Maxwell propagator in Kaku's QFT book

[hyungrokkim]

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:
\mathcal L=\frac12 A^\mu P_{\mu\nu}\partial^2A^\nu
where
P_{\mu\nu}=g_{\mu\nu}-\partial_\mu\partial_\nu/(\partial)^2
The problem with this operator is that it is not invertible. [...]​

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

When factoring out A^\mu, how does he get the \partial^2?

 

[RedX]

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:
\mathcal L=\frac12 A^\mu P_{\mu\nu}\partial^2A^\nu
where
P_{\mu\nu}=g_{\mu\nu}-\partial_\mu\partial_\nu/(\partial)^2
The problem with this operator is that it is not invertible. [...]​

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

When factoring out A^\mu, how does he get the \partial^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.

 

[haushofer]

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

<br /> \phi Y \phi<br />

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;

<br /> \partial A \partial B = \partial(A \partial B) - A \partial^2 B<br />

The first term gives a boundary condition after integration and can be discarted after suitable boundary conditions.

 

[haushofer]

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

<br /> \int d^n x \phi(\partial^2 + m^2)\phi<br />

So Y = \partial^2 + m^2. You can do the same thing for the vector field A.

 

[hyungrokkim]

I did figure it out, thanks for the replies. What threw me off was the (unusual, but still possible) occurrence of (\partial^2)^{-1}, i.e., the inverse of the d'Alembertian.