- #1
maverick280857
- 1,789
- 4
Hi everyone
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.
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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