# 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 – physshyp Nov 23 '20 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 Nov 23 '20 at 22:55
• former term $t\to\infty(1-i\epsilon)$ – physshyp Nov 24 '20 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 Nov 24 '20 at 19:38

## 1 Answer

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? – Dani Nov 26 '20 at 9:44
• As mentioned by @physshyp you also need to change the limit when you want to do it that way. – Oбжорoв Nov 27 '20 at 16:01