I know that the sum of vacuum bubbles can be related to the Vacuum energy, but I'm trying to understand how this follows from the Gell-Mann Low theorem/equation. My question will use equations from Peskin and Schroeder's (1995) text.

I start from the equation just below (4.30) on page 87, reproduced here

$$1=\langle\Omega|\Omega\rangle=\big(\big|\langle 0|\Omega\rangle\big|^2 e^{-iE_0(2T)}\big)^{-1}\langle0|U(T,t_0)U(t_0,-T)|0\rangle\,,$$


  1. $T$ is a large time, and $t_0$ is arbitrary reference time.
  2. $|\Omega\rangle$ is ground state of interacting theory with energy $E_0$.
  3. $|0\rangle$ is the ground state of free theory.

I would now like to solve for $E_0$, the ground state energy. By first taking the Log of both sides. I get almost the desired result.

$$E_0=\frac{i}{2T}\ln \langle0|U(T,t_0)U(t_0,-T)|0\rangle-\frac{i}{2T}\ln\big|\langle 0|\Omega\rangle\big|^2\,.$$

The first term is what I'm looking for: this generates the vacuum bubbles. But then there's the extra second term which I don't know how to interpret. How do I understand this second term? or how can I justify dropping it?


1 Answer 1


It looks like P&S have been a little bit dirty at this point. Go back to equation 4.29... you really need to take the limit $T\to \infty(1-i\epsilon)$ of this expression. Physically what's happening is that the true vacuum $|\Omega\rangle$ is not the perturbative vacuum $|0\rangle$. You are extracting the true vacuum contribution by evolving to large imaginary times. Equality only holds in the limit. Your first term is an increasing phase divided by $T$ so it goes like a constant, but the second term is a $T$ independent constant divided by $T$ so it disappears.


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