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QED propagator

  1. Nov 20, 2007 #1
    1. The problem statement, all variables and given/known data

    A. Zee Quantum Field theory in a nutshell, p. 31. There is painfully little explanation on this page.

    I'm okay with the action:

    [itex]S(A) = \int d^4 x \mathcal{L} = \int d^4 x\{ \frac{1}{2}A_\mu [(\partial^2 +m^2)g^{\mu \nu}-\partial^\mu\partial^\nu]A_\nu + A_\mu J^\mu \}[/itex]

    where I'm assuming [itex]\partial^2 = \partial^\mu\partial_\mu[/itex], but how does he jump from that to the Green's function

    [itex][(\partial^2+m^2)g^{\mu\nu}-\partial^\mu\partial^\nu]D_{\nu\lambda}(x) = \delta^\mu_\lambda \delta^{(4)}(x)[/itex]?

    Where did the Kronecker delta and the subscript \lambda come from??
     
  2. jcsd
  3. Nov 20, 2007 #2
    It happens that the propagator is the inverse of the operator appearing in the quadratic term in the Lagrangian. I haven't seen a proof, but i believe there exists one. Thus, what you simply have in your equation is

    Quadratic term * Propagator = 1

    the delta and the lambda are there to satisfy this assumption.
     
  4. Nov 20, 2007 #3
    Thanks for your reply.

    I understand where the equation comes from but I don't understand the presence of the Kronecker delta [itex]\delta^\mu_\lambda[/itex] nor why the green's function is indexed by [itex]\nu,\lambda[/itex].
     
    Last edited: Nov 20, 2007
  5. Nov 20, 2007 #4
    If we put [itex]\lambda = \mu[/itex] we get

    [itex][(\partial^2+m^2)g^{\mu\nu}-\partial^\mu\partial^\nu]D_{\nu\mu}(x) = \delta^{(4)}(x)[/itex]

    which makes a little more sense since we're now summing over all indices. Is there any reason for summing over the [itex]\nu[/itex] index as opposed to [itex]\mu[/itex] when [itex]\lambda\neq\mu[/itex].

    Further question: fourier transforming gives

    [itex][-(k^2-m^2)g^{\mu\nu}+k^\nu k^\mu]D_{\nu\lambda}(k) = \delta^\mu_\lambda[/itex].

    I don't see how to get from this to Eq. (3):

    [itex]D_{\nu \lambda}(k)= -\frac{g_{\nu\lambda} + k_\nu k_\lambda /m^2}{k^2-m^2}[/itex].

    Am I missing something obvious?
     
    Last edited: Nov 20, 2007
  6. Nov 20, 2007 #5

    nrqed

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    To find [itex]D_{\nu \lambda}[/itex] from the equation, write the most general possible form, which is [tex] A(k) k_{\nu} k_{\lambda} + B(k) g_{\nu \lambda} [/tex] and solve for A and B.
     
  7. Nov 20, 2007 #6
    *Head explodes*. Is there some way you can justify that [tex]D_{\nu\lambda} = A(k) k_{\nu} k_{\lambda} + B(k) g_{\nu \lambda} [/itex]??

    I wish I knew more about Green's functions...

    Is it just because there's some theorem which says that the Green's function must be of the same general form as the operator?
     
    Last edited: Nov 20, 2007
  8. Nov 20, 2007 #7

    nrqed

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    It's simply that [tex] D_{\nu \lambda} [/tex] must be a tensor with two indices (a second rank tensor). the only things you have at your disposal to build D from are k and the metric. So that's the the most general thing you can write down, that's all.
     
  9. Nov 20, 2007 #8
    Hey nrqed,

    Thanks. That makes things much clearer. I'm still troubled about how we're supposed to find A(k) and B(k), we have

    [itex][-(k^2-m^2)g^{\mu\nu}+k^\nu k^\mu](A(k) k_{\nu} k_{\lambda} + B(k) g_{\nu \lambda}) = \delta^\mu_\lambda[/itex]

    But now what? Expanding things out doesn't help much. It seems like we need some conditions on the behaviour of A and B for various k, but we aren't told anything? Where is the extra equation?
     
  10. Nov 20, 2007 #9
    Let [itex]D_{\nu\lambda}(k) = A(k) g_{\nu\lambda} + B(k) k_\nu k_\lambda[/itex].

    [itex][-(k^2-m^2)g^{\mu\nu}+k^\nu k^\mu](A(k) g_{\nu\lambda} + B(k) k_\nu k_\lambda) = \delta^\mu_\lambda[/itex]
    [itex][-(k^2-m^2)g^{\mu\nu}+k^\nu k^\mu](A(k) g_{\nu\lambda} + B(k) k_\nu k_\lambda) = 1[/itex]
    [itex]-(k^2-m^2)g^{\mu\nu}g_{\nu\mu}A(k) + k^\mu k^\nu g_{\nu\mu}A(k) -(k^2 -m^2)g^{\mu\nu}k_\nu k_\mu B(k) + k^\mu k^\nu k_\nu k_\mu B(k)=1[/itex]
    [itex]-(k^2-m^2)A(k) + k^2A(k) -(k^2 -m^2)k^2 B(k) + k^4 B(k)=1 \implies[/itex]
    [itex]A(k) = \frac{1}{m^2}- k^2 B(k)\implies[/itex]

    [itex]D_{\nu\lambda} = \frac{g_{\nu\lambda}}{m^2} - g_{\nu\lambda}k^\nu k_\nu B(k) + B(k) k_\nu k_\lambda = \frac{g_{\nu\lambda}}{m^2}[/itex]

    What am I missing here?
     
    Last edited: Nov 20, 2007
  11. Nov 20, 2007 #10

    Avodyne

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    You are multiplying 2 matrices together, the first with indices [itex]\mu\nu[/itex] and the second with indices [itex]\nu\lambda[/itex], and you are summing over [itex]\nu[/itex]. You should not set [itex]\mu=\lambda[/itex], because you lose information when you do this.

    Then, in your last line, you illegally use the index nu twice in the second term; [tex]g_{\nu\lambda}k^\nu k_\nu[/tex] should be [tex]g_{\nu\lambda}k^\rho k_\rho[/tex].
     
    Last edited: Nov 20, 2007
  12. Nov 20, 2007 #11
    Second attempt:

    Assume [itex]A(k) = -\frac{1}{k^2-m^2}[/itex]. Then

    [itex][-(k^2-m^2)g^{\mu\nu}+k^\nu k^\mu](A(k) g_{\nu\lambda} + B(k) k_\nu k_\lambda) = \delta^\mu_\nu - \frac{k^\mu k_\lambda}{k^2-m^2} + Bk_\nu k_\lambda[-(k^2-m^2)g^{\mu\nu}+ k^\mu k^\nu][/itex]

    [itex][-(k^2-m^2)g^{\mu\nu}+k^\nu k^\mu](A(k) g_{\nu\lambda} + B(k) k_\nu k_\lambda) = \delta^\mu_\nu - \frac{k^\mu k_\lambda}{k^2-m^2} + B[-(k^2-m^2)k^\mu k_\lambda+ k^2k^\mu k_\lambda][/itex]

    [itex][-(k^2-m^2)g^{\mu\nu}+k^\nu k^\mu](A(k) g_{\nu\lambda} + B(k) k_\nu k_\lambda) = \delta^\mu_\nu - \frac{k^\mu k_\lambda}{k^2-m^2} + B[k^\mu k_\lambda m^2][/itex]

    Therefore [itex]B = \frac{1/m^2}{k^2-m^2}[/itex].

    But I hate the fact that I needed to assume the form of A(k). Is there are more elegant way to derive this?
     
  13. Nov 20, 2007 #12

    Avodyne

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    You don't have to assume it.

    [tex][-(k^2-m^2)g^{\mu\nu}+k^\mu k^\nu](A g_{\nu\lambda} + B k_\nu k_\lambda)=-(k^2-m^2)(A \delta^\mu{}_\lambda +Bk^\mu k_\lambda)+k^\mu(Ak_\lambda+Bk^2k_\lambda)=-(k^2-m^2)A \delta^\mu{}_\lambda +(A+Bm^2)k^\mu k_\lambda[/tex]

    This is supposed to equal [tex]\delta^\mu{}_\lambda[/tex], so we must have [tex]-(k^2-m^2)A =1 [/tex] and [tex]A+Bm^2=0.[/tex]
     
  14. Nov 21, 2007 #13

    nrqed

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    There was no reason to replace the [itex] \delta^\mu_\lambda [/itex] by one on the rhs there!
    And then you set lambda = mu but you should not have done that.

    See Avodyne's post for the full derivation.


    As an aside, it is instructive to do the same calculation for a massless A_\mu. Then you find out that there is no solution A(k) and B(k) satisfying the equation! This is because of gauge invariance; one must add a gauge-fixing term to get a solution. Here, it's not a problem because the massive field case does not have that gauge invariance.
     
  15. Nov 21, 2007 #14

    dextercioby

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    IF you put [itex] \lambda=\mu [/itex] and sum, then [itex] \delta_{\mu}^{\mu} = 4 [/itex]
     
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