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Question about non-relativistic limit of QFT

  1. Sep 12, 2013 #1
    In pages 41-42 of these notes: http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf , it is said that [itex]|\vec{p}|\ll m[/itex] implies [itex]|\ddot{\tilde{\phi}}|\ll m|\dot{\tilde{\phi}}|[/itex]

    Why is this so?
     
    Last edited: Sep 12, 2013
  2. jcsd
  3. Sep 12, 2013 #2

    Avodyne

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    Take a plane-wave solution of the KG equation,
    [tex]\phi=\exp(i\vec p\cdot\vec x - i E t)[/tex]
    where
    [tex] E =\sqrt{\vec p^2+m^2}[/tex]
    Now assume ##|\vec p|\ll m##. Then we have
    [tex] E \simeq m + {\vec p^2\over 2m}[/tex]
    and the solution can be written as
    [tex]\phi=\exp(-imt)\tilde\phi[/tex]
    where
    [tex]\tilde\phi=\exp\bigl[i\vec p\cdot\vec x - i(\vec p^2\!/2m)t\bigr][/tex]
    Now we can check that
    [tex]\left|\ddot{\tilde\phi}\right|\ll m\left|\dot{\tilde\phi}\right|[/tex]
    as claimed. This will also apply to superpositions of different plane waves, provided that only plane waves with ##|\vec p|\ll m## are included in the superposition.
     
  4. Sep 14, 2013 #3
    Thanks, and another question

    Thanks for that! I've got another question though. In the same document, a bit later, he says that one can also derive the Schrodinger Lagrangian by taking the non-relativistic limit of the (complex?) scalar field Lagrangian. And for that he uses the condition [itex]\partial_{t} \Psi \ll m \Psi[/itex], which in fact I suppose he means [itex]|\partial_{t} \tilde{\Psi}| \ll |m \tilde{\Psi}|[/itex], otherwise I don't get it. In any case, starting with the Lagrangian:

    [itex]\mathcal{L}=\partial^{\mu}\tilde{\psi} \partial_{\mu} \tilde{\psi}^{*} -m^{2}\tilde{\psi}\tilde{\psi}^{*}[/itex]

    Using the inequationI think it's correct, I can only get to:

    [itex]\mathcal{L}=-\nabla\tilde{\psi} \nabla \tilde{\psi}^{*} -m^{2}\tilde{\psi}\tilde{\psi}^{*}[/itex]

    And from that I've tried relating [itex]\tilde{\psi}[/itex] or [itex]\psi[/itex] (as we can write the above Lagrangian with both, as it's invariant under multiplying by a pure phase), to [itex]\dot{\psi}[/itex]
     
    Last edited: Sep 14, 2013
  5. Sep 16, 2013 #4

    Avodyne

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    Yes, start with the lagrangian for a complex field,
    [tex]{\cal L}=\partial^\mu\psi^*\partial_\mu\psi-m^2\psi^*\psi[/tex] Let
    [tex]\psi=e^{-imt}\tilde\psi[/tex] Then we have
    [tex]\partial_t\psi=e^{-imt}(-im\tilde\psi+\partial_t\tilde\psi) \quad\hbox{and}\quad \partial_t\psi^*=e^{+imt}(+im\tilde\psi^*+\partial_t\tilde\psi^*)[/tex] Multiply these together, and drop the [itex]\partial_t\tilde\psi^*\partial_t\tilde\psi[/itex] term as "small", but do not drop the cross terms. If you like, then integrate by parts to move the time derivative off [itex]\tilde\psi^*[/itex] and onto [itex]\tilde\psi[/itex].
     
    Last edited: Sep 16, 2013
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