# Issue with path integrals for the partition function

I was going through Kapusta and Gale "Finite temperature Field theory" In chap 2, Eq. 2.24, they need to do the path integral $$Z = Lim_{N-> \infty} \left (\prod_{i=1}^{N} \int_{-\infty}^{\infty} \frac{d\pi_i}{2\pi} \int_{periodic} d\phi_i \right )\\ \times \exp \Bigg ( \sum_{j=1}^{N}\int d^3x \Big \{ i\pi_j(\phi_{j+1} - \phi_J) -\Delta \tau [\frac{1}{2}\pi_j^2 + \frac{1}{2}(\nabla \phi_j)^2 + \frac{1}{2}m^2\phi^2 + U(\phi)] \Big \} \Bigg )$$

Then they divide the space into small cubes with $$V = L^3, L = aM, a\rightarrow 0, M \rightarrow \infty$$.

Then, they write $$\pi_j = A_j/(a^3\Delta \tau)^{1/2}$$. So far, so good. Then they write the $$\pi$$ integral, in Eq. 2.25 as $$\int_{-\infty}^{+\infty} \frac{dA_j}{2\pi} \exp \Big [ -\frac{1}{2}A_j^2 + i\left (\frac{a^3}{\Delta \tau} \right )^{1/2} (\phi_{j+1} - \phi_j)A_j \Big ]$$ = $$(2\pi)^{-1/2} \exp\left (\frac{-a^3(\phi_{j+1} - \phi_j)^2}{2\Delta \tau} \right )$$

This integration is fine, but they completely miss out the Jacobian $$\frac{1}{(a^3\Delta \tau)^{1/2}}$$. I say this, because, subsequently, in Eq. 2.26, they write: $$Z = Lim_{M,N-> \infty} (2\pi)^{-M^3N/2} \int \left (\prod_{i=1}^{N} d\phi_i \right )\\ \times \exp \Bigg \{ \Delta \tau \sum_{j=1}^{N}\int d^3x \Big [ -\frac{1}{2}\left ( \frac{\phi_{j+1} - \phi_j}{\Delta \tau} \right )^2 - \frac{1}{2}(\nabla \phi_j)^2 - \frac{1}{2}m^2\phi^2 -U(\phi_j)\Big ] \Bigg \}$$

There is no Jacobian. My question is, have they actually missed out on the Jacobian, or am I missing something here? And this missing out of the Jacobian does not seem trivial. They subsequently call the constants outside the $$d\phi$$ integrals, a trivial normalization constant that is irrelevant, and will not affect the thermodynamics. Actually, in the Jacobian, one can substitute $$\Delta \tau = \frac{\beta}{N}$$ and $$a^3 = \frac{V}{M^3}$$. This Jacobian then becomes, $$\left ( \frac{M^3N}{V\beta} \right)^{1/2}$$. Since this depends on $$\beta$$ and $$V$$, it is not a trivial normalization constant, and it will definitely affect the thermodynamics.

These derivations should be pretty standard. Am I missing something here?

I'm thinking of the below reason. Please let me know if someone sees an issue with it.

There are two Jacobians to consider

a) $$\sqrt{\frac{1}{\Delta \tau a^3}}$$ and

b)The Jacobian $$\sqrt{\beta^3 V}$$, which appears after converting $$\phi$$ into its fourier components using the transform $$\phi(x,\tau) \rightarrow \sqrt{\frac{\beta}{V}} \exp \left [ i\omega_n \tau \right ] \exp \left [ ip.x \right ]$$.

The above transformation can be written as $$\sqrt{\beta^3V} \left ( \frac{1}{\beta}\exp \left [ i\omega_n \tau \right ] \frac{1}{V} \exp \left [ ip.x \right ] \right )$$. The term inside the () brackets is unitary, which gives the Jacobian as $$\sqrt{\beta^3V}$$.

After substituting $$a^3 = \frac{V}{M^3}$$ and $$\Delta \tau = \frac{\beta}{N}$$, we can combine the two Jacobians to get $$\beta\sqrt{ NM^3}$$. This means the partition function is of the form $$Z = \beta \sqrt{NM^3}\times (.....)$$.

Then, $$\ln Z = \ln \beta + ......$$ And energy = $$\langle E \rangle = \frac{\partial \ln Z}{\partial \beta} = \frac{1}{\beta} + .....$$

Now comes the crucial thing. $$\frac{1}{\beta}$$ is independent of momentum, $$p$$ or mass,$$m$$. So, when momentum and mass are zero, we would expect the energy to be 0, but the contribution from $$\frac{1}{\beta}$$ remains. This contribution has no physical significance. Hence subtract it.

Since the $$\frac{1}{\beta}$$ is coming from the Jacobians, the Jacobians were never considered in the first place.