# QED propagator

1. Nov 20, 2007

### jdstokes

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:

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

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

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

Where did the Kronecker delta and the subscript \lambda come from??

2. Nov 20, 2007

### PhysiSmo

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.

3. Nov 20, 2007

### jdstokes

Thanks for your reply.

I understand where the equation comes from but I don't understand the presence of the Kronecker delta $\delta^\mu_\lambda$ nor why the green's function is indexed by $\nu,\lambda$.

Last edited: Nov 20, 2007
4. Nov 20, 2007

### jdstokes

If we put $\lambda = \mu$ we get

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

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

Further question: fourier transforming gives

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

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

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

Am I missing something obvious?

Last edited: Nov 20, 2007
5. Nov 20, 2007

### nrqed

To find $D_{\nu \lambda}$ from the equation, write the most general possible form, which is $$A(k) k_{\nu} k_{\lambda} + B(k) g_{\nu \lambda}$$ and solve for A and B.

6. Nov 20, 2007

*Head explodes*. Is there some way you can justify that $$D_{\nu\lambda} = A(k) k_{\nu} k_{\lambda} + B(k) g_{\nu \lambda} [/itex]?? I wish I knew more about Green's functions... Is it just because there's some theorem which says that the Green's function must be of the same general form as the operator? Last edited: Nov 20, 2007 7. Nov 20, 2007 ### nrqed It's simply that [tex] D_{\nu \lambda}$$ 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.

8. Nov 20, 2007

### jdstokes

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

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

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?

9. Nov 20, 2007

### jdstokes

Let $D_{\nu\lambda}(k) = A(k) g_{\nu\lambda} + B(k) k_\nu k_\lambda$.

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

$D_{\nu\lambda} = \frac{g_{\nu\lambda}}{m^2} - g_{\nu\lambda}k^\nu k_\nu B(k) + B(k) k_\nu k_\lambda = \frac{g_{\nu\lambda}}{m^2}$

What am I missing here?

Last edited: Nov 20, 2007
10. Nov 20, 2007

### Avodyne

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

Then, in your last line, you illegally use the index nu twice in the second term; $$g_{\nu\lambda}k^\nu k_\nu$$ should be $$g_{\nu\lambda}k^\rho k_\rho$$.

Last edited: Nov 20, 2007
11. Nov 20, 2007

### jdstokes

Second attempt:

Assume $A(k) = -\frac{1}{k^2-m^2}$. Then

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

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

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

Therefore $B = \frac{1/m^2}{k^2-m^2}$.

But I hate the fact that I needed to assume the form of A(k). Is there are more elegant way to derive this?

12. Nov 20, 2007

### Avodyne

You don't have to assume it.

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

This is supposed to equal $$\delta^\mu{}_\lambda$$, so we must have $$-(k^2-m^2)A =1$$ and $$A+Bm^2=0.$$

13. Nov 21, 2007

### nrqed

There was no reason to replace the $\delta^\mu_\lambda$ 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.

14. Nov 21, 2007

### dextercioby

IF you put $\lambda=\mu$ and sum, then $\delta_{\mu}^{\mu} = 4$

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