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?


1 Answer 1


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.


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