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Peskin Page 284

  1. Sep 8, 2013 #1
    Hi,

    I am studying Peskin's An Introduction To Quantum Field Theory. On the beginning of page 284, the authors say We can turn the field [itex]\phi_S(x_1)|\phi_1\rangle=\phi_1(x_1)|\phi_1\rangle[/itex]. I tried hard to prove this relation but still can't get it right. Could anyone give me some hints? Thanks.
     
    Last edited: Sep 8, 2013
  2. jcsd
  3. Sep 9, 2013 #2

    vanhees71

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    It's a bit strangely formulated. The generalized kets [itex]|\varphi \rangle[/itex] are defined as generalized eigenstates of the field operators (here in the Schrödinger picture of time evolution), i.e.,
    [tex]\hat{\phi}_S(\vec{x}_1) |\varphi \rangle = \varphi(x) |\varphi \rangle.[/tex]
    Note that [itex]\hat{\phi}_S[/itex] is a field operator in the Schrödinger picture while [itex]\varphi[/itex] is a (complex or real-valued, depending on whether you describe charged or strictly neutral scalar bosons) c-number field.
     
    Last edited: Sep 9, 2013
  4. Sep 9, 2013 #3

    Bill_K

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    This is difficult to read. Please format your LaTeX correctly, or use UTF. :wink:
     
  5. Sep 9, 2013 #4

    vanhees71

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    Done (butt no UTF, which is hard to read either ;-)).
     
  6. Sep 9, 2013 #5
    Thanks! That makes sense but I still don't understand why this given state is an eigenstate of field operators with eigenvalue being the field amplitudes at some specific position.
     
  7. Sep 9, 2013 #6
    Because that's how we're defining the state [itex]|\phi_1\rangle[/itex]. We're trying to pick out the state in the Hilbert space that, when you hit it with the field operator at any position [itex]x[/itex], will give you an eigenvalue equal to the c-number [itex]\phi(x)[/itex]. It's a way of going from states in a Hilbert space to simple c-number functions, so that you can perform the functional integral over them.
     
    Last edited: Sep 9, 2013
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