5Propagators and Green functions 5.1 The causality issue Quantum field theory is buit over the same postulates as quantum mechanics, but with fields and conjugate canonical momenta replacing positions and ordinary momenta. A key ingredient, a priori absent in QM, is causality; i.e. the statement that information cannot propagate faster than light. A good way analyzing this issue is by considering the Green function or propagator G(~x, ~x 0 ; t). This is defined as the probability amplitude fot a particle that is initially localized at point ~x to be at point ~x0 after a time t has elapsed G(~x, ~x0 ; t) = h~x0 |e−iHt |~xi. (199) H is assumed to be time-independent (this is almost invariably the case in field theory) and, in fact we shall assume it is the free hamiltonian. This is also the relevant case for field theory since perturbation theory is the analytical tool that is commonly used. Between interactions particles are assumed to propagate freely. It is a well known reault in QM that G0 (~x, ~x0 ; t) = h~x0 |e−iH0 t |~xi (200) 3 m 2 im |~x0 −~x|2 2t = d ph~x |e |~ pih~ p|~xi = (201) e 2πit From this expression it is quite apparent that the propagator is non-zero bewtween any two Z 3 0 2 P −i 2m t space-time points regardless whether one is in the causal cone of the other or not. In fact, nothing special happens for points on the boundary of the light-cone itself. Moving to a relativistic theory, but still sticking to the one-particle, or first quantized, interpretation, we would change (201) to G0 (~x, ~x0 ; t) = Z d3 ph~x0 |e−i √ p ~2 +m2 t |~ pih~ p|~xi. (202) A direct calculation shows that G0 (~x, ~x0 ; t) ∼ e−m √ |~ x0 −~ x|2 −t2 . (203) This expresion, although relativistic does not really solve the causality issue present in (201); the propagation amplitude is small but non zero for |~x 0 − ~x|2 t2 (well outside the light cone). 36 h0|φ(x)φ(y)|0i = d3 p (2π)3 2Ep Z = Z Z d3 k e−ipx+iky h0|ap a†k |0i (2π)3 2Ek d3 p e−ip(x−y) . y) = h0|φ(x)φ(y)|0i. i. (204) Using elementary QM techniques.2 Propagators in field theory We shall present the discussion in terms of a scalar field and in the next section we shall discuss the modifications for fermions and vector fields. a function of x − y. When we ask whether an observation made at space-time point x can influence another at y. the analogy with (201) is manifest. 0)e−iP 0 (x0 −y 0 ) φ(~y .e. 5. ~r = ~x − ~y . we will find that the amplitudes due to particle and antiparticle propagation exactly cancel. 0)|0i.Quantum field theory solves this problems by introducing antiparticles. To fulfill Green’s equation (see below) the factor θ(x 0 − y 0 ) is included in quantum mechanics in the propagator. A simple calculation shows that D(x. y) is a c-number and. (2π)3 2Ep (206) (207) Take for instance a space-like separation. Let us introduce the following twopoint function in field theory D(x. with (x − y)2 ≤ 0. in fact. We shall see that the culprit are the negative energy modes and that causality-violating propagation of negative energy modes is turned into causality compliant propagation thanks to antiparticles. the propagation seems to be non causal. this function can be written as h0|φ(~x. x 0 = y 0 . we work with a normal ordered hamiltonian H = P 0 . 37 (208) . then Z d3 p e−i~p~r ∼ e−mr . (2π)3 2Ep Again. Taking into account that we go from QM to field theory by the replacement ~x → φ(~x). (205) where we have defined the vacuum energy to be zero. if (x − y) 2 < 0 (y is outside the lightcone of x. it is actually an hermitian operation acting in the Hilbert space of the system. and that eipx is by itself invariant. since it would reverse the time ordering of physical events. the commutator is zero. φ(y)]|0i. Therefore we conclude that if (x − y)2 < 0. (2π)3 2Ep (210) (211) We see that indeed if x0 = y 0 . Actually we know that this is is always the case if x0 = y 0 . Now we can pose the causality issue in very clear terms. φ(y)] = 0 (209) a measurement made at x cannot affect another made at y. as this is actually the content of one of the canonical commutation relations. [φ(x). It is more interesting to note that D(x − y) is a Lorentz invariant. Notice that the cancellation takes place between positive and negative energy solutions. If [φ(x). and changing p~ → −~ p in the second integral. Now. ak e−ikx + a†k eikx ] (2π)3 2Ek d3 p (e−ip(x−y) − e−ip(y−x) ) = D(x − y) − D(y − x). we can write [φ(x). but it is so by noticing that Z d3 p = (2π)3 2Ep Z d4 p δ(p2 − m2 )θ(p0 ) (2π)3 (212) which is manifestly invariant under proper gauge tranformations (those connected to the identity). Since [φ(x). Thus antiparticles are basic to restore causality. φ(y)] = 0 (214) Causality in the sense above indicated is now manifest. or viceversa) we can find a Lorentz transformation such that x − y → −(x − y). as expected. Let us now compute the above commutator for unequal times [φ(x). This is not totally obvious. φ(y)] = h0|[φ(x). (213) This is obviously impossible if x − y is time-like. φ(y)] = = Z Z d3 p (2π)3 2Ep Z d3 k [ap e−ipx + a†p eipx .Taking into account that φ(x) is real field. 38 (215) . Thus φ(x) is an observable. Physically it means that we can set our initial data for a constant time slice. φ(y)] is a c-number. φ(y)}|0i. (216) (217) This is called the “retarded” propagator. φ(y)]|0i = Z C i d4 e−ip(x−y) . (218) We can check that the retarded propagator verifies the following differential equation (2 + m2 )DR (x − y) = −iδ (4) (x − y). Feynman.. The result is thus 0 0 DR (x − y) ≡ θ(x − y )h0|[φ(x).) ¯ − y) = −iδ 4 (x − y). 0 0 0 Z 0 0 = θ(x − y ) Z = θ(x0 − y 0 ) Z 0 θ(x − y )h0|[φ(x). (219) This is proposed as an exercise. (2 + m2 )D(x (220) and thus all of them in momentum space obey ¯ (−p2 + m2 )D(p) = −i. but is of no particular relevance to us. In fact all propagators are related to solutions of this partial diferential equations (Green’s ¯ − y) a generic equation) and they are therefore called Green’s functions. Another way of enforcing the integration circuit is by adding a small and negative imaginary part. declaring that the poles lie at p 0 = ±Ep − i and integrating p0 over the real line. To be precise. 2 2πi p − m2 where C is a circuit containing the two poles at p 0 = ±Ep closed through the lowest half of the complex plane (p0 = −i∞) since x0 − y 0 > 0. Likewise we can consider an “advanced” propagator with poles at p0 = ±Ep + i which is non-zero for y 0 > x0 . Denoting by D(x propagator (retarded.. it is usually defined as θ(y 0 − x0 )h0|{φ(x). 4 2 (2π) p − m2 DR (x − y) = θ(x0 − y 0 )(D(x − y) − D(y − x)). φ(y)]|0i = θ(x − y ) d3 p (e−ip(x−y) − eip(x−y) ) (2π)3 2Ep d3 p (2π)3 d3 p (2π)3 1 −ip(x−y) 1 e |p0 =Ep + e−ip(x−y) |p0 =−Ep 2Ep −2Ep Z C −1 dp0 e−ip(x−y) . and ¯ − y) = D(x Z i d4 p e−ip(x−y) 4 2 (2π) p − m2 39 (221) (222) ! .Let us assume that x0 > y 0 . In short DF (x − y) = θ(x0 − y 0 )D(x − y) + θ(y 0 − x0 )D(y − x). in this case the non-zero contribution comes from the negative energy solution.The difference between the different Green functions (D R (x − y). 4 2 (2π) p − m2 + i (224) Now. as we have seen.) lies in the prescription to handle the poles appearing in the p 0 integral. Then DF = Z i d4 p e−ip(x−y) . (226) This is denoted as The different prescriptions for the poles are summarized in the following figure Advanced Retarded Feynman The so-called Dyson prescription consists in taking p2 i − m2 − i with positive. and we pick only the contribution from the pole at p0 = +Ep . − m2 + i (223) Thus. includes both of them in the contour. 40 (227) .. if x0 > y 0 we have to close the circuit in the lower half-plane. that is D(x − y) . The retarded prescription. The opposite happens if y 0 > x0 . . Feyman prescription amounts to considering p2 i .. poles are in p0 = Ep − i and p0 = −Ep + i. DR (x − y). (225) h0|T φ(x)φ(y)|0i. Green’s functions are solutions of the partial differential equation ¯ − y) = iδ (4) (x − y). (231) We define and where the minus sign appears on account of the anticommuting character of the variables. (235) We have as a non-vanishing contribution Z X d3 k d3 p e−ipx+iky u(p. σ)¯ u(k. and also in order to discuss the peculiarities of spin 1/2 fields. σ)c† (k. the result is What this actually means is i . − m2 + i (234) An alternative way of deriving the previous expressions is by directly inserting the mode expansion.5. 6k − m (232) 6k + m i =i 2 . (229) ¯ S− (y − x) = h0|ψ(y)ψ(x)|0i. σ)e−ip(x−y) . We can easily solve Green’s equation in momentum space. For this reason. σ 0 )h0|c(p. (230) ¯ ¯ SF (x − y) = θ(x0 − y 0 )h0|ψ(x)ψ(y)|0i − θ(y 0 − x0 )h0|ψ(y)ψ(x)|0i.σ0 = Z X d3 p u(p. 6k − m k − m2 (233) Feynman propagator is obtained by considering i k2 6k + m . (i 6 ∂x − m)S(x (228) ¯ S+ (x − y) = h0|ψ(x)ψ(y)|0i. for instance in ¯ S+ (x − y) = h0|ψ(x)ψ(y)|0i. The basic definitions are the same. σ 0 )|0i (2π)3 2Ep (2π)3 2Ek σ.3 The fermion and gauge-boson propagators The contribution from particles and antiparticles it is best discussed in a case where the distinction matters. σ)¯ u(k. (2π)3 2Ep σ 41 (236) (237) . σ)eip(x−y) . ψ(y)} (239) vanishes for space-like separations. bosons have their own antiparticles. We are thus led to the following hypothesis due to Feynman: “The emission (absorption) of an antiparticle of 4-momentum p µ is physically equivalent to the absorption (emission) of a particle of 4-momentum −p µ ” 42 . The Feynman propagator contains both contributions too SF (x − y) = θ(x0 − y 0 )S+ (x − y) − θ(y 0 − x0 )S− (y − x). Working in an arbitrary gauge.Likewise ¯ ψ(x)|0i ¯ S− (x − y) = h0|ψ(y) = Z X d3 p v(p. k 2 + i (242) where we have already adopted Feynman prescription.4 Feynman interpretation of negative energy states and antiparticles It should be clear by now that in spite of its brilliant success for spin 1/2 particles. σ)¯ v (p. We shall be brief in this case as we have already discussed most of the subtle issues. both propagating forward in time. We finally turn to gauge fields. is now the statement that ¯ {ψ(x). the corresponding Green’s equation is 1 (2g µν − (1 − )∂ µ ∂ ν )Dνα (x − y) = iδαµ δ (4) (x − y). particles and antiparticles. ξ whose solution is. 3 (2π) 2Ep σ (238) The sum over spinors gives 6 k + m (6 k − m) in the first (second) case. but with reversed momentum. This can be easily proven by using the same methods as for the bosonic case and it necessitates both both contributions. ¯ We now see that h0|ψ(x)ψ(y)|0i propagates particle (positive energy solutions) from y to x ¯ while h0|ψ(y)ψ(x)|0i propagates antiparticles from x to y. 5. in momentum space (241) ν α g να − (1 − ξ) k kk2 (−i) . Dirac theory of holes and particles cannot be applied to bosons. Causality. while for y 0 > x0 we have antiparticle propagation. Yet. (240) but not simultaneosuly! If x0 > y 0 we have particle propagation.