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More correlation functions (Chapter 4, Peskin and Schroeder)

  1. Sep 8, 2009 #1
    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.
     
    Last edited: Sep 8, 2009
  2. jcsd
  3. Sep 9, 2009 #2
    I've attached another (wrong) attempt to get the final result, along with a description of why I think its wrong. Curiously it gives the right answer. I'd appreciate some help with this.
     

    Attached Files:

  4. Sep 9, 2009 #3

    Avodyne

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    You have to be careful with the [itex]i\epsilon[/itex]'s when taking the hermitian conjugate. I don't like the way P&S do this (or the way they do much of anything else, for that matter). See problem 9.5 in Srednicki for a clearer (IMO) explanation.
     
  5. Sep 9, 2009 #4
    Ok, I had a look at it. Seems like you don't have to expand the ket (or bra) in terms of [itex]|n\rangle[/itex] the way P&S have done. Nice clean way.

    But can you please point out the mistake in the P&S-based approach? I have detailed all my steps in the pdf file attached in my last post.

    EDIT -- Okay, even in the step in Srednick's book where he asks to prove that

    [tex]\langle 0|U^{\dagger}(T,0) = \langle 0|\phi\rangle\langle\phi|[/tex]

    there is a similar problem I face when trying to simplify to get the right hand side. First, how do contributions from the excited states vanish, and second, how does the first term [itex]\langle 0|\phi\rangle\langle\phi[/itex] appear without the exponential terms involving H and H_0?
     
    Last edited: Sep 9, 2009
  6. Sep 10, 2009 #5

    Avodyne

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    In Srednicki's approach,

    [tex]U^\dagger(T,0)=e^{iHT}e^{-iH_0 T}.[/tex]

    Then, [itex]\langle 0|H=0,[/itex] so [itex]\langle 0|e^{iHT}=\langle 0|[/itex], and so [itex]\langle 0|U^\dagger(T,0)=\langle 0|e^{-iH_0 T}.[/itex] Now replace [itex]H_0[/itex] with [itex](1-i\epsilon)H_0[/itex], so [itex]e^{-iH_0 T}[/itex] becomes [itex]e^{- H_0 \epsilon T}e^{-iH_0 T}[/itex]. Now insert a complete set of eigenstates of [itex]H_0[/itex] to the left of this operator; then we can replace [itex]H_0[/itex] with its eigenvalue, which is positive or zero. If it's positive, the factor of [itex]e^{-E_n \epsilon T}[/itex] goes to zero as [itex]T\to +\infty[/itex].
     
  7. Sep 10, 2009 #6
    Yeah, got it..[itex]t_0 = 0[/itex].
     
  8. Sep 10, 2009 #7
    In general, if an operator definition consists of limits or integral limits that are themselves complex, then does the Hermitian adjoint affect the limits?

    The way I thought about this is that [itex]T\rightarrow\infty(1-i\epsilon)[/itex] is equivalent to the substitution [itex]T = \Lambda(1-i\epsilon)[/itex] and [itex]\Lambda\rightarrow\infty[/itex], so that now, the limit is over a purely real number. But if one has an integral with complex limits, it may not always be convenient or correct to reparametrize the measure (e.g. volume) in the complex plane, right?
     
  9. Aug 14, 2010 #8
    I'm reading this too recently.
    I think you proposed a good point.
    It's dangerous to directly take hermitian conjugate of (4.28) to obtain (4.29).
    Because, we will face the expression
    [tex] \langle 0| e^{iH2T} [/tex]
    which we can't deal with.
    However, I can't explain the coincidence that if we take the above expression to be
    [tex] \langle 0| e^{iE_0 2T} [/tex]
    we get the (4.29).

    The strict derivation of (4.29) should start from the equation above (4.27), take its conjugate, and take the appropriate limit to get rid of higher modes:
    [tex] \langle \Omega | = \lim_{T\rightarrow\infty(1-i\epsilon)} \langle 0|e^{-iHT}\left( e^{-iE_0T}\langle0|\Omega\rangle \right)^{-1} \\
    = \lim_{T\rightarrow\infty(1-i\epsilon)}\langle 0 | U(T,t_0) \left( e^{-iE_0(T-t_0)}\langle0|\Omega\rangle \right)^{-1} = (4.29)[/tex]
     
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