Hi everyone(adsbygoogle = window.adsbygoogle || []).push({});

I am trying to get equation 4.29 of Peskin and Schroeder from equation 4.28. This is what I did

[tex]|\Omega\rangle = \lim_{T\rightarrow\infty(1-i\epsilon)}\left(e^{-iE_{0}(t_{0}-(-T))}\langle\Omega|0\rangle\right)U(t_0, -T)|0\rangle[/tex]

Take the Hermitian Adjoint of both sides.

[tex]\langle \Omega| = \lim_{T\rightarrow\infty(1-i\epsilon)}\langle 0|U^{\dagger}(t_0, -T)\left(e^{iE_0(t_0-(-T))}\langle 0|\Omega\rangle\right)^{-1}[/tex]

Make the transformation [itex]t_0 \rightarrow -t_0[/itex].

[tex]\langle \Omega| = \lim_{T\rightarrow\infty(1-i\epsilon)}\langle 0|U^{\dagger}(-t_0, -T)\left(e^{iE_0(-t_0+T))}\langle 0|\Omega\rangle\right)^{-1}[/tex]

Equation 4.25 is

[tex]U(t, t') = e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}[/tex]

So,

[tex][U(t, t')]^{\dagger} = e^{iH_0(t'-t_0)}e^{-iH(t'-t)}e^{-iH_0(t-t_0)} = U(t', t)[/tex]

using which, the third equation from top becomes

[tex]\langle \Omega| = \lim_{T\rightarrow\infty(1-i\epsilon)}\langle 0|U(-T,-t_0)\left(e^{iE_0(-t_0+T))}\langle 0|\Omega\rangle\right)^{-1}[/tex]

How to proceed further?

I have to show that

[tex]\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}[/tex]

Also, isn't [itex]U(a, b)[/itex] defined only when [itex]a \geq b[/itex]? Strictly, [itex][U(t, t')]^{\dagger} = U(t', t)[/itex] shouldn't even be a valid statement.

Thanks in advance.

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# More correlation functions (Chapter 4, Peskin and Schroeder)

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