I'm reading chapter 4 of Introduction to Quantum Field Theory by Peskin & Schroeder. In the $\phi^4$ theory, the authors state that the ground state of the interaction theory $|\Omega\rangle$ can be written as ($\hbar=1$ )

$$|\Omega\rangle=\lim_{T\to\infty(1-i\epsilon)}\left( e^{-iE_{0}T} \langle \Omega | 0 \rangle \right)^{-1}e^{-iHT}|0\rangle$$

where $E_{0}=\langle\Omega|H|\Omega\rangle$,$|0\rangle$ is the free theory vacuum and $E_n, |n\rangle$ are the eigenvalues and eigenstates of the hamiltonian $H$ .I'm trying to understand that letting the time be "slightly ($\epsilon$ small in some sense) imaginary" is the way to get rid of $n\neq0$ terms in the following series, via a decaying real exponential:

$$e^{-iHT}|0\rangle=e^{-iE_{0}T}|\Omega\rangle\langle\Omega|0\rangle\ + \sum_{n\neq0}e^{-iE_{n}T}|n\rangle\langle n|0\rangle $$

$$ \sum_{n\neq0}e^{-iE_{n}T(1-i\epsilon)}|n\rangle\langle n|0\rangle= \sum_{n\neq0}e^{-iE_{n}T}|n\rangle\langle n|0\rangle+ \sum_{n\neq0}e^{-\epsilon E_{n} T}|n\rangle\langle n|0\rangle$$

Question 1) I don't really see the oscillating term can be neglected as $T \to \infty$

Question 2) How can be physically justified the following substitution $T\to T-i\epsilon$ ?

Thanks for your time

  • $\begingroup$ I haven't seen it in any books, but for those who are interested in the mathematical well definiteness, I'm pretty sure that this small imaginary part is related to contour integrals in the complex plane and some Cauchy theorem. $\endgroup$ – Noix07 Jan 24 '14 at 14:34

Question 2 :

In quantum mechanics or quantum field theory, you use, for calculus of transition probabilities, path integrals as :

$$Z = \int d\Phi \, e^{i\, S(\Phi)} \, or Z = \int d\Phi \, e^{i\, \int \, H(\Phi) \, dt } $$

To prove that this kind of integral is convergent, you cannot stay with a imaginary unit - times the real action term - in the coefficient of the exponential. You need to have a negative real quantity in the coefficient of the exponential such as to be sure that the integral is convergent. So you can go to an euclidean time, or at least add an imaginary part to the standard time (this is your case). So this is the justification of the following substitution $T→T−iϵ$

Question 1 :

However, in your case, you will have $e^{- E_n T\epsilon} <<< e^{- E_0 T\epsilon}$, when $T$ goes to infinity, because $E_n > E_0$.

  • $\begingroup$ Hi, thank you. Does Euclidean time mean $t\to it$? $\endgroup$ – Jorge Lavín May 22 '13 at 11:44
  • $\begingroup$ Yes, your are right. $\endgroup$ – Trimok May 22 '13 at 12:24

Maybe I am missing something but

\begin{equation} \sum_{n \neq 0} e^{-i E_n (T-i\epsilon)} |n \rangle \langle n|0\rangle =\sum_{n \neq 0} e^{-i E_n T} e^{-\epsilon E_n } |n \rangle \langle n|0\rangle  \end{equation} and not a sum of series. This really allows you then to kill non zero $n$ as $T$ goes to infinity.

As for the physical meaning of this trick, you can imagine it (I imaginie it like that at least) as an account for vanishing mixing instead of purely zero mixing. In other word you add some kind of temperature to your system that you know tend to select out the ground state and then you take the limit when the temperature goes to zero.

Essentially what happens is that the initial state $|0\rangle$ is not an eigenstate of the hamiltonian $H$. Because of that the state evolution is best expressed by expanding it in the eigenstate basis set $|n\rangle$. Now, you can imagine all these states being visited every now and then during the evolution (that's what the above equation states).

A system is always coupled in some way to a thermostat even if its temperature is vanishing. Over a long period, the frequency with which the quantum oscillations will allow each state $|n\rangle$ to be visited will be roughly given by a Boltzmann weight. That's what I meant with the term "mixing", your initial state is "shared" among all the eigenstates and eventually, one can estimate the fraction of it that goes to each $|n\rangle$. In the case of a vanishing temperature, the ground state is greatly favored which explains the result.

Instead of a trick, it is actually more a reminder of what should be the true calculation if it were to be done with the density matrix instead of pure states.

  • $\begingroup$ Hi, thank you Indeed, the first question was that, I really need to learn to read, sorry. THe temperature point I don't really get it. I was googling and there seems to be some connection between QFT and Statistical Mechanics via this kind of imaginary time but I don't really see the details. $\endgroup$ – Jorge Lavín May 21 '13 at 9:35
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    $\begingroup$ @Nivalth Maybe this question will help: physics.stackexchange.com/questions/44726/… $\endgroup$ – twistor59 May 21 '13 at 11:02

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