3

I know that there has already been many questions related to this question, such as in Differentiating Propagator, Green's function, Correlation function, etc. However, that question mainly discriminate the Green function and kernel, just briefly discuss the propagator as we often know it. Now I don’t mean to duplicate other questions related to this question, if you find other related, please inform me and I will delete it, I just haven’t found out a satisfying answer. To be more specific, what I mean by propagator is the following:

$$ \Delta (x,t;x’,t’) = \langle x | U(t, t’) | x’ \rangle $$

Or in QFT settings $$ \Delta (x,t;x’,t’) = \langle 0| \mathcal{T} [\phi^{(H)}(x’,t’) \phi ^{\dagger(H)} (x,t)]| 0 \rangle. $$

I want to know how to connect this to the green function or correlation function, which is defined to be (two-point)

$$G(x1,x2) = \langle \phi (x1) \phi (x2) \rangle = \frac{\int D \phi e^{-S[\phi]}\phi(x1) \phi(x2)}{Z}.$$

In my own try to understand this, we could try to write the green function as the following. (In QFT settings)

$$G(x1,t1;x2,t2) = \langle \mathcal{T} [\phi ^{(H)}(x1,t1) \phi^{\dagger (H)} (x2,t2)] \rangle = \langle \mathcal{T} [e^{i H t_1}\phi (x1) e^{-i H(t_1-t_2)} \phi^{\dagger} (x2)e^{-i H t_2}] \rangle. $$

Now it seems to feel like the evolution function in the propagator, but how can one deal with the “expectation value” part of the green function definition, which is missing in the propagator definition?

I also know that partition function $Z$ could be related to the integral of imaginary time propagator, but couldn’t really get all these fuzzy things in place at once.

Jiahao Fan
  • 503
  • 1
    Your definition of two-point function is given in path-integral language, whereas expression for $\Delta$ is written in operator language. Both languages are the same, they describe the same quantities. Only one difference you should keep into account: Green function solves equation $Lf=\text{RHS}$, where RHS can contain delta function or not. – Artem Alexandrov Mar 15 '20 at 08:29
  • 1
    Look up the first chapter of Quantum field theory on a lattice by Montvay and Munster, you'll find a mathematically rigorous treatment of the subject – Davide Morgante Mar 15 '20 at 09:37
  • @ Davide Morgante. Oh thanks! I have just read the book, it is excellent! I think I miss two crucial point here: 1) the expectation value is “ground state expectation” not “thermal expectation”; 2) when we work in path integral language, we have to take $\tau$ to infinity so as to make the ground state expectation value become the dominant term in the trace. Is my understanding correct? – Jiahao Fan Mar 15 '20 at 10:19
  • I think J. Zinn-Justin’s book Path integral in quantum mechanics also gives an excellent description of the subject, to whom it may concern. – Jiahao Fan Mar 15 '20 at 10:55
  • Spelling out the time ordered operator you can write what you call the (causal) propagator, in terms of correlators. There's no black magic. – lcv Mar 16 '20 at 19:14

1 Answers1

0

All right so after days of looking textbooks I finally get a feel of how things are arranged, I’ll try to put all things together to give a clear distinction for the people who are also confused by this.

So basically it is the difference between the operator language and the path integral language, and it uses the fact that the real-time green function is defined on zero temperature.

In the path integral formulation, we tend to talk about the expectation value, so in this language, we write the green function in terms of expectation value of “pure function” or “correlation function”, there is no operator anymore:

$ G( x_1,x_2) = \langle \phi(x_1) \phi(x_2) \rangle $

In the operator formulation, we tend to care how operator operates on the states and what is its outcome. In this language, we write green function in expectation value of operators’ matrix elements.

$ G(x_1,x_2) = \langle \mathcal{T} [\phi(x_1,t_1) \phi^{\dagger} (x_2,t_2) ]\rangle $

While doing this expectation value calculation, we actually face two situations, finite temperature or zero-temperature. In the zero-temperature scenario, the ground state contributions dominate and we could write the operators expectation value as:

$ G(x_1,x_2) = \langle 0| \mathcal{T} [\phi(x_1, t_1) \phi^{\dagger} (x_2,t_2) ]| 0 \rangle $

And that is what we usually call “propagator”.

Jiahao Fan
  • 503