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In many textbooks the notions Green function and propagator are used interchangeably. But are they really the same thing?

This popular answer argues that a retarded propagator function $D_R(x,t,x',t')$ in quantum field theory is a Green function since it can be understood as the product of the kernel $K(x,t,x',t')$ and a Heaviside function $\theta(t-t')$: $$ D_R(x,t,x',t') = K(x,t,x',t') \Theta(t-t') . \tag{1} $$ The defining property of a Green function is that $$ DD_R(x,t,x',t') = \delta(t-t') \delta(x-x') , \tag 2 $$ where $D$ is the differential operator in question. Moreover, the defining condition of a kernel is $$ DK(x,t,x',t') = 0 .$$ Therefore, we can check whether or not Eq. 1 is correct: \begin{align} D D_R(x,t,x',t')&= D\Theta(t-t') K(x,t,x',t') \\ &= \Big( D\Theta(t-t')\Big) K(x,t,x',t') + \Theta(t-t') \Big( D K(x,t,x',t')\Big)\\ &= \delta(t-t') K(x,t,x',t'). \end{align} Thus we can see that this is equal to Eq. 2 only if $$K(x,t,x',t)=\delta(x-x').\tag{3}$$

In quantum field theory, however, the kernel is given by the Wightman function $$W(x,t,x^\prime,t') = \langle0| \varphi(x,t) \varphi(x^\prime,t') |0\rangle\,, $$ which is not equal to a delta distribution for $t=t'$. Instead, we have $$ \langle 0| \varphi(x,t) \varphi(x',t)|0\rangle=\frac{m}{4\pi^2 r} K_1(mr),$$ where $K_1$ is the modified Bessel function.

Thus it seems as if a QFT propagator, in general, is not necessarily a Green function. (It's still possible that some propagator (e.g. the Feynman propagator) is a Green function. However, so far I haven't found a source which clarifies which propagators are actually Green functions and which are not. There is the additional complication that there are different definitions e.g. for the retarded and advanced propagator which may explain some of the confusion.)

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    $\begingroup$ Short answer: they are (essentially) the same thing only for quadratic, integrable models. Otherwise there is an infinite hierarchy of equations. Differentiating the two point function you get higher order point functions. $\endgroup$ – lcv Oct 29 '19 at 19:06
  • $\begingroup$ n.b. They're called "Green's functions" because they were first studied by a mathematician named Green, so they're "his" functions. en.wikipedia.org/wiki/Green%27s_function $\endgroup$ – Sean E. Lake Oct 30 '19 at 8:38
  • $\begingroup$ @SeanE.Lake see the comment at the beginning of this answer physics.stackexchange.com/a/20812/37286 why it makes sense to talk about Green functions. We also don't say Feynman's propagator or Dirac's equation. (I really don't care about this. But I don't think it's wrong to talk about Green functions.) $\endgroup$ – jak Oct 30 '19 at 8:49
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No, propagator is sometimes used to mean Green function, sometimes used to mean kernel, as explicitly written in the answer cited in the question itself.

EDIT. Also, the argumentation in the question has an issue: Eq. (1) from the cited answer is incorrect in general. By the obvious reason that when the operator $D$ is multiplied by 2, a kernel remains a kernel (whatever boundary conditions are) but a Green function cannot remain a Green function. A related issue is replacing $D\Theta(t-t')$ by $\delta(t-t')$. They need not be equal; this depends on a particular differential operator $D$.

But Eq.~(1) is correct e.g. for wave equation, see this answer for a detailed proof and further references.

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  • $\begingroup$ Thanks. Since I'm interested in QFT and for simplicity, we can restrict ourselves to the Klein-Gordon equation. In this particular case, I think, the cited answer does not really answer the question which propagator is a Green function. In particular, the retarded propagator $D_R(x,x^\prime) = \Theta(x^0 - x^{\prime\,0})\,\langle0| \varphi(x) \varphi(x^\prime) |0\rangle\,$ given in the answer is not a Green function as far as I see. This follows if we plug it into the KG equation. $\endgroup$ – jak Oct 30 '19 at 9:59
  • $\begingroup$ The answer explains that some propagators are kernels while others are Green function but is somewhat vague about which is which. So any elaboration would be much appreciated. $\endgroup$ – jak Oct 30 '19 at 10:00
  • $\begingroup$ I asked a new question physics.stackexchange.com/questions/511001/… which maybe explains a bitter what I'm trying to get at. $\endgroup$ – jak Oct 30 '19 at 10:19

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