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Maxwell propagator in Kaku's QFT book

  1. Dec 19, 2009 #1
    In Michio Kaku's QFT book, p. 106, he writes:

    [To illustrate problems with direct quantization due to gauge invariance]
    let us write down the action [of the Maxwell theory] in the following form:
    [tex]\mathcal L=\frac12 A^\mu P_{\mu\nu}\partial^2A^\nu[/tex]
    The problem with this operator is that it is not invertible. [...]​

    I don't understand his notation. Normally, the same Lagrangian is written
    [tex]\mathcal L=-\frac14F^2[/tex]

    When factoring out [tex]A^\mu[/tex], how does he get the [tex]\partial^2[/tex]?
  2. jcsd
  3. Dec 20, 2009 #2
    Kaku wrote everything in terms of the vector potential 'A', rearranged the resulting "expression" to get another "expression + 4-divergence" , and threw away the 4-divergence (you can throw away a 4-divergence in a Lagrangian). This is a typical technique, not unique to Kaku's book.
  4. Dec 20, 2009 #3


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    In these kind of expression you often want to get an expression like

    \phi Y \phi

    where Y is an expression involving a second order derivative, and with some contraction depending on what phi exactly is. The reason is that these kind of expressions give you Gaussian integrals which you know how to handle. The way to get them is via partial integration and using Stokes;

    \partial A \partial B = \partial(A \partial B) - A \partial^2 B

    The first term gives a boundary condition after integration and can be discarted after suitable boundary conditions.
  5. Dec 20, 2009 #4


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    Maybe you should try it for the scalar field; the action of the scalar field can be rewritten as

    \int d^n x \phi(\partial^2 + m^2)\phi

    So [itex]Y = \partial^2 + m^2 [/itex]. You can do the same thing for the vector field A.
  6. Dec 20, 2009 #5
    I did figure it out, thanks for the replies. What threw me off was the (unusual, but still possible) occurrence of [tex](\partial^2)^{-1}[/tex], i.e., the inverse of the d'Alembertian.
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