Interaction term in path integral

[QuantumDevil]

In some QFT books it is written that the generating functional
Z[J]=\int \mathcal{D}\phi e^{i\int d^{4}x(\mathcal{L}_{o} +V(\phi) +J\phi) }
can be expressed in equivalent form:

Z[J]=e^{i\int d^{4}xV(\phi)} \int \mathcal{D}\phi e^{i\int d^{4}x(\mathcal{L}_{o} +J\phi )}.
The only argument supporting this statement I found is that V(\phi) does not depend on J. But I'm still suspicious about it because we have still to integrate over all possible paths \mathcal{D}\phi, which is ommited in the second definition of the generating functional.

So...can anybody explain me why these two froms of Z[J] are equivalent?

 

[blechman]

something's very wrong with this! You should not be able to take out the potential from the path integral unless it's constant (in phi). What textbooks are you referring to? Are you sure it's not a typo?

 

[QuantumDevil]

http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

page 61, equations 6.21 and 6.22
page 72, equation 9.6

 

[blechman]

These formulae are very different than what you wrote!

As they are in the text, they are correct. The easiest way to see it is to go ahead and expand both the LHS and the RHS in powers of V, and they match.

This is (roughly) nothing more than a formal derivation of perturbation theory.

 

[QuantumDevil]

Can't see it's really a proof because one still have to integrate powers of V over measure \mathcal{D}\phi ...and V depends still on \phi.
There is also question: how to interpret integrals over d^4x of interaction term parametrized by functional derivative over current J:
V (\frac{\delta}{\delta J(x)})?
How e^{i\int d^4 x V (\frac{\delta}{\delta J(x)})} actually acts on Z_{0}(J)?
For me the description given by Srednicki is ambiguous.

 

[blechman]

check out Zee's "QFT in a Nutshell" - it has a very nice explanation of how this all works.

You need the d^4x integral because these are functional derivatives. To get an intuitive idea of what's happening, replace

\int d^4x\frac{\delta}{\delta J(x)} \rightarrow \frac{d}{dJ}

To see how the exponential operator acts, Taylor Expand.

 

[QuantumDevil]

Thanks for reference. It was very helpful.
The reason of my problems was that I assumed V (\frac{\delta}{\delta J(x)}) to be potential parametrized by functional derivative not just product of potential and the "derivative".

 

[blechman]

Thanks for reference. It was very helpful.
The reason of my problems was that I assumed V (\frac{\delta}{\delta J(x)}) to be potential parametrized by functional derivative not just product of potential and the "derivative".

V (\frac{\delta}{\delta J(x)}) is referring to the potential, where you plug the functional derivative into the argument. So for example, if V(x)=x^2, then this would be the SECOND FUNCTIONAL DERIVATIVE operator. Not the product.

 

[QuantumDevil]

You're right.