Exploring Zee's Quantum Field Theory: Unraveling 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:

[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??

 

[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 [itex]\delta^\mu_\lambda[/itex] nor why the green's function is indexed by [itex]\nu,\lambda[/itex].

 

[jdstokes]

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?

 

[nrqed]

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?

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.

 

[jdstokes]

*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?

 

[nrqed]

*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?

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.

 

[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

[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?

 

[jdstokes]

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?

 

[Avodyne]

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].

 

[jdstokes]

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?

 

[Avodyne]

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]

 

[nrqed]

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]

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

[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]

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 [itex]\lambda = \mu[/itex] we get

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