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Pertubation Theory

  1. Feb 23, 2010 #1
    Again, I am having difficulty deciphering my class notes - in this case there are missing lines of explanation. If we consider a system of particles that approach and interact, the Heisenberg representation of the interacting field is:

    [tex]\phi(\vec{x} , t) = U^{-1} (t) \phi_{a} (\vec{x} , t) U(t)[/tex]

    (where [tex]\phi_a[/tex] is the free field before the interaction.

    Why is it that we can write:

    [tex]\frac{\partial}{\partial t} \phi_{a}= \frac{\partial}{\partial t} U \phi U^{-1}=[\frac{\partial}{\partial t} UU^{-1},\phi_{a}]+iU[H,\phi]U^{-1} [/tex]

    where the square brackets in the third equality are commutators?

    I don't understand where the third expression comes from?

    Last edited: Feb 23, 2010
  2. jcsd
  3. Feb 23, 2010 #2
    \frac{\partial \phi_a}{\partial t} =
    \frac{\partial U}{\partial t} \phi U^{-1}
    + U\phi\frac{\partial U^{-1}}{\partial t}
    + U\frac{\partial\phi}{\partial t}U^{-1}
    where the last term of RHS involves the derivative of time with respect to the field [tex]\phi[/tex] whose equation of motion is well known, the Heisenberg's EoM.

    For the first two terms of eq(*), note that,
    [tex] \frac{\partial U^{-1}}{\partial t} = -U^{-1}\frac{\partial U}{\partial t} U^{-1}
    then you will see why they can be grouped into
    \frac{\partial U}{\partial t}U^{-1} , \phi_a
  4. Feb 23, 2010 #3


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    I do not recall this identity .. can you provide a brief derivation/proof/justification? It seems quite useful ...
  5. Feb 24, 2010 #4
    Take time derivative of both sides of the equality

    [tex]1 = UU^{-1} [/tex]

  6. Feb 24, 2010 #5


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    That'll do it ... and it certainly was brief. :redface: Thanks!
  7. Feb 24, 2010 #6
    Thank you ever so much ismaili - spent ages trying to see this!
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