4
$\begingroup$

I'm currently reading Peskin's "An introduction to Quantum Field Theory", but I'm stuck on page 87; I don't understand why he gets such a Bra for the vaccum state of the interacting theory, say $\langle \Omega |$: $$\langle \Omega | = \lim_{t\rightarrow\infty(1-i\epsilon)} \langle 0 | U(t,t_0)\left(e^{-iE_0(t-t_0)}\langle 0|\Omega\rangle \right)^{-1} $$ If i start from the previous definition of the Ket: $$|\Omega\rangle = \lim_{t\rightarrow\infty(1-i\epsilon)}\left(e^{-iE_0(t+t_0)}\langle \Omega|0\rangle \right)^{-1}U(t_0,-t)|0\rangle $$ Just applying the rules for the adjoint (or dual), I end up with the following: $$\langle \Omega | = \lim_{t\rightarrow\infty(1-i\epsilon)} \langle 0 | U(-t,t_0)\left(e^{iE_0(t+t_0)}\langle 0|\Omega\rangle \right)^{-1} $$ Where I've used the property $U^\dagger(t,t')=U(t',t)$.

From here, the author deduces the rest of the expressions from his previous result, but I cannot understand why he gets that.

$\endgroup$
4
  • $\begingroup$ be careful with the imaginary term in limit $\endgroup$
    – physshyp
    Nov 23, 2020 at 21:55
  • $\begingroup$ what do you mean, which term do you refer to? are you talking about $t\rightarrow \infty(1-i\epsilon)$? or about the exponential term. $\endgroup$
    – Dani
    Nov 23, 2020 at 22:55
  • $\begingroup$ former term $t\to\infty(1-i\epsilon)$ $\endgroup$
    – physshyp
    Nov 24, 2020 at 11:36
  • $\begingroup$ Okay, but I could redefine things in a different way. If I just transform temporal parameter as $ t\rightarrow t(1-i\epsilon)$ and then take the real limit $t\rightarrow \infty$, Things are right up to the Ket: $$|\Omega\rangle = \left(e^{-iE_0(t+t_0)}\langle \Omega|0\rangle \right)^{-1}U(t_0,-t)|0\rangle $$ which must be valid for $t\gg 1$. But taking the dual of this expression I obtain $$\langle \Omega | = \langle 0 | U(-t,t_0)\left(e^{iE_0(t+t_0)}\langle 0|\Omega\rangle \right)^{-1}\qquad (t\gg 1)$$ $\endgroup$
    – Dani
    Nov 24, 2020 at 19:38

1 Answer 1

2
$\begingroup$

Start from $$ \langle 0|e^{-iHT} = \sum_n \langle 0|e^{-iHT} |n\rangle \langle n|= \sum_n \langle 0|e^{-iE_nT} |n\rangle \langle n| $$ from which $$ \lim_{T\rightarrow \infty(1-i\varepsilon)}\langle 0|e^{-iHT}= e^{-iE_\Omega T} \langle 0 |\Omega \rangle \langle \Omega| $$ and \begin{align} \langle \Omega| =&\, \lim_{T\rightarrow \infty(1-i\varepsilon)}\left( e^{-iE_\Omega T} \langle 0 |\Omega \rangle \right)^{-1}\langle 0|e^{-iHT}\\ =&\, \lim_{T\rightarrow \infty(1-i\varepsilon)}\left( e^{-iE_\Omega (T-t_0)} \langle 0 |\Omega \rangle \right)^{-1}\langle 0|e^{-iH(T-t_0)} \\ =&\, \lim_{T\rightarrow \infty(1-i\varepsilon)}\left( e^{-iE_\Omega (T-t_0)} \langle 0 |\Omega \rangle \right)^{-1}\langle 0|e^{iH_0(T-t_0)}e^{-iH(T-t_0)}\\ =&\, \lim_{T\rightarrow \infty(1-i\varepsilon)}\left( e^{-iE_\Omega (T-t_0)} \langle 0 |\Omega \rangle \right)^{-1}\langle 0|U(T,t_0) \end{align}

$\endgroup$
3
  • $\begingroup$ That's right, but how could you derive the same final expression by taking the dual of ket?: $$|\Omega\rangle = \lim_{t\rightarrow\infty(1-i\epsilon)}\left(e^{-iE_0(t+t_0)}\langle \Omega|0\rangle \right)^{-1}U(t_0,-t)|0\rangle$$ Shouldn't you get the same? $\endgroup$
    – Dani
    Nov 26, 2020 at 9:44
  • $\begingroup$ As mentioned by @physshyp you also need to change the limit when you want to do it that way. $\endgroup$ Nov 27, 2020 at 16:01
  • $\begingroup$ Directly conjugation simply won't work. At least not directly. Because it would be equivalent to expanding $\langle 0|e^{iHt}=\sum_n \langle 0|e^{iHt}|n\rangle\langle n|$ and sending $T$ to $\infty(1+i\epsilon)$; which is the same as expanding $\langle 0|e^{-iHt}=\sum_n \langle 0|e^{-iHt}|n\rangle\langle n|$ and sending $T$ to $\infty(-1-i\epsilon)$. This will give a different expression. As the answer suggested, you need to send $T$ to $\infty(1-i\epsilon)$ in $\langle 0|e^{-iHt}=\sum_n \langle 0|e^{-iHt}|n\rangle\langle n|$ to get Peskin and Schroeder's equation. $\endgroup$ Aug 19, 2022 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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