PHYSICS IIDr. Ing. Valeric˘ a D. Ninulescu 2010 Contents 1 The 1.1 1.2 1.3 1.4 1.5 1.6 1.7 experimental foundations of quantum mechanics Thermal radiation . . . . . . . . . . . . . . . . . . . . . . . . Photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . Compton effect . . . . . . . . . . . . . . . . . . . . . . . . . . Atomic spectra . . . . . . . . . . . . . . . . . . . . . . . . . . Bohr model of the hydrogen atom . . . . . . . . . . . . . . . Experimental confirmation of stationary states . . . . . . . . Einstein’s phenomenological theory of radiation processes . . 1.7.1 Relations between Einstein coefficients . . . . . . . . . 1.7.2 Spontaneous emission and stimulated emission as competing processes . . . . . . . . . . . . . . . . . . . . . 1.8 Correspondence principle . . . . . . . . . . . . . . . . . . . . 1.9 Wave-particle duality . . . . . . . . . . . . . . . . . . . . . . . 1.10 Heisenberg uncertainty principle . . . . . . . . . . . . . . . . 1.10.1 Uncertainty relation and the Bohr orbits . . . . . . . . 1.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Principles of quantum mechanics and applications 2.1 First postulate . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Schr¨odinger equation . . . . . . . . . . . . . . . . . . . . . . . 2.3 Probability conservation . . . . . . . . . . . . . . . . . . . . . 2.4 Constraints on the wavefunction . . . . . . . . . . . . . . . . 2.5 Time-independent Schr¨odinger equation . . . . . . . . . . . . 2.6 Potential wells . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 The one-dimensional infinite well . . . . . . . . . . . . . . . . 2.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 2.7.2 Energy eigenvalues and eigenfunctions . . . . . . . . . 2.7.3 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 The rectangular potential barrier. Potential barrier penetration 1 1 4 4 4 6 10 11 13 15 16 17 17 18 19 21 21 21 23 25 25 27 28 28 29 33 34 2 Case E > V0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .CONTENTS iv 2.10. . . . . . 2.1 Case 0 < E < V0 . . . Non-rectangular potential barriers . . . 2. . . . . .8. . science. 34 37 37 38 39 39 40 41 47 48 51 B Greek letters used in mathematics. The quantum harmonic oscillator . A Fundamental physical constants .12 2. Three-dimensional Schr¨odinger equation Problems . 2. . . . . .10. . . . . . . . .11 2. . . and engineering 55 . . . . . . . . 2. . . . . . . . . .13 2. . Applications of tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9 2. . . . . .1 Field emission of electrons . . . . .2 Alpha-particle emission . . . . . . . . . .8. . . . . . . . . . . . . . . . . . .10 2. . . . . . . . .8. . . . . . . . . . . . .3 Discussion . . . . . . . . . poorly approximates Boltzmann constant centre of mass time-dependent Schr¨odinger equation time-independent Schr¨odinger equation .i r Aˆ† v ∼ kB CM TDSE TISE exponential function.) R z∗ Rh. ez = exp(z) complex unity. i2 = −1 imaginary part of a complex number the set of real numbers complex conjugated of z average integral over the whole three-dimensional space adjoint of Aˆ vector roughly similar.Glossary of notations e i Im(. . it seemed that physics was able to explain all physical processes. 1.Chapter 1 The experimental foundations of quantum mechanics At the end of the 19th century. For example. this coefficient is dependent on wavelength. According to the ideas of that time. the matter motion could be studied by Newton’s laws and the radiation was described by Maxwell’s equations. Examples: • the infrared radiation of a household radiator • solar radiation The radiation incident on the surface of a body is partially reflected and the other part is absorbed. dark bodies absorb most of the incident radiation and light bodies reflect most of the radiation.1 Thermal radiation It is well known that every body with non-zero temperature emits electromagnetic radiation with a continuous spectrum that contains all wavelengths. . the Universe was composed of matter and radiation. The absorption coefficient of a surface is defined as the fraction of the incident radiation energy that is absorbed. This confidence began to disintegrate. For an observer placed outside. this is why it is called a “displacement law”. For example. when iron is heated up in a fire. • The emissive power (energy radiated from a body per unit area per unit time). then yellow. and white at very high temperatures. Kirchhoff stated that the ratio of the emissivity power (power emitted by unit area at a given wavelength) and the absorption coeficient at that wavelength is the same for all bodies at a given temperature. R.1) where σ ≈ 5. The entrance hole behaves as a blackbody. a ray that enters the cavity is absorbed as it is scattered by the interior walls. (1. Consider a cavity with a small entrance hole and maintained at a constant temperature. otherwise its temerature can not remain constant. Such a body emits and absorbs the same energy in unit time. of a blackbody at temperature T grows as T 4 : P = σT 4 . This result is known as Stefan–Boltzmann law. In 1859 G. λmax shifts with the temperature. λmax T = b . It follows that: • the blackbody is not only the best absorber. A blackbody is an object that absorbs all electromagnetic radiation falling on it. Blackbody radiation laws: • The blackbody radiation is isotrope and nonpolarized.67×10−8 W m−2 K−4 is Stefan–Boltzmann constant. P .2 1 The experimental foundations of quantum mechanics Suppose a body at thermal equilibrium with its surroundings.898 × 10−3 m K . but it is also the best emitter • the emissivity power of a blackbody is a universal function A good approximation of a blackbody can be done as follows. signifying that all the visible frequencies are being emitted equally. • The wavelength λmax for maximum emissive power from a blackbody is inversely proportional to the absolute temperature. . Further increase in temperature causes the colour to change to orange. The constant b is called Wien’s displacement constant and its value is b ≈ 2. The radiation emitted by a body at thermal equilibrium is termed thermal radiation. (1. the first visible radiation is red.2) which is called Wien’s displacement law. λ5 exp(hc/λkB T ) − 1 (1.6) ρ(ν.7) and . T ) given by Eq. Classical physics furnished a result in good agreement with the experiment only at small frequencies.5). Another form of the Planck’s law gives the spectral energy density ρ(λ.1 presents the graph of the spectral energy density ρ(λ. 3 c exp(hν/kB T ) − 1 (1.3) The constant h is called Planck’s constant and has the value h ≈ 6. A straightforward calculation gives for the spectral energy density the expressions ρ(ω.1 Thermal radiation 3 • For a complete quantitative characterization of the blackbody radiation.5) Figure 1. By use of the relationship ρ(λ. 5 λ exp(hc/λkB T ) − 1 (1. Planck radiation formula can be expressed in terms of angular frequency or the frequency. we should give the thermal radiation power per unit of area and unit of wavelength. hν. T ) .1. T ) (1. T ) = 2 hc2 1 . T ) = 8 hc 1 . T ) of a blackbody. (1. Planck’s radiation formula contains all the information previously obtained. In 1900 Max Planck determined a formula which agrees with the experiment at whatever temperature based on a new and revolutionary idea: the exchange of energy between a body and its surroundings can be performed only in discrete portions. T ) = ω2 ~ω 2 c3 exp(~ω/k T ) − 1 B (1. the minimum energy implied in the exchange being proportional to the frequency.4) we find ρ(λ. T ) = (4/c)R(λ. T ) = hν 8 ν2 . denoted here R(λ.626 × 10−34 J s . Planck radiation law reads R(λ. • On passing through a very thin slit and then through a prism the electromagnetic radiation can be separated into its component frequencies.3 Compton effect 1. 1. • Infrared thermography: a fast non-distructive inspection method that maps the temperature differences of any object in a range from −50 ◦ C to 1500 ◦ C.2 Photoelectric effect 1. 103J m−4 5 4 1750 K 3 2 1500 K 1 1000 K 0 0 1 2 λ. 1.T). µm 3 4 5 Fig. Applications • Pyrometer: device for measuring relatively high temperatures by measuring radiation from the body whose temperature is to be measured. the following observations concerning the atomic emission spectra had been done: • When a gas of an element at low pressure is subjected to an input of energy. such as from an electric discharge. the gas emits electromagnetic radiation.1 The experimental foundations of quantum mechanics 4 6 2000 K ρλ(λ.1 Spectral energy density ρ(λ. T ) of a blackbody at a few temperatures.4 Atomic spectra By the early 1900s. . n1 = 2. 3. .10) Tn = R/n2 . λn 2 n λn = C where the new constant R is nowadays called Rydberg constant. For a given series. ν˜n1 n2 = λn 1 n 2 n1 n2 n1 n2 (1. the spectral lines in a series get closer together with increasing frequency. (1. the wavenumber approaches the limit R/n21 . . 1 1 1 ν˜n = = R 2 − 2 . . . . • Each element has its own unique emission spectrum. however. a spectral term has a more complicated form. 4. (1. . • The emission spectrum is made up of spectral series in the different regions (infrared.6 nm . n = 3. . . According to Eq. Here. and generalized in the form 1 1 1 R R = R 2 − 2 = 2 − 2 . . . Its four spectral lines in the visible have wavelengths that can be represented accurately by the Johann Balmer formula (1885) n2 n = 3.1. n2 = 3. an infinite number of lines lie at the series limit. The simpliest line spectrum is that of the hydrogen atom.9) the wavenumber of any line of the hydrogen spectrum is the difference of two spectral terms: ν˜n1 n2 = Tn1 − Tn2 . . .11) with It proves empirically that the lines of other chemical elements can also be expressed as the difference of two terms. and ultraviolet) of the spectrum of electromagnetic radiation. .9) and n1 < n2 . (1. None of the above quantitative result could be explained satisfactorily in the frame of clasical physics. . . visible. The separation of consecutive wavenumbers of a given series decreases so that the wavenumber can not exceed the series limit. (1.8) n2 − 4 where C ≈ 364. 4. with increasing values of n2 . In principle. This formula was written by Johannes Rydberg (1888) for wavenumbers.4 Atomic spectra 5 It was found that a gas at low pressure emits only discrete lines on the frequency scale. . 4. n1 defines the spectral series. . .12) where E1 and E2 denote the energy of the atom in the two stationary states. (1. E2 . • Emission and absorption of radiation are always associated to a transition of the atom from one stationary state to another. According to the electromagnetic theory. . with the nucleus playing the role of the Sun and the electrons that of the planets bound in their orbits by Coulomb attraction to the nucleus. The energies of the stationary states. The frequency of the radiated energy is the same as the orbiting one. Ernest Rutherford. . This conclusion completely disagrees with experiment. Hans Geiger and Ernest Marsden (1909) directed -particles from radioactive radium at thin gold foil. The frequency ν of the radiation emitted or absorbed respectively during such a transition is given by the equation hν = |E2 − E1 | . . To study the internal structure of atoms. form a discrete set of values. Electrons lose energy and they should spiral into the nucleus in a time of the order of 10−8 s . the discrete emission spectrum of atoms can not be explained. As the orbiting frequency can take a continuous range of values. Rutherford (1911) furnished the nuclear atom model (planetary model of the atom): the atom consists of a positively charged heavy core (nucleus) of radius on the order of 10−15 m and a cloud of negatively charged electrons. Niels Bohr (1913) improved the Rutherford model of the atom by introducing the following postulates: • For every atom there is an infinite number of stationary states in which the atom can exist without emitting radiation.5 Bohr model of the hydrogen atom Atoms have radii on the order of 10−10 m . Based on the results of the -particles scattering by the gold atoms. To overcome these deficiencies. About as soon as the model was published it was realized that the atom model suffers from two serious deficiencies: 1. Orbital motion is an accellerated motion and electrons are charged particles. E1 . 2. the electrons should radiate energy in the form of electromagnetic waves. since the atoms are stable. . Atomic structure was pictured as analogous to the solar system.1 The experimental foundations of quantum mechanics 6 1. E3 . 17) me2 is called the Bohr radius.14). 4 ǫ0 r (1. (1. .29 × 10−11 m (1.13) We are now in a position to solve the problem. Newton’s second law applied for the electron motion on a circular orbit of radius r with the velocity v under the attractive Coulomb force exerted by the nucleus yields m v2 e2 = . 4 ǫ0 ~ n (1. The potential energy of the electron in the hydrogen atom is given by a0 = V (r) = − e2 .2 presents the diagram of . . we find rn = 4 ǫ 0 ~2 2 n = a 0 n2 me2 (1. (1.19) The integer n determines the energy of the bound state of the atom and it is called principal quantum number. 2.1. . .16) and vn = The radius 4 ǫ 0 ~2 ≈ 5.5 Bohr model of the hydrogen atom 7 Bohr applied these postulates for the hydrogen atom. The selection of the allowed stationary states is performed by the following quantization rule for the angular momentum: the angular momentum of the electron in a stationary state is an integer multiple of ~: mvr = n~ . n2 n = 1. where n = 1. . (1. 3. 2. Figure 1.18) The energy of the atom in state n is e2 1 En = Tn + Vn = mvn2 − 2 4 ǫ 0 rn that gives m En = − 2 2~ e2 4 ǫ0 2 1 . . This value sets the scale for atomic dimension.15) e2 1 . r 4 ǫ0 r 2 (1.14) Combining Eqs.13) and (1. The electron of mass m and electric charge −e is assumed to move around the nucleus of electric charge e in a circular orbit. . . 3. has the energy m E1 = − 2 2~ e2 4 ǫ0 2 ≈ 13. The ionization energy of the hydrogen atom.21) Let us now calculate the frequencies of the hydrogen spectrum. (1. called ground-state. In fact. a number of phenomena in spectroscopy can be dealt with by making use of Bohr theory alone. (1. . the subscript reminds us the premise that the nucleus is exceedingly massive compared with the electron.22) The wavelength λn1 n2 = c/νn1 n2 of the radiation is given by 1 λn 1 n 2 m = 4 c~3 e2 4 ǫ0 2 1 1 − n21 n22 where R∞ m = 4 c~3 e2 4 ǫ0 2 = R∞ 1 1 − n21 n22 ≈ 1. states of positive energies refers to the ionized atom.1 The experimental foundations of quantum mechanics 8 energy levels for the hydrogen atom.20) This value sets the scale for atomic energy.097 × 107 m−1 .24) is the Rydberg constant for hydrogen. (1. (1.23) (1. The states of higher energy are called excited states. The great success of the Bohr model had been in explaining the spectra of hydrogen-like (single electron around a positive nucleus) atoms. All energies of the bound atom are negative. its knowledge considerably helps the understanding of new theories.6 eV . the frequency of the radiation emitted or absorbed is νn1 n2 = (En2 − En1 )/h and the calculus yields ν n1 n2 m = 4 ~3 e2 4 ǫ0 2 1 1 − 2 2 n1 n2 . Even though the Bohr theory is now extended and altered in some essential respects by quantum mechanics. In the transition between the states n1 and n2 > n1 . defined as the amount of energy required to force the electron from its lowest energy level entirely out of the atom is EI = −E1 ≈ 13. The state of minimum energy.6 eV . 6 Fig.2 nm 1875 nm 1282 nm ? ? 3 ? ? ? ? Paschen series Balmer series −3.51 1 ? ? ? ? Lyman series −13.1 nm 434.85 Brackett series −1.3 nm 486.3 nm 95. and Paschen series.5 Bohr model of the hydrogen atom n ∞ 9 E/eV 0. Balmer.1.6 nm 102.0 nm 2 −0. 1.00 Ionized atom 5 4 656.40 121.2 Energy level diagram for hydrogen and a few spectral lines of the Lyman.6 nm 97.0 nm 410. .54 ? ? −0. The minimum kinetic energy of the electron for the atom excitation is practically E2 − E1 (see Problem 1. .5 V is applied so that only those electrons above an energy threshold will reach it. • If the electron has a kinetic energy T < E2 − E1 it is not able to excite a mercury atom. Electrons emitted from a hot cathode C are accelerated toward the mesh grid G through a low pressure gas of Hg vapour by means of an adjustable voltage U ..8). 1. Due to the interaction between electron and atom. The experimental result is explained in terms of electron–mercury atom collisions. At this point. U′ Experimental confirmation of stationary states The idea of stationary states had been introduced to explain the discrete spectrum of atomic systems.1 The experimental foundations of quantum mechanics 10 C Hg vapour G A A U 1.9 V.9 eV is interpreted as the energy required to excite a mercury atom to its first excited state. 1. This explains the sequence of peaks. The first experimental confirmation of this hypothesis was provided by J. Hence. as would be the case for a vacuum tube. E2 − E1 . Hertz (1914). 1.9 eV.6 Fig.9 eV of kinetic energy and cause a second excitation event before reaching the anode.3. Franck and G. the kinetic energy is conserved) and the electron moves through the vapor losing energy very slowly (see Problem 1. Let us denote E1 and E2 the energy of the atom in its ground state and first excited state. a sudden decrease of the current through the tube is noticed.4) does not increase monotonically. • If the voltage is more than twice 4.e. The anodic current as a function of voltage (Fig.9). • If the kinetic energy of an electron has reached the value of 4. The collision is elastic (i. there is an exchange of energy. Between the grid and the anode A a small retarding voltage U ′ ∼ 0. but rather displays a series of peaks at multiples of 4. respectively. it is able to transfer its energy to a mercury atom in a collision. the energy of 4.3 Franck-Hertz set-up. the electron is able to regain 4.9 V. Their experimental set-up is presented schematically in Fig. 1. The Physics Nobel Prize for the year 1925 has been awarded to Professor James Franck and Professor Gustav Hertz for their discovery of the laws governing the impact of an electron upon an atom. The two atomic levels are allowed to be multiplets with degeneracies g1 and g2 . Assuming that all the atoms are in these states.7 Einstein’s phenomenological theory of radiation processes 11 Current 6 0 Fig. E2 > E1 (Fig. The mean numbers of atoms per unit volume in the two multiplet states are denoted by N1 and N2 .25) Einstein’s phenomenological theory of radiation processes In 1917 Einstein explained phenomenologically the radiation-matter interaction based on the quantum ideas of that time: Planck’s quantum hypothesis and Bohr’s planetary model of the hydrogen atom. N1 + N2 = N . when excited.7 (1. each atom having a pair of bound-state energy levels E1 and E2 . Einstein’s postulates could all be justified by the later developed quantum mechanical treatments of the interaction processes.5). V It is known from spectroscopy that.26) The atomic medium is considered in a radiation field whose spectral energy density at angular frequency ω given by . This is a confirmation of the stationary states hypothesis. Radiation of this wavelength is observed during the passage of the electron beam through the Hg vapour.7 nm. (1. 0 5 10 15 Accelerating voltage. mercury vapor. λ = h/p 1. Suppose N identical atoms in unit volume. emits radiation whose wavelength is λ = 253.4 Current through a tube of Hg vapor versus accelerating voltage in the Franck–Hertz experiment.9 eV.1. 1. the energy of the corresponding photons is hc/λ ≈ 4. g 1 . The total rate of spontaneous emissions is −N˙ 2 (t) = −A21 N2 (t) and the integration with the initial condition at t = 0 gives N2 (t) = N2 (0) exp(−A21 t) . Einstein considers three basic interaction processes between radiation and atoms (Fig.1 The experimental foundations of quantum mechanics 12 6 ~ω ? E2 . 1.27) is ρ(ω) . g 2 . 1. Spontaneous emission An atom in state 2 spontaneously performes a transition to state 1 and a photon of frequency ω is emitted. 1.5). N 2 6 B12 ρ(ω) A21 ? Spontaneous emission Absorption B21 ρ(ω) ? Stimulated emission E1 . ~ω = E2 − E1 (1. N 1 Fig. In the absence of a radiation field the transition 1 → 2 is impossible due to violation of the energy conservation law. The time in which the population falls to 1/e of its initial value is τ21 = 1/A21 (1.5 Radiative transitions between to energy levels. . The probability per unit time for occurence of this process is denoted by A21 and is called Einstein coefficient for spontaneous emission.28) and is called lifetime of level 2 with respect to the spontaneous transition to level 1. The photon is emitted in a random direction with arbitrary polarization. energy conservation law asks for the emission of a photon of frequency ω. Absorption In the presence of a radiation field an atom initially in state 1 can jump to state 2 by absorption of a photon of frequency ω. (1.29) where B12 is called Einstein coefficient for absorption. This special case will lead us to establish relations between Einstein coefficients. They depend only on the properties of the two atomic states. the total rate of absorptions is −N˙ 1 = B12 ρ(ω)N1 . the spectral energy density ρ(ω) is given now by Planck’s formula [Eq. direction of propagation.1 N1 + N2 = N . (1.7.2 of the two energy levels change in time according to the equations N˙ 1 = −N˙ 2 = −B12 ρ(ω) N1 + [A21 + B21 ρ(ω)]N2 .31) Relations between Einstein coefficients Let us consider the atomic system defined above in equilibrium with thermal radiation. the populations N1. Einstein coefficients defined above are independent of the radiation field properties and are treated here as phenomenological parameters. Due to all radiative transitions presented above. 3.7 Einstein’s phenomenological theory of radiation processes 13 2. (1.6)].30) where B21 is called Einstein coefficient for stimulated emission. The reason for its introduction by Einstein will be made clear during next section. polarization and phase as the radiation that forces the emission process. The probability per unit time for this process is assumed to be proportional to the spectral energy density at frequency ω. this means that radiation created through stimulated emission has the same frequency.1. the total rate of stimulated emissions is −N˙ 2 = B21 ρ(ω)N2 . The radiation produced during the stimulated emission process adds coherently to the existing one. 1. (1. The stimulated emission process was unknown before 1917. The probability per unit time for this process is assumed to be proportional to the spectral energy density at frequency ω. Stimulated emission Einstein postulates that the presence of a radiation field can also stimulate the transition 2 → 1 of the atom. . the spectral energy density of the thermal field can be expressed as ρ(ω. (g1 B12 /g2 B21 ) exp(~ω/kB T ) − 1 (1. T ) gives ρ(ω. For the atomic system.32) would not be consistent with Planck’s formula. T ) = A21 /B21 . (1. T ) = g2 A21 exp(−~ω/kB T ) . This subject will be treated below. g1 B12 As this formula verifies experimentally at high frequencies.6). From Eq. it is expected that stimulated emission of atoms is an unlikely event at high frequencies. These relations permit to express the transition rates between a pair of levels in terms of a single Einstein coefficient. Eq. (1. . so g1 B12 = g2 B21 (1. Let us notice that without the introduction of the stimulated emission process.14 1 The experimental foundations of quantum mechanics The equilibrium condition for the radiation-matter interaction expresses as follows. atomic energy levels are occupied according to the Maxwell–Boltzmann statistics. T )]N2 = 0 . c B21 (1.33a) and ω2 A21 = 2 3 ~ω .32) This formula should be Planck’s radiation formula (1.33) are known as Einstein’s relations.31) at equilibrium. (B12 /B21 )(N1 /N2 ) − 1 In thermal equilibrium.33b) Relations (1.32) becomes ρ(ω. T )N1 + [A21 + B21 ρ(ω. These conditions are equivalent. for B21 = 0 Eq. the equilibrium condition writes N˙ 1 = N˙ 2 = 0 . −B12 ρ(ω. (1. For the radiation field the equilibrium condition means equality of emitted quanta and absorbed quanta. The use of this ratio into the expression of ρ(ω. T ) = A21 /B21 . Indeed. In exact terms the ratio of the population densities of the two levels is N1 /N2 = (g1 /g2 ) exp[−(E1 − E2 )/kB T ] = (g1 /g2 ) exp(~ω/kB T ) . The ratio is R = 1 for ν = (kB T /h) ln 2 ≈ 4.6. exp(~ω/kB T ) − 1 For numerical evaluations we choose a typical temperature T = 300 K. −40 10 −60 10 10 10 1.7 Einstein’s phenomenological theory of radiation processes 15 0 10 −20 10 R Fig. R ≈ 624. we consider the ratio R of the stimulated emission rate and spontaneous emission one. The use of Planck’s radiation formula and the second Einstein equation gives R= 1 . so the stimulated emission in more likely than the spontaneous emission.1.2 m) in infrared region.6 Ratio of stimulated emission probability and spontaneous emission probability for a two-level atom in a thermal radiation field of temperature T = 300 K. This ratio is R= B21 ρ(ω) A21 and is dependent on the radiation field and frequency. For smaller frequencies spontaneous emission becomes negligible with respect to stimulated emission. Which process is most likely? To ask this question. 1. • In the near-infrared and visible region the ratio R takes on very small values. • In the microwave region. To be more specific. The dependence R versus frequency is presented in Fig.2 11 10 12 13 10 10 ν (Hz) 14 10 15 10 Spontaneous emission and stimulated emission as competing processes An atom in an excited state can jump to a lower energy state through a spontaneous emission or a stimulated one.7. 1. . so the spontaneous emission is the dominant process. for example at ν = 1010 Hz. let us consider the thermal radiation field.33×1012 Hz (λ ≈ 69. it follows that the classical continuum description of the spectrum should correspond to transitions between two such states. 3 2 rn 2 ~ 4 ǫ0 n3 where Eqs. i.22) the frequency of the radiation in the transition between states n − 1 and n is νn−1.n m = 4 ~3 e2 4 ǫ0 2 2 2 1 1 2n − 1 e m − 2 = . By use of Eq. the comparison of the frequency expressions gives νn−1.8 Correspondence principle One of the guiding principles used in the development of quantum theory was Bohr’s correspondence principle (1920) which indicates that quantum theory should give results that approach the classical physics results for large quantum numbers. The stimulated emission dominates over spontaneous for a laser. (1.16) were used.e. More that that. The quantum description of the atom is performed in the Bohr theory framework..15) and (1. (1.n ≈ νn in agreement to the correspondence principle. To illustrate this principle let us discuss the hydrogen atom spectrum. n ≫ 1 .n m ≈ 4 ~3 e2 4 ǫ0 2 m 2n = 4 n 2 ~3 e2 4 ǫ0 2 1 .1 The experimental foundations of quantum mechanics 16 Can stimulated emission dominate over spontaneous emission for a classical source of visible radiation? It can be proven that the spectral energy density of a spectroscopic lamp is not sufficient to ensure a ratio R > 1 . The frequency of the quantum transition becomes νn−1. 2 3 2 (n − 1) n 4 ~ 4 ǫ0 n (n − 1)2 We consider now large quantum numbers. 1. . for large values of the quantum number n the hydrogen energy levels lie so close together that they form almost a continuum. n3 On the other hand the motion of the electron on the nth orbit has the frequency 2 2 vn m 1 e νn = = . According to classical theory the atom emits radiation at the frequency νn . The wavelength of the wave associated to the electrons is given by Eq. according to the theory of the diffraction produced by a rectangular slit. refers to the simultaneous measurement of the position and the momentum of a particle and states that the uncertainty ∆x involved in the measurement of a coordonate of the particle and the uncertainty ∆px involved in the measurement of the momentum in the same direction are related by the relationship ∆x ∆px ≥ ~/2 .25).10 17 Wave-particle duality Heisenberg uncertainty principle An interesting interpretation of the wave-particle duality of all physical entities has been given by Werner Heisenberg (1927). the angle θ is given by d sin θ = λ .9 Wave-particle duality 1. we consider a thought experiment. d in agreement with Eq.9 1. (1.34). This slit produces a wider central maximum in the diffraction pattern. a larger uncertainty in the determination of the y-component of the momentum of the particle. 1. a diffraction pattern is observed on the photographic plate. The uncertainty principle. a narrower slit is required.1. (1.34) To illustrate the uncertainty principle. i. the uncertainty of the y-position is ∆y ∼ d . Suppose a parallel beam of monoenergetic electrons that passes through a narrow slit and is then recorded on a photographic plate (Fig.e.7). Note that: • To reduce the uncertainty in the determination of the coordinate y of the particle. or indeterminacy principle. The precision with which we know the y-position of an electron is determined by the size of the slit. The uncertainty in the knowledge of the y-component of electron momentum after passing through the slit is determined by the angle θ corresponding to the central maximum of the diffraction pattern.. Reducing the width of the slit. . (1. where p designate the momentum of the electrons. We have ∆py ∼ p sin θ = p h λ = d d and ∆y ∆py ∼ d h = h. if d is the width of the slit. 1 Uncertainty relation and the Bohr orbits In Sect.7 Diffraction through a single slit. The above example clearly shows that in contrast to the classical situation. at the atomic level. (1. because the uncertainties implied by this principle are too small to be observed. Let us investigate the compatibility of this picture with the uncertainty relation [Eq. measurement inevitably introduces a significant perturbation in the system. 1.1 The experimental foundations of quantum mechanics 18 y 6 Incoming beam.34)].? θ Screen with a single slit O Observing screen Fig. 1. It follows that for such particles the description of the motion needs a different picture. For everyday macroscopic objects the uncertainty principle plays a negligible role in limiting the accuracy of measurements. It is common to consider in everyday life that a measurement can be performed without changing the state of the physical system. • To improve the precision in the determination of the y-component of the momentum of the particle. The uncertainty principle implies that the concept of trajectory for particles of atomic dimensions is meaningless.10.35) . The classical treatment of the electron motion is justified for small uncertainties in the position and momentum. 1.6 . which means a larger uncertainty in the y-coordinate of the particle.d . ∆x ≪ r and ∆px ≪ p .5 the electron of the hydrogen atom is supposed in a circular motion around the nucleus. This requires a larger slit. the width of the central maximum in the diffraction pattern must be reduced. (1. 1. find the velocity of the fastest electrons ejected and the stopping voltage. The transmitter of (a) radiates a power of 100 W.36) for small values of the quantum number n . (c) The Sun. . It follows that the classical motion picture of the electron on circular orbits must be rejected. The use of Heisenberg uncertainty relation yields ∆x ∆px 1 ≥ .1. The penetration depth of the radiation is approximately equal to the radiation wavelength. (a) Determine the threshold wavelength for the photoelectric effect. Calculate the change in energy of the scattered photon and the momentum of the recoiling electron.3 eV.11 Problems 19 from which we infer ∆x ∆px ≪ 1. 1.4 The work function of lithium is 2. the quantization rule gives pr = n~ . In the classical model of radiation-matter interaction.11 Problems 1.1 Calculate the energy in eV for photons typical of: (a) A long-wave radio transmitter (λ = 1500 m). (1. r p 2n It is clear now that this result is in contradiction with Eq. 1. (b) The 2. the radiation energy is equally shared by all free electrons. Free electron density in zinc is n = 1029 m−3 and the work function is Φ = 4 eV. Assume hν = kB T .36) r p For the motion on the Bohr orbit n . Calculate the minimum irradiation time for an electron to accumulate sufficient energy to escape from the metal.2 Derive Wien’s displacement law from Planck’s radiation formula. Assume T = 5800 K. (b) Supposing that UV radiation having a wavelength of 300 nm falls on a lithium surface. How many photons per second are emitted? 1. 1.3 A zinc plate is irradiated at a distance R = 1 m from a mercury lamp that emits through a spectral filter P = 1 W radiation power at λ = 250 nm. (1.7 K cosmic background radiation.5 A photon from the cosmic background radiation collides with a stationary electron and is directly back scattered. required for the atom excitation to the first excited state of energy E2 . (1 + M/m)2 . 1. 1. and the fringes were observed by means of an electron microscope at a distance of 35 cm. when it collides with a particle of mass M initially at rest in the laboratory frame of reference is ∆T = − Discuss the case M ≫ m .20 1 The experimental foundations of quantum mechanics 1. with initial kinetic energy T .8 An electron is bombarding an atom at rest in its ground state of energy E1 . Prove that the threshold kinetic energy of the electron. 4M/m T. is approximately equal to E2 − E1 .9 Show that the change in kinetic energy of a particle of mass m.7 Use position-momentum uncertainty relation to estimate the ground state energy of the hydrogen atom. Calculate the fringe separation. 1.6 In a two-slit interference experiment using 50 keV electrons the slits were separated by 2 m. 1) We extend the treatment for the case of microparticles. t) = A exp[i(k · r − ωt)] .2) In case of microparticles. t) = A exp[i(kx x − ωt)] . In general.Chapter 2 Principles of quantum mechanics and applications 2. (2.3) . is proportional to the probability density of the particle at that point and time. That wavefunction determines everything that can be known about the particle and is a single-valued function of the space coordinates and the time. evaluated at a particular point and time. In case of light. 2. it is a complex function whose squared modulus. The expression of a wave of constant kx and angular frequency ω is ψ(x.1 First postulate Postulate 1 For a particle moving under the influence of an external potential. (2.2 Schr¨ odinger equation Consider the two-slit experiment. there is an associated wavefunction. kx = px ~ and ω= E p2 = x . the interference pattern is explained using waves of form ψ(r. ~ 2m~ (2. we consider the one-dimensional case. For simplicity. ψ(x. ∂x (2. 2.7? We associate operators to the dynamical variables E and px of the particle: ˆ = i~ ∂ E ∂t and pˆx = −i~ ∂ .9) The energy of the particle is now given by E = p2x /2m + V (x.7 in the form ˆ = Tˆψ Eψ where Tˆ = pˆ2x /2m (2.12) is the kinetic energy operator. ∂t 2m (2. Let us extend the equation for the non-free particle. t) = A exp (px x − Et) = A exp (px x − ~ ~ 2m 22 We have ∂ i ψ = − Eψ . t) . in the one-dimensional case. Eψ .11) By writing Eq. t) . ∂t ~ ∂ i ψ = px ψ . ∂x ~ Combining these equations.8) A(px ) exp (px x − E(px )t) dpx . 2. We do this for the case of a force Fx deriving from a potential energy V (x.7) As the differential equation is linear and homogeneous. i~ ∂ p2 ψ = Eψ = x ψ . t) = ~ −∞ Equation 2.7 is called the Schr¨odinger equation for the free particle. the natural extension of the equation is ˆ = (Tˆ + Vˆ )ψ . t): Fx = − ∂ V (x.5) ∂2 1 ψ = − 2 p2x ψ . (2. the general solution is Z ∞ i (2. ∂x2 ~ (2. (2.4) ψ(x.6) i~ ∂ ~2 ∂ 2 ψ=− ψ.10) How the extend Eq. ∂t 2m ∂x2 (2.2 Principles of quantum mechanics and applications i p2x i t) . ∂x (2. 13) ∂t 2m ∂x2 The extension to the three-dimensional case is i~ ∂ ~2 2 ψ(r. The 1D-Schr¨odinger equation for a particle in a potential field is now ∂ ~2 ∂ 2 i~ ψ = − ψ + V (x. d dt Z Z ∂ ∂ ∗ |ψ(r. t) ψ(r. ∂t 2m (2. t) = −i ~ [ψ ∗ (r. t) ψ (r. t) − ψ(r.2. The rate of change of this quantity can be calculated from the rate of change of ψ as given by TDSE (2. t) + V (r. t)] . t)∇2 ψ ∗ (r.3 Probability conservation 23 We take Vˆ = V . I Z d 2 3 (2. t)∇ψ ∗ (r.16) Applying the divergence theorem.14). t)∇2 ψ(r. t) · n dS . t)∇ψ(r. The probability interpretation of the wavefunction only makes sense if the normalization condition remains satisfied at all subsequent times. t)∇ψ ∗ (r. (2. t) d3 r ∂t ∂t V V Z ∗ i~ ψ (r. t)] d3 r = 2m V Z = − ∇ · j(r. t)| d r = − j(r. dt V Σ .14) Postulate 2 The time evolution of the wavefunction is given by the timedependent Schr¨ odinger equation (TDSE). t) = − ∇ ψ(r. t)|2 d3 r . 2m (2. t) d3 r .15) 2 3 ∗ V where j(r. (2. t) . The probability R of locating the particle within a volume V at time t is proportional to V |ψ(r. 2. The wavefunction evolves in time according to the time-dependent Schr¨odinger equation. t) + ψ(r. t)∇ψ(r. t) . t) d3 r = 2m V Z i~ ∇ · [ψ ∗ (r. which relates the volume integral of the divergence of a vector to the surface integral of the vector. t)| d r = ψ (r. t) − ψ(r. t) ψ(r. it is proven below that this is the case.17) |ψ(r.3 Probability conservation Consider the wavefunction ψ(r) is normalized at some time t0 . t) − ψ(r. t) = A exp[i(px − Et)/~] .19) This is the continuity equation or the local law of probability conservation. t) . We have j(x. This is analogous to electricity. in which the current density is the charge density multiplied by the velocity of charge carriers. 2m ~ ~ m (2. the wavefunction must approach zero to ensure the posibility of normalization. t)|2 d3 r = 0 . the probability current density (2. t)| + ∇ · j(r. Now. On the left side we have the rate of change of the probability that the particle is located within V. the integrand must vanish everywhere: ∂ P (r. Let us analyze the physical content of Eq. t) d3 r = 0 . t) = 0 . t) ψ(x. t) − ψ (x. t) ψ (x. the surface integral must vanish.17) is the integral form of the law of probability conservation.16) acquires the form ~ ∂ ∂ ∗ ∗ j(x. Z d |ψ(r. this quantity is the analog of the current density encountered in electricity. (2. t) + ∇ · j(r.18) dt r so the norm of the wavefunction does not change in time.24 2 Principles of quantum mechanics and applications where Σ is the bounding surface of V and n is the surface’s outward unit normal. (2. If we extend the surface Σ to infinity. the probability current density is equal to the probability density |A|2 multiplied by the particle velocity p/m.17).20) ψ(x. Eq. (2.15) and arrange it as Z ∂ 2 |ψ(r. ∂t V Since the volume V is arbitrary. ∂t (2. It follows that rhs can be interpreted as a current of probability and j is the probability current density. Let us come back to Eq. (2. t) = i p|A|2 p|A|2 p ~ −i −i = |A|2 .21) Thus. 2m ∂x ∂x Let us calculate the probability current density for the free-particle wavefunction ψ(x. t) = i (2. In one dimension. . To ensure that the probability for finding the particle in any region of space do not exceed unity. 2. r This implies that the wavefunction approaches zero at infinity.5 Time-independent Schr¨ odinger equation Suppose the potential V does not have explicit time dependence. t) = u(r) T (t) .4 Constraints on the wavefunction 2. The wavefunction may be a complex valued function. T (t) dt u(r) 2m .4 25 Constraints on the wavefunction The probability amplitude interpretation of the wavefunction and its time evolution equation imply some constraints on the wavefunction: 1. 2. the method of separation of variables. It is shown below that TDSE can be simplified to a form not containing the time variable. Sometimes functions that violate this condition are used and Eq. first derivative has to be continuous (the wavefunction is smooth). (2. For the existence of second spatial derivative. t) d3 r < ∞ . After dividing both sides by u(r) T (t) we get 1 d 1 ~2 2 i~ T (t) = − ∇ u(r) + V (r) u(r) . To guarantee a unambiguous value of probability of finding the particle at a particular position and time.2. gives one such function. the wavefunction must be square integrable.e.. The wavefunction must be continuous everywhere. 3. This requirement can not be fulfilled Having set up these constraints on the wavefunction we are now ready to consider the solutions to the Schr¨odinger equation in some specific cases. the quantity |ψ|2 must be single valued. an electron in an atom experiences an electrostatic force directed towards the nucleus. 4. since it is only |ψ|2 that has physical significance. Z |ψ|2(r.22) This is substituted into the TDSE (2. For example. i.14). ψ(r. this force derive from a potential dependent only on the distance between electron and nucleus. We look for a solution in the form of a product of spatial and temporal terms. 26). This equation is an eigenvalue one for the operator ~2 2 Hˆ = − ∇ + V (r) 2m (2. and a set of corresponding functions u(r). . as the dimension of the constant is that of energy. − T (t) dt u(r) 2m The first of these differential equations can be solved immediately.26) The time-dependence of this solution is contained entirely in an overall phase with no physical implications. (2. in the latter case we say that the eigenvalue is degenerate. T (t) = exp(−iEt/~) .24) is the time-independent Schr¨ odinger equation (TISE). (2. arranged as − ~2 2 ∇ u(r) + V (r) u(r) = E u(r) . The general solution of TDSE is a linear combination of functions given by Eq. the constant is denoted by E: 1 d 1 ~2 2 i~ T (t) = E and ∇ u(r) + V (r) u(r) = E . Most of the effort in quantum mechanics goes into solving the TISE for various systems. (2. (2. Based on our previous discussion on the individual terms of the TDSE. Let E be an energy eigenvalue and u(r) an energy eigenfunction for E. a few simple examples in the one-dimensional case follow. while u(r) is referred to as energy eigenfunction (eigenstate). the corresponding solution of TDSE is ψ(r. and the rhs only on r. the only way this equation can be true is if both sides equal a constant. E. 2m (2. this property justifies the name of stationary state for the state given by Eq. t) = u(r) exp(−iEt/~) .25) called energy operator . E is called energy eigenvalue.26). E is identified as the energy of the system.23) The second equation.2 Principles of quantum mechanics and applications 26 Since the lhs depends only on t. TISE is a linear second-order differential equation in which one has to solve simultaneously for a set of energy values. To each eigenvalue E it corresponds one ore more eigenfunctions. 6 Potential wells 27 V (x) 6 Fig. a particle having the potential energy V (x) is subjected to the force Fx (x) = − d V (x) . The interaction between particles in such cases is represented through a potential energy that possesses a local minimum. dx (2. electrons in atoms. Consider for simplicity the one-dimensional case (Fig. E xmin 2. the state of the particle is called unbound state. 2.e. the force tends to bring the particle back toward this position. The energy of the particle is E ≥ Vmin and the motion takes place between the abscissas xmin and xmax determined as solutions of the equation V (x) = E . xmax ] .2. and so on.1 A one-dimensional potential well. x0 .27) The force is zero for the local minimum position.. The region surrounding the local minimum of potential energy is called potential well. Classically. this position is an equilibrium one. It is said that the particle is in a bound state.6 x0 xmax x Potential wells There are many examples of particles whose motion is confined to a limited region of space: electrons in metals. The classical mechanics treatment of the motion allows the particle to be found in whatever position on the axis.1). . 2. if the particle is displaced from x0 . The classical motion of a particle of energy E takes place in the coordinate interval [xmin . nucleons in nuclei. Suppose now a potential which is finite along the whole axis and a particle of energy E such that V (x) ≤ E for all x-values. i. the probability per unit length of finding the particle is constant inside the well and the normalization condition gives Pcl (x) = 1/a . (2. 2. 2. The energy of the particle is E = 0 in the first case and E > 0 in the latter. (2.7.1 The one-dimensional infinite well Introduction Let us imagine a particle of mass m is moving freely inside a closed tube of length a. Classically. Since the particle can not leave the tube. 2. nor can they change infinitely fast.29) . the potential experienced by the particle can be approximated as V (x) = ( 0 0≤x≤a ∞ otherwise. collisions between the particle and the ends of the tube are assumed to be elastic. the constant value of the potential inside the tube is chosen V (x) = 0 .2 The infinitely deep one-dimensional square potential well of width a. the approximation is justified for reasons of mathematical simplicity. we say that the particle has a continuous spectrum of energy. For the moving particle. Whatever value E > 0 is possible.7 2.28) where the axis Ox is chosen with the origin at one end of the tube and the other end is positioned at x = a (Fig. It is clear that real potentials can not be infinite.28 2 Principles of quantum mechanics and applications V (x) 6 V (x) = ∞ V (x) = ∞ a 0 -x Fig.2). the particle inside such a well can be at rest or it bounces back and forth due to ellastic collisions with the walls. In the interval [0. dx2 ~ (2. The solution ( A exp(κx) + B exp(−κx) 0 ≤ x ≤ a u(x) = 0 otherwise. Defining κ = −2mE/~ the solution is u(x) = A exp(κx) + B exp(−κx) . 2m dx2 that can be expressed as d2 2m u(x) + 2 E u(x) = 0 .2 29 Energy eigenvalues and eigenfunctions In the quantum mechanical framework. we need to solve the one dimensional time-independent Schr¨odinger equation with the potential given by Eq.7 The one-dimensional infinite well 2. Continuity of the wavefunction at the points x = 0 and x = a gives A + B = 0 and A exp(κa) + B exp(−κa) = 0 . The result is that expected as the energy is the sum of the kinetic energy.2.28). The result is A = B = 0 . a] the time-independent equation is that of a free particle of zero potential energy. where A and B are constants that are to be determined. so u(x) = 0 everywhere. and the potential one which is also positive here. We conclude that the particle can not have a negative energy. a positive defined quantity.7. 0 ≤ x ≤ a. − ~ 2 d2 u(x) = E u(x) .30) The form of the solution depends on the sign of the energy. This is not an acceptable wavefunction. √ Let us first look for negative energies. must be a continuous function. In the regions x < 0 and x > 0 we have − ~ 2 d2 u(x) + ∞ u(x) = E u(x) 2m dx2 and the only possible solution is u(x) = 0 which means that the probability of finding the particle outside the well is zero. (2. . 30) has the solution u(x) = Ax + B . (2. . The result is that expected. .. a unacceptable solution. gives B = 0 and Aa + B = 0 with the solution A = B = 0 .34) Now. . . The same condition at x = 0 reads A sin ka = 0 .e.33) This means that the wavenumber k is quantized (takes on discrete values) and can have only the particular values kn = n /a . . n = 1. (2. Since k > 0 the solution is ka = n . . . 0≤x≤a and the wavefunction is u(x) = ( A sin kx + B cos kx 0 0≤x≤a otherwise. 2. a]. n = 1. i. (2. (2. . Defining √ k = 2mE/~ > 0 . We turn now to searching for positive energies. . . from Eq.35) . 0≤x≤a and the continuity condition of the wavefunction ( Ax + B 0 ≤ x ≤ a u(x) = 0 otherwise. (2.31) the energy of the particle is also quantized: En = 2 ~2 2ma2 n2 . as a state of zero energy would imply an exact position determination and also an exact value (zero) of the momentum in contradiction with the Heisenberg uncertainty principle. n = 1. . Again u(x) = 0 everywhere.30 2 Principles of quantum mechanics and applications Next we look for an energy E = 0 .32) The continuity condition at x = 0 gives B = 0 . 2.30) has the general solution u(x) = A sin kx + B cos kx . the condition sin ka = 0 must be satisfied. (2. 2. (2. Not to get u(x) = 0 everywhere. . this equation can be recognized as the condition for a standing wave in the interval [0.31) Eq. The equation (2. . .36) un (x) = 0 otherwise. (2. In the ground state (n = 1) the energy is positive. 25 31 n 5 20 4 15 10 3 5 2 Ground state 0 Fig. Quantification of the energy arises naturally by imposing boundary conditions on solutions of Schr¨odinger equation as opposed to its ad-hoc introduction in its model of the hydrogen atom. the valleys correspond to positions of low probability.2. Since the phase of the constant remains arbitrary. The probability per unit length of finding the particle at x in a state n is ( (2/a) sin2 n x/a 0 ≤ x ≤ a Pn (x) = |un (x)|2 = n = 1. 1 The energy level diagram of the particle is shown in Fig. . 2.37) 0 otherwise. .e. 2. .. Figure 2. . the peaks correspond to positions of high probability of finding the particle.7 The one-dimensional infinite well E/E1 6 .4 shows the plot of first few wavefunctions and probability densities. Normalization condition determines the value of the constant A: Z a Z ∞ n a 2 |An |2 sin2 |u(x)| dx = x dx = |An |2 = 1 a 2 0 −∞ p hence |An | = 2/a . In a plot of the probability density.3. 2.3 Energy level diagram for the particle in the one-dimensional infinite well. The wavefunctions can now be written in the final form (p 2/a sin n x/a 0 ≤ x ≤ a n = 1. . i. arg An = 0 .. . 2. we take the simpliest choice. . . (2. 2.5 x/a 1 1 5 0 0.0 0. each successive eigenfunction has one more node.5 x/a 1. .2 Principles of quantum mechanics and applications 32 n 1 1 1 0 −1 0 1 1 2 0 −1 0 1 1 3 0 −1 0 1 1 4 0 0 (a/2)|u(x)|2 (a/2)1/2u(x) −1 1 0 −1 0 0.4 The lowest five energy eigenfunctions for the infinite potential well (left) and the corresponding probability densities (right).0 Fig. With increasing n. 35) gives n ≈ 4.55 eV. 4. classical behaviour is expected for large values of the quantum number n . the probability density (2. This is a manifestation of the uncertainty principle since the particle is localized within ∆x = a/2 so ∆px ≥ ~/2∆x = ~/a and E= 1 2 1 1 ~2 hpx i = hp2x i − hpx i2 = (∆px )2 ≥ . The energy spacing to adjacent levels n ± 1 is |En±1 − En | ≈ 2 ~2 2ma2 2n = 2 En ≈ 4. Eq. We consider an interval of length ∆x inside the well. The quantum description shows that the minimum energy is strictly positive. 13. m and a is the same as that of E1 . the classical result.7 The one-dimensional infinite well 2. Suppose an electron trapped in an atomic-scale potential well of width a = 0. Equation (2. the probability per unit length has a number of peaks of the order of 1033 .3 33 Discussion 1.2. Let us derive this result as a limit case of the quantum description. The uncertainty principle is often used to provide a quick order of magnitude estimate for the ground state energy. (2.7. sin = a a a2 a i. Further. the next energy levels being positioned at 5.7 × 10−34 J .3 × 1033 . for large values of n . Supposing the system in an energy eigenstate. and so on.37) has many oscillations on the interval and its average value is 2 D 2 n xE 2 1 1 = . 2. note that the dependence on ~. Classically a particle in an infinite potential well can have zero velocity. According to the correspondence principle. energy appears to take on a continuous range of values. 2m 2m 2m 2ma2 The actual ground state energy E1 is 2 larger than the lower bound.e. the significant . 3..29) gives the probability per unit length of finding the particle inside the well in the classical description. Let us consider an example of system we are dealing with. and hence unresolvable from each other.02 eV.51 eV. The energy of the ground state is E1 = 2 ~2/2ma2 ≈ 1. The peak spacing is ∼ 10−33 m. To make clear why we do not detect the quantum nature of matter at a macroscopic scale let us consider a mass of 1 kg trapped in a potential well of width 1 m. n much less than the energy of the system and impossibly small to detect. so zero energy.5 nm. 5.2 Principles of quantum mechanics and applications 34 probability density is the averaged one and this coincides with the classical one.39) V0 0 ≤ x ≥ a . that is. 2. the probability density is the same at x and −x. The energy of a particle in this potential is E > 0 . (2. The energy eigenfunctions for x ∈ [−a/2. the potential is an even-parity function [V (−x) = V (x) for all x]. Potential barrier penetration We now consider the scattering of a particle by a potential barrier such as the one shown in Fig. 2. 2.38) These wavefunctions are alternatively even and odd.1 Case 0 < E < V0 Classical mechanics predicts that a particle coming from the left with an energy 0 < E < V0 is reflected back at x = 0 . The potential barrier in Fig.5 can be written mathematically ( 0 x < 0.8 The rectangular potential barrier. The physical background of this is that there is no reason for the particle to distinguish between the two sides of a symmetrical well. The property of energy eigenfunctions to have a definite parity is a general one in case of a symmetric well and useful to simplify the determination of the time-independent Schr¨odinger equation solution. By solving the TISE we will find an important quantum phenomenon—barrier penetration or tunnelling. We shall consider the cases 0 < E < V0 and E > V0 separately. a/2] reads un (x) = r 2 n (x + a/2) sin = a a r n x n 2 sin + a a 2 and by a phase change of in case of need (p 2/a cos n x/a n odd un (x) = p 2/a sin n x/a n even.8. . Let us choose the axis origin in the middle of the well. Now. so |un (−x)|2 = |un (x)|2 for all n and x. x > a V (x) = (2. 5. 2. 0 a -x We can immediately write the general solution: C1 exp(ikx) + C2 exp(−ikx) x < 0 u(x) = C3 exp(κx) + C4 exp(−κx) 0≤x≥a exp[ik(x − a)] x > a.8 The rectangular potential barrier. For simplicity. (2. the energy of the particle is not .43c) (2. By applying the boundary conditions at x = 0 and x = a. 2. (2. (2. κ[C3 exp(κa) − C4 exp(−κa)] = ik . (2.5 A 1-D rectangular potential barrier of width a and height V0 . the amplitude of the wave in the region x > a has modulus 1 . (2.41) κ = [2m(V0 − E))]1/2 /~ .43a) ik(C1 − C2 ) = κ(C3 − C4 ) .43b) C3 exp(κa) + C4 exp(−κa) = 1 . Potential barrier penetration 35 V (x) 6 V0 Fig. In other words.2. It means that the energy has a continuous spectrum. A particle is incident from the left.43d) The solution for C1 is C1 = cosh κa + (i/2)(κ/k − k/κ) sinh κa . (2.42) and For a particle incident from the left we anticipate that in the region x > a there is no wave propagating to the left.44) There is solution whatever the value of E . we get the set of algebraic equations C1 + C2 = C3 + C4 .40) where k = (2mE)1/2 /~ (2. The parameter ε expresses the energy in units of the barrier height. m We define the transmission coefficient of the barrier as j(x) = T = Here. (2. we introduce the dimensionless parameters r 2mV0 α= a and ε = E/V0 .50) can be approximated: √ √ [(1/2) exp(α 1 − ε )]2 exp(2κa) [(1/2) exp(α 1 − ε)]2 −1 = = T ≈ 4ε(1 − ε) 4ε(1 − ε) 16ε(1 − ε) so T ≈ 16ε(1 − ε) exp(−2κa) (2. (2. The use of these parameters in the expression of the transmission coefficient gives √ sinh2 α 1 − ε −1 T =1+ .2) ~k |C1 |2 − |C2 |2 . The probability current density in this region is (see Problem 2. |C1 |2 The expression of C1 is introduced here and it yields ! r 2 2 κ 1 V 2m(V − E) k 1 0 0 sinh2 κa = 1+ + sinh2 a .49) ~2 T = The parameter α characterizes the potential barrier and it will be called strength parameter. |jtransmitted | .51) . (2.2 Principles of quantum mechanics and applications 36 quantized.50) 4ε(1 − ε) √ Special case: For α 1 − ε ≫ 1 the transmission coefficient is small and Eq. |jincident | (2.46) (2.45) (2. T −1 = 1+ 4 k κ 4 E(V0 − E) ~2 (2. In region x < a there are two counterpropagating waves: the incident wave and the reflected wave. Let us consider now the probability flux density. x < 0 m while the probability current density in the region x > a is j(x) = ~k .48) In order to write the transmission coefficient in a simpler form.47) 1 . 8. T =1+ 4ε(ε − 1) (2. E = 1 eV .72.78 . Potential barrier penetration 37 1 (a) 0 1 (b) 0 1 (c) 0 1 T (d) 0 0 1 2 E /V0 3 4 Fig.3 Discussion • 0 < ε < 1 .2. Exercise: √ sin2 α ε − 1 −1 . ε = 0.52) Figure 2.8 The rectangular potential barrier. 2. 2. a = 0. The transmission coefficient is an increasing function of energy and it can take significant values for sufficiently small values of the strength parameter α .6 Transmission coefficient of a rectangular potential barrier versus normalized energy of the incident particle for some values of the barrier parameter α: (a) α = 1 . 2.1 nm . (c) α = 5 . there is a significant .6 shows the transmission coefficient of a barrier as a function of the normalized energy for various values of the barrier strength parameter.5 and T ≈ 0.1 × 10−31 kg).2 Case E > V0 Not treated during the lecture. Electron (me ≈ 9. We get α ≈ 0. and (d) α = 10 . V0 = 2 eV.8. Examples: 1. (b) α = 2 . (2. We get α ≈ 31.55) In other words. V0 = 2 eV. We divide the barrier into a large number of back-to-back rectangular barriers.50) of the transmission probability is restricted to rectangular barriers only. In reality. (2. Barrier penetration can also be explained in terms of energy–time uncertainty relation. (2. Figure 2.5 × 10−19 . we are not likely to encounter such a simpleshaped barrier. E = 1 eV . It can be shown that the transmission coefficient of the barrier can be approximated as ) ( Z p x2 2m[V (x) − E] dx . A fluctuation giving the energy E + ∆E > V0 can enable the particle to pass over the barrier.54) the condition for maxima becomes k′ a = n . The graphs exhibit oscillations and the transmission is unity for √ α ε − 1 = n . ε = 0.2 Principles of quantum mechanics and applications 38 probability of barrier penetration. a = 0. The energy of the particle fluctuates.9 Non-rectangular potential barriers The expression (2. 2. • ε > 1 . .53) By introducing the constant of the wave inside the barrier k ′ = [(2m(E − V0 )1/2 /~ (2. Proton (mp ≈ 1 836me ).5 and T ≈ 3. in the study of the Fabry–Perot interferometer. the width of the barrier is an integral number of half wavelengths. . .56) T = exp −2 ~ x1 . n = 1.7 shows a barrier of arbitrary shape towards which a particle of mass m and energy E is directed from the left. The probability of barrier penetration is not significant. . This condition was encountered in optics. 2.1 nm . 3. We give below an approximate formula for the transmission coefficient of an arbitrarily-shaped potential barrier. 2. the magnitude of the fluctuation is ∆E on a timescale ∆t ∼ ~/∆E . The maxima are explained as a constructive interference of waves. 1. By approximating the linear part V (x) = V0 − eEx by a series of square barriers the transmission coefficient for electrons is .7 A potential barrier of arbitrary shape. The electrons faces now a potential barrier and its penetration is expected. The energy difference V0 − EF (V0 is the height of the well) is the electron work function Φ of the metal. another method that gives rise to the emission of electrons from a metal is the application of an intense electric field. The application of a uniform electric field E directed towards the metal surface x = 0 produces a change in the potential field obeying dV = −eEdx . The potential experienced by metal electrons is shown in Fig.57) V0 − eEx x ≥ 0 .10.8(b).10 Applications of tunneling 2. 2. shown in Fig. 2.1 Field emission of electrons Electrons in a metal are bound by a potential that may be approximated by a finite potential well.8(a). Electrons can be extracted from the metal in the photoelectric effect (see Sect. Electrons fill up these energy levels up to an energy EF called Fermi energy. 2. 2. The coordinates x1 and x2 > x1 are defined by V (x1 ) = V (x2 ) = E . By integration with the condition V (0) = V0 we find the modified potential ( 0 x<0 V (x) = (2. We consider the metal situated in the region x < 0 .2. a process referred to as field emission. The barrier is decomposed into a sequence of rectangular barriers. where e denotes the magnitude of the electron charge.2).10 Applications of tunneling 39 V (x) 6 E x1 x2 -x Fig. Here it experiences strong attractive nuclear forces. A straightforward calculation gives ! √ 4 2m Φ3/2 T = exp − . 2. i. an STM can thus be used to construct a very accurate contour map of the surface under investigation.58) It can be seen that the probability of emission increases strongly as the electric field strength near the surface of the metal increases. Assuming that the potential difference is held constant. 2. electric field strengths as high as 1010 V/m are required. -emission is the process wherein a nucleus spontaneously decays by emitting an particle..e. To obtain a high probability of tunneling. The half-life of this process is about 4.40 2 Principles of quantum mechanics and applications calculated by help of Eq.56). A good example is provided by the decay of the isotope uranium238 (parent nucleus) into thorium-234 (daughter nucleus). Electrons tunneling between the surface and the probe tip give rise to a weak electric current whose magnitude is proportional to the tunneling probability given by Eq.58). (2. For r > r1 the -particle is repelled by the Coulomb force between it and the daughter nucleus. The field emission of electrons from a metal surface is used for imaging surfaces at the atomic level in the scanning tunneling microscope (STM). (2. A very sharp conducting probe is brought close to the surface of a solid conducting medium and a large voltage difference is applied between the probe and the surface.10. x2 = Φ/eE. (2. We apply this formula for electrons at energy E = EF . The explanation of this process is based on the assumption that an particle has a separate existence inside the nucleus.2 Alpha-particle emission An -particle is a synonymous term for a helium nucleus.9). 4 ǫ0 r . the electrostatic potential energy is V (r) = 2(Z − 2)e2 . the resulting potential can be modeled as a deep rectangular well of radius r1 that is interpreted as the nucleus radius (Fig. 3~eE (2.56) are x1 = 0 and x2 given by V0 − eEx2 = EF . the current is an extremely sensitive function of the spacing between the tip and the surface. the integration ends in Eq.2 MeV. The tunneling is also advantaged by a small work function (see Table ??).47 billion years and the kinetic energy of the ejected -particle is about 4. (2. The restoring force is the elastic force F given by Hooke’s law F = −kx . 4 ǫ 0 r2 The -decay can be modeled as a penetration through this nuclear-plusCoulomb barrier.59) .11 The quantum harmonic oscillator 41 V (x) V0 EF = V 0 − Φ 6 METAL (a) VACUUM -x V (x) V0 EF = V 0 − Φ 6 E (b) - x1 = 0 x2 x Fig. 2. A typical example of a harmonic oscillator is that of a mass attached to a spring. a harmonic oscillator (also known as linear oscillator or simple oscillator) is a physical system that is bound to a position of stable equilibrium by a restoring force proportional to the displacement from this position. (b) In the presence of an electric field E the potential step is changed into a triangular potential barrier that can be penetrated by electrons.8 Field emission of electrons: (a) The electrons in a conductor placed in the region x < 0 are restrained by a potential step of minimal height equal to the work function Φ. where Z denotes the atomic number of the parent nucleus. 2.2.11 The quantum harmonic oscillator In classical mechanics. For a particle of energy E smaller than V (r1 ) the barrier extends to the radius r2 given by 2(Z − 2)e2 =E . 2 Principles of quantum mechanics and applications 42 V (r) 6 ← Coulomb potential E 0 V0 r1 r2 -r ↑ Nuclear potential Fig. 2.9 The potential energy experienced by an α-particle inside the atomic nucleus (0 ≤ r < r1 ) and its vicinity. where x is the displacement and k is the spring constant. The motion of a body of mass m attached to the spring is governed by Newton’s second law m d2 x(t) = −kx dt2 (2.60) whose general solution is x(t) = A cos(ωt + φ) . (2.61) Here, ω= p k/m (2.62) is the natural oscillating frequency, A is the amplitude of the oscillation, and φ is the phase constant; both A and φ are constants determined by the initial condition (initial displacement and velocity). The work of force F between two positions x1 and x2 is Z x2 1 1 W = F (x) dx = kx21 − kx22 (2.63) 2 2 x1 and this do not depend on the path between them, so a potential energy V (x) can be defined. As a general rule, W = −∆V = V (x1 ) − V (x2 ) , (2.64) 2.11 The quantum harmonic oscillator 43 so V (x) = (1/2)kx2 + const; the simple choice V (0) = 0 gives V (x) = (1/2)kx2 = (1/2)mω 2 x2 . (2.65) The harmonic oscillator is one of the most important models in mechanics because any potential V (x) can be approximated as a harmonic potential in the vicinity of a stable equilibrium point. We now turn our attention to the quantum description of the harmonic oscillator. The 1-D TISE with the potential given by Eq. (2.65) is − ~ 2 d2 1 u(x) + mω 2 x2 u(x) = E u(x) . 2m dx2 2 (2.66) As first step in this equation solving, dimensionless coordinate and energy are introduced; the substitutions ξ = (mω/~)1/2 x = αx give the equation and ε = E/(~ω/2) d2 u(ξ) + (ε − ξ 2 )u(ξ) = 0 . dξ 2 (2.67) (2.68) There is not a straightforward method to solve this equation. We first investigate the asymptotic behavior of u(y). For large values of |ξ|, the parameter ε is negligible compared to ξ 2 and the above equation becomes d2 u(ξ) − ξ 2 u(ξ) = 0 . dξ 2 The approximate solutions of this equation are u(ξ) = C exp(±ξ 2/2) , Indeed, C = const . d2 u(ξ) = (ξ 2 ± 1)u(ξ) ≈ ξ 2 u(ξ) dξ 2 for large |ξ| . The function defined by u(ξ) = exp(ξ 2/2) is not satisfactory because it becomes infinite as |ξ| → ∞ and the function defined by u(ξ) = exp(−ξ 2/2) approaches 0 at infinity, so it is well-behaved. Retaining the asymptotic behavior given by exp(−ξ 2/2) , we are led for searching solutions of the form u(ξ) = H(ξ) exp(−ξ 2/2) , where H(ξ) is a function to be determined. (2.69) 2 Principles of quantum mechanics and applications 44 Substitution of Eq. (2.69) into Eq. (2.68) gives d d2 H(ξ) − 2ξ H(ξ) + (ε − 1)H(ξ) = 0 . 2 dξ dξ (2.70) The solution of this differential equation is searched in the form of a series H(ξ) = a0 + a1 ξ + a2 ξ 2 + · · · . (2.71) This form of H is inserted into Eq. (2.70); the coefficient of ξ s (s ∈ N) in left-hand side of the equation is (s + 1)(s + 2)as+2 − 2sas + (ε − 1)as and it must be zero, so as+2 = 2s + 1 − ε as , (s + 1)(s + 2) s ∈ N. (2.72) In the limit of large s , as+2 /as = 2/s , and this ratio can also be encountered in the Taylor expansion of exp(ξ 2 ): 2 eξ = c0 + c1 ξ + c2 ξ 2 + c3 ξ 3 + · · · = 1 + ξ 2 + ξ4 ξ6 + + ··· . 2! 3! Indeed, the ratio of adjacent coefficients for large s is c2s+2 1 1 2 = ≈ = . c2s s+1 s 2s It follows that H(ξ) and exp(ξ 2 ) have the same asymptotic behaviour. Coming back to the solution (2.69) of Schr¨odinger equation, this behaves like exp(ξ 2/2) for large ξ. This function diverges and is not an acceptable solution. The only way to avoid the divergence of the wave function is to terminate the series (2.71) after a finite number of terms. Examining Eq. (2.72) we find that an (n ∈ N) is the last term in the series if 2n + 1 − ε = 0 , which means ε = 2n + 1 ≡ εn , n ∈ N . By use of Eq. (2.67), the possible values of the energy are given by En = (~ω/2)εn = ~ω(n + 1/2) , n ∈ N. (2.73) The energy of the quantum oscillator is thus quantified. The minimum energy is positive, ~/2 , and the energy levels are equally spaced (Fig 2.10). . . where Nn is the normalization constant.1 First six Hermite polynomials. −∞ (2. (2. n 0 1 2 3 4 5 Hn (x) 1 2x 4x2 8x3 − 12x 16x4 − 48x2 + 12 32x5 − 160x3 + 120x We now focus on the energy eigenfunctions determination.2.11 The quantum harmonic oscillator E/~ω 6 5 45 n .76) (2.75) Table 2. dξ 2 dξ (2.74) The solution of this equation is known as Hermite polynomial of order n and is usually expressed in the form Hn (ξ) = (−1)n eξ 2 dn −ξ2 e . By imposing Z ∞ |un (x)|2 dx = 1 . dξ n (2. The solution of Schr¨odinger equation for the energy En reads now un (x) = Nn exp(−α2 x2/2) Hn (αx) . The use of εn into Eq. 3 2 2 1 1 Ground state 0 0 Table 2.70) gives d2 d H(ξ) − 2ξ H(ξ) + 2nH(ξ) = 0 .10 Energy level diagram for the quantum harmonic oscillator. 4 4 3 Fig. 2.77) .1 lists first few Hermite polynomials. 1 0.1 0. the energy level separation is ~ω.5 0.3 0.3 0.3 0. For a radiation field of angular frequency ω.5 0.0 −0.3 0. 2.2 0.3 0.0 0.1 0.5 −4 −2 0 αx 2 4 0.5 0. it is found Nn = r α √ .0 0.0 −0.1 0.2 0.2 0.5 0.11 shows first few energy eigenfunctions and corresponding probability per unit length.11 The lowest five energy eigenfunctions for the quantum harmonic oscillator (left) and the corresponding probability densities (right).5 0.78) Figure 2.1 0.0 −4 1 2 3 4 5 −2 0 αx 2 4 Fig.5 0.5 0.0 −0.0 0.2 0.5 0.2 Principles of quantum mechanics and applications 46 n 0.5 0. the energy of the field is n~ω above the ground state. One can associate a set of n .0 0.0 α−1|ψn(x)|2 α−1/2ψn(x) −0.2 0. Suppose the radiation field in the energy eigenstate n.0 −0. 2n n! (2. Applications of the quantum oscillator • Study of the oscillations of the atoms of a molecule about their equilibrium position • Study of the oscillations of atoms of a crystalline lattice • Study of a the electromagnetic field of radiation One mode of the radiation field is treated as a quantum oscillator. the interaction processes are the stimulated (induced) emission and absorption. the fluctuations of the field induce the dezexcitation of the atom. these particles are called photons.24)] is a partial differential equation which gives rise to considerable mathematical complications in general. dividing by X(x)Y(y)Z(z) and rearranging. (2. (2. . solving of the 3-D TISE will be limited below to those cases where the separation of variables technique can be used.12 Three-dimensional Schr¨odinger equation 47 identical particles to the radiation field.7. each one carrying the energy ~ω. This subject was treated phenomenologically in Sect. 2.2.24). each of which is a function of only one of the three Cartesian coordinates: V (r) = Vx (x) + Vy (y) + Vz (z) . (2. The interaction between the radiation field and an atom can produce jumps of the atom from one state to another.81) 2m Z(z) dz 2 This equation says that a function of x alone (the term inside the first square bracket).79) We look for wavefunctions that express as a product of three one-dimensional functions.80) into Eq. plus one of y alone. (2. (2. If the field is in an excited state (n ≥ 1). 1. there is no radiation field. If the radiation field is in its ground state1 and the atom in an excited state.80) Substituting Eqs. 3-D TISE [Eq. Let us consider the case where the potential energy V (r) can be written as a sum of three terms. (2. The transition of the radiation field from one energy eigenstate to another corresponds to the creation (emission) or destruction (absorption) of a number of photons. we get ~2 1 d2 X(x) ~2 1 d2 Y (y) − + Vx (x) + − + Vy (y) 2m X(x) dx2 2m Y (y) dy 2 ~2 1 d2 Z(z) + − + Vz (z) = E . plus one of z alone.79) and (2. equals a constant 1 Classically. The quantum theory confirms Einstein’s theory. u(r) = X(x) Y (y) Z(z) . the phenomenon is the spontaneous emission.12 Three-dimensional Schr¨ odinger equation Unlike the 1-D case. 24)] are given by Eq. 2m dy 2 ~2 d2 Z(z) − + Vz (z)Z(z) = Ez Z(z) .83) 2. a = 0.26). Hint: Apply the probability conservation condition for the wavefunction (2.2 Calculate the probability current density for a particle of mass m and energy E described in a region of the Ox axis by the wavefunction ψ(x. Ey . The equation can only be true if each square bracket is independently constant.4 An electron is trapped in an infinite potential well of 1 cm width. (2. 2.2 Principles of quantum mechanics and applications 48 E. 2. t) = u(x) exp[−i(E/~)t] .81) split into three one-dimensional TISEs: ~2 d2 X(x) + Vx (x)X(x) = Ex X(x) . Give a direct proof. 2m dx2 ~2 d2 Y (y) − + Vy (y)Y (y) = Ey Y (y) .82c) After solving these three problems. A and B are complex constants and κ is a positive quantity.3 Determine the ground state energy of an electron confined to an infinite square well of width equal to the diameter of a hydrogen atom. Is quantum theory required for system description? . (2.11 nm. Interpret the results. 2m dz 2 − (2. 2. (2. where (a) u(x) = A exp(ikx) + B exp(−ikx) . Eq. (2.82a) (2.1 In Eq.24) it is understood that E is a real quantity.13 Problems 2. (2.82b) (2. and Ez the constants. A and B are complex constants and k is a positive quantity. Denoting Ex .80) and the energy eigenvalues are given by E = Ex + Ey + Ez . the energy eigenfunctions of the 3-D TISE [Eq. (b) u(x) = A exp(κx) + B exp(−κx) . 8 Use position-momentum uncertainty relation to estimate the ground state energy of the quantum oscillator of frequency ω.7 The surface of an electrode lies in the plane x = 0 and the potential energy outside it is approximated in the form ( V0 (1 − x2/a2 ) 0 ≤ x ≤ a V (x) = 0 otherwise. Determine the probability of barrier penetration. (b) Determine the probabilities of the energy eigenstates. where α = (mω/~)1/2 .9 The ground state energy eigenfunction of an oscillator of mass m and frequency ω is of form u(x) = C exp(−α2x2/2) .6 An electron of energy E = (1/2)V0 is incident on a potential barrier given by V (x) = V0 exp(−x2/x20 ).13 Problems 49 2. 2.5 A particle in an infinite square well is described by the initial wavefunction ( N x(a − x) 0 ≤ x ≤ a ψ(x) = 0 otherwise. where N is a complex constant. . V0 = 1 eV and x0 = 0.2. (a) Normalize the wavefunction.1 nm. 2. Calculate the probability of penetration of this barrier for an electron of energy 0 < E < V0 . 2. Determine the normalized eigenfunction. 2. 50 2 Principles of quantum mechanics and applications . . The standard uncertainty in the last two digits is given in parenthesis. 7].Appendix A Fundamental physical constants The tables give values of some basic physical constants recommended by the Committee on Data for Science and Technology (CODATA) based on the 2006 adjustment [6. 109 382 15(45) × 10−31 kg 0.272 013(23) MeV 1. p = 101.152 672 47(80) 7.135 667 33(10) × 10−15 eV s 1.817 940 2894(58) × 10−15 m 927.605 691 93(34) eV h/2π ~ electron mass energy equivalent in MeV electron charge to mass quotient proton mass mp = Ar (p) u energy equivalent in MeV neutron mass mn = Ar (n) u energy equivalent in MeV proton-electron mass ratio fine-structure constant e2/4πǫ0 ~c inverse fine-structure constant Rydberg constant α2 me c/2h R∞ hc in eV Wien displacement law constants b = λmax T b′ = νmax /T Stefan-Boltzmann constant (π2/60)k 4/~3 c2 Bohr radius α/4πR∞ = 4πǫ0 ~2/me e2 Compton wavelength h/me c classical electron radius α2 a0 Bohr magneton e~/2me me me c 2 −e/me mp b b′ 2.400 915(23) × 10−26 J T−1 5.413 996(39) × 10−3 m3 mol−1 2.1 An abbreviated list of the CODATA recommended values of the fundamental constants of physics and chemistry. × 10−12 F m−1 6.565 346(23) MeV 1836.510 998 910(13) MeV −1.854 187 817.617 343(15) × 10−5 eV K−1 Vm n0 e F h 22.672 621 637(83) × 10−27 kg 1.426 310 2175(33) × 10−12 m 2.670 400(40) × 10−8 W m−2 K−4 0.626 068 96(33) × 10−34 J s 4.897 7685(51) × 10−3 m K 5.050 783 24(13) × 10−27 J T−1 3.035 999 679(94) 10 973 731.3399(24) C mol−1 6. c0 µ0 ǫ0 G NA R k molar volume of ideal gas (T = 273.15 K.568 527(73) m−1 13.297 352 5376(50) × 10−3 137.152 451 2326(45) × 10−8 eV T−1 mp c 2 mn mn c 2 mp /me α 1/α R∞ .380 6504(24) × 10−23 J K−1 8..674 927 211(84) × 10−27 kg 1.A Fundamental physical constants 52 Table A.529 177 208 59(36) × 10−10 m 2.054 571 628(53) × 10−34 J s 6.314 472(15) J mol−1 K−1 1.022 141 79(30) × 1023 mol−1 8.007 276 466 77(10) u 938..686 7774(47) × 1025 m−3 1.788 381 7555(79) × 10−5 eV T−1 5.758 820 150(44) × 1011 C kg−1 1.008 664 915 97(43) u 939.325 kPa) Loschmidt constant NA /Vm elementary charge Faraday constant NA e Planck constant Value 299 792 458 m s−1 (exact) 4π × 10−7 N A−2 8. Quantity speed of light in vacuum magnetic constant electric constant 1/µ0 c2 Newtonian constant of gravitation Avogadro constant molar gas constant Boltzmann constant R/NA Symbol c.878 933(10) × 1010 Hz K−1 σ a0 λC re µB nuclear magneton e~/2mp µN 5.674 28(67) × 10−11 m3 kg−1 s−2 6.602 176 487(40) × 10−19 C 96 485.582 118 99(16) × 10−16 eV s 9. 53 Table A. .660 538 782(83) × 10−27 kg a The electronvolt is the kinetic energy acquired by an electron in passing through a potential difference of one volt in vacuum.602 176 487(40) × 10−19 J b (unified) atomic mass unit 1 u = mu = (1/12)m(12 C) = 10−3 kg mol−1/NA u 1. Name of unit Symbol Value in SI units ˚ ˚ angstr¨om A 0. at rest and in its ground state.1 nm = 100 pm = 10−10 m a electron volt : (e/C) J eV 1. b The unified atomic mass unit is equal to 1/12 times the mass of a free carbon 12 atom.2 The values in SI units of some non-SI units. 54 A Fundamental physical constants . . and engineering Appendix B Greek letters used in mathematics. ε ζ η θ. ϑ ι κ. ς τ υ φ. ϕ χ ψ ω .56 B Greek letters used in mathematics. and engineering Name of letter alpha beta gamma delta epsilon zeta eta theta iota kappa lambda mu nu xi omicron pi rho sigma tau upsilon phi chi psi omega Capital letter A B Γ ∆ E Z H Θ I K Λ M N Ξ O Π P Σ T Υ Φ X Ψ Ω Lower-case letter α β γ δ ǫ. science. ̺ σ. science. ̟ ρ. κ λ µ ν ξ o π. Principles of Quantum Mechanics (2nd ed. Newell. N. Vol. The article can also be found at http://physics. Ref. [7] CODATA Internationally recommended values of the fundamental physical constants. Popescu. Editura Universit˘a¸tii din Bucure¸sti. Vols. Kluwer Academic. Fizic˘ a.Bibliography [1] I. 633–730 (2008). Mod. New York (1994). B. [6] P.).html. Prentice Hall (1995). and F. Bucure¸sti (2007).nist. II. Phys. Data 37(3). John Wiley & Sons. Rev. M. and D.gov/cuu/Constants. 1187–1284 (2008). [3] C.gov/cuu/Constants/papers. http://physics. Griffiths. Diu. B. Mohr. Shankar. . CODATA recommended values of the fundamental physical constants: 2006. Lect¸ii de mecanic˘ a cuantic˘ a. [5] R. B. Phys 80(2). Taylor. Bucure¸sti (1983). [4] D.nist. Chem. Editura Didactic˘a ¸si Pedagogic˘a Bucure¸sti. J. Cohen-Tannoudji. Quantum Mechanics. J. Lalo¨e. New York (1977). Introduction to Quantum Mechanics. I and II. J. Florescu. [2] V.