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.

  • $\begingroup$ be careful with the imaginary term in limit $\endgroup$ – physshyp Nov 23 '20 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 '20 at 22:55
  • $\begingroup$ former term $t\to\infty(1-i\epsilon)$ $\endgroup$ – physshyp Nov 24 '20 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 '20 at 19:38

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}

  • $\begingroup$ 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 '20 at 9:44
  • $\begingroup$ As mentioned by @physshyp you also need to change the limit when you want to do it that way. $\endgroup$ – Oбжорoв Nov 27 '20 at 16:01

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