I'm learning quantum field theory. In Peskin & Schroeder, when they derive $$\int {D\phi(x)\phi ({x_1})\phi ({x_2})\exp [i\int {{d^4}x\mathcal{L(x)}] = \left\langle {{\phi _b}|{e^{ - iHT}}T\{ \phi ({x_1})\phi ({x_2})\} {e^{ - iHT}}|{\phi _a}} \right\rangle } } $$ and what to make it nearly equal to two-point correlation function, they says:

...Just as in Section 4.2, this trick projects out the vacuum state $ |\Omega\rangle$ from $|\phi_a\rangle$and $|\phi_b\rangle$ (provided that these states have some overlap with $ |\Omega\rangle$, which we assume). For example, decomposing $|\phi_a\rangle$ into eigenstates $|n\rangle$ of $H$, we have $${e^{ - iHT}}|{\phi _a}\rangle = \sum\limits_n {{e^{ - i{E_n}T}}|n\rangle \left\langle {n|{\phi _a}} \right\rangle } \rightarrow\left\langle {\Omega |{\phi _a}} \right\rangle {e^{ - i{E_0}\infty (1 - i\epsilon )}}|\left. \Omega \right\rangle, $$ cf. eqs. (4.27)-(4.28). Thus we obtain a simple formula $$\left\langle {\Omega |T\{ \phi ({x_1})\phi ({x_2})\} |\Omega } \right\rangle = \lim_{T \to \infty (1 - i\epsilon )} \frac{{\int {D\phi (x)\phi ({x_1})\phi ({x_2})} \exp [i\int {{d^4}x{{\mathcal L}}} {\rm{ }}]}}{{\int {D\phi (x)} \exp [i\int {{d^4}x\mathcal L]}}}. $$

In Section 4.2 p.86 they illustrate this track as below:

We must assume that $|\Omega \rangle$ has some overlap with $|0\rangle$, that is, $\langle \Omega | 0 \rangle\ne0$(if this were not the case, $H_I$ would in no sense be a small perturbation).

Actually I haven't get what that means.

  1. Does it mean if we are working in non-perturbative situation, the $|\Omega\rangle$ is orthogonal to $| 0 \rangle$ in the Hilbert space?

  2. Why when it comes to a non-pertubative situation, the formula between path integral and correlation function still works? Do we require a stricter proof?


1 Answer 1


Haag's theorem is the precise mathematical statement that non-perturbatively, $\langle \Omega | 0 \rangle = 0$. People sometimes take this to mean that there is something fundamentally wrong with the interaction picture, but from my point of view

  • it's clear that our calculations using the interaction picture have been very successful at describing physics experiments
  • there isn't even a construction of any interacting quantum field theories in 4 spacetime dimensions, so somehow it's not so worrying. "Stick to what works".

The only place where path-integrals (for four-dimensional interacting theories) seem to be well defined is in lattice-discretisations, e.g. Lattice-QCD. Since everything is regularized and finite-dimensional, you can formally prove relationships between correlation functions and path integrals.

  • $\begingroup$ Thanks. "People sometimes take this to mean that there is something fundamentally wrong with the interaction picture" Are there any reference? $\endgroup$
    – Errorbar
    Commented Mar 28 at 2:27

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