3
$\begingroup$

In An Introduction to Quantum Field Theory by Peskin and Schroeder chapter 4, it has discussed about the ground state $|\Omega\rangle$ (where $|0\rangle$ is the ground state in free field theory) in interacting field theory. As (4.27) shows, $$|\Omega\rangle = \lim_{T\rightarrow\infty(1-i\epsilon)}\left(e^{iE_0T}\langle\Omega|0\rangle\right)^{-1}e^{-iHT}|0\rangle\tag{4.27}$$ And then there is a confusing derivation in (4.28), $$|\Omega\rangle = \lim_{T\rightarrow\infty(1-i\epsilon)}\left(e^{iE_0T}\langle\Omega|0\rangle\right)^{-1}e^{-iH(T+t_0)}|0\rangle$$ $$= \lim_{T\rightarrow\infty(1-i\epsilon)}\left(e^{iE_0T}\langle\Omega|0\rangle\right)^{-1}e^{-iH(t_0-(-T))}e^{-iH_0(-T-t_0)}|0\rangle$$ $$ = \lim_{T\rightarrow\infty(1-i\epsilon)}\left(e^{iE_0T}\langle\Omega|0\rangle\right)^{-1}U(t_0, -T)|0\rangle\tag{4.28}$$ I got very confused about the derivation.

  1. First, I'm not sure about add $t_0$ is valid or not, but I think if $T$ goes to infinity, add this term $t_0$ would be okay.
  2. I got confused with the second line of the derivation, why it is accpectable to add $e^{-iH_0(-T-t_0)}$ in the equation, is the reason same as 1.?
  3. I got confused with the apperance of $U(t_0, -T)$ and I think the order in second line is wrong, because in the previous text in the book, that is, eq(4.17), we define $$U(t, t_0) = e^{iH_0(t-t_0)}e^{-iH(t-t_0)}.\tag{4.17}$$ But in second line the order of $H$ and $H_0$ is different and the sign is a little bit different, too. Please help me figure out the confusing points above.

Related questions:

Derivation of Peskin & Schroeder eq. (4.29)

Imaginary time in QFT

$\endgroup$
4
  • $\begingroup$ Wow this derivation is so confusing that there are 4 separate questions about it in stack exchange. Working on an answer. $\endgroup$
    – JohnA.
    Dec 31, 2023 at 14:47
  • 1
    $\begingroup$ More on eq. (4.27) in P&S. $\endgroup$
    – Qmechanic
    Dec 31, 2023 at 15:19
  • $\begingroup$ Well, because I didn't expect the ground state for interacting field theory can be derived by this way, which is kind of reasonable from the point of veiw that it must relate to ground state in free theory, but I thought there would be more straight forward way to find it at the first sight. Many thanks! @JohnA. $\endgroup$ Dec 31, 2023 at 15:43
  • $\begingroup$ Yeah, I am mostly criticizing Peskin, not the learner. Go Srednicki ! If you have a question about 4.29 (the dual), checkout the linked questions. $\endgroup$
    – JohnA.
    Dec 31, 2023 at 15:44

1 Answer 1

3
$\begingroup$
  1. First, I'm not sure about add 𝑑0 is valid or not, but I think if 𝑇 goes to infinity, add this term 𝑑0 would be okay.

This is correct. Usually you see it used the opposite way, i.e. $x + X \approx X$ when $X>>x$ but here we are just doing $T \approx T +t_0$ since $T>>T_0$.

I got confused with the second line of the derivation, why it is accpectable to add $𝑒^{βˆ’π‘–π»_0(βˆ’π‘‡βˆ’π‘‘_0)}$ in the equation, is the reason same as 1.?

No, you can add this because $H_0|0\rangle = 0$, i.e. $e^{xH_0}|0\rangle = |0\rangle .$

I got confused with the apperance of $π‘ˆ(𝑑_0,βˆ’π‘‡)$

We can't use $U(t, t_0) = e^{iH_0(t-t_0)}e^{-iH(t-t_0)}$ here, it simply doesn't apply because we can't change $t_0$. Use Eq. (4.25) where we can set both times freely

$$U(t,t') = e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}$$ so for $$U(t_0, -T) = e^{iH_0(t_0-t_0)}e^{-iH(t_0-(-T))}e^{-iH_0((-T)-t_0)}$$ and we obtain whats needed.

$\endgroup$

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.