Srednicki's normalization of path integral

[geoduck]

In Srednicki's QFT textbook, equation 9.6, he writes:

$$Z[J]=e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} \int \, \mathcal D\phi e^{i \int d^4x \, \mathcal [\mathcal L_0+J\phi]} \propto <br /> e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} Z_0[J]$$

and said that we have to use "proportional to" rather than equals because the ε-trick does not give the correct overall normalization.

I don't quite understand this. Aside from physics, aren't the two sides equal mathematically? If you were to discretize the path integral of $Z_0[J]$, and take the limit of infinite number of time slices, then you would get $Z_0[J]=e^{i\int \int J(y)\Delta(x-y) J(x)}$. If you then wanted to calculate

$$\int \, \mathcal D\phi e^{i \int d^4x \, \mathcal [\mathcal L_0+\mathcal L_I+J\phi]}$$

then it would be equal mathematically to $e^{i \int d^4x \, \mathcal L_I\left(\frac{1}{i}\frac{\delta}{\delta J(x)} \right)} Z_0[J]$ would it not?

 

[andrien]

He is saying that in a general case it may not be possible to get the normalization factor which is rather easy in case of a harmonic oscillator. So to avoid this problem, you need to enforce a normalization convention.

 

[Avodyne]

I don't have the book in front of me, but I believe he defines $Z[J]$ so that $Z[0]=1$. This implies a certain normalization of ${\cal D}\varphi$, which is different from the normalization in the free theory. It is this difference in the normalization of ${\cal D}\varphi$ that leads to proportionality rather than equality.

You can also just check that the final right-hand side does not equal 1 for J=0; it is 1 plus a sum of vacuum diagrams.

 

[geoduck]

I don't have the book in front of me, but I believe he defines $Z[J]$ so that $Z[0]=1$. This implies a certain normalization of ${\cal D}\varphi$, which is different from the normalization in the free theory. It is this difference in the normalization of ${\cal D}\varphi$ that leads to proportionality rather than equality.

You can also just check that the final right-hand side does not equal 1 for J=0; it is 1 plus a sum of vacuum diagrams.

I guess this is where I'm getting confused. If you imagine a discretized path integral instead of a continuous one, then ${\cal D}\varphi$ should be the same whether your theory is interacting or not. The integral over the (discrete) conjugate momenta should yield the same normalization factors whether you have an interacting theory or not, because the interacting part factors away from the free part as far as conjugate momenta is concerned, i.e.,

$$\int [\Pi dp_i] \, e^{i px-ip^2/2m}=\text{normalization factor, free and interacting, since independent of potential }$$

 

[andrien]

Ok, I see where is the problem which is troubling you. You are right that the normalization factor which you will get in both cases are same (caution) and that is the problem.

In free field theory case, you get the normalization as something like $\int D\phi[...]$ ( Srednicki has not mentioned it anywhere probably)and it is independent of source term $J$, so you divide by this term and when $J=0$ you get $Z_0(0)=1$.

In the interacting case, you have same normalization factor coming out if you take the interaction part apart but you see that the field $\phi$ is a functional derivative in the interacting lagrangian and it will act on $Z_0(J)$ to give some other factors which you have not got by free field action alone and if you set $J=0$, your normalization will differ because it will contain some vacuum diagram along with free field case which has come because of functional differentiation. So you have to divide by this whole normalization so that when you put $J=0$, you get $Z(0)=1$.

So in some way you can say that normalization is different in interacting case from the free field case.(I guess you will find it rather trivial)