This is the action for the 1+1 dimensional interacting electron system;

$$S_{cl}[\theta , \phi]= \frac{1}{2\pi} \int dxd\tau \left(g^{-1}v(\partial_x \theta)^2 + gv(\partial_x \phi)^2 + 2i\partial_{\tau} \theta \partial_x \phi \right).$$

I want to integrate out the Gaussian field $\phi$. This book says that it is just an "elementary" Gaussian integration. So, I tried some modification to this action;

$$S_{cl}[\theta , \phi]= \frac{1}{2\pi} \int dxd\tau \left(g^{-1}v(\partial_x \theta)^2 + (\sqrt{gv}\partial_x \phi + \frac{i}{\sqrt{gv}}\partial_{\tau} \theta)^2 + \frac{1}{gv}(\partial_{\tau} \theta)^2 \right).$$

For this action, partition function is given by

$$\int D\theta D\phi \exp[-S_{cl}].$$

Maybe, the second term in the action is related to Gaussian integral. But, I don't know how to calculate it.

How can I calculate this?


1 Answer 1


OP has already completed the square in the second term

$$\tag{1} (\sqrt{gv}\partial_x \phi + \frac{i}{\sqrt{gv}}\partial_{\tau} \theta)^2 ~=~ gv (\partial_x \phi + \frac{i}{gv}\partial_{\tau} \theta)^2 ~=~ gv \left(\partial_x( \phi + \frac{i}{gv}\partial_{\tau} \Theta)\right)^2$$

of the action. Here we defined the antiderivative (aka. primitive or indefinite integral)

$$\tag{2} \Theta(x,t)~:=~ \int_0^x \!dx^{\prime}~\theta(x^{\prime},t).$$

So the Gaussian integration over $\phi$ removes the second term in the classical action, even for an imaginary shift (1).

Quantum mechanically, there will also appear a multiplicative Van Vleck-Morette determinantal factor

$$\tag{3} ({\rm Det}^{\prime}(-\Delta))^{-\frac{1}{2}}$$

in front of the remaining path integral over $\theta$. Here $\Delta:=\partial_x^2$. The prime in eq. (3) indicates that a zeromode should be excluded.


  1. A. Altland and B. Simons, Condensed Matter Field Theory, 2010, p. 180-191.

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