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The Heisenberg picture and causality

Schrödinger picture

In the Schrödinger picture of the Klein-Gordon field, the state of the system at any one time t is described by a vector |ψ(t) in a Hilbert space. As time varies, the state vector evolves according to the Schrödinger equation: iddt|ψ(t)=H|ψ(t), where H is the quantum Hamiltonian operator and we are working in units where =1. The solution of this equation with initial state |ψ(0) is |ψ(t)=eiHt|ψ(0), assuming the system is autonomous, that is the Hamiltonian has no time dependence.

The Heisenberg picture

The Heisenberg picture emphasises instead the operators A as dynamical objects which vary in time according to the Heisenberg equation ddtA(t)=i[H,A(t)]. If H is autonomous then a solution is given by A(t)=eiHtA(0)eiHt. Note that if |ψ(t) solves the Schrödinger equation and A(t) solves the Heisenberg equation, then ψ(0)|A(t)|ψ(0)=ψ(t)|A(0)|ψ(t), so the two pictures are equivalent as far as computing expectation values is concerned.

Our favourite operators in the Heisenberg picture

For the Klein-Gordon system, the creation and annihilation operators, ap and ap, satisfy the following commutation relations with the Hamiltonian [H,ap]=ωpap,[H,ap]=ωpap. The Heisenberg equation becomes ddtap(t)=iωpap,ddtap(t)=iωpap and the solutions are ap(t)=eiωmathbfpap(0),ap(t)=eiωmathbfpap(0). The operators ϕ(x)=ϕ(x,t) and π(x)=π(x,t) are therefore given by ϕ(x)=d3p(2π)312ωp(ap(0)eipx+ap(0)eipx)π(x)=id3p(2π)3ωp2(ap(0)eipxap(0)eipx) where p=(ωp,p) and px=ωptpx.

Causality

Recall that [ap,aq]=(2π)3δ(pq). Using this commutation relation, we have [ϕ(x),ϕ(y)]=d3p(2π)312ωpd3q(2π)312ωq([ap,aq]eiqypx+[ap,aq]eipxqy)=d3p(2π)312ωp(eip(yx)eip(xy))=D(xy)D(yx), where we have defined D(z):=d3p(2π)312ωpeipz. The function D(z) is Lorentz invariant (the integration measure d3p/2ωp is the Lorentz invariant measure on the mass shell and the integrand eipz is certainly Lorentz invariant). If x and y are spacelike separated ((xy)2<0), then, after a Lorentz transformation, we can assume that they lie on the spacelike slice t=0. In that case the integrals D(xy) and D(yx) are equal when we make the change of variables pp, so, if (xy)2<0 we get [ϕ(x),ϕ(y)]=0. The quantum Klein-Gordon field is therefore causal: local operators at spacelike separations commute.

Note that the amplitude 0|ϕ(x)ϕ(y)|0 for the process "particle is created at y and moves to x where it is destroyed" is equal to D(xy). This is actually nonzero (exponentially decaying) outside the lightcone and oscillatory inside the lightcone. However, when (xy)2<0, this amplitude is cancelled by the process "particle is created at y and moves to x where it is destroyed", as the expectation value of the commutator [ϕ(x),ϕ(y)] is zero.

The propagator

Consider the operators ϕ(x) and ϕ(y).

Tϕ(x)ϕ(y)=:ϕ(x)ϕ(y):+ΔF(xy).
We have ϕ(x)=ϕ+(x)+ϕ(x) where ϕ+(x)=d3p(2π)312ωpapeipxϕ(x)=d3p(2π)312ωpapeipx. If we assume that x0>y0 then Tϕ(x)ϕ(y)=ϕ(x)ϕ(y)=(ϕ+(x)+ϕ(x))(ϕ+(y)+ϕ(y))=ϕ+(x)ϕ+(y)+ϕ(x)ϕ+(y)+ϕ+(x)ϕ(y)+ϕ(x)ϕ(y). All of the terms here are normal-ordered except ϕ+(x)ϕ(y). We have ϕ+(x)ϕ(y)=[ϕ+(x),ϕ(y)]+ϕ(y)ϕ+(x). The term ϕ(y)ϕ+(x) is normal-ordered; moreover, we have [ϕ+(x),ϕ(y)]=d3p(2π)312ωpd3q(2π)312ωq[ap,aq]ei(qypx)=d3p(2π)312ωpd3q(2π)312ωq(2π)3δ(pq)=d3p(2π)312ωpeip(yx)=D(yx). A similar calculation when y0x0 gives D(xy).

The Feynman propagator can be written as ΔF(z)=d3p(2π)3Cdp02πieipzp2m2, where the p0-integral is taken along the following contour C in the complex p0-plane:

where the two dots indicate the poles p0=±ωp.
If x0>y0 then we can close the contour C with a "semicircle at infinity" in the lower half p0-plane, p0Reiθ for large R, as the factor eip0(x0y0) in the integrand will tend to zero along this semicircle. The value of the line integral along C is therefore equal to the sum of residues of the integrand in the region of the plane below C, in other words, the residue of the integrand at p0=ωp. This is 1/2ωp, giving d3p(2π)3Cdp02πieipzp2m2=D(yx) when x0>y0. Similarly, if x0y0 then we can close the contour in the upper half p0-plane and we obtain d3p(2π)3Cdp02πieipzp2m2=D(xy).

The integral in the lemma is often written as ΔF(z)=d4p(2π)4ieipzp2m2+iϵ, as the poles of this integrand are at p0=±m2+|p|2iϵ, which are slightly above/below the real axis. We may send ϵ0 and deform the real axis to the contour C without crossing the poles to recover the original integral.

Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.