# Dual Bra of the ground state of interacting theory

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.

• be careful with the imaginary term in limit Commented Nov 23, 2020 at 21:55
• 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.
– Dani
Commented Nov 23, 2020 at 22:55
• former term $t\to\infty(1-i\epsilon)$ Commented Nov 24, 2020 at 11:36
• 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 $$\langle \Omega | = \langle 0 | U(-t,t_0)\left(e^{iE_0(t+t_0)}\langle 0|\Omega\rangle \right)^{-1}\qquad (t\gg 1)$$
– Dani
Commented Nov 24, 2020 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}
• 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?
• 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. Commented Aug 19, 2022 at 4:14