ 2δ (3) ∂ 2 S ∂x 2 = i¯ h 2δ (4) The quantum-mechanical Hamilton-Jacobi equation for a free particle is ∂S ∂t + 1 2m ¸ ∂S ∂x 2 − i¯ h 2m ∂ 2 S ∂x 2 = 0 5 Inserting (??), (??), and (??) into this equation, we find − ¯ h 2 4mδ − ¯ h 2 β 2 8mδ 2 + ¯ h 2 β 2 8mδ 2 + ¯ h 2 4mδ = 0 so, sure enough, the equation is satisfied. Problem 2.3 Consider a wave function that initially is the superposition of two well-separated narrow wave packets: Ψ 1 (x, 0) + Ψ 2 (x, 0) chosen so that the absolute value of the overlap integral γ(0) = +∞ −∞ Ψ ∗ 1 (x)Ψ 2 (x)dx is very small. As time evolves, the wave packets move and spread. Will |γ(t)| increase in time, as the wave packets overlap? Justify your answer. It seems to me that the answer to this problem depends entirely on the specifics of the particular problem. One could well imagine a situation in which the overlap integral would not increase with time. Consider, for example, the neutron wave packets plotted in the figure from problem 2.1 If one of those wave packets were centered in Chile and another in China, the overlap integral would be tiny, since the wave packets only have appreciable value within a few angstroms of their centers. Furthermore, even if the neutrons are initially moving toward each other, their wave packets spread out on a time scale of ≈ 100 fs, long before their centers ever come close to each other. On the other hand, if the two neutron wave packets were each centered, say, 20 angstroms apart, then they would certainly overlap a little before collapsing entirely. Problem 2.4 A high resolution neutron interferometer narrows the energy spread of thermal neutrons of 20 meV kinetic energy to a wavelength dis- persion level of ∆λ/λ= 10 −9 . Estimate the length of the wave packets in the direction of motion. Over what length of time will the wave packets spread appreciably? 6 First let’s compute the average momentum of the neutrons. p 0 = [ 2mE ] 1/2 ≈ [ 2 · (940Mev · c −2 ) · (20mev) ] 1/2 = 6.1 kev / c We’re given the fractional wavelength dispersion level; what does this tell us about the momentum dispersion level? p = 2π¯ h λ dp = −2π¯ h λ 2 dλ so dp p = dλ λ = 10 −9 so the momentum uncertainty is ∆p = p 0 · 10 −9 = 6.1 · 10 −6 ev. This implies a position uncertainty of ∆x ≈ ¯ h ∆p = 6.6 · 10 −16 ev · s 6.1 · 10 −6 ev / c ≈ c · 10 −10 s ≈ 30 cm. This is HUGE! So the point is, if we know with this precision how quickly our thermal neutrons are moving, we have only the most rough indication of where in the room they might be. To estimate the time scale of spreading of the wave packets, we can imagine that they are Gaussian packets. In this case we start to get appreciable spreading when ¯ h 2m t ≈ (∆x) 2 or t ≈ 2m(∆x) 2 /¯ h ≈ 2 · 9.4 · 10 8 ev · (0.3 m) 2 c 2 · 6.6 · 10 −16 ev · s ≈ 32 ks ≈ 10 hours. 7 Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid March 8, 1999 Chapter 3 Problem 3.1 If the state Ψ(r) is a superposition, Ψ(r) = c 1 Ψ 1 (r) + c 2 Ψ 2 (r) where Ψ 1 (r) and Ψ 2 (r) are related to one another by time reversal, show that the probability current density can be expressed without an interference term involving Ψ 1 and Ψ 2 . I found this to be a pretty cool problem! First of all, we have the probability conservation equation: d dt ρ = − ∇ · J. To show that J contains no cross terms, it suffices to show that its divergence has no cross terms, and to show this it suffices (by probability conservation) to show that dρ/dt has no cross terms. We have ρ = Ψ ∗ Ψ = [c ∗ 1 Ψ ∗ 1 + c ∗ 2 Ψ ∗ 2 ] · [c 1 Ψ 1 + c 2 Ψ 2 ] = |c 1 ||Ψ 1 | + |c 2 ||Ψ 2 | + c 1 c ∗ 2 Ψ 1 Ψ ∗ 2 + c ∗ 1 c 2 Ψ ∗ 1 Ψ 2 (1) 1 Problem 3.2 For a free particle in one dimension, calculate the variance at time t, (∆x) 2 t ≡ (x − x t ) 2 t = x 2 t −x 2 t without explicit use of the wave function by applying (3.44) repeatedly. Show that (∆x) 2 t = (∆x) 2 0 + 2 m ¸ 1 2 xp x + p x x 0 − x 0 p x t + (∆p x ) 2 m 2 t 2 and (∆p x ) 2 t = (∆p x ) 2 0 = (∆p x ). 2 I find it easiest to use a slightly different notation: w(t) ≡ (∆x) 2 t . (The w reminds me of “width.”) Then w(t) = w(0) + t dw dt t=0 + 1 2 t 2 d 2 w dt 2 t=0 + · · · (2) We have dw dt = d dt x 2 − x 2 = d dt x 2 − 2 x d dt x (3) d 2 w dt 2 = d 2 dt 2 x 2 − 2 ¸ d dt x 2 − 2 x d 2 dt 2 x (4) We need to compute the time derivatives of x and x 2 . The relevant equation is d dt F = 1 i¯ h FH − HF + ∂F ∂t for any operator F. For a free particle, the Hamiltonian is H = p 2 /2m, and the all-important commutation relation is px = xp − i¯ h. We can use this to calculate the time derivatives: d dt x = 1 i¯ h [x, H] = 1 2im¯ h xp 2 − p 2 x = 1 2im¯ h xp 2 − p(xp − i¯ h) = 1 2im¯ h xp 2 − pxp + i¯ hp 2 = 1 2im¯ h xp 2 − (xp − i¯ h)p + i¯ hp = 1 2im¯ h 2i¯ hp = p m (5) d 2 dt 2 x = 1 m d dt p = 0 (6) d dt x 2 = 1 i¯ h [x 2 , H] = 1 2im¯ h x 2 p 2 − p 2 x 2 = 1 2im¯ h x 2 p 2 − p(xp − i¯ h)x = 1 2im¯ h x 2 p 2 − pxpx + i¯ hpx = 1 2im¯ h x 2 p 2 − (xp − i¯ h) 2 + i¯ h(xp − i¯ h) = 1 2im¯ h x 2 p 2 − xpxp + 2i¯ hxp + ¯ h 2 + ¯ h 2 + i¯ hxp = 1 2im¯ h x 2 p 2 − x(xp − i¯ h)p + 3i¯ hxp + 2¯ h 2 = 1 2im¯ h 2¯ h 2 + 4i¯ hxp = − i¯ h m + 2 m xp (7) d 2 dt 2 x 2 = 2 m d dt xp = 2 i¯ hm [xp, H] = − 1 i¯ hm 2 xp 3 − p 2 xp = − 1 i¯ hm 2 xp 3 − p(xp − i¯ h)p = − 1 i¯ hm 2 xp 3 − pxp 2 + i¯ hp 2 = − 1 i¯ hm 2 xp 3 − (xp − i¯ h)p 2 + i¯ hp 2 = 1 i¯ hm 2 2i¯ hp 2 = 2 m 2 p 2 (8) d 3 dt 3 x 2 = 2 i¯ hm 2 [p 2 , H] = 0 (9) Now that we’ve computed all time derivatives of x and x 2 , it’s time to 3 plug them into (3) and (4) to compute the time derivatives of w. dw dt = d dt x 2 − 2 x d dt x = − i¯ h m + 2 m xp − 2 m x p = 2 m − i¯ h 2 + xp − 2 m x p = 2 m px − xp 2 + xp − 2 m x p = 2 m px + xp 2 − 2 m x p (10) d 2 w dt 2 = d 2 dt 2 x 2 − 2 ¸ d dt x 2 − 2 x d 2 dt 2 x = 2 m 2 p 2 − 2 m 2 p 2 = 2 m 2 (∆p) 2 (11) Finally, we plug these into the original equation (2) to find w(t) = w(0) + 2 m ¸ 1 2 px + xp − x p t + (∆p) 2 m 2 t. 2 The other portion of this problem, the constancy of (∆p) 2 , is trivial, since (∆p) 2 contains expectation values of p and p 2 , which both commute with H. 4 Problem 3.3 Consider a linear harmonic oscillator with Hamiltonian H = T + V = p 2 2m + 1 2 mω 2 x 2 . (a) Derive the equation of motion for the expectation value x t , and show that it oscillates, similarly to the classical oscillator, as x t = x 0 cos ωt + p 0 mω sinωt. (b) Derive a second-order differential equation of motion for the expectation value T − V t by repeated application of (3.44) and use of the virial theorem. Integrate this equation and, remembering conservation of energy, calculate x 2 t . (c) Show that (∆x) 2 t ≡ x 2 t − x 2 t = (∆x) 2 0 cos 2 ωt + (∆p) 2 0 m 2 ω 2 sin 2 ωt + ¸ 1 2 xp + px 0 − x 0 p 0 sin 2ωt mω Verify that this reduces to the result of Problem 2 in the limit ω → 0. (d) Work out the corresponding formula for the variance (∆p) 2 t . (a) Again I like to use slightly different notation: e(t) = x t . Then d dt e(t) = 1 i¯ h xH − Hx = 1 2i¯ hm xp 2 − p 2 x = 1 2i¯ hm xp 2 − p(xp − i¯ h) = 1 2i¯ hm xp 2 − (xp − i¯ h)p + i¯ hp = 1 2i¯ hm 2i¯ hp = p m . d 2 dt 2 e(t) = d dt p m 5 = 1 i¯ hm pH − Hp = ω 2 2i¯ h px 2 − x 2 p = ω 2 2i¯ h (xp − i¯ h)x − x 2 p = ω 2 2i¯ h x(xp − i¯ h) − i¯ hx − x 2 p = ω 2 2i¯ h −2i¯ hx = −ω 2 x . So we have d 2 dt 2 e(t) = −ω 2 e(t) with general solution e(t) = Acos ωt +Bsin ωt. The coefficients are determined by the boundary conditions: e(0) = x 0 → A = x 0 e (0) = p 0 m → B = p 0 mω . (b) Let’s define v(t) = T − V t . Then d dt v(t) = 1 i¯ h (T − V )H − H(T − V ) = 1 i¯ h (T − V )(T + V ) − (T + V )(T − V ) = 2 i¯ h TV − V T = ω 2 2i¯ h p 2 x 2 − x 2 p 2 . We already worked out this commutator in Problem 2: p 2 x 2 − x 2 p 2 = − 4i¯ hxp + 2¯ h 2 so d dt v(t) = −2ω 2 xp + i¯ hω 2 . = −2ω 2 xp + ω 2 xp − px = −ω 2 xp + px (12) Next, d 2 dt 2 v(t) = − 2ω 2 i¯ h xpH − Hxp = − 2ω 2 i¯ h ¸ 1 2m xp 3 − p 2 xp + mω 2 2 xpx 2 − x 3 p (13) 6 The bracketed expressions are xp 3 − p 2 xp = xp 3 − p(xp − i¯ h)p = xp 3 − (xp − i¯ h)p 2 + i¯ hp 2 = 2i¯ hp 2 xpx 2 − x 3 p = x(xp − i¯ h)x − x 3 p = x 2 (xp − i¯ h) − i¯ hx 2 − x 3 p = −2i¯ hx 2 and plugging these back into (13) gives d 2 dt 2 v(t) = = −4ω 2 ¸ p 2 m − mω 2 2 x 2 ¸ = −4ω 2 v(t) with solution v(t) = Acos 2ωt + B sin 2ωt. (14) Evaluating at t = 0 gives A = T 0 − V 0 . Also, we can use (12) evaluated at t = 0 to determine B: −ω 2 xp + px 0 + i¯ hω 2 = 2ωB so B = − ω xp + px 0 2 . The next task is to compute x 2 t : x 2 t = 2 mω 2 V t = 1 mω 2 H − (T − V ) t = 1 mω 2 [H t − v(t)] . Since H does not depend explicitly on time, H is constant in time. For v(t) we can use (14): x 2 t = 1 mω 2 T 0 + V 0 − [T 0 − V 0 ] cos 2ωt + ω xp + px 0 2 sin 2ωt = 1 mω 2 2 T 0 sin 2 ωt + 2 V 0 cos 2 ωt + ω xp + px 0 2 sin 2ωt = p 2 0 m 2 ω 2 sin 2 ωt + x 2 0 cos 2 ωt + 1 2 xp + px 0 sin 2ωt mω . (15) 7 (c) Earlier we found that x t = x 0 cos ωt + p 0 mω sin ωt x 2 t = x 2 0 cos 2 ωt + p 2 0 m 2 ω 2 sin 2 ωt + x 0 p 0 sin 2ωt. Subtracting from (15) gives (∆x) 2 t = x 2 − x 2 = x 2 0 − x 2 0 cos 2 ωt + 1 m 2 ω 2 p 2 0 − p 2 0 + ¸ 1 2 xp + px 0 − x 0 p 0 sin 2ωt = (∆x) 2 0 cos 2 ωt + (∆p) 2 0 m 2 ω 2 sin 2 ωt + ¸ 1 2 xp + px 0 − x 0 p 0 sin 2ωt mω . As ω → 0, cos 2 ωt → 1, (sin 2 ωt/ω 2 ) → 1, and (sin 2ωt/ω) → 2, as needed to ensure matchup with the result of Problem 2. Problem 3.4 Prove that the probability density and the probability current den- sity at position r 0 can be expressed in terms of the operators r and p as expectation values of the operators ρ(r 0 ) → δ(r − r 0 ) j(r 0 ) → 1 2m [pδ(r − r 0 ) + δ(r − r 0 )p] . Derive expressions for these densities in the momentum represen- tation. The first one is trivial: δ(r − r 0 ) = Ψ ∗ (r)δ(r − r 0 )Ψ(r)dr = Ψ ∗ (r 0 )Ψ(r 0 ) = ρ(r 0 ). For the second one, 1 2m pδ(r − r 0 ) + δ(r − r 0 )p = − i¯ h 2m [Ψ ∗ ∇δ(r − r 0 )Ψ + Ψ ∗ δ(r − r 0 )∇Ψ] dr The gradient operator in the first term operates on everything to its right: = − i¯ h 2m [Ψ ∗ Ψ∇δ(r − r 0 ) + 2δ(r − r 0 )Ψ ∗ ∇Ψ] dr. 8 Here we can use the identity f(x)δ (x − a)dx = −f (a) : = − i¯ h 2m |−∇(Ψ ∗ Ψ) + 2Ψ ∗ ∇Ψ| r=r0 = i¯ h 2m |Ψ∇Ψ ∗ − Ψ ∗ ∇Ψ| r=r0 = j(r 0 ). Problem 3.5 For a system described by the wave function Ψ(r ), the Wigner distribution function is defined as W(r , p ) = 1 (2π¯ h) 3 exp(−ip ·x /¯ h)Ψ ∗ r − r 2 Ψ r + r 2 dr . (a) Show that W(r , p ) is a real-valued function, defined over the six-dimensional “phase space” (r , p ). (b) Prove that W(r , p )dp = |Ψ(r )| 2 and that the expectation value of a function of the operator r in a normalized state is f(r) = f(r )W(r , p )dr dp . (c) Show that the Wigner distribution function is normalized as W(r , p )dr dp = 1. (d) Show that the probability density ρ(r 0 ) at position r 0 is ob- tained from the Wigner distribution function with ρ(r 0 ) → f(r) = δ(r − r 0 ). (a) 9 Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid June 24, 2000 Chapter 5 Problem 5.1 Calculate the matrix elements of p 2 x with respect to the energy eigenfunctions of the harmonic oscillator and write down the first few rows and columns of the matrix. Can the same result be obtained directly by matrix algebra from a knowledge of the matrix elements of p x ? For the harmonic oscillator, we have H = 1 2m p 2 x + 1 2 mω 2 x 2 so p 2 x = 2mH −m 2 ω 2 x 2 and < Ψ n |p 2 x |Ψ k >= 2m¯ hω(n + 1 2 )δ nk −m 2 ω 2 < Ψ n |x 2 |Ψ k > . (1) The nth eigenfunction is Ψ(x) = 1 2 n n! 1/2 mω ¯ hπ 1/4 exp(− mω 2¯ h x 2 )H n ( mω ¯ h x). The matrix element of x 2 is then < Ψ n |x 2 |Ψ k >= 1 2 n+k n!k! 1/2 mω ¯ hπ 1/2 ∞ −∞ x 2 exp( mω ¯ h x 2 )H n ( mω ¯ h x)H k ( mω ¯ h x) dx. 1 Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 2 The obvious substitution is u = (mω/¯ h)x, with which we obtain < Ψ n |x 2 |Ψ k >= 1 2 n+k n!k!π 1/2 ¯ h mω ∞ −∞ u 2 e −u 2 H n (u)H k (u) du. (2) The integral is what Merzbacher calls I nkp with p = 2. The useful formula is ¸ n,k,p I nkp s n t k (2λ) p n! k! p! = √ π e λ 2 +2λ(s+t)+2st . = √ π 1 +λ 2 + 1 2 λ 4 + · · · 1 + 2λ(s +t) + 1 2 (2λ) 2 (s +t) 2 + · · · 1 + (2st) + 1 2 (2st) 2 + · · · (3) There are two ways to get a λ 2 term out of this. One way is to take the λ 2 term from the first series and the 1 from the second series, together with any term from the last series. The second way is to take the 1 from the first series and the λ 2 term from the second series, along with any term from the last series. Writing down only terms obtainable in this way, we have = · · · + √ πλ 2 1 + 2(s +t) 2 1 + (2st) + 1 2 (2st) 2 + · · · + · · · = · · · + √ πλ 2 1 + 2s 2 + 2t 2 + 4st ∞ ¸ j=0 1 j! (2st) j + · · · = · · · + √ πλ 2 ∞ ¸ j=0 2 j j! s j t j + 2 j+1 j! s j+2 t j + 2 j+1 j! s j t j+2 + 2 j+2 j! s j+1 t j+1 + · · · Comparing termwise with (3), we can read off I nk2 = (n + 2)!2 n √ π , n = k − 2 n!2 n−1 √ π(1 + 2n) , n = k 0 , otherwise. Plugging this into (2), we have < Ψ n+2 |x 2 |Ψ n > = 1 2 [(n + 2)(n + 1)] 1/2 ¯ h mω < Ψ n |x 2 |Ψ n > = 1 2 (2n + 1) ¯ h mω . Finally, from (1), < Ψ n+2 |p 2 |Ψ n > = − 1 2 [(n + 2)(n + 1)] 1/2 (mω¯ h) < Ψ n |x 2 |Ψ n > = 1 2 (2n + 1)(mω¯ h). Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 3 I find it kind of confusing that the matrix element for p 2 comes out negative in the first case. It would be absurd for the expectation value (i.e., diagonal matrix element) of the square of an observable operator to come out negative. In this case it is less absurd since there’s no classical interpretation of the off- diagonal matrix elements of an operator, but it’s still weird. However, in another sense it seems inescapable that p 2 should have a negative off-diagonal matrix element here, because the off-diagonal matrix elements of H must vanish in the energy eigenfunction basis, but x 2 has a nonvanishing matrix element, and H is just a sum of x 2 and p 2 terms, so p 2 must have a negative matrix element to cancel out the positive matrix element of x 2 . Problem 5.2 Calculate the expectation values of the potential and kinetic energies in any station- ary state of the harmonic oscillator. Compare with the results of the virial theorem. The potential energy operator is U = mω 2 x 2 /2. We found the expectation values of x 2 in the last problem, so U = 1 2 mω 2 x 2 = ¯ hω 2 (n + 1 2 ) which is just half the energy expectation value. The kinetic energy expectation value must of course make up the difference, so we have T = U. On the other hand, the virial theorem is supposed to be saying 2 T = x d dx V (x) . In this case, d dx V (x) = mω 2 x, so the virial theorem says that T = 1 2 mω 2 x 2 = U in accord with what we concluded earlier. Problem 5.3 Calculate the expectation value of x 4 for the nth energy eigenstate of the harmonic oscillator. Ψ n x 4 Ψ n = 1 n! 2 n mω ¯ hπ 1/2 ∞ −∞ x 4 exp(− mω ¯ h x 2 )H 2 n ( mω ¯ h x) dx Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 4 = 1 n! 2 n √ π ¯ h mω 2 ∞ −∞ u 4 e −u 2 H 2 n (u) du (4) For the integral we want to use (3) again, but this time we’ll need to write out the expansion a little further than before. ¸ n,k,p I nkp s n t k (2λ) p n! k! p! = √ π e λ 2 +2λ(s+t)+2st . (5) = √ π 1 +λ 2 + 1 2 λ 4 + · · · 1 + · · · + 1 2 (2λ) 2 (s +t) 2 + · · · + 1 4! (2λ) 4 (s +t) 4 + · · · ∞ ¸ j=0 (2st) j j! = √ π 1 +λ 2 + 1 2 λ 4 + · · · 1 + · · · + 4λ 2 st + · · · + 4λ 4 (st) 2 + · · · ∞ ¸ j=0 (2st) j j! = · · · + √ π ∞ ¸ j=0 (λ 4 )(st) j 2 j−1 j! + 4 2 j−1 (j − 1)! + 4 2 j−2 (j − 2)! + · · · = · · · + √ π ∞ ¸ j=0 (λ 4 )(st) j 2 j j! 1 2 + 2j +j(j − 1) = · · · + √ π ∞ ¸ j=0 (λ 4 )(st) j 2 j j! 1 2 +j +j 2 In the first line, I only wrote out terms that can be combined to give a factor of λ 4 . In the second line, I further limited it to terms that also contain the same number of powers of s as t. Equating powers in (5), I nn4 = 3 2 2 n n! √ π( 1 2 +n +n 2 ), so (4) is Ψ n x 4 Ψ n = 3 2 ¯ h mω 2 ( 1 2 +n +n 2 ). Problem 5.4 For the energy eigenstates with n=0, 1, and 2, compute the probability that the coordinate of a linear harmonic oscillator in its ground state has a value greater than the amplitude of the classical oscillator of the same energy. The classical amplitude is A = (2E)/(mω 2 ). The probability of finding the particle with coordinate greater than this is P(|x| > A) = −A −∞ Ψ 2 n (x) dx + ∞ A Ψ 2 n (x) dx Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 5 = 2 ∞ A Ψ 2 n (x) dx = 2 n! 2 n mω ¯ hπ 1/2 ∞ A exp(− mω ¯ h x 2 )H 2 n ( mω ¯ h x) dx = 2 n! 2 n √ π ∞ √ 2E/¯ hω e −u 2 H 2 n (u) du = 1 n! 2 n−1 √ π ∞ √ 2n+1 e −u 2 H 2 n (u) du In going from the first line to the second we invoked the fact that Ψ n has either even or odd parity, so Ψ 2 n has even parity. In going from the second to last line to the last line, we noted that the energy of the nth eigenstate is ¯ hω(n + 1/2). In particular, P n=0 (|x| > A) = 1 √ π ∞ 1 e −u 2 du = 1 √ 2π ∞ 0 e −p 2 /2 dp − √ 2 0 e −p 2 /2 dp ¸ = 1 √ 2π π 2 − √ 2π · erf( √ 2) = 1 2 − erf( √ 2) ≈ 0.31 Problem 5.5 Show that if an ensemble of linear harmonic oscillators is in thermal equilibrium, governed by the Boltzmann distribution, the probability per unit length of finding a particle with displacement x is a Gaussian distribution. Plot the width of the distribution as a function of temperature. Check the results in the classical and low-temperature limits. [Hint: Equation (5.43) may be used.] Suppose we denote the number of oscillators in the nth energy state by N n . If the ensemble is in thermal equilibrium, the ratio of the number of oscillators in the n th state to the number of oscillators in the nth state is N n N n = e −(n −n)¯ hω/kT . In particular, for any n, N n = N 0 e −n¯ hω/kT . Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 6 The probability of finding a particle between x and dx is P(x)dx = ∞ ¸ n=0 C n |Ψ n (x)| 2 dx = C 0 mω ¯ hπ 1/2 exp(− mω ¯ h x 2 ) ∞ ¸ n=0 e −n¯ hω/kT n! 2 n H 2 n ( mω ¯ h x)dx This can be summed using the Mehler formula with t = exp(−¯ hω/kT) : P(x) = C 0 mω ¯ hπ 1/2 exp(− mω ¯ h x 2 ) 1 √ 1 −t 2 exp ¸ 2t 1 +t mω ¯ h x 2 = C 0 mω ¯ hπ 1/2 1 √ 1 −t 2 exp ¸ − 1 −t 1 +t mω ¯ h x 2 This is a Gaussian distribution with variance σ 2 = ¯ h 2mω 1 +t 1 −t = ¯ h 2mω 1 +e −¯ hω/kT 1 −e −¯ hω/kT = ¯ h 2mω coth ¯ hω 2kT Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid June 24, 2000 Chapter 6 Problem 6.1 Obtain the transmission coefficient for a rectangular potential barrier of width 2a if the energy exceeds the height V 0 of the barrier. Plot the transmission coefficient as a function E/V 0 (up to E/V 0 = 3), choosing (2ma 2 V 0 ) 1/2 = (3π/2). In the text, Merzbacher treats this problem for the case where the particle’s energy is less than the potential barrier. He obtains the result M 11 = cosh2κa + i 2 sinh 2κa e 2ika (1) where κ = 2m(V 0 −E) 2 and = κ k − k κ . (2) We can re-use the result (1) for the case where the energy is greater than the potential barrier. To do this we note that κ becomes imaginary in this case, and we write κ = iβ = i 2m(E −V 0 ) 2 so that (2) becomes = iβ k − k iβ = i β k + k β ≡ iλ 1 Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 2 Plugging into (1) and noting that coshix = cos x, sinhix = i sinx we have M 11 = cos 2βa − iλ 2 sin 2βa e 2ika . and |M 11 | = cos 2 2βa + λ 2 4 sin 2 βa 1/2 = ¸ 1 + λ 2 4 −1 sin 2 βa 1/2 = ¸ 1 + β 4 +k 4 4β 2 k 2 − 1 2 sin 2 βa 1/2 = ¸ 1 + (β 2 −k 2 ) 2 4β 2 k 2 sin 2 βa 1/2 = ¸ 1 + V 2 0 4E(E −V 0 ) sin 2 βa 1/2 = ¸ 1 + 1 4γ(γ −1) sin 2 βa 1/2 = ¸ 4γ(γ −1) + sin 2 βa 4γ(γ −1) 1/2 where γ = E/V 0 . We have βa = ¸ 2m h 2 (E −V 0 ) 1/2 a = 2mV 0 a 2 h 2 1/2 (γ −1) 1/2 = 3π 2 (γ −1) 1/2 so the transmission coefficient is T = 1 |M 11 | 2 = ¸ 4γ(γ −1) 4γ(γ −1) + sin 2 3π 2 (γ −1) 1/2 ¸ This is plotted in Figure 1. Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.5 1 1.5 2 2.5 3 3.5 4 T r a n s m i s s i o n C o e f f i c i e n t E/V0 Figure 1: Transmission coefficient versus E/V 0 for Problem 6.1. Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 4 Problem 6.2 Consider a potential V = 0 for x > a, V = −V 0 for a ≥ x ≥ 0, and V = +∞ for x < 0. Show that for x > a the positive energy solutions of the Schr¨ odinger equation have the form e i(kx+2δ) −e −ikx Calculate the scattering coefficient |1 − e 2iδ | 2 and show that it exhibits maxima (resonances) at certain discrete energies if the potential is sufficiently deep and broad. We have Ψ(x) = 0, x ≤0 Ae ik1x +Be −ik1x , 0 ≤x ≤ a Ce ik2x +De −ik2x , a ≤ x with k 1 = 2m(E +V 0 ) 2 k 2 = 2mE 2 . Applying the requirement that Ψ be continuous at x = 0, we see we must take A = −B, so Ψ(x) = γ sin k 1 x for 0 ≤ x ≤ a. The other standard requirement, that the derivative of Ψ also be continuous, does not hold at x = 0 because the potential is infinite there. Hence γ is undetermined as yet. Eventually, we could apply the normalization condition on Ψ to find γ if we wanted to. Next applying continutity of Ψ and its derivative at x = a, we obtain γ sin k 1 a = Ce ik2a +De −ik 2 a k 1 γ cos k 1 a = ik 2 [Ce ik2a −De −ik 2 a ] Combining these yields C = 1 2i γe −ik2a ¸ k 1 k 2 cos k 1 a +i sink 1 a (3) D = − 1 2i γe +ik2a ¸ k 1 ik 2 cos k 1 a −i sink 1 a (4) It’s now convenient to write k 1 k 2 cos k 1 a +i sink 1 a = αe iδ (5) where α 2 = k 1 k 2 2 cos 2 k 1 a + sin 2 k 1 a (6) Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 5 and δ = tan −1 ¸ k 2 k 1 tank 1 a so that α contains magnitude information, while δ represents phase information. Then we can rewrite (3) and (4) as C = γα 2i e −ik2a e iδ D = − γα 2i e +ik2a e −iδ Then the expression for the wavefunction to the left of x = a becomes Ψ(x) = Ce ik2x +De −ik2x (x > a) = γα 2i e ik2(x−a) e iδ −e −ik2(x−a) e −iδ = γα 2i e −iδ e ik2(x−a) e 2iδ −e −ik2(x−a) . Using (5) and (6), the scattering coefficient is |1 −e 2iδ | 2 = 1 − k1 k2 2 cos 2 k 1 a + 2i k1 k2 cos k 1 a sin k 1 a −sin 2 k 1 a k1 k2 2 cos 2 k 1 a + sin 2 k 1 a 2 = 2 sink 1 a sin k 1 a −i k1 k2 cos k 1 a k1 k2 2 cos 2 k 1 a + sin 2 k 1 a 2 = 4 sin 2 k 1 a k1 k2 2 cos 2 k 1 a + sin 2 k 1 a (7) We have k 1 k 2 2 = E +V 0 E = 1 + 1 λ k 1 a = 2ma 2 (E +V 0 ) 2 = β(λ + 1) 1/2 where λ = E/V 0 and β = (2ma 2 V 0 / 2 ) 1/2 in Merzbacher’s notation. Then the scattering coefficient (7) is scattering coefficient = 4 sin 2 [β(λ + 1) 1/2 ] (1 + 1 λ ) 2 cos 2 [β(λ + 1) 1/2 ] + sin 2 [β(λ + 1) 1/2 ] In Figure 6.2 I have plotted this for β = 25. Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 6 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 2 2.5 3 S c a t t e r i n g c o e f f i c i e n t E/V0 Figure 2: Scattering coefficient versus E/V 0 for Problem 6.2. Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 7 Problem 6.3 A particle of mass m moves in the one-dimensional double well potential V (x) = −gδ(x −a) −gδ(x +a). If g > 0, obtain transcendental equations for the bound-state energy eigenvalues of the system. Compute and plot the energy levels in units of 2 /ma 2 as a function of the dimensionless parameter mag/ 2 . Explain the features of the plot. In the limit of large separation, 2a, between the wells, obtain a simple formula for the splitting ∆E between the ground state (even parity) energy level, E + , and the excited (odd parity) energy level, E − . In this problem, we can divide the x axis into three regions. In each region, the wavefunction is just the solution to the free-particle Schr¨ odinger equation, but with energy E < 0 since we’re looking for bound states. Putting k = 2mE/ 2 , we have Ψ(x) = Ae kx +Be −kx , x ≤ −a Ce kx +De −kx , −a ≤ x ≤ a Ee kx +Fe −kx , a ≤ x. Now, first of all, the wavefunction can’t blow up at infinity, so B = E = 0. Also, since the potential in this problem has mirror-reversal symmetry, the wavefunction will have definite parity. Considering first the even parity solution, Ψ(x) = Ae kx , x ≤ −a Bcosh(kx), −a ≤ x ≤ a Ae −kx , a ≤ x. (8) Matching the value of the wavefunction at x = −a gives Ae −ka = Bcosh(−ka). (9) Since the potential becomes infinite at x = −a, the normal derivative-continuity condition doesn’t hold there. Instead, we can write down the Schr¨ odinger equa- tion, d 2 dx 2 Ψ(x) = 2m 2 V (x)Ψ(x) − 2m 2 EΨ(x), then integrate from −a − to −a + and take the limit as →0. This gives dΨ dx −a+ −a− = − 2mg 2 Ψ(−a). (10) Applying this condition to the wavefunction (8) yields kBsinh(−ka) −kAe −ka = − 2mg 2 Bcosh(−ka). Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 8 Substituting from (9), kBsinh(−ka) −kBcosh(−ka) = − 2mg 2 Bcosh(−ka) or tanh(ka) = 2mg 2 k −1 = β ka −1 with β = 2mag/ 2 . This equation determines the energy eigenvalue of the even- parity state, which will be the ground state. On the other hand, the odd parity state looks like Ψ(x) = Ae kx , x ≤ −a Bsinh(kx), −a ≤ x ≤ a −Ae −kx , a ≤ x. (11) Matching values at x = −a gives Ae −ka = Bsinh(−ka) and applying condition (10) gives kBcosh(−ka) −kAe −ka = − 2mg 2 Bsinh(−ka) kBcosh(ka) +kBsinh(ka) = 2mg 2 Bsinh(ka) coth(ka) = β ka −1 so this is the condition that determines the energy of the odd parity state. In Figure (3) I have plotted tanh(ka), coth(ka), and β/(ka) −1 for the case β = 3. As expected, the coth curve crosses the β/(ka) −1 curve at a lower value of ka than the tanh curve; that means that the energy eigenvalue for the odd parity state is smaller in magnitude (less negative) than the even parity state. Problem 6.4 Problem 3 provides a primitive model for a one-electron linear diatomic molecule with interatomic distance 2a = |X|, if the potential energy of the “molecule” is taken as E ± (|X|), supplemented by a repulsive interaction λg/|X| between the wells (“atoms”). Show that, for a sufficiently small value of λ, the system (“molecule”) is stable if the particle (“electron”) is in the even parity state. Sketch the total potential energy of the system as a function of |X|. Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 9 0 0.5 1 1.5 2 1 1.2 1.4 1.6 1.8 2 P S f r a g r e p l a c e m e n t s tanh(ka) coth(ka) β/(ka) −1 ka Figure 3: Graphical determination of energy levels for Problem 6.3 with β = 3. Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 10 Problem 6.5 If the potential in Problem 3 has g < 0 (double barrier), calculate the transmission coefficient and show that it exhibits resonances. (Note the analogy between the system and the Fabry-Perot ´etalon in optics.) Now we’re assuming that the energy E is positive, so Ψ(x) = Ae ikx +Be −ikx x ≤ −a Ce ikx +De −ikx −a ≤ x ≤ a Ee ikx +Fe −ikx a ≤ x (12) with k = 2mE/ 2 . Matching values at x = −a, we have Ae −ika +Be ika = C −ika +D ika (13) Also, as before, we have the derivative condition dΨ dx x=−a+ x=a− = − 2mg 2 Ψ(−a) where g is now negative. Applying this to the wavefunction in (12), we have ik[Ce −ika −De ika −Ae −ika +Be ika ] = − 2mg 2 [Ae −ika +Be ika ]. (14) Combining (13) and (14) yields C = 1 + β ika A+ β ika e 2ka B (15) D = − β ika e −2ka A+ 1 − β ika B (16) with β = mag/ 2 as before. Now, applying the matching conditions to the wavefunction at x = +a will give two equations exactly like (13) and (14), but with the substitutions A →C, B →D, C →E, D →F, and a →−a. Making these substitutions in (15) and (16) we obtain E = 1 − β ika C − β ika e 2ka D (17) F = + β ika e 2ka C + 1 + β ika D (18) Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 11 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 15 20 25 30 P S f r a g r e p l a c e m e n t s ka T ( k ) Figure 4: Transmission coefficient in Problem 6.4 with β = 15. Combining equations (15) through (18), we have E = ¸ 1 + β ika 2 (e −4ika −1) ¸ A+ + ¸ 2β ka 1 − β ika sin2ka B F = ¸ 2β ka 1 + β ika sin 2ka A+ ¸ 1 + β ika 2 (e 4ika −1) ¸ B This is the M matrix, and the transmission coefficient is given by T = 1/|M 11 | 2 , or T = 1 1 + 2 β ka 2 ¸ β ka 2 + 1 (1 −cos(4ka)) In Figure 4 I have plotted this for β = 15. Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 12 Problem 6.6 A particle moves in one dimension with energy E in the field of a potential defined as the sum of a Heaviside step function and a delta function: V (x) = V 0 η(x) +gδ(x) (with V 0 and g > 0) The particle is assumed to have energy E > V 0 . (a) Work out the matrix M, which relates the amplitudes of the incident and reflected plane waves on the left of the origin (x < 0) to the amplitudes on the right (x > 0). (b) Derive the elements of the matrix S, which relates incoming and outgoing amplitudes. (c) Show that the S matrix is unitary and that the elements of the S matrix satisfy the properties expected from the applicable symmetry considerations. (d) Calculate the transmission coefficient for particles incident from the right and for particles incident from the left, which have the same energy (buf different velocities). We have Ψ(x) = Ae ik1x +Be −ik1x , x ≤ 0 Ce ik2x +De −ik2x , x ≥ 0 with k 1 = 2m 2 E k 2 = 2m 2 (E −V 0 ). Matching values at x = 0 gives C +D = A+B (19) Also, the delta function at the origin gives rise to a discontinuity in the derivative of the wavefunction as before: dΨ dx 0+ 0− = 2mg 2 Ψ(0) so ik 2 (C −D) −ik 1 (A−B) = 2mg 2 (A+B) or C −D = k 1 k 2 (A−B) + 2mg ik 2 2 (A+B). (20) Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 13 Adding and subtracting (19) and (20), we can read off C = 1 2 ¸ 1 + k 1 k 2 + 2mg ik 2 2 A+ 1 2 ¸ 1 − k 1 k 2 + 2mg ik 2 2 B D = 1 2 ¸ 1 − k 1 k 2 − 2mg ik 2 2 A 1 2 ¸ 1 + k 1 k 2 − 2mg ik 2 2 B. We could also write this as C D = M 11 M 12 M ∗ 12 M ∗ 11 A B Or instead of the M matrix we could use the S matrix, which is defined by B C = S 11 S 12 S 21 S 22 A D Since we already know the M coefficients, we can calculate the elements of the S matrix from the formula S 11 = − M ∗ 12 M ∗ 11 S 12 = 1 M ∗ 11 S 21 = 1 M ∗ 11 S 22 = + M12 M ∗ 11 However, this is tedious and long and boring and I don’t want to do it. Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid April 5, 2001 Chapter 7 Before starting on these problems I found it useful to review how the WKB approximation works in the first place. The Schr¨ odinger equation is − 2 2m d 2 dx 2 Ψ(x) +V (x)Ψ(x) = EΨ(x) or d 2 dx 2 Ψ(x) +k 2 (x)Ψ(x) = 0, k(x) ≡ 2m 2 [E −V (x)]. We postulate for Ψ the functional form Ψ(x) = Ae iS(x)/ in which case the Schr¨ odinger equation becomes iS (x) = [S (x)] 2 − 2 k 2 (x). (1) This equation can’t be solved directly, but we obtain guidance from the obser- vation that, for a constant potential, S(x) = ±kx, so that S vanishes. For a nonconstant but slowly varying potential we might imagine S (x) will be small, and we may take S = 0 as the seed of a series of successive approximations to the exact solution. To be specific, we will construct a series of functions S 0 (x), S 1 (x), · · · , where S 0 is the solution of (1) with 0 on the left hand side; S 1 is a solution with S 0 on the left hand side; and so on. In other words, at the nth step in the approximation sequence (by which point we have computed S n (x)), we compute S n (x) and use that as the source term on the LHS of (1) to calculate S n+1 (x). Then we compute the second derivative of S n+1 (x) and use this as the source term for calculating S n+2 , and so on ad infinitum. In 1 Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 2 symbols, 0 = [S 0 (x)] 2 − 2 k 2 (x) (2) iS 0 = [S 1 (x)] 2 − 2 k 2 (x) (3) iS 1 = [S 2 (x)] 2 − 2 k 2 (x) (4) · · · · Equation (2) is clearly solved by taking S 0 (x) = ±k(x) ⇒ S 0 (x) = S 00 ± x −∞ k(x )dx (5) for any constant S 00 . Then S 0 (x) = ±k (x), so (3) is S 1 (x) = ± k 2 (x) ±ik (x). With the two ± signs here, we appear to have four possible choices for S 1 . But let’s think a little about the ± signs in this equation. The ± sign under the radical comes from the two choices of sign in (5). But if we chose, say, the plus sign in that equation, so that S 0 > 0, we would also expect that S 1 > 0. Indeed, if we choose the plus sign in (5) but the minus sign in (3), then S 0 and S 1 have opposite sign, so S 1 differs from S 0 by an amount at least as large as S 0 , in which case our approximation sequence S 0 , S 1 , · · · has little hope of converging. So we choose either both plus signs or both minus signs in (3), whence our two choices are S 1 = + k 2 (x) +ik (x) or S 1 = − k 2 (x) −ik (x). (6) If V (x) is constant, k(x) is constant, and, as we observed before, the sequence of approximations terminates at 0th order with S 0 being an exact solution. By extension, if V (x) is not constant but changes little over one particle wavelength, we have k (x)/k 2 (x) 1, so we may expand the radicals in (6): S 1 ≈ k(x) 1 + ik (x) 2k 2 (x) or S 1 ≈ −k(x) 1 − ik (x) 2k 2 (x) or S 1 ≈ ±k(x) + ik (x) 2k(x) . (7) Integrating, S 1 (x) = S 1 (a) ± x a k(u)du + i 2 x a k (u) k(u) dx = S 1 (a) ± x a k(u)du + i 2 ln k(x) k(a) where a is some point chosen such that the approximation (7) is valid in the full range a < x < x. We could go on to compute S 2 , S 3 , etc., but in practice it seems the approximation is always terminated at S 1 . Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 3 The wavefunction at this order of approximation is Ψ(x) = exp(iS 1 (x)/) = e iS1(a)/ e ±i R x a k(u)du e ln k(x)/k(a) −1/2 = Ψ(a) k(a) k(x) e ±i R x a k(u)du = Ψ(a)G ± (x; a) where (8) G ± (x; a) ≡ k(a) k(x) e ±i R x a k(u)du . (9) We have written it this way to illustrate that the function G(x, a) is kind of like a Green’s function or propagator for the wavefunction, in the sense that, if you know what Ψ is at some point a, you can just multiply it by G ± (x; a) to find out what Ψ is at x. But this doesn’t seem quite right: Schr¨ odinger’s equation is a second-order differential equation, but (8) seems to be saying that we need only one initial condition—the value of Ψ at x = a—to find the value of Ψ at other points. To clarify this subtle point, let’s investigate the equations leading up to (8). If the approximation (7) makes sense, then there are two solutions of Schr¨ odinger’s equation at x = a, one whose phase increases with increasing x (positive derivative), and one whose phase decreases. Equation (8) seems to be saying that we can use either G + or G − to get to Ψ(x) from Ψ(a); but the requirement the dΨ/dx be continuous at x = a means that only one or the other will do. Indeed, in using (8) to continue Ψ from a to x we must choose the appropriate propagator—either G + or G − , according to the derivative of Ψ at x = a; otherwise the overall wave function will have a discontinuity in its first derivative at x = a. So to use (8) to obtain values for Ψ at a point x, we need to know both Ψ and Ψ at a nearby point x = a, as should be the case for a second-order differential equation. If we want to de-emphasize this nature of the solution with the propagator we may write Ψ(x) = C 1 k(x) e ±i R x a k(u)du (10) where C = Ψ(a) k(a). In regions where V (x) > E, k(x) is imaginary, so it’s useful to define κ(x) = −ik(x) = 2m 2 [V (x) −E] (11) and Ψ(x) = C 1 κ(x) e ± R x a κ(u)du . (12) Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 4 If we have a region of space in which the WKB approximation is valid, knowing the value of Ψ (and its derivative) at one point within the region is equivalent to knowing it everywhere, because we can use the propagator (9) to get from that one point to every other point within the region. The WKB method, however, gives us no way of determining the value of Ψ at that one starting point. Furthermore, even if we know Ψ at one point within a region of validity, we can’t use (8) to determine Ψ in other, nonadjacent regions, because we can’t carry the propagator across regions of invalidity. So basically what we need is a way of finding one starting value for Ψ(x) in every region of validity of the WKB approximation. How do we find such points? Well, one sure-fire way to get starting points in regions of validity is to identify regions of invalidity, of which there will be at least one adjacent to each region of validity, and then get values of Ψ at the boundaries of the regions of invalidity—which will also count as values in the regions of validity. So we need to identify the regions of invalidity of the WKB approximation and do a more accurate solution of the Schr¨ odinger equation there. The WKB approximation breaks down when k /k 2 1 ceases to hold, which is true when k ≈ 0 but k = 0, which happens near a classical turning point of the motion—i.e., a point x 0 at which V (x 0 ) = E. But near such a point we may expand V (x) − E in a Taylor series around the point x 0 ; if we keep only the first (linear in x) term in the series, we arrive at a Schr¨ odinger equation which we can solve exactly in the vicinity of x 0 . To do this, suppose the point x 0 is a classical turning point of the motion, so that V (x 0 ) = E. In the neighborhood of x 0 we may expand V (x): V (x) = E + (x −x 0 )V (x 0 ) +· · · (13) Then the Schr¨ odinger equation becomes d 2 dx 2 Ψ(x) − 2m 2 V (x 0 )(x −x 0 )Ψ(x) = 0. (14) The useful substitution here is u(x) = γ(x −x 0 ) γ ≡ ¸ 2m 2 V (x 0 ) 1/3 so x(u) = u γ +x 0 . If we define Φ(u) = Ψ(x(u)) then dΦ du = dΨ dx dx du = 1 γ Ψ (x(u)) d 2 Φ du 2 = 1 γ 2 Ψ (x(u)) Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 5 so (14) becomes γ 2 d 2 du 2 Φ(u) −γ 3 (x −x 0 )Φ(u) = 0 or d 2 du 2 Φ(u) −uΦ(u) = 0. The solution to this differential equation is Φ(u) = β 1 Ai(u) +β 2 Bi(u) (15) so the solution to the Schr¨ odinger equation (14) is Ψ(x) = β 1 Ai γ(x −x 0 ) +β 2 Bi γ(x −x 0 ) . For γ(x − x 0 ) 1 we have the asymptotic expression Ψ(x) ≈ π −1/2 [γ(x −x 0 )] −1/4 β 1 2 e − 2 3 |γ(x−x0)| 3/2 +β 2 e + 2 3 |γ(x−x0)| 3/2 (16) and for γ(x −x 0 ) −1 we have Ψ(x) ≈ π −1/2 |γ(x −x 0 )| −1/4 β 1 cos 2 3 |γ(x −x 0 )| 3/2 − π 4 −β 2 sin 2 3 |γ(x −x 0 )| 3/2 − π 4 . (17) To simplify these, we need to consider two possible kinds of turning point. Case 1: V (x 0 ) > 0. In this case the potential is increasing through the turning point at x 0 , which means that V (x) < E for x < x 0 , and V (x) > E for x > x 0 . Hence the region to the left of the turning point is the classically accessible region, while the right of the turning point is classically forbidden. Since V (x 0 ) > 0, γ > 0, so for x < x 0 (??) holds. For points close to the turning point on the left side, k(x) = 2m 2 [E −V (x)] ≈ ¸ 2m 2 V 1/2 (x 0 −x) 1/2 = γ 3/2 (x 0 −x) 1/2 so |γ(x −x 0 )| −1/4 = γ k(x) (18) and x0 x k(u)du = γ 3/2 x0 x (x 0 −x) 1/2 du = 2 3 γ 3/2 (x 0 −x) 3/2 = 2 3 |γ(x −x 0 )| 3/2 . (19) On the other hand, for points close to the turning point on the left side we have x > x 0 , so γ(x −x 0 ) > 0. In this region, Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 6 κ(x) = 2m 2 V (x) −E ≈ ¸ 2m 2 V (x 0 )(x −x 0 ) 1/2 = γ 3/2 (x −x 0 ) 1/2 (20) so, for x near x 0 , x x0 κ(u)du = γ 3/2 x x0 (u −x 0 ) 1/2 du = 2 3 γ 3/2 (x −x 0 ) 3/2 (21) and also |γ(x −x 0 )| −1/4 = γ κ(x) . (22) Using (18) and (19) in (17), and (21) and (22) in (16), the solutions to the Schr¨ odinger equation on either side of a classical turning point x 0 at which V (x 0 ) > 0 are Ψ(x) = 1 k(x) 2β 1 cos x0 x k(u)du − π 4 −β 2 sin x0 x k(u)du − π 4 , x < x 0 (23) Ψ(x) = 1 κ(x) β 1 e R x x 0 κ(u)du +β 2 e − R x x 0 κ(u)du , x < x 0 (24) (we redefined the β constants slightly in going to this equation). Case 2: V (x 0 ) < 0. In this case the potential is decreasing through the turning point, so the classically accessible region is to the right of the turning point, and the forbidden region to the left. Since V (x 0 ) < 0, γ < 0. That means that the regions of applicability of (16) and (17) are on opposite sides of the turning points as they were in the previous case. The solutions to the Schr¨ odinger equation on either side of the turning point are Ψ(x) = 1 κ(x) β 1 e R x 0 x κ(u)du +β 2 e − R x x 0 κ(u)du , x < x 0 (25) Ψ(x) = 1 k(x) 2β 1 cos x x0 k(u)du − π 4 −β 2 sin x x0 k(u)du − π 4 , x > x 0 (26) (27) So, to apply the WKB approximation to a given potential V (x), the first step is to identify the classical turning points of the motion, and to divide space Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 7 up into regions bounded by turning points, within which regions the WKB approximation (7) is valid. Then, for each turning point, we write down (23) and (24) (or (25) and (26)) at nearby points on either side of the turning point, and then use (10) to evolve the wavefunction from those points to other points within the separate regions. We should probably quantify the meaning of “nearby” in that last sentence. Suppose x 0 is a classical turning point of the motion, and we are looking for points x 0 ± at which to make the “handoff” from approximations (16) and (17) to the WKB approximation These points must satisfy several conditions. First, the approximate Schr¨ odinger equation (14) is only valid as long as we can neglect the quadratic and higher-order terms in the expansion (13), so we must have |V (x 0 )| |V (x 0 )| ⇒ V (x 0 ) V (x 0 ) . (28) But at the same time, γ must be sufficiently greater than 1 to justify the approximation (16) (or sufficiently less than -1 to justify (17)); the condition here is 2m 2 V (x 0 ) 1/3 1 ⇒ 2m 2 V (x 0 ) −1/3 . (29) Finally, the points x± must be sufficiently far away from the turning points that the approximation (7) is valid for the derivative of the phase of the wavefunction; the condition for this to be the case was k (x) k 2 (x) 1 ⇒ 1 2 2 2m V (x ±) [E −V (x ±)] 3/2 1. (30) If there are no points x 0 ± satisfying all three conditions, the WKB approxi- mation cannot be used. To apply all of this to the problem of bound states in a potential well, consider a potential like that shown in Figure 1, with two classical turning points at x = a and x = b. Although there are no discontinuities in the potential here, the problem may be analyzed in a manner similar to that used in the consideration of one-dimensional piecewise constant potentials, as in Chapter 6: we divide space into a number of distinct regions, obtain solutions of the Schr¨ odinger equation in each region, and then match values and derivatives at the region boundaries. To divide space into distinct regions in this case, we begin by identifying narrow regions around the turning points a and b in which the linear approx- imation (13) is valid. In the narrow region around x = a, we may use (25) and (26); around x = b we may use (23) and (24). Let the narrow such region around a be a− 1 < x < a+ 1 , and that around b be b − 2 < x < b + 2 . Then space divides naturally into five regions: (a) x < a − In this region we are far enough to the left of the turning point that the WKB approximation is valid, and the wavefunction takes the form (10). However, we Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 8 P S f r a g r e p l a c e m e n t s a b E V (x) Figure 1: A potential V (x) with two classical turning points for an energy E. Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 9 must throw out the term that grows exponentially as x → −∞, so we are left with Ψ(x) = A 1 κ(x) e − R (a−) x κ(u)du , x < a −. (31) (b) a − < x < a + In this region we are close enough to the turning point that (13) is valid, so (25) and (26) may be used. Ψ(x) = 1 κ(x) β 1 e − R a x κ(u)du +β 2 e + R a x κ(u)du , x < (a −). (32) From (31) and (32) we see that continuity of both the value and first derivative of Ψ(x) at x = a− requires taking β 1 = A, β 2 = 0. With this choice of constants, we achive continuity not only of the value and first derivative of Ψ but also of all higher derivatives, as must be the case since there is no discontinuity in the potential. But now that we know the value of Ψ at x = a − , we also know it at x = a + , because of course the solution of the Schr¨ odinger equation in the narrow strip around a (to which (32) is an asymptotic approximation for x < a) is valid throughout the strip; the same solution that’s valid at x = a − is valid at x = a +. With β 1 = A and β 2 = 0, (26) becomes Ψ(x) = 2A 1 k(x) cos x a k(u)du − π 4 = A 1 k(x) e +i( R x a k(u)du−π/4) +e −i( R x a k(u)du−π/4) , x = (a +)− (33) (c) a + < x < b − In this region the WKB approximation (7) is valid, so we may use (8) to find the wavefunction at any point within the region. Using the expression (33) for the wavefunction at x = a+, integrating from a+ to x in the propagator (9), and using G + and G − , respectively, to propagate the first and second terms in Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 10 (33), we obtain for the wavefunction at a point x in this region Ψ(x) = A 1 k(x) ¸ e +i “ R (a+) a k(u)du+ R x (a+) k(u)du−π/4 ” +e −i “ R (a+) a k(u)du+ R x (a+) −π/4 ” = A 1 k(x) e +i( R x a k(u)du−π/4) +e −i( R x a k(u)du−π/4) = 2A 1 k(x) cos x a k(u)du − π 4 = 2A 1 k(x) cos b a k(u)du − b x k(u)du − π 4 , a + < x < b − (34) Okay, I have now carried this analysis far enough to see for myself exactly where the Bohr-Sommerfeld quantization condition b a k(u)du = n + 1 2 π, n = 1, 2, · · · (35) comes from, which was my original goal, so I am now going to stop this exercise and proceed directly to the problems. Problem 7.1 Apply the WKB method to a particle that falls with acceleration g in a uniform gravitational field directed along the z axis and that is reflected from a perfectly elastic plane surface at z = 0. Compare with the rigorous solutions of this problem. We’ll start with the exact solution to the problem. The requirement of perfect elastic reflection at z = 0 may be imposed by taking V (x) to jump suddenly to infinity at z = 0, i.e. V (x) = mgz, z > 0 ∞, z ≤ 0. For z > 0, the Schr¨ odinger equation is 0 = d 2 dx 2 Ψ(x) + 2m 2 [E −mgz] Ψ(x) = d 2 dx 2 Ψ(x) + 2m 2 g 2 ¸ E mg −z Ψ(x) = d 2 dx 2 Ψ(x) − 2m 2 g 2 [z −z 0 ] Ψ(x) (36) Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 11 where z 0 = E/mg. With the substitution u = γ(z −z 0 ) γ = 2m 2 g 2 1/3 and taking Φ(u) = Ψ(x(u)), we find that (36) is just the Airy equation for Φ(u), d 2 du 2 Φ(u) −uΦ(u) = 0 with solutions Φ(u) = β 1 Ai(u) +β 2 Bi(u). Since we require a solution that remains finite as z →∞, we must take β 2 = 0. The solution to (36) is then Ψ(x) = β 1 Ai γ(z −z 0 ) , (z > 0). (37) For z < 0, I wasn’t quite sure how to account for the infinite potential jump at z = 0, so instead I supposed the potential for z < 0 to be a constant, V (z) = V 0 , where eventually I’ll take V 0 → ∞. Then the Schr¨ odinger equation for z < 0 is d 2 dz 2 Ψ(z) − 2m 2 [V 0 − E] Ψ(z) = 0 with solution Ψ(z) = Ae −kz , k = 2m 2 [V 0 −E]. (38) Matching values and derivatives of (37) and (38) at z = 0, we have β 1 Ai(−γz 0 ) = A γβ 1 Ai (−γz 0 ) = −kA Dividing, we obtain 1 γ Ai(−γz 0 ) Ai (−γz 0 ) = − 1 k Now taking V 0 → ∞, we also have k → ∞, so the RHS of this goes to zero; thus the condition is that −γz 0 be a zero of the Airy function, which means the energy eigenvalues E n are given by 2m 2 g 2 1/3 E n mg = x nm ⇒E n = mg 2 2 2 1/3 x n (39) where x n is the nth root of the equation Ai(−x n ) = 0. Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 12 So that’s the exact solution. In the WKB approximation, the spectrum of energy eigenvalues is determined by the condition (35). In this case the classical turning points are at z = 0 and z = z 0 , so we have n + 1 2 π = z0 0 k(z) dz = 2m 2 z0 0 [E −mgz] 1/2 dz = 2m 2 g 2 z0 0 [z 0 −z] 1/2 du = 2m 2 g 2 − 2 3 (z 0 −z) 3/2 z0 0 = 2 3 2m 2 g 2 E mg 3/2 so the nth eigenvalue is given by E n = mg 2 2 2 1/3 ¸ 3 2 n + 1 2 π 2/3 . Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition Homer Reid May 13, 2001 Chapter 8 1 Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 2 Problem 8.1 Apply the variational method to estimate the ground state energy of a particle confined in a one-dimensional box for which V = 0 for −a < x < a, and Ψ(±a) = 0. (a) First, use an unnormalized trapezoidal trial function which vanishes at ±a and is symmetric with respect to the center of the well: Ψ t (x) = (a −|x|), b ≤ x ≤ a (a −b), |x| ≤ b. (b) A more sophisticated trial function is parabolic, again vanishing at the end points and even in x. (c) Use a quartic trial function of the form Ψ t (x) = (a 2 −x 2 )(αx 2 +β), where the ratio of the adjustable parameters α and β is determined variation- ally. (d) Compare the results of the different variational calculations with the exact ground state energy, and, using normalized wave functions, evaluate the mean square deviation a −a |Ψ(x) −Ψ t (x)| 2 dx for the various cases. (e) Show that the variational procedure produces, in addition to the approximation to the ground state, an optimal quartic trial function with nodes between the endpoints. Interpret the corresponding stationary energy value. First let’s observe that the exact expressions for the ground state wavefunction and energy are Ψ n (x) = 1 √ a cos(k n x), k n = nπ 2a , E n = n 2 2 π 2 8ma 2 ≈ 1.23 2 ma 2 . (a) We need first to normalize the trial wavefunction. Taking Ψ t (x) = γ(a −|x|), b ≤ |x| ≤ a γ(a −b), x| ≤ b. Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 3 we have a −a Ψ 2 t (x)dx = 2 a 0 Ψ 2 t (x)dx = 2γ 2 (a −b) 2 b 0 dx + a b (a −x) 2 dx ¸ = 2γ 2 b(a −b) 2 + 1 3 (a −b) 3 = 2γ 2 b(a 2 +b 2 −2ab) + 1 3 (a 3 −b 3 ) −a 2 b +b 2 a = γ 2 2 3 a 3 + 2b 3 −3ab 2 so Ψ t is normalized by taking γ 2 = 3 2 1 a 3 + 2b 3 −3ab 2 . (1) Now we can compute the energy expectation value of Ψ t : < Ψ t |H|Ψ t > = − 2 2m a −a Ψ t (x) d 2 dx 2 Ψ t (x) dx Integrating by parts, = − 2 2m Ψ t (x)Ψ t (x) a −a − a −a Ψ 2 t (x) dx (the first integral vanishes since Ψ t vanishes at the endpoints) = + 2 m a 0 Ψ 2 t (x) dx = 2 m γ 2 a b dx = 2 m γ 2 (b −a). Using (1), this is < H >= 3 2 2m (b −a) a 3 + 2b 3 −3ab 2 . To find the optimal value of b, we zero the derivative of this with respect to b: 0 = 1 (a 3 + 2b 3 −3ab 2 ) − 6b 2 (b −a) (a 3 + 2b 3 −3ab 2 ) 2 + 6ab(b −a) (a 3 + 2b 3 −3ab 2 ) 2 = −4b 3 + 9b 2 a −6a 2 b +a 3 Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 4 (b) For a parabolic trial function we take Ψ t (x) = γ(a 2 −x 2 ). The normalization integral is a −a Ψ 2 t (x) dx = 2γ 2 a 0 (a 2 −x 2 ) 2 dx = 2γ 2 a 0 (a 4 +x 4 −2a 2 x 2 )dx = 2γ 2 a 5 + 1 5 a 5 − 2 3 a 5 = 16 15 γ 2 a 5 so Ψ t (x) is normalized by taking γ 2 = 15 16a 5 . The expectation value of the energy is < H > = − 2 2m a −a Ψ t (x) d 2 dx 2 Ψ t (x) dx = 2 2 m γ 2 a 0 (a 2 −x 2 )dx = 4 3 2 m γ 2 a 3 = 5 4 2 ma 2 ≈ 1.25 2 ma 2 . So this is in good agreement with the exact ground state energy. (c) In this case we have Ψ t (x) = γ(a 2 −x 2 )(αx 2 +β) = γ[−αx 4 + (αa 2 −β)x 2 +βa 2 ] The kinetic energy is − 2 2m d 2 dx 2 Ψ t (x) = γ 2 m [6αx 2 −(αa 2 −β)]. The expectation value of the energy is Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 5 Problem 8.2 Using scaled variables, as in Section 5.1, consider the anharmonic oscillator Hamil- tonian, H = 1 2 p 2 ξ + 1 2 ξ 2 +λξ 4 where λ is a real-valued parameter. (a) Estimate the ground state energy by a variational calculation, using as a trial function the ground state wave function for the harmonic oscillator H 0 (ω) = 1 2 p 2 ξ + 1 2 ω 2 ξ 2 where ω is an adjustable variational parameter. Derive an equation that relates ω and λ. (b) Compute the variational estimate of the ground state energy of H for various positive values of the strength λ. (c) Note that the method yields answers for a discrete energy eigenstate even if λ is slightly negative. Draw the potential energy curve to judge if this result makes physical sense. Explain. (a) To find the ground state eigenfunction of the Hamiltonian Merzbacher pro- poses, it’s convenient to rewrite it: H 0 (ω) = 1 2 p 2 ξ + 1 2 ω 2 ξ 2 = − 1 2 ∂ 2 ∂ξ 2 + 1 2 ω 2 ξ 2 Upon substituting u = ω 1/2 ξ we obtain = ω − 1 2 ∂ 2 ∂u 2 + 1 2 u 2 and now this is just the ordinary harmonic oscillator Hamiltonian, scaled by a constant factor ω, with ground-state eigenfunction Ψ(ω) = Ce −u 2 /2 = Ce −ωξ 2 /2 . Adding the normalization constant, Ψ(ω) = ω π 1/4 e −ωξ 2 /2 . Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 6 Now we want to treat ω as a parameter and vary it until the energy expectation value of Ψ(ω) is minimized. The energy expectation value is Ψ| H |Ψ = Ψ| T |Ψ +Ψ| V |Ψ where T = p 2 ξ /2 and V = ξ 2 /2 +λξ 4 . Let’s compute the two expectation values separately. First of all, to compute the expectation value of T, we need to know the result of operating on Ψ(ω) with p 2 ξ : p 2 ξ Ψ(ω) = − ω π 1/4 ∂ ∂ξ ¸ ∂ ∂ξ e −ωξ 2 /2 = − ω π 1/4 ∂ ∂ξ −ωξe −ωξ 2 /2 = − ω π 1/4 −ω +ω 2 ξ 2 e −ωξ 2 /2 Then for the expectation value of T we have Ψ| T |Ψ = 1 2 ∞ −∞ Ψ(ξ)p 2 ξ Ψ(ξ)dξ = − 1 2 ω π ∞ −∞ −ω +ω 2 ξ 2 e −ωξ 2 dξ = − 1 2 ω π ¸ −ω π ω + 1 2 ω 2 π ω 3 = ω 4 . (2) On the other hand, for the expectation value of V we have (3) exptwoΨV Ψ = ω π 1 2 ∞ −∞ ξ 2 e −ωξ 2 dξ +λ ∞ −∞ ξ 4 e −ωξ 2 dξ = ω π 1 4 π ω 3 + 3 4 λ π ω 5 = 1 4ω + 3λ 4ω 2 . (4) Adding (2) and (4), exptwoΨHΨ = 1 2 ¸ ω + 1 ω + 3λ ω 2 . (5) To minimize this with respect to ω we equate its ω derivative to 0: 0 = 1 − 1 ω 2 − 6λ ω 3 Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 7 or ω 3 −ω −6λ = 0. (6) We could then solve this equation for ω in terms of λ to obtain the energy- minimizing value of ω for a given perturbing potential strength λ. But writing down the full solution would be tedious. Instead let’s see what happens when λ is small. Evidently, when λ = 0 the Hamiltonian in this problem degenerates to the normal harmonic oscillator Hamiltonian, for which the energy is minimized by the (unscaled) ground state harmonic oscillator wavefunction, i.e. Ψ(ω) with ω = 1. We can thus imagine that, for small λ, the energy-minimizing value of ω will be close to 1, and we may write ω(λ) ≈ 1 + for some small . Inserting this in (6), (1 + 3 + 3 2 + 3 ) −(1 +) = 6λ Keeping only terms of zeroth or first order in the small quantity (which is equivalent to keeping terms of lowest order in the perturbing potential strength λ) we obtain from this ≈ 3λ, so for λ 0 the minimizing value of ω is ω ≈ 1 + 3λ. Inserting this estimate into (5) and again keeping only terms of lowest order in λ we find (7) exptwoΨHΨ = 1 4 (1 +) + (1 +) −1 + 3λ(1 +) −2 ≈ 1 4 (1 + 3λ) + (1 + 3λ) −1 + 3λ(1 + 3λ) −2 ≈ 1 4 [(1 + 3λ) + (1 −3λ) + 3λ(1 −6λ)] ≈ 1 2 + 3 4 λ. (8) Since the 1/2 term is the normal (unperturbed) energy of the state, the energy shift caused by the perturbing potential is ∆E = 3 4 λ. (9) Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 8 Problem 8.3 In first-order perturbation theory, calculate the change in the energy levels of a linear harmonic oscillator that is perturbed by a potential gx 4 . For small values of the coefficient, compare the result with the variational calculation in Problem 2. The energy shift to first order is ∆E = exptwoΨ n (x)|gx 4 |Ψ n (x) = g Ψ n | x 4 |Ψ n . I worked out this expectation value in Problem 5.3: ∆E = gexptwoΨ n x 4 Ψ n = 3g 2 mω 2 1 2 +n +n 2 In particular, the energy shift of the ground state is ∆E 0 = 3g 4 mω 2 which agrees with () (the difference in the factor (/mω) 2 just represents the fact that in Problem 8.2 we used scaled variables, whereas in this problem we inserted the units explicitly). Problem 8.4 Using a Gaussian trial function, e −λx 2 , with an adjustable parameter, make a vari- ational estimate of the ground state energy for a particle in a Gaussian potential well, represented by the Hamiltonian H = p 2 2m −V 0 e −αx 2 (V 0 > 0, α > 0). For notational simplicity, I like to put β/2 = λ. Then Ψ(x) = Ce −βx 2 /2 and the normalization constant is determined by 1 = C 2 ∞ −∞ e −βx 2 dx ⇒ C = β π 1/4 . Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 9 The kinetic energy operator operating on this state yields TΨ = p 2 2m Ψ(x) = − 2 2m ∂ ∂x ¸ ∂ ∂x Ψ(x) = − β π 1/4 2 2m ∂ ∂x −βxe −βx 2 /2 = − β π 1/4 2 2m −β +β 2 x 2 e −βx 2 /2 and its expectation value is T = β π 1/2 2 2m ¸ β π β − β 2 2 π β 3 = 2 β 4m . (10) The expectation value of the potential energy is V = −V 0 ∞ −∞ Ψ 2 (x)e −αx 2 dx = −V 0 β π ∞ −∞ e −βx 2 e −αx 2 dx = −V 0 β π π (α +β) = −V 0 β (α +β) . (11) Combining (10) and (11), exptwoΨ|H|Ψ = Ψ| T |Ψ +Ψ| V |Ψ = 2 β 2m −V 0 β (α +β) . (12) To minimize with respect to β we equate the first β derivative of this to zero: 0 = 2 2m − V 0 2 ¸ 1 β(α +β) − √ β (α +β) 3 ¸ = 2 2m − V 0 2 ¸ α 2 β(α +β) 3 1/2 = β(α +β) 3 − mV 0 α 2 2 = β 4 + 3β 3 α + 3β 2 α 2 +βα 3 − mV 0 α 2 2 = x 4 + 3x 3 + 3x 2 +x − mV 0 2 α 2 Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 10 where I put x = β/α. In theory we could write down an explicit expression for the roots of this quartic in terms of mV 0 / 2 α, and then insert said expression into (12) to obtain the lowest energy attainable with this form of trial wave function. In practice, however, this would be a mess, and I can’t see any way to proceed other than numerically. Am I missing some kind of trick here? Problem 8.5 Show that as inadequate a variational trial function as Ψ(x) = C 1 − |x| a |x| ≤ a 0 |x| > a yields, for the optimum value of a, an upper limit to the ground state energy of the linear harmonic oscillator, which lies within less than 10 percent of the exact value. The first task is to evaluate the normalization constant C. 1 = C 2 a −a Ψ(x) 2 dx = 2C 2 a 0 1 − x a 2 dx = 2C 2 a 0 1 −2 x a + x 2 a 2 = 2C 2 a −a + a 3 so C = 3 2a . The harmonic oscillator hamiltonian is E = T +V = p 2 2m + mω 2 x 2 2 . (13) exptwoΨTΨ = − 2 2m a −a Ψ(x) ∂ 2 ∂x 2 Ψ(x) dx (14) Integrating by parts, = − 2 2m Ψ(x) ∂ ∂x Ψ(x) a −a − a −a ¸ ∂ ∂x Ψ(x) 2 dx ¸ Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 11 The first term vanishes... = 2 2m 3 2a a −a 1 a 2 dx = 3 2 2ma 2 (15) (16) exptwoΨV Ψ = mω 2 2 a −a x 2 Ψ(x) 2 dx = mω 2 a 0 x 2 Ψ(x) 2 dx = mω 2 3 2a a 0 x 2 − 2x 3 a + x 4 a 2 dx = mω 2 3 2a x 3 3 − x 4 2a + x 5 5a 2 a 0 = mω 2 a 2 20 (17) exptwoΨ|H|Ψ = Ψ| T |Ψ +Ψ| V |Ψ = 3 2 2ma 2 + mω 2 a 2 20 . (18) To minimize with respect to a we set the a derivative of this to zero: 0 = − 3 2 ma 3 + mω 2 a 10 or a 4 = 30 2 m 2 ω 2 a 2 = √ 30 mω . Inserting into (18), exptwoΨHΨ = 3 √ 30 ω ≈ 0.547 · ω. Of course the actual ground state energy is 0.5 · ω, so the fractional error is 0.047/0.5 < 10%. Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 12 Problem 8.6 A particle of mass m moves in a potential V (r). The n −th discrete energy eigen- function of this system, Ψ n (r), corresponds to the energy eigenvalue E n . Apply the variational principle by using as a trial function, Ψ t (r) = Ψ n (λr), where λ is a variational (scaling) parameter, and derive the virial theorem for sta- tionary states. (I have dropped the x subscripts, and I write k0 instead of k). To proceed we need to complete the square in the exponent: x2 + i(k0 − k)x 4(∆x)2 x2 − i(k0 − k)x − (k0 − k)2 (∆x)2 + (k0 − k)2 (∆x)2 4(∆x)2 2 − = − = − x − i(k0 − k)∆x − (k0 − k)2 (∆x)2 2(∆x) 1 2 x − 2i(k0 − k)(∆x)2 − (k0 − k)2 (∆x)2 = − 4(∆x)2 (2) Now we plug (??) into (??) to find: ∞ Φ(k) = (2π)−1/2 C exp[−(k0 −k)2 (∆x)2 ] −∞ exp − 1 [x − 2i(k0 − k)(∆x)2 ]2 dx 4(∆x)2 In the integral we can make the shift x → x − 2i(k0 − k)(∆x)2 and use the ∞ standard formula −∞ exp(ax2 )dx = (π/a)1/2 . The result is Φ(k) = √ 2C∆x exp −(k0 − k)2 (∆x)2 To put this into direct correspondence with the form of the wave packet in configuration space, we can write Φ(k) = √ 2C∆x exp − (k0 − k)2 4(∆k)2 where ∆k = 1/(2∆x). This is the minimum possible k width attainable for a wave packet with x width ∆x, which is why the Gaussian wave packet is sometimes referred to as a minimum uncertainty wave packet. The next step is to compute Ψ(x, t) for t > 0. Since we are talking about a free particle, we know that the momentum eigenfunctions are also energy eigenfunctions, which makes their time evolution particularly simple to write down. In the above work we have expressed the initial wave packet Ψ(x, 0) as a linear combination of momentum eigenfunctions, i.e. as a sum of terms exp(ikx), with the kth term weighted in the sum by the factor Φ(k). The wave packet at a later time t > 0 will be given by the same linear combination, but now with the kth term multiplied by a phase factor exp[−iω(k)t] describing its time evolution. In symbols we have ∞ Ψ(x, t) = −∞ Φ(k)ei[kx−ω(k)t] dk. For a free particle the frequency and wave number are connected through ω(k) = 2 hk 2 ¯ . 2m Using our earlier expression for Φ(k), we find √ 2C∆x ∞ −∞ Ψ(x, t) = exp −(k0 − k)2 (∆x)2 + ikx − i hk 2 ¯ t dk. 2m (3) Again we complete the square in the exponent: −(k0 − k)2 (∆x)2 + ikx − i hk 2 ¯ i¯ h 2 t = − [(∆x)2 + t]k 2 − [2k0 (∆x)2 + ix]k + k0 (∆x)2 2m 2m β β2 β2 k+ 4 − 4 2 α 4α 4α 2 β β = −α2 [k − 2 ]2 + 2 − γ 2α 4α = −α2 k 2 − where we have defined some shorthand: α2 = (∆x)2 + i¯ h t 2m β = 2k0 (∆x)2 + ix 2 γ = k0 (∆x)2 . = − α2 k 2 − βk + γ −γ Using this in (??) we find: Ψ(x, t) = √ 2C∆x exp β2 −γ 4α2 ∞ −∞ exp −α2 (k − β 2 ) dk. 2α2 The integral evaluates to π 1/2 /α. We have √ √ 1 α β2 −γ 4α2 1/2 Ψ(x, t) = = 2πC∆x exp (∆x)2 2πC i¯ h (∆x)2 + 2m t e−k0 (∆x) exp 2 2 (ix + 2k0 (∆x)2 )2 i¯ h 4[(∆x)2 + 2m t] This is pretty ugly, but it does display the relevant features. The important point is that term i¯ t/2m adds to the initial uncertainty (∆x)2 , so that the h wave packet spreads out with time. In the figure, I’ve plotted this function for a few values of t, with the following parameters: m=940 Mev (corresponding to a proton or neutron), ∆x=3 ˚, A k0 =0.8 ˚−1 . This value of k0 corresponds, for a neutron, to a velocity of about A 5 · 104 m/s; and note that, sure enough, the center of the wave packet travels about 5 nm in 100 fs. The time scale of the spread of this wave packet is ≈ 100 fs. 3 Gaussian wave packet example 350 t=0 s t=30 fs t=70 fs t=100 fs 300 Wavefunction (arbitrary units) 250 4 200 150 100 50 0 -1e-08 -5e-09 0 Distance (meters) 5e-09 1e-08 Problem 2.2 Express the spreading Gaussian wave function Ψ(x, t) obtained in Problem 1 in the form Ψ(x, t) = exp[iS(x, t)/¯ ]. Identify the h function S(x, t) and show that it satisfies the quantum mechanical Hamilton-Jacobi equation. In the last problem we found ∆x β2 exp −γ α 4α2 √ β2 ∆x + 2 −γ 2πC = exp ln α 4α 2πC √ Ψ(x, t) = (1) where we’ve again used the shorthand we defined earlier: α2 = (∆x)2 + h i¯ t 2m β = 2k0 (∆x)2 + ix 2 γ = k0 (∆x)2 . From (??) we can identify (neglecting an unimportant additive constant): S(x, t) = β2 h ¯ − ln α + 2 . i 4α Things are actually easier if we define δ = α2 . Then δ = (∆x)2 + i¯ h t 2m S(x, t) = 1 ¯ β2 h − ln δ + i 2 4δ Now computing partial derivatives: ∂S ∂t h ¯ 1 β 2 ∂δ − − 2 i 2δ 4δ ∂t β2 h2 1 ¯ + 2 2m 2δ 4δ h β ∂β ¯ i 2δ ∂x hβ ¯ 2δ i¯ h 2δ (2) = = − ∂S ∂x = = ∂2S ∂x2 = (3) (4) The quantum-mechanical Hamilton-Jacobi equation for a free particle is 1 ∂S ∂S + ∂t 2m ∂x 2 − 5 i¯ ∂ 2 S h =0 2m ∂x2 long before their centers ever come close to each other. the overlap integral would be tiny. we find h2 ¯ h2 β 2 ¯ h2 β 2 ¯ h2 ¯ − =0 + + 2 2 4mδ 8mδ 8mδ 4mδ so. say.4 A high resolution neutron interferometer narrows the energy spread of thermal neutrons of 20 meV kinetic energy to a wavelength dispersion level of ∆λ/λ= 10−9 . 0) chosen so that the absolute value of the overlap integral +∞ γ(0) = −∞ Ψ∗ (x)Ψ2 (x)dx 1 is very small. − Problem 2. (??). the equation is satisfied. Furthermore. their wave packets spread out on a time scale of ≈ 100 fs. and (??) into this equation. even if the neutrons are initially moving toward each other.3 Consider a wave function that initially is the superposition of two well-separated narrow wave packets: Ψ1 (x. sure enough. 20 angstroms apart. As time evolves.1 If one of those wave packets were centered in Chile and another in China. Estimate the length of the wave packets in the direction of motion. On the other hand. then they would certainly overlap a little before collapsing entirely. Will |γ(t)| increase in time.Inserting (??). for example. since the wave packets only have appreciable value within a few angstroms of their centers. the neutron wave packets plotted in the figure from problem 2. if the two neutron wave packets were each centered. as the wave packets overlap? Justify your answer. One could well imagine a situation in which the overlap integral would not increase with time. 0) + Ψ2 (x. Over what length of time will the wave packets spread appreciably? 6 . the wave packets move and spread. It seems to me that the answer to this problem depends entirely on the specifics of the particular problem. Problem 2. Consider. if we know with this precision how quickly our thermal neutrons are moving. p0 = [ 2mE ]1/2 ≈ [ 2 · (940Mev · c−2 ) · (20mev) ]1/2 = 6. To estimate the time scale of spreading of the wave packets.4 · 108 ev · (0.3 m)2 ≈ c2 · 6. 7 or . what does this tell us about the momentum dispersion level? p = dp = so dp p so the momentum uncertainty is ∆p = p0 · 10−9 = 6.1 · 10−6 ev / c = 2π¯ h λ −2π¯ h dλ λ2 dλ = 10−9 λ ≈ c · 10−10 s ≈ 30 cm. This is HUGE! So the point is. we have only the most rough indication of where in the room they might be.1 · 10−6 ev. In this case we start to get appreciable spreading when h ¯ t ≈ (∆x)2 2m t ≈ 2m(∆x)2 /¯ h 2 · 9.6 · 10−16 ev · s 6.6 · 10−16 ev · s ≈ 32 ks ≈ 10 hours.First let’s compute the average momentum of the neutrons.1 kev / c We’re given the fractional wavelength dispersion level. This implies a position uncertainty of ∆x ≈ = h ¯ ∆p 6. we can imagine that they are Gaussian packets. show that the probability current density can be expressed without an interference term involving Ψ1 and Ψ2 . I found this to be a pretty cool problem! First of all.1 If the state Ψ(r) is a superposition.Solutions to Problems in Merzbacher. Ψ(r) = c1 Ψ1 (r) + c2 Ψ2 (r) where Ψ1 (r) and Ψ2 (r) are related to one another by time reversal. we have the probability conservation equation: d ρ=− dt · J. 1999 Chapter 3 Problem 3. and to show this it suffices (by probability conservation) to show that dρ/dt has no cross terms. To show that J contains no cross terms. Quantum Mechanics. Third Edition Homer Reid March 8. We have ρ = Ψ∗ Ψ = [c∗ Ψ∗ + c∗ Ψ∗ ] · [c1 Ψ1 + c2 Ψ2 ] 1 1 2 2 = |c1 ||Ψ1 | + |c2 ||Ψ2 | + c1 c∗ Ψ1 Ψ∗ + c∗ c2 Ψ∗ Ψ2 2 2 1 1 (1) 1 . it suffices to show that its divergence has no cross terms. 2 For a free particle in one dimension. For a free particle. The relevant equation is d 1 F = F H − HF + dt i¯ h ∂F ∂t for any operator F . (The w t reminds me of “width.44) repeatedly.2 t 0 2 1 xpx + px x m 2 0− x 0 px t+ (∆px )2 2 t m2 I find it easiest to use a slightly different notation: w(t) ≡ (∆x)2 . the Hamiltonian is H = p2 /2m. and the all-important commutation relation is px = xp − i¯ . calculate the variance at time 2 t.Problem 3. Show that (∆x)2 = (∆x)2 + t 0 and (∆px )2 = (∆px )2 = (∆px ). H] i¯ h 1 xp2 − p2 x 2im¯ h 1 xp2 − p(xp − i¯ ) h 2im¯ h 1 xp2 − pxp + i¯ p h 2im¯ h 2 = = = = . We can use this to h calculate the time derivatives: d x dt 1 [x.”) Then w(t) = w(0) + t We have dw dt = = d2 w dt2 = d 2 x2 − x dt d d x2 − 2 x x dt dt d2 d x2 − 2 x 2 dt dt dw dt + t=0 1 2 d2 w t 2 dt2 t=0 +··· (2) (3) 2 −2 x d2 x dt2 (4) We need to compute the time derivatives of x and x2 . (∆x)2 ≡ (x − x t )2 t = x2 t − x t without explicit use of the t wave function by applying (3. = = = d2 x dt2 d x2 dt = = = = = = = = = = d2 x2 dt2 = = = = = = = = d3 x2 dt3 = 1 xp2 − (xp − i¯ )p + i¯ p h h 2im¯ h 1 2i¯ p h 2im¯ h p m 1 d p =0 m dt 1 [x2 . H] i¯ m h 1 − xp3 − p2 xp i¯ m2 h 1 − xp3 − p(xp − i¯ )p h i¯ m2 h 1 − xp3 − pxp2 + i¯ p2 h i¯ m2 h 1 − xp3 − (xp − i¯ )p2 + i¯ p2 h h i¯ m2 h 1 2i¯ p2 h i¯ m2 h 2 p2 m2 2 [p2 . it’s time to 3 . H] i¯ h 1 x2 p 2 − p 2 x2 2im¯ h 1 x2 p2 − p(xp − i¯ )x h 2im¯ h 1 x2 p2 − pxpx + i¯ px h 2im¯ h 1 x2 p2 − (xp − i¯ )2 + i¯ (xp − i¯ ) h h h 2im¯ h 1 x2 p2 − xpxp + 2i¯ xp + h2 + h2 + i¯ xp h ¯ ¯ h 2im¯ h 1 x2 p2 − x(xp − i¯ )p + 3i¯ xp + 2¯ 2 h h h 2im¯ h 1 2¯ 2 + 4i¯ xp h h 2im¯ h i¯ h 2 − + xp m m 2 d xp m dt 2 [xp. H] = 0 i¯ m2 h (5) (6) (7) (8) (9) Now that we’ve computed all time derivatives of x and x2 . m2 The other portion of this problem. is trivial.plug them into (3) and (4) to compute the time derivatives of w. 4 . since (∆p)2 contains expectation values of p and p2 . dw dt d d x x2 − 2 x dt dt i¯ h 2 2 − + xp − x p m m m i¯ h 2 2 − + xp − x p m 2 m 2 px − xp 2 + xp − x p m 2 m 2 2 px + xp − x p m 2 m d2 d2 d x x x2 − 2 −2 x dt2 dt dt2 2 2 2 p2 − 2 p 2 = 2 (∆p)2 m2 m m 2 = = = = = (10) d2 w dt2 = = (11) Finally. which both commute with H. the constancy of (∆p)2 . we plug these into the original equation (2) to find w(t) = w(0) + 2 1 px + xp − x p m 2 t+ (∆p)2 2 t. Integrate this equation and.3 Consider a linear harmonic oscillator with Hamiltonian H =T +V = p2 1 + mω 2 x2 . x t = x 0 cos ωt + mω (b) Derive a second-order differential equation of motion for the expectation value T − V t by repeated application of (3. calculate x2 t . 2m 2 (a) Derive the equation of motion for the expectation value x t .44) and use of the virial theorem. m d p dt m 5 . remembering conservation of energy.Problem 3. similarly to the classical oscillator. (c) Show that (∆x)2 ≡ x2 t − x t 2 t = (∆x)2 cos2 ωt + 0 + (∆p)2 0 sin2 ωt m2 ω 2 sin 2ωt 1 xp + px 0 − x 0 p 0 2 mω Verify that this reduces to the result of Problem 2 in the limit ω → 0. and show that it oscillates. (d) Work out the corresponding formula for the variance (∆p)2 . as p0 sin ωt. t (a) Again I like to use slightly different notation: e(t) = x t . Then d e(t) = dt = = = = = d2 e(t) = dt2 1 xH − Hx i¯ h 1 xp2 − p2 x 2i¯ m h 1 xp2 − p(xp − i¯ ) h 2i¯ m h 1 xp2 − (xp − i¯ )p + i¯ p h h 2i¯ m h 1 2i¯ p h 2i¯ m h p . d2 2ω 2 v(t) = − xpH − Hxp dt2 i¯ h 2ω 2 1 mω 2 = − xp3 − p2 xp + xpx2 − x3 p i¯ 2m h 2 6 → → A= x 0 p0 B= . Then 1 (T − V )H − H(T − V ) i¯ h 1 = (T − V )(T + V ) − (T + V )(T − V ) i¯ h 2 = TV − V T i¯ h 2 ω p 2 x2 − x 2 p 2 . = 2i¯ h We already worked out this commutator in Problem 2: d v(t) dt = p2 x2 − x2 p2 = − 4i¯ xp + 2¯ 2 h h so d v(t) = −2ω 2 xp + i¯ ω 2 .= = = = = = So we have 1 i¯ m h ω2 2i¯ h ω2 2i¯ h ω2 2i¯ h ω2 2i¯ h −ω 2 pH − Hp px2 − x2 p (xp − i¯ )x − x2 p h x(xp − i¯ ) − i¯ x − x2 p h h −2i¯ x h x . d2 e(t) = −ω 2 e(t) dt2 with general solution e(t) = A cos ωt + B sin ωt. h dt 2 = −2ω xp + ω 2 xp − px = −ω 2 xp + px Next. mω (12) (13) . The coefficients are determined by the boundary conditions: e(0) = x 0 p0 e (0) = m (b) Let’s define v(t) = T − V t . we can use (12) evaluated at t = 0 to determine B: −ω 2 xp + px so B=− 0 + i¯ ω 2 = 2ωB h ω xp + px 0 . H is constant in time. Also. mω 2 t Since H does not depend explicitly on time. Evaluating at t = 0 gives A= T 0 p2 mω 2 2 − x m 2 (14) − V 0 . mω (15) .The bracketed expressions are xp3 − p2 xp = = = xpx2 − x3 p = = = xp3 − p(xp − i¯ )p h 2i¯ p2 h h h xp3 − (xp − i¯ )p2 + i¯ p2 x(xp − i¯ )x − x3 p h −2i¯ x2 h x2 (xp − i¯ ) − i¯ x2 − x3 p h h and plugging these back into (13) gives d2 v(t) = = −4ω 2 dt2 = −4ω 2 v(t) with solution v(t) = A cos 2ωt + B sin 2ωt. 2 The next task is to compute x2 t : x2 t = = = 2 V t mω 2 1 H − (T − V ) mω 2 1 [ H t − v(t)] . For v(t) we can use (14): x2 t = = = 1 mω 2 1 mω 2 ω xp + px 0 sin 2ωt 2 ω xp + px 0 2 T 0 sin2 ωt + 2 V 0 cos2 ωt + sin 2ωt 2 T 0 + V 0 −[ T 0 − V 0 ] cos 2ωt + p2 0 sin2 ωt + x2 m2 ω 2 0 cos2 ωt + 7 1 xp + px 2 0 sin 2ωt . 8 . 2m Derive expressions for these densities in the momentum representation. Subtracting from (15) gives (∆x)2 = x2 − x t 2 = + + x2 1 0 − x 2 0 cos2 ωt 0 m2 ω 2 p2 − p 0 2 0 1 xp + px 2 − x 0 p 0 sin 2ωt 0 = (∆x)2 cos2 ωt + 0 1 (∆p)2 0 sin2 ωt + xp + px m2 ω 2 2 − x 0 p 0 sin 2ωt . and (sin 2ωt/ω) → 2. The first one is trivial: δ(r − r0 ) = For the second one. mω As ω → 0. (sin2 ωt/ω 2 ) → 1. [Ψ∗ δ(r − r0 )Ψ + Ψ∗ δ(r − r0 ) Ψ] dr The gradient operator in the first term operates on everything to its right: =− i¯ h 2m [Ψ∗ Ψ δ(r − r0 ) + 2δ(r − r0 )Ψ∗ Ψ] dr. as needed to ensure matchup with the result of Problem 2.4 Prove that the probability density and the probability current density at position r0 can be expressed in terms of the operators r and p as expectation values of the operators ρ(r0 ) → δ(r − r0 ) j(r0 ) → 1 [pδ(r − r0 ) + δ(r − r0 )p] . 1 i¯ h pδ(r − r0 ) + δ(r − r0 )p = − 2m 2m Ψ∗ (r)δ(r − r0 )Ψ(r)dr = Ψ∗ (r0 )Ψ(r0 ) = ρ(r0 ).(c) Earlier we found that x x t 2 t = = p0 sin ωt mω 2 p x 2 cos2 ωt + 2 02 sin2 ωt + x 0 m ω x 0 cos ωt + 0 p 0 sin 2ωt. cos2 ωt → 1. Problem 3. (b) Prove that W (r . p ) = 1 r exp(−ip ·x /¯ )Ψ∗ r − h 3 (2π¯ ) h 2 Ψ r + r 2 dr . Problem 3. the Wigner distribution function is defined as W (r . p )dr dp . p )dp = |Ψ(r )|2 and that the expectation value of a function of the operator r in a normalized state is f (r) = f (r )W (r . p ). (d) Show that the probability density ρ(r0 ) at position r0 is obtained from the Wigner distribution function with ρ(r0 ) → f (r) = δ(r − r0 ).5 For a system described by the wave function Ψ(r ). p ) is a real-valued function. (c) Show that the Wigner distribution function is normalized as W (r . defined over the six-dimensional “phase space” (r . p )dr dp = 1.Here we can use the identity = − f (x)δ (x − a)dx = −f (a) : i¯ h |− (Ψ∗ Ψ) + 2Ψ∗ Ψ|r=r0 2m i¯ h = |Ψ Ψ∗ − Ψ∗ Ψ|r=r0 2m = j(r0 ). (a) 9 . (a) Show that W (r . Third Edition Homer Reid June 24. Can the same result be obtained directly by matrix algebra from a knowledge of the matrix elements of px ? For the harmonic oscillator.1 Calculate the matrix elements of p2 with respect to the energy eigenfunctions of the x harmonic oscillator and write down the first few rows and columns of the matrix. 2000 Chapter 5 Problem 5. Quantum Mechanics. h ¯ The matrix element of x2 is then < Ψn |x2 |Ψk >= 1 n+k n!k! 2 1/2 mω hπ ¯ 1/2 ∞ x2 exp( −∞ mω 2 x )Hn ( h ¯ mω x)Hk ( h ¯ mω x) dx. we have H= so p2 = 2mH − m2 ω 2 x2 x and 1 < Ψn |p2 |Ψk >= 2m¯ ω(n + )δnk − m2 ω 2 < Ψn |x2 |Ψk > . h x 2 The nth eigenfunction is Ψ(x) = 1 n n! 2 1/2 1 2 1 p + mω 2 x2 2m x 2 (1) mω hπ ¯ 1/4 exp(− mω 2 x )Hn ( 2¯ h mω x). h ¯ 1 .Solutions to Problems in Merzbacher. from (1).Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 2 The obvious substitution is u = (mω/¯ )x. we have Comparing termwise with (3). 1 1/2 h < Ψn+2 |p2 |Ψn > = − [(n + 2)(n + 1)] (mω¯ ) 2 1 < Ψn |x2 |Ψn > = (2n + 1)(mω¯ ). together with any term from the last series. n! k! p! 1 1 + 2λ(s + t) + (2λ)2 (s + t)2 + · · · 2 1 1 + (2st) + (2st)2 + · · · 2 (3) There are two ways to get a λ2 term out of this.k. we can read off  √ . with which we obtain h < Ψn |x2 |Ψk >= 1 2n+k n!k!π 1/2 h ¯ mω ∞ u2 e−u Hn (u)Hk (u) du.p √ 2 sn tk (2λ)p = π eλ +2λ(s+t)+2st . Writing down only terms obtainable in this way. h 2 . n=k−2  (n + 2)!2n π √ n!2n−1 π(1 + 2n) . we have = = ··· + = ··· + = ··· + √ πλ2 1 + 2(s + t)2 1 1 + (2st) + (2st)2 + · · · + · · · 2 ∞ √ 1 π 1 + λ 2 + λ4 + · · · 2 √ 2 πλ 1 + 2s2 + 2t2 + 4st √ ∞ j=0 1 (2st)j + · · · j! +··· πλ2 j=0 2 j j 2 2j+1 j j+2 2j+2 j+1 j+1 s t + sj+2 tj + s t + s t j! j! j! j! j j+1 Plugging this into (2). One way is to take the λ2 term from the first series and the 1 from the second series. along with any term from the last series. n = k Ink2 =  0 . The useful formula is Inkp n. The second way is to take the 1 from the first series and the λ2 term from the second series. 2 mω h ¯ mω Finally. −∞ 2 (2) The integral is what Merzbacher calls Inkp with p = 2. < Ψn+2 |x2 |Ψn > = < Ψn |x2 |Ψn > = 1 1/2 [(n + 2)(n + 1)] 2 h ¯ 1 (2n + 1) . otherwise. In this case it is less absurd since there’s no classical interpretation of the offdiagonal matrix elements of an operator. Problem 5. but it’s still weird. dx 1 mω 2 x2 2 so the virial theorem says that T = = U in accord with what we concluded earlier. Problem 5. so we have T = U . dx d V (x) = mω 2 x. We found the expectation values of x2 in the last problem. x d V (x) . the virial theorem is supposed to be saying U = 2 T = In this case. diagonal matrix element) of the square of an observable operator to come out negative.3 Calculate the expectation value of x4 for the nth energy eigenstate of the harmonic oscillator. in another sense it seems inescapable that p2 should have a negative off-diagonal matrix element here. The kinetic energy expectation value must of course make up the difference.2 Calculate the expectation values of the potential and kinetic energies in any stationary state of the harmonic oscillator. However. It would be absurd for the expectation value (i. Compare with the results of the virial theorem. The potential energy operator is U = mω 2 x2 /2. 1 n! 2n mω hπ ¯ 1/2 ∞ Ψ n x4 Ψ n = x4 exp(− −∞ mω 2 2 x )Hn ( h ¯ mω x) dx h ¯ . and H is just a sum of x2 and p2 terms.. so p2 must have a negative matrix element to cancel out the positive matrix element of x2 . On the other hand.e. so 1 hω ¯ 1 mω 2 x2 = (n + ) 2 2 2 which is just half the energy expectation value. but x2 has a nonvanishing matrix element. because the off-diagonal matrix elements of H must vanish in the energy eigenfunction basis.Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 3 I find it kind of confusing that the matrix element for p2 comes out negative in the first case. In the second line. 1.p √ 2 sn tk (2λ)p = π eλ +2λ(s+t)+2st .4 For the energy eigenstates with n=0. The probability of finding the particle with coordinate greater than this is −A ∞ P (|x| > A) = −∞ Ψ2 (x) dx + n A Ψ2 (x) dx n . 2 2 so (4) is Ψ n x4 Ψ n = 3 2 h ¯ mω 2 1 ( + n + n2 ). Equating powers in (5). I further limited it to terms that also contain the same number of powers of s as t.k. and 2. but this time we’ll need to write out the expansion a little further than before. √ 1 3 Inn4 = 2n n! π( + n + n2 ). Inkp n.Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 2 4 = 1 √ n! 2n π h ¯ mω ∞ 2 u4 e−u Hn (u) du −∞ 2 (4) For the integral we want to use (3) again. compute the probability that the coordinate of a linear harmonic oscillator in its ground state has a value greater than the amplitude of the classical oscillator of the same energy. The classical amplitude is A = (2E)/(mω 2 ). 2 Problem 5. I only wrote out terms that can be combined to give a factor of λ4 . n! k! p! (5) ∞ = = √ √ 1 π 1 + λ 2 + λ4 + · · · 2 1 π 1 + λ 2 + λ4 + · · · 2 √ √ √ ∞ 1 1 1 + · · · + (2λ)2 (s + t)2 + · · · + (2λ)4 (s + t)4 + · · · 2 4! ∞ j=0 (2st)j j! 1 + · · · + 4λ2 st + · · · + 4λ4 (st)2 + · · · j−1 j=0 (2st) j! j = ··· + = ··· + = ··· + π j=0 ∞ (λ4 )(st)j (λ4 )(st)j j=0 ∞ 2 2j j! 2j j! j! +4 2 2 +4 (j − 1)! (j − 2)! j−1 j−2 +··· π π j=0 1 + 2j + j(j − 1) 2 1 + j + j2 2 (λ4 )(st)j In the first line. Check the results in the classical and low-temperature limits.] Suppose we denote the number of oscillators in the nth energy state by Nn .5 Show that if an ensemble of linear harmonic oscillators is in thermal equilibrium. [Hint: Equation (5. h Nn = N0 e−n¯ ω/kT . governed by the Boltzmann distribution. 1 √ π 1 √ 2π ∞ 2 Pn=0 (|x| > A) = = = = e−u du 1 ∞ e 0 −p2 /2 √ 2 dp − e−p 0 2 /2 dp √ π √ 1 √ − 2π · erf( 2) 2 2π √ 1 − erf( 2) ≈ 0. ¯ In particular. so Ψ2 has even parity. the ratio of the number of oscillators in the n th state to the number of oscillators in the nth state is Nn h = e−(n −n)¯ ω/kT . In going from the second to last line n to the last line.43) may be used. . we noted that the energy of the nth eigenstate is hω(n + 1/2).Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 ∞ 5 = 2 A Ψ2 (x) dx n mω x) dx h ¯ = = = mω 1/2 ∞ 2 mω 2 2 exp(− x )Hn ( n n! 2 hπ ¯ h ¯ A ∞ 2 2 2 √ √ e−u Hn (u) du n! 2n π 2E/¯ ω h n! 2 √ n−1 1 ∞ π √ 2n+1 2 e−u Hn (u) du 2 In going from the first line to the second we invoked the fact that Ψn has either even or odd parity.31 2 Problem 5. If the ensemble is in thermal equilibrium. for any n. Plot the width of the distribution as a function of temperature. the probability per unit length of finding a particle with displacement x is a Gaussian distribution. Nn In particular. Homer Reid’s Solutions to Merzbacher Problems: Chapter 5 6 The probability of finding a particle between x and dx is ∞ P (x)dx = n=0 Cn |Ψn (x)|2 dx mω hπ ¯ 1/2 = C0 exp(− h e−n¯ ω/kT 2 mω 2 x ) Hn ( h ¯ n! 2n n=0 ∞ mω x)dx h ¯ This can be summed using the Mehler formula with t = exp(−¯ ω/kT ) : h P (x) = C0 = C0 mω hπ ¯ mω hπ ¯ 1/2 exp(− √ mω 2 x ) h ¯ 1/2 1 1 − t2 h ¯ 2mω 1 2t exp 2 1+t 1−t 1 − t mω 2 exp − x 1+t h ¯ √ mω 2 x h ¯ This is a Gaussian distribution with variance σ2 = = = 1+t 1−t h 1 + e−¯ ω/kT h ¯ h 2mω 1 − e−¯ ω/kT hω ¯ h ¯ coth 2mω 2kT . choosing (2ma2 V0 )1/2 = (3π/2) . Plot the transmission coefficient as a function E/V0 (up to E/V0 = 3). Quantum Mechanics. He obtains the result M11 = where κ= and 2m(V0 − E) 2 cosh 2κa + i sinh 2κa e2ika 2 (1) κ k − .1 Obtain the transmission coefficient for a rectangular potential barrier of width 2a if the energy exceeds the height V0 of the barrier. Merzbacher treats this problem for the case where the particle’s energy is less than the potential barrier.Solutions to Problems in Merzbacher. In the text. 2000 Chapter 6 Problem 6. and we write 2m(E − V0 ) κ = iβ = i 2 = so that (2) becomes = iβ k − =i k iβ 1 k β + k β ≡ iλ . (2) k κ We can re-use the result (1) for the case where the energy is greater than the potential barrier. Third Edition Homer Reid June 24. To do this we note that κ becomes imaginary in this case. We have βa = = λ2 − 1 sin2 βa 4 1 β4 + k4 − 4β 2 k 2 2 (β 2 − k 2 )2 4β 2 k 2 V02 4E(E − V0 ) 1 4γ(γ − 1) 1/2 1/2 sin2 βa 1/2 sin2 βa 1/2 sin2 βa 1/2 sin2 βa 1/2 4γ(γ − 1) + sin2 βa 4γ(γ − 1) 2m (E − V0 ) h2 1/2 1/2 a 2mV0 a2 (γ − 1)1/2 h2 3π (γ − 1)1/2 = 2 so the transmission coefficient is T = 1 4γ(γ − 1) = 2 |M11 | 4γ(γ − 1) + sin2 3π (γ − 1)1/2 2 This is plotted in Figure 1. sinh ix = i sin x we have M11 = and |M11 | = cos2 2βa + λ2 sin2 βa 4 1/2 cos 2βa − iλ sin 2βa e2ika .Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 2 Plugging into (1) and noting that cosh ix = cos x. . 2 = 1+ = 1+ = 1+ = 1+ = 1+ = where γ = E/V0 . 8 0.6 0.1.4 0.9 0.3 0.Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 3 1 0.5 3 3.2 0.5 0.5 2 E/V0 2.7 Transmission Coefficient 0.5 1 1.1 0 0 0. .5 4 Figure 1: Transmission coefficient versus E/V0 for Problem 6. we obtain γ sin k1 a = Ceik2 a + De−ik 2 a 2 k1 γ cos k1 a = ik2 [Ceik2 a − De−ik a ] Combining these yields 1 −ik2 a k1 γe cos k1 a + i sin k1 a 2i k2 1 k1 D = − γe+ik2 a cos k1 a − i sin k1 a 2i ik2 C= It’s now convenient to write k1 cos k1 a + i sin k1 a = αeiδ k2 where α2 = k1 k2 2 (3) (4) (5) cos2 k1 a + sin2 k1 a (6) .   ik2 x Ce + De−ik2 x . V = −V0 for a ≥ x ≥ 0. does not hold at x = 0 because the potential is infinite there. Eventually. that the derivative of Ψ also be continuous. Next applying continutity of Ψ and its derivative at x = a. we see we must take A = −B. k1 = 2m(E + V0 ) 2 x ≤0 0 ≤x ≤ a≤ .2 Consider a potential V = 0 for x > a. so Ψ(x) = γ sin k1 x for 0 ≤ x ≤ a.  Ψ(x) = Aeik1 x + Be−ik1 x .Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 4 Problem 6. a x with k2 = 2mE 2 Applying the requirement that Ψ be continuous at x = 0. we could apply the normalization condition on Ψ to find γ if we wanted to. The other standard requirement. and V = +∞ for x < 0. Hence γ is undetermined as yet. Show that for x > a the positive energy solutions of the Schr¨dinger o equation have the form ei(kx+2δ) − e−ikx Calculate the scattering coefficient |1 − e2iδ |2 and show that it exhibits maxima (resonances) at certain discrete energies if the potential is sufficiently deep and broad. We have   0. 2 I have plotted this for β = 25. the scattering coefficient is k1 k2 2 k1 cos2 k1 a + 2i k2 cos k1 a sin k1 a − sin2 k1 a k1 k2 2 2 |1 − e2iδ |2 = 1 − cos2 k1 a + sin2 k1 a 2 = k1 2 sin k1 a sin k1 a − i k2 cos k1 a k1 k2 2 cos2 k1 a + sin2 k1 a (7) = k1 k2 4 sin2 k1 a 2 cos2 k1 a + sin2 k1 a We have k1 k2 2 = E + V0 = E 2 1+ 1 λ = β(λ + 1)1/2 k1 a = 2ma2 (E + V0 ) where λ = E/V0 and β = (2ma2 V0 / 2 )1/2 in Merzbacher’s notation.Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 5 and δ = tan−1 k2 tan k1 a k1 so that α contains magnitude information. Then we can rewrite (3) and (4) as γα −ik2 a iδ e e 2i γα D = − e+ik2 a e−iδ 2i C= Then the expression for the wavefunction to the left of x = a becomes Ψ(x) = Ceik2 x + De−ik2 x (x > a) γα ik2 (x−a) iδ −ik2 (x−a) −iδ e e −e e = 2i γα −iδ ik2 (x−a) 2iδ e = e e − e−ik2 (x−a) . Then the scattering coefficient (7) is scattering coefficient = 1 2 λ) (1 + 4 sin2 [β(λ + 1)1/2 ] cos2 [β(λ + 1)1/2 ] + sin2 [β(λ + 1)1/2 ] In Figure 6. . 2i Using (5) and (6). while δ represents phase information. 5 1 1.5 3 Figure 2: Scattering coefficient versus E/V0 for Problem 6.5 2 1.5 E/V0 2 2.2.5 0 0 0. .5 3 Scattering coefficient 2.5 1 0.Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 6 4 3. 2 dx then integrate from −a − to −a + and take the limit as → 0. In each region.Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 7 Problem 6. (10) Applying this condition to the wavefunction (8) yields kB sinh(−ka) − kAe−ka = − 2mg 2 B cosh(−ka). 2a. we can divide the x axis into three regions. Explain the features of the plot. so B = E = 0. o but with energy E < 0 since we’re looking for bound states. Compute and plot the energy levels in units of 2 /ma2 as a function of the dimensionless parameter mag/ 2 . a ≤ x. obtain a simple formula for the splitting ∆E between the ground state (even parity) energy level. In this problem. we can write down the Schr¨dinger equao tion.3 A particle of mass m moves in the one-dimensional double well potential V (x) = −gδ(x − a) − gδ(x + a). a ≤ x. −a ≤ x ≤ a Ψ(x) = (8)  Ae−kx . Also. E− . we have  x ≤ −a  Aekx + Be−kx . Now. Cekx + De−kx . This gives dΨ dx −a+ −a− =− 2mg 2 Ψ(−a). B cosh(kx). and the excited (odd parity) energy level. E+ . the wavefunction is just the solution to the free-particle Schr¨dinger equation. the wavefunction can’t blow up at infinity. Considering first the even parity solution. If g > 0. between the wells. first of all. since the potential in this problem has mirror-reversal symmetry. (9) Since the potential becomes infinite at x = −a. Matching the value of the wavefunction at x = −a gives Ae−ka = B cosh(−ka). Instead. In the limit of large separation. . −a ≤ x ≤ a Ψ(x) =  Eekx + F e−kx . the normal derivative-continuity condition doesn’t hold there. Putting k = 2mE/ 2 .  x ≤ −a  Aekx . the wavefunction will have definite parity. obtain transcendental equations for the bound-state energy eigenvalues of the system. d2 2m 2m Ψ(x) = 2 V (x)Ψ(x) − 2 EΨ(x). Sketch the total potential energy of the system as a function of |X|. Matching values at x = −a gives Ae−ka = B sinh(−ka) and applying condition (10) gives kB cosh(−ka) − kAe−ka = − kB cosh(ka) + kB sinh(ka) = coth(ka) = β −1 ka 2mg 2 2 B sinh(−ka) 2mg B sinh(ka) so this is the condition that determines the energy of the odd parity state. −a ≤ x ≤ a (11) Ψ(x) =  −Ae−kx . coth(ka). On the other hand. the coth curve crosses the β/(ka) − 1 curve at a lower value of ka than the tanh curve. if the potential energy of the “molecule” is taken as E± (|X|).4 Problem 3 provides a primitive model for a one-electron linear diatomic molecule with interatomic distance 2a = |X|. kB sinh(−ka) − kB cosh(−ka) = − or tanh(ka) = β 2mg −1= −1 2k ka 2mg 2 B cosh(−ka) with β = 2mag/ 2 . that means that the energy eigenvalue for the odd parity state is smaller in magnitude (less negative) than the even parity state. Problem 6. supplemented by a repulsive interaction λg/|X| between the wells (“atoms”). and β/(ka) − 1 for the case β = 3. a ≤ x. This equation determines the energy eigenvalue of the evenparity state. the odd parity state looks like  x ≤ −a  Aekx .Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 8 Substituting from (9). for a sufficiently small value of λ. B sinh(kx). the system (“molecule”) is stable if the particle (“electron”) is in the even parity state. As expected. Show that. . which will be the ground state. In Figure (3) I have plotted tanh(ka). 3 with β = 3.6 1.8 2 Figure 3: Graphical determination of energy levels for Problem 6. .5 1 0.Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 9 PSfrag replacements 2 tanh(ka) coth(ka) β/(ka) − 1 1.4 ka 1.5 0 1 1.2 1. Matching values at x = −a. D → F . so  ikx −ikx x ≤ −a  Ae + Be  ikx −ikx Ψ(x) = Ce + De −a≤x≤a   ikx −ikx a≤x Ee + F e 2mE/ 2 . C → E. as before. Applying this to the wavefunction in (12). applying the matching conditions to the wavefunction at x = +a will give two equations exactly like (13) and (14).Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 10 Problem 6. B → D. we have ik[Ce−ika − Deika − Ae−ika + Beika ] = − Combining (13) and (14) yields β 2ka e B ika β −2ka β D=− B e A+ 1− ika ika C= 1+ β ika A+ (15) (16) 2mg 2 [Ae−ika + Beika ]. Making these substitutions in (15) and (16) we obtain E= β 2ka e D ika β 2ka β F =+ D e C + 1+ ika ika 1− C− β ika (17) (18) . Now.) e Now we’re assuming that the energy E is positive. (14) with β = mag/ 2 as before. calculate the transmission coefficient and show that it exhibits resonances. (Note the analogy between the system and the Fabry-Perot ´talon in optics. but with the substitutions A → C. we have the derivative condition dΨ dx x=−a+ x=a− (12) with k = (13) =− 2mg 2 Ψ(−a) where g is now negative. and a → −a.5 If the potential in Problem 3 has g < 0 (double barrier). we have Ae−ika + Beika = C −ika + Dika Also. Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 11 PSfrag replacements 1 0.6 T (k) 0.8 0. or 1 T = 2 2 β β + 1 (1 − cos(4ka)) 1 + 2 ka ka In Figure 4 I have plotted this for β = 15.5 0.4 0. .4 with β = 15.9 0. we have E = 1+ F = 2β ka β ika 1+ 2 (e−4ika − 1) A + + β ika sin 2ka A + 1 + 2β ka β ika 1− 2 β ika sin 2ka B (e4ika − 1) B This is the M matrix. Combining equations (15) through (18).1 0 0 5 10 15 ka 20 25 30 Figure 4: Transmission coefficient in Problem 6.2 0. and the transmission coefficient is given by T = 1/|M11 |2 .7 0.3 0. which relates the amplitudes of the incident and reflected plane waves on the left of the origin (x < 0) to the amplitudes on the right (x > 0). (c) Show that the S matrix is unitary and that the elements of the S matrix satisfy the properties expected from the applicable symmetry considerations. (a) Work out the matrix M . the delta function at the origin gives rise to a discontinuity in the derivative of the wavefunction as before: dΨ dx so ik2 (C − D) − ik1 (A − B) = or C −D = 0+ = 0− 2mg 2 Ψ(0) 2mg 2 (A + B) (20) k1 2mg (A − B) + (A + B). x≤0 x≥0 E k2 = 2m 2 (E − V0 ). which have the same energy (buf different velocities). Ce ik2 x + De −ik2 x .Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 12 Problem 6. Matching values at x = 0 gives C +D =A+B (19) Also. (b) Derive the elements of the matrix S. which relates incoming and outgoing amplitudes. k2 ik2 2 .6 A particle moves in one dimension with energy E in the field of a potential defined as the sum of a Heaviside step function and a delta function: V (x) = V0 η(x) + gδ(x) (with V0 and g > 0) The particle is assumed to have energy E > V0 . (d) Calculate the transmission coefficient for particles incident from the right and for particles incident from the left. We have Ψ(x) = with k1 = 2m 2 Aeik1 x + Be−ik1 x . Homer Reid’s Solutions to Merzbacher Problems: Chapter 6 13 Adding and subtracting (19) and (20). which is defined by B C = S11 S21 S12 S22 A D Since we already know the M coefficients. we can read off 1 2mg 2mg 1 k1 k1 A+ B + + 1+ 1− 2 k2 ik2 2 2 k2 ik2 2 2mg 2mg 1 1 k1 k1 − − D= A 1+ B. . 1− 2 k2 ik2 2 2 k2 ik2 2 C= We could also write this as C D = M11 ∗ M12 M12 ∗ M11 A B Or instead of the M matrix we could use the S matrix. we can calculate the elements of the S matrix from the formula S11 S21 = − M12 ∗ 11 1 = M∗ 11 M∗ 1 S12 = M ∗ 11 M S22 = + M12 ∗ 11 However. this is tedious and long and boring and I don’t want to do it. we will construct a series of functions S0 (x). Third Edition Homer Reid April 5. where S0 is the solution of (1) with 0 on the left hand side. To be specific. we compute Sn (x) and use that as the source term on the LHS of (1) to calculate Sn+1 (x). S1 (x). S1 is a solution with S0 on the left hand side. In 1 . but we obtain guidance from the observation that. In other words. · · · . 2001 Chapter 7 Before starting on these problems I found it useful to review how the WKB approximation works in the first place. The Schr¨dinger equation is o − or d2 Ψ(x) + V (x)Ψ(x) = EΨ(x) 2m dx2 2 d2 Ψ(x) + k 2 (x)Ψ(x) = 0. Ψ(x) = AeiS(x)/ in which case the Schr¨dinger equation becomes o i S (x) = [S (x)]2 − 2 2 k (x). at the nth step in the approximation sequence (by which point we have computed Sn (x)). Then we compute the second derivative of Sn+1 (x) and use this as the source term for calculating Sn+2 . Quantum Mechanics. (1) This equation can’t be solved directly.Solutions to Problems in Merzbacher. S(x) = ±kx. so that S vanishes. and we may take S = 0 as the seed of a series of successive approximations to the exact solution. for a constant potential. and so on ad infinitum. and so on. For a nonconstant but slowly varying potential we might imagine S (x) will be small. dx2 We postulate for Ψ the functional form k(x) ≡ 2m 2 [E − V (x)]. so that S0 > 0. in which case our approximation sequence S0 . S3 . the sequence of approximations terminates at 0th order with S0 being an exact solution. 2k(x) ik (x) 2k 2 (x) (7) k(u)du + where a is some point chosen such that the approximation (7) is valid in the full range a < x < x. if we choose the plus sign in (5) but the minus sign in (3). With the two ± signs here. But if we chose. but in practice it seems the approximation is always terminated at S1 . 0 = [S0 (x)]2 − 2 2 2 2 k (x) k (x) k (x) (2) (3) (4) i S0 = [S1 (x)] − 2 2 i S1 = [S2 (x)] − ···· 2 2 Equation (2) is clearly solved by taking x S0 (x) = ± k(x) ⇒ S0 (x) = S00 ± k(x )dx −∞ (5) for any constant S00 . So we choose either both plus signs or both minus signs in (3). k(x) is constant. S1 (x) = S1 (a) ± = S1 (a) ± x k (u) i dx 2 a k(u) a x i k(x) k(u)du + ln 2 k(a) a x ik (x) 2k 2 (x) or S1 ≈ − k(x) 1 − i k (x) . We could go on to compute S2 . S1 . · · · has little hope of converging. and. But let’s think a little about the ± signs in this equation.Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 2 symbols. Indeed. etc. whence our two choices are S1 = + k 2 (x) + ik (x) or S1 = − k 2 (x) − ik (x). say. By extension.. The ± sign under the radical comes from the two choices of sign in (5). Then S0 (x) = ± k (x). (6) If V (x) is constant. so (3) is S1 (x) = ± k 2 (x) ± ik (x). if V (x) is not constant but changes little over one particle wavelength. we would also expect that S1 > 0. we have k (x)/k 2 (x) 1. the plus sign in that equation. so S1 differs from S0 by an amount at least as large as S0 . we appear to have four possible choices for S1 . . then S0 and S1 have opposite sign. as we observed before. so we may expand the radicals in (6): S1 ≈ k(x) 1 + or S1 ≈ ± k(x) + Integrating. according to the derivative of Ψ at x = a. one whose phase increases with increasing o x (positive derivative). you can just multiply it by G± (x.Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 3 The wavefunction at this order of approximation is Ψ(x) = exp(iS1 (x)/ ) = eiS1 (a)/ = Ψ(a) e±i Rx a k(u)du eln k(x)/k(a) −1/2 k(a) ±i R x k(u)du e a k(x) = Ψ(a)G± (x. we need to know both Ψ and Ψ at a nearby point x = a. So to use (8) to obtain values for Ψ at a point x. but the requirement the dΨ/dx be continuous at x = a means that only one or the other will do. a) ≡ k(a) ±i R x k(u)du e a . Equation (8) seems to be saying that we can use either G+ or G− to get to Ψ(x) from Ψ(a). let’s investigate the equations leading up to (8). But this doesn’t seem quite right: Schr¨dinger’s equation o is a second-order differential equation. If the approximation (7) makes sense. if you know what Ψ is at some point a. κ(x) (12) 2m 2 [V (x) − E] (11) . otherwise the overall wave function will have a discontinuity in its first derivative at x = a. as should be the case for a second-order differential equation. Indeed. in using (8) to continue Ψ from a to x we must choose the appropriate propagator—either G+ or G− . then there are two solutions of Schr¨dinger’s equation at x = a. so it’s κ(x) = −ik(x) = and Ψ(x) = C 1 ± R x κ(u)du e a . and one whose phase decreases. a) to find out what Ψ is at x. If we want to de-emphasize this nature of the solution with the propagator we may write 1 ±i R x k(u)du Ψ(x) = C e a (10) k(x) where C = Ψ(a) useful to define k(a). a) where (8) G± (x. k(x) (9) We have written it this way to illustrate that the function G(x. k(x) is imaginary. but (8) seems to be saying that we need only one initial condition—the value of Ψ at x = a—to find the value of Ψ at other points. To clarify this subtle point. a) is kind of like a Green’s function or propagator for the wavefunction. in the sense that. In regions where V (x) > E. γ . gives us no way of determining the value of Ψ at that one starting point. The WKB method. So we need to identify the regions of invalidity of the WKB approximation and do a more accurate solution of the Schr¨dinger equation o there. Furthermore. In the neighborhood of x0 we may expand V (x): V (x) = E + (x − x0 )V (x0 ) + · · · Then the Schr¨dinger equation becomes o d2 2m Ψ(x) − 2 V (x0 )(x − x0 )Ψ(x) = 0. which is true when k ≈ 0 but k = 0. because we can’t carry the propagator across regions of invalidity.. How do we find such points? Well. even if we know Ψ at one point within a region of validity. of which there will be at least one adjacent to each region of validity.e. if we keep only the first (linear in x) term in the series. So basically what we need is a way of finding one starting value for Ψ(x) in every region of validity of the WKB approximation. which happens near a classical turning point of the motion—i. we can’t use (8) to determine Ψ in other. But near such a point we may expand V (x) − E in a Taylor series around the point x0 . so that V (x0 ) = E.Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 4 If we have a region of space in which the WKB approximation is valid. The WKB approximation breaks down when k /k 2 1 ceases to hold. because we can use the propagator (9) to get from that one point to every other point within the region. one sure-fire way to get starting points in regions of validity is to identify regions of invalidity. knowing the value of Ψ (and its derivative) at one point within the region is equivalent to knowing it everywhere. suppose the point x0 is a classical turning point of the motion. however. and then get values of Ψ at the boundaries of the regions of invalidity—which will also count as values in the regions of validity. a point x0 at which V (x0 ) = E. To do this. we arrive at a Schr¨dinger equation which o we can solve exactly in the vicinity of x0 . 2 dx The useful substitution here is u(x) = γ(x − x0 ) so x(u) = If we define Φ(u) = Ψ(x(u)) then dΦ dΨ dx 1 = = Ψ (x(u)) du dx du γ 1 d2 Φ = 2 Ψ (x(u)) du2 γ γ≡ 2m 2 1/3 (13) (14) V (x0 ) u + x0 . nonadjacent regions. Case 1: V (x0 ) > 0. (17) To simplify these. For points close to the turning point on the left side. du2 The solution to this differential equation is Φ(u) = β1 Ai(u) + β2 Bi(u) so the solution to the Schr¨dinger equation (14) is o Ψ(x) = β1 Ai γ(x − x0 ) + β2 Bi γ(x − x0 ) . . γ > 0. while the right of the turning point is classically forbidden. for points close to the turning point on the left side we have x > x0 . Since V (x0 ) > 0. 3 On the other hand. so for x < x0 (??) holds. which means that V (x) < E for x < x0 . we need to consider two possible kinds of turning point. so γ(x − x0 ) > 0. Hence the region to the left of the turning point is the classically accessible region. and V (x) > E for x > x0 . In this region.Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 5 so (14) becomes γ2 or d2 Φ(u) − γ 3 (x − x0 )Φ(u) = 0 du2 d2 Φ(u) − uΦ(u) = 0. For γ(x − x0 ) 1 we have the asymptotic expression −1/4 3/2 β1 − 2 |γ(x−x0 )|3/2 2 e 3 + β2 e+ 3 |γ(x−x0 )| 2 (15) Ψ(x) ≈ π −1/2 [γ(x − x0 )] and for γ(x − x0 ) (16) −1 we have −1/4 Ψ(x) ≈ π −1/2 |γ(x − x0 )| β1 cos 2 π 3/2 |γ(x − x0 )| − 3 4 π 2 3/2 − β2 sin |γ(x − x0 )| − 3 4 . In this case the potential is increasing through the turning point at x0 . k(x) = so |γ(x − x0 )| and x0 x0 −1/4 2m 2 [E − V (x)] ≈ 2m 2 1/2 V (x0 − x)1/2 = γ 3/2 (x0 − x)1/2 = γ k(x) 2 3/2 γ (x0 − x)3/2 3 (19) (18) k(u)du = γ 3/2 x x (x0 − x)1/2 du = 2 = |γ(x − x0 )|3/2 . κ(x) (22) Using (18) and (19) in (17). x > x0 (26) (27) − β2 sin k(u)du − So. x < x0 x < x0 (23) (24) − β2 sin Ψ(x) = k(u)du − Rx R 1 κ(u)du − x κ(u)du β1 e x 0 + β 2 e x0 . and to divide space . the first step is to identify the classical turning points of the motion. Since V (x0 ) < 0.Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 6 κ(x) = 2m 2 V (x) − E ≈ 2m 2 1/2 V (x0 )(x − x0 ) = γ 3/2 (x − x0 )1/2 (20) so. In this case the potential is decreasing through the turning point. κ(x) (we redefined the β constants slightly in going to this equation). the solutions to the Schr¨dinger equation on either side of a classical turning point x0 at which o V (x0 ) > 0 are Ψ(x) = 1 2β1 cos k(x) x0 x k(u)du − x0 x π 4 π 4 . for x near x0 . The solutions to the Schr¨dinger equation on either o side of the turning point are Ψ(x) = Ψ(x) = R Rx 1 0 − x κ(u)du β1 e x κ(u)du + β2 e x0 . to apply the WKB approximation to a given potential V (x). x x0 κ(u)du = γ 3/2 x x0 (u − x0 )1/2 du = 2 3/2 γ (x − x0 )3/2 3 (21) and also |γ(x − x0 )| −1/4 = γ . and (21) and (22) in (16). so the classically accessible region is to the right of the turning point. γ < 0. Case 2: V (x0 ) < 0. κ(x) x < x0 (25) 1 2β1 cos k(x) x x0 k(u)du − x x0 π 4 π 4 . and the forbidden region to the left. That means that the regions of applicability of (16) and (17) are on opposite sides of the turning points as they were in the previous case. and then use (10) to evolve the wavefunction from those points to other points within the separate regions. we may use (25) and (26). Then space divides naturally into five regions: (a) x < a − In this region we are far enough to the left of the turning point that the WKB approximation is valid. and we are looking for points x0 ± at which to make the “handoff” from approximations (16) and (17) to the WKB approximation These points must satisfy several conditions. around x = b we may use (23) and (24).Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 7 up into regions bounded by turning points. the condition for this to be the case was k (x) k 2 (x) 1⇒ 1 2 2 2m [E − V (x ± )] V (x ± ) 3/2 1. Then. In the narrow region around x = a. and the wavefunction takes the form (10). Although there are no discontinuities in the potential here. First. Suppose x0 is a classical turning point of the motion. the condition here is 1/3 −1/3 2m 2m 1 ⇒ . we . (29) V (x0 ) V (x0 ) 2 2 Finally. We should probably quantify the meaning of “nearby” in that last sentence. Let the narrow such region around a be a − 1 < x < a + 1. obtain solutions of the Schr¨dinger equation in each region. the problem may be analyzed in a manner similar to that used in the consideration of one-dimensional piecewise constant potentials. within which regions the WKB approximation (7) is valid. To divide space into distinct regions in this case. and then match values and derivatives at o the region boundaries. with two classical turning points at x = a and x = b. we begin by identifying narrow regions around the turning points a and b in which the linear approximation (13) is valid. γ must be sufficiently greater than 1 to justify the approximation (16) (or sufficiently less than -1 to justify (17)). for each turning point. and that around b be b − 2 < x < b + 2. However. (30) If there are no points x0 ± satisfying all three conditions. we write down (23) and (24) (or (25) and (26)) at nearby points on either side of the turning point. the points x± must be sufficiently far away from the turning points that the approximation (7) is valid for the derivative of the phase of the wavefunction. To apply all of this to the problem of bound states in a potential well. (28) |V (x0 )| |V (x0 )| ⇒ V (x0 ) But at the same time. the approximate Schr¨dinger equation (14) is only valid as long as we can o neglect the quadratic and higher-order terms in the expansion (13). consider a potential like that shown in Figure 1. as in Chapter 6: we divide space into a number of distinct regions. the WKB approximation cannot be used. so we must have V (x0 ) . PSfrag replacements V (x) E a b .Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 8 Figure 1: A potential V (x) with two classical turning points for an energy E. But now that we know the value of Ψ at x = a − . κ(x) x < (a − ). so (25) and (26) may be used. x<a− . the same solution that’s valid at x = a − is valid at x = a + . With this choice of constants. so we may use (8) to find the wavefunction at any point within the region. respectively. we also know it at x = a + .Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 9 must throw out the term that grows exponentially as x → −∞. to propagate the first and second terms in . (31) Ψ(x) = A κ(x) (b) a − < x < a + In this region we are close enough to the turning point that (13) is valid. so we are left with 1 − R (a− ) κ(u)du e x . (32) From (31) and (32) we see that continuity of both the value and first derivative of Ψ(x) at x = a − requires taking β1 = A. Ψ(x) = Ra Ra 1 β1 e− x κ(u)du + β2 e+ x κ(u)du . Using the expression (33) for the wavefunction at x = a + . (26) becomes 1 cos k(x) 1 e+i( k(x) Rx a x a Ψ(x) = 2A =A k(u)du − π 4 Rx a k(u)du−π/4) + e−i( k(u)du−π/4) . we achive continuity not only of the value and first derivative of Ψ but also of all higher derivatives. because of course the solution of the Schr¨dinger equation in the o narrow strip around a (to which (32) is an asymptotic approximation for x < a) is valid throughout the strip. as must be the case since there is no discontinuity in the potential. With β1 = A and β2 = 0. integrating from a + to x in the propagator (9). x = (a + )− (33) (c) a + < x < b − In this region the WKB approximation (7) is valid. β2 = 0. and using G+ and G− . Compare with the rigorous solutions of this problem. · · · (35) comes from. the Schr¨dinger equation is o 0= d2 2m Ψ(x) + 2 [E − mgz] Ψ(x) 2 dx d2 2m2 g E = 2 Ψ(x) + − z Ψ(x) 2 dx mg d2 2m2 g = 2 Ψ(x) − [z − z0 ] Ψ(x) 2 dx (36) . The requirement of perfect elastic reflection at z = 0 may be imposed by taking V (x) to jump suddenly to infinity at z = 0. we obtain for the wavefunction at a point x in this region Ψ(x) = A =A = 2A = 2A 1 +i e k(x) “R (a+ ) a k(u)du+ Rx (a+ ) k(u)du−π/4 ” +e −i “R (a+ ) a k(u)du+ Rx (a+ ) −π/4 ” Rx Rx 1 e+i( a k(u)du−π/4) + e−i( a k(u)du−π/4) k(x) 1 cos k(x) 1 cos k(x) x a b a k(u)du − k(u)du − π 4 b x k(u)du − π 4 .1 Apply the WKB method to a particle that falls with acceleration g in a uniform gravitational field directed along the z axis and that is reflected from a perfectly elastic plane surface at z = 0. V (x) = mgz. For z > 0. We’ll start with the exact solution to the problem. which was my original goal. z ≤ 0. I have now carried this analysis far enough to see for myself exactly where the Bohr-Sommerfeld quantization condition b k(u)du = a n+ 1 2 π. a+ <x <b− (34) Okay.Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 10 (33). i. Problem 7. n = 1.e. so I am now going to stop this exercise and proceed directly to the problems. z > 0 ∞. 2. With the substitution u = γ(z − z0 ) γ= 2m2 g 2 1/3 and taking Φ(u) = Ψ(x(u)). where eventually I’ll take V0 → ∞. thus the condition is that −γz0 be a zero of the Airy function. we must take β2 = 0. I wasn’t quite sure how to account for the infinite potential jump at z = 0. Then the Schr¨dinger equation o for z < 0 is d2 2m Ψ(z) − 2 [V0 − E] Ψ(z) = 0 2 dz with solution 2m [V0 − E]. d2 Φ(u) − uΦ(u) = 0 du2 with solutions Φ(u) = β1 Ai(u) + β2 Bi(u). (z > 0). which means the energy eigenvalues En are given by 2m2 g 2 1/3 En = xnm ⇒ En = mg mg 2 2 2 1/3 xn (39) where xn is the nth root of the equation Ai(−xn ) = 0. we obtain 1 Ai(−γz0 ) 1 =− γ Ai (−γz0 ) k Now taking V0 → ∞. we also have k → ∞. The solution to (36) is then Ψ(x) = β1 Ai γ(z − z0 ) . (37) For z < 0. we find that (36) is just the Airy equation for Φ(u). V (z) = V0 . .Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 11 where z0 = E/mg. k= 2 Matching values and derivatives of (37) and (38) at z = 0. (38) Ψ(z) = Ae−kz . Since we require a solution that remains finite as z → ∞. so instead I supposed the potential for z < 0 to be a constant. we have β1 Ai(−γz0 ) = A γβ1 Ai (−γz0 ) = −kA Dividing. so the RHS of this goes to zero. .Homer Reid’s Solutions to Merzbacher Problems: Chapter 7 12 So that’s the exact solution. so we have n+ 1 2 z0 π= 0 k(z) dz 2m 2 0 z0 = = = = 2 3 [E − mgz]1/2 dz z0 2m2 g 2 0 [z0 − z]1/2 du z0 0 2m2 g 2 2 − (z0 − z)3/2 3 E mg 3/2 2m2 g 2 so the nth eigenvalue is given by En = mg 2 2 2 1/3 3 2 n+ 1 2 2/3 π . In this case the classical turning points are at z = 0 and z = z0 . the spectrum of energy eigenvalues is determined by the condition (35). In the WKB approximation. 2001 Chapter 8 1 . Quantum Mechanics. Third Edition Homer Reid May 13.Solutions to Problems in Merzbacher. again vanishing at the end points and even in x. an optimal quartic trial function with nodes between the endpoints. evaluate the mean a square deviation −a |Ψ(x) − Ψt (x)|2 dx for the various cases.Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 2 Problem 8. (d) Compare the results of the different variational calculations with the exact ground state energy. and. (b) A more sophisticated trial function is parabolic. where the ratio of the adjustable parameters α and β is determined variationally. Interpret the corresponding stationary energy value. (c) Use a quartic trial function of the form Ψt (x) = (a2 − x2 )(αx2 + β).23 . 2a En = n 2 2 π ≈ 1. use an unnormalized trapezoidal trial function which vanishes at ±a and is symmetric with respect to the center of the well: Ψt (x) = (a − |x|). 8ma2 ma2 2 2 (a) We need first to normalize the trial wavefunction. Taking γ(a − |x|). (a − b). Ψt (x) = . a kn = nπ . using normalized wave functions. γ(a − b). (a) First.1 Apply the variational method to estimate the ground state energy of a particle confined in a one-dimensional box for which V = 0 for −a < x < a. b ≤ |x| ≤ a x| ≤ b. and Ψ(±a) = 0. b≤x≤a |x| ≤ b. First let’s observe that the exact expressions for the ground state wavefunction and energy are 1 Ψn (x) = √ cos(kn x). in addition to the approximation to the ground state. (e) Show that the variational procedure produces. this is m γ2 Ψt2 (x) dx dx 2 a b m 2 m γ 2 (b − a). < H >= To find the optimal value of b. 2m Ψt (x) −a d2 Ψt (x) dx dx2 2 =− 2m Ψt (x)Ψt (x) a −a a − Ψt2 (x) dx −a (the first integral vanishes since Ψt vanishes at the endpoints) 2 a 0 =+ = = Using (1). 3 2 2m (b − a) a3 + 2b3 − 3ab2 .Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 3 we have a −a Ψ2 (x)dx = 2 t a 0 Ψ2 (x)dx t (a − b)2 b a = 2γ 2 = 2γ 2 = 2γ 2 = γ2 dx + 0 b (a − x)2 dx 1 b(a − b)2 + (a − b)3 3 1 b(a2 + b2 − 2ab) + (a3 − b3 ) − a2 b + b2 a 3 2 3 a + 2b3 − 3ab2 3 3 2 1 a3 + 2b3 − 3ab2 a so Ψt is normalized by taking γ2 = . (1) Now we can compute the energy expectation value of Ψt : 2 < Ψt |H|Ψt > = − Integrating by parts. we zero the derivative of this with respect to b: 0= (a3 1 6b2 (b − a) 6ab(b − a) − 3 + 3 3 − 3ab2 ) 3 − 3ab2 )2 + 2b (a + 2b (a + 2b3 − 3ab2 )2 = −4b3 + 9b2 a − 6a2 b + a3 . 2 t 2m dx m 2 = γ[−αx4 + (αa2 − β)x2 + βa2 ] The expectation value of the energy is .25 .Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 4 (b) For a parabolic trial function we take Ψt (x) = γ(a2 − x2 ). The normalization integral is a −a a 0 a 0 Ψ2 (x) dx = 2γ 2 t = 2γ = 2γ = 2 (a2 − x2 )2 dx (a4 + x4 − 2a2 x2 )dx 2 1 2 a5 + a5 − a5 5 3 16 2 5 γ a 15 so Ψt (x) is normalized by taking γ2 = The expectation value of the energy is 2 a 15 . (c) In this case we have Ψt (x) = γ(a2 − x2 )(αx2 + β) The kinetic energy is − 2 d2 Ψ (x) = γ [6αx2 − (αa2 − β)]. = 4 ma2 ma2 So this is in good agreement with the exact ground state energy. 16a5 <H >=− =2 = 2m 2 Ψt (x) −a a 0 d2 Ψt (x) dx dx2 m γ2 (a2 − x2 )dx 4 2 2 3 γ a 3m 2 5 2 ≈ 1. 1. (a) Estimate the ground state energy by a variational calculation. Draw the potential energy curve to judge if this result makes physical sense. .Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 5 Problem 8. with ground-state eigenfunction Ψ(ω) = Ce−u Adding the normalization constant. as in Section 5. e−ωξ 2 /2 . (c) Note that the method yields answers for a discrete energy eigenstate even if λ is slightly negative. consider the anharmonic oscillator Hamiltonian. Ψ(ω) = ω π 1/4 2 /2 = Ce−ωξ 2 /2 . (a) To find the ground state eigenfunction of the Hamiltonian Merzbacher proposes. scaled by a constant factor ω. Explain. 1 1 H = p2 + ξ 2 + λξ 4 2 ξ 2 where λ is a real-valued parameter. Derive an equation that relates ω and λ. it’s convenient to rewrite it: H0 (ω) = 1 2 1 2 2 p + ω ξ 2 ξ 2 1 1 ∂2 + ω2 ξ 2 =− 2 2 ∂ξ 2 Upon substituting u = ω 1/2 ξ we obtain =ω − 1 ∂2 1 + u2 2 ∂u2 2 and now this is just the ordinary harmonic oscillator Hamiltonian.2 Using scaled variables. (b) Compute the variational estimate of the ground state energy of H for various positive values of the strength λ. using as a trial function the ground state wave function for the harmonic oscillator H0 (ω) = 1 2 1 2 2 p + ω ξ 2 ξ 2 where ω is an adjustable variational parameter. exptwoΨHΨ = π ω5 (4) 1 3λ 1 ω+ + 2 . 2 ω ω (5) To minimize this with respect to ω we equate its ω derivative to 0: 0=1− 1 6λ − 3 ω2 ω . The energy expectation value is Ψ| H |Ψ = Ψ| T |Ψ + Ψ| V |Ψ where T = p2 /2 and V = ξ 2 /2 + λξ 4 . for the expectation value of V we have (3) exptwoΨV Ψ = ω π ω π 1 2 ∞ ∞ ξ 2 e−ωξ dξ + λ −∞ −∞ 2 ξ 4 e−ωξ dξ 2 3 1 π = + λ 3 4 ω 4 1 3λ = + . Let’s compute the two expectation values ξ separately. 4ω 4ω 2 Adding (2) and (4).Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 6 Now we want to treat ω as a parameter and vary it until the energy expectation value of Ψ(ω) is minimized. First of all. 4 On the other hand. to compute the expectation value of T . we need to know the result of operating on Ψ(ω) with p2 : ξ p2 Ψ(ω) = − ξ ω π ω =− π ω =− π 1/4 1/4 ∂ ∂ −ωξ2 /2 e ∂ξ ∂ξ 2 ∂ −ωξe−ωξ /2 ∂ξ −ω + ω 2 ξ 2 e−ωξ 2 1/4 /2 Then for the expectation value of T we have Ψ| T |Ψ = 1 2 1 2 1 2 ∞ Ψ(ξ)p2 Ψ(ξ)dξ ξ −∞ =− =− = ω π ∞ −∞ −ω + ω 2 ξ 2 e−ωξ dξ π 1 + ω2 ω 2 π ω3 (2) 2 ω −ω π ω . so for λ 0 the minimizing value of ω is ω ≈ 1 + 3λ. (6) We could then solve this equation for ω in terms of λ to obtain the energyminimizing value of ω for a given perturbing potential strength λ. for which the energy is minimized by the (unscaled) ground state harmonic oscillator wavefunction. the energy-minimizing value of ω will be close to 1.e. Ψ(ω) with ω = 1. Evidently. Inserting this in (6). But writing down the full solution would be tedious. 2 4 (8) Since the 1/2 term is the normal (unperturbed) energy of the state. the energy shift caused by the perturbing potential is ∆E = 3 λ.Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 7 or ω 3 − ω − 6λ = 0. 4 (9) . when λ = 0 the Hamiltonian in this problem degenerates to the normal harmonic oscillator Hamiltonian. and we may write ω(λ) ≈ 1 + for some small . We can thus imagine that. (1 + 3 + 3 2 + 3 ) − (1 + ) = 6λ Keeping only terms of zeroth or first order in the small quantity (which is equivalent to keeping terms of lowest order in the perturbing potential strength λ) we obtain from this ≈ 3λ. Instead let’s see what happens when λ is small. Inserting this estimate into (5) and again keeping only terms of lowest order in λ we find (7) exptwoΨHΨ = ≈ ≈ ≈ 1 (1 + ) + (1 + )−1 + 3λ(1 + )−2 4 1 (1 + 3λ) + (1 + 3λ)−1 + 3λ(1 + 3λ)−2 4 1 [(1 + 3λ) + (1 − 3λ) + 3λ(1 − 6λ)] 4 1 3 + λ. i. for small λ. make a variational estimate of the ground state energy for a particle in a Gaussian potential well. For small values of the coefficient.3 In first-order perturbation theory. I like to put β/2 = λ.3: ∆E = gexptwoΨn x4 Ψn = 3g 2 2 mω 1 + n + n2 2 In particular. I worked out this expectation value in Problem 5. compare the result with the variational calculation in Problem 2. α > 0). Problem 8. Then Ψ(x) = Ce−βx 2 /2 and the normalization constant is determined by ∞ 1 = C2 −∞ e−βx dx 2 ⇒ C= β π 1/4 .Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 8 Problem 8. calculate the change in the energy levels of a linear harmonic oscillator that is perturbed by a potential gx4 . represented by the Hamiltonian H= 2 p2 − V0 e−αx 2m 2 (V0 > 0. the energy shift of the ground state is ∆E0 = 3g 4 2 mω which agrees with () (the difference in the factor ( /mω)2 just represents the fact that in Problem 8. The energy shift to first order is ∆E = exptwoΨn (x)|gx4 |Ψn (x) = g Ψn | x4 |Ψn . whereas in this problem we inserted the units explicitly).4 Using a Gaussian trial function. For notational simplicity. with an adjustable parameter. e−λx .2 we used scaled variables. . Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 9 The kinetic energy operator operating on this state yields TΨ = 2 ∂ p2 ∂ Ψ(x) = − Ψ(x) 2m 2m ∂x ∂x =− =− and its expectation value is T = = β π 2 1/2 β π β π 2 1/4 2 ∂ −βxe−βx /2 2m ∂x 2 1/4 2 2m −β + β 2 x2 e−βx 2 /2 2m β β2 π − β 2 π β3 (10) β . (α + β) (12) To minimize with respect to β we equate the first β derivative of this to zero: √ 2 V0 1 β − − 0= 2m 2 β(α + β) (α + β)3 2 = 2m − α2 V0 2 β(α + β)3 mV0 α 2 1/2 2 = β(α + β)3 − = β 4 + 3β 3 α + 3β 2 α2 + βα3 − = x4 + 3x3 + 3x2 + x − mV0 2α mV0 α 2 2 2 . Ψ2 (x)e−αx dx −∞ 2 β π β π ∞ e−βx e−αx dx −∞ 2 2 π (α + β) (11) β . 4m The expectation value of the potential energy is ∞ V = −V0 = −V0 = −V0 = −V0 Combining (10) and (11). (α + β) exptwoΨ|H|Ψ = Ψ| T |Ψ + Ψ| V |Ψ = β − V0 2m 2 β . for the optimum value of a.5 Show that as inadequate a variational trial function as Ψ(x) = C 1− 0 |x| a |x| > a |x| ≤ a yields. and then insert said expression into (12) to obtain the lowest energy attainable with this form of trial wave function.Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 10 where I put x = β/α. an upper limit to the ground state energy of the linear harmonic oscillator. 1 = C2 = 2C 2 = 2C 2 0 a Ψ(x)2 dx 1− a −a a 0 x a 2 dx 1−2 a 3 x x2 + 2 a a = 2C so C= 2 a−a+ 3 . 2m 2 (13) The harmonic oscillator hamiltonian is E =T +V = 2 a exptwoΨT Ψ = − Integrating by parts. In practice. In theory we could write down an explicit expression for the roots of this quartic in terms of mV0 / 2 α. and I can’t see any way to proceed other than numerically. which lies within less than 10 percent of the exact value. this would be a mess. however. Am I missing some kind of trick here? Problem 8. 2m Ψ(x) −a ∂ Ψ(x) dx ∂x2 2 (14) 2 =− 2m Ψ(x) ∂ Ψ(x) ∂x a −a a − −a ∂ Ψ(x) ∂x 2 dx . 2a p2 mω 2 x2 + . The first task is to evaluate the normalization constant C. . 3 exptwoΨHΨ = √ ω ≈ 0. 30 Of course the actual ground state energy is 0.5 < 10%. 2 2ma 20 To minimize with respect to a we set the a derivative of this to zero: exptwoΨ|H|Ψ = Ψ| T |Ψ + Ψ| V |Ψ = mω 2 a 3 2 + ma3 10 0=− or a4 = 30 2 m2 ω 2 √ .. a2 = 30 mω (17) (18) Inserting into (18). so the fractional error is 0.5 · ω.Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 11 The first term vanishes.047/0. 2 = 2m 3 2a a −a 1 dx a2 (15) 3 2 = 2ma2 (16) exptwoΨV Ψ = mω 2 2 a −a a x2 Ψ(x)2 dx = mω 2 0 x2 Ψ(x)2 dx 3 2a 3 2a a 0 = mω 2 = mω 2 x2 − 2x3 x4 + 2 a a a 0 dx x3 x4 x5 − + 2 3 2a 5a mω 2 a2 = 20 3 2 mω 2 a2 + . .547 · ω. and derive the virial theorem for stationary states. where λ is a variational (scaling) parameter. Ψn (r).Homer Reid’s Solutions to Merzbacher Problems: Chapter 8 12 Problem 8. Ψt (r) = Ψn (λr). The n − th discrete energy eigenfunction of this system. corresponds to the energy eigenvalue En . Apply the variational principle by using as a trial function.6 A particle of mass m moves in a potential V (r). .
Copyright © 2021 DOKUMEN.SITE Inc.