How does Zee derive the Green's function in Quantum Field Theory on p. 31?

[jdstokes]

Homework Statement

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??

 

[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.

 

[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.

 

[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?

 

[nrqed]

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?

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.

 

[jdstokes]

*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]??<br /> <br /> I wish I knew more about Green&#039;s functions...<br /> <br /> Is it just because there&#039;s some theorem which says that the Green&#039;s function must be of the same general form as the operator?

 

[nrqed]

*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]??<br /> <br /> I wish I knew more about Green&#039;s functions...<br /> <br /> Is it just because there&#039;s some theorem which says that the Green&#039;s function must be of the same general form as the operator?

<br /> <br /> It&#039;s simply that 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&#039;s the the most general thing you can write down, that&#039;s all.

 

[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?

 

[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?

 

[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.

 

[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?

 

[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.

 

[nrqed]

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

There was no reason to replace the \delta^\mu_\lambda by one on the rhs there!

-(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

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.

 

[dextercioby]

If we put \lambda = \mu we get

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