Say I have a path integral $\int D \phi \exp(i S_0)$. $S_0$ is the usual free action $$S_0=\frac{1}{2}\int\phi (-\Box-m^2) \phi=\frac{1}{2}\int \phi G^{-1} \phi,$$ and at the moment I'm not interested in interactions. When I do this integration, I just get a term like $(\det G)^{-1/2}$, which is the zero point energy. Normally I don't care about it and it is absorbed in the normalization of the path integral.

But say that this propagator depends on a parameter, and I would like to actually keep this determinant. How do I actually compute it? I know I can do the trick

$$(\det G)^{-1/2}=\exp(-\frac{1}{2}\text{Tr} \log G) $$

but then I would have to compute something like (after Wick rotating and going to momentum space)

$$\int \frac{d^d k}{(2 \pi)^d} \log(\frac{1}{k^2+m^2})$$

which of course diverges (because I have infinitely many harmonic oscillators) and requires some kind of renormalization. Hence my question: how do you compute such a determinant?

  • $\begingroup$ Consider trading the inverse of the determinant for a path integral over the fermionic scalar field (which violates spin-statistics). This will give you extra Feynman rules and thus keep the desired dependence on the parameter. Works well for gauge theories. $\endgroup$ Nov 28, 2015 at 15:13
  • $\begingroup$ But I will still be in the same situation, won't I? I don't have interactions so I will trade a scalar free theory for a ghost free theory, but I still would need to compute the zero point energy $\endgroup$
    – user22710
    Nov 28, 2015 at 16:13
  • $\begingroup$ Yes, you are correct. But I can't imagine how your determinant can be physically significant if it does not depend on the fields (which would yield the interaction with ghosts). You could always add a constant term to the lagrangian, right? This term will yield a constant undetermined factor in the path integral, which absorbs your determinant. $\endgroup$ Nov 28, 2015 at 17:14
  • $\begingroup$ Determining this consant would be my aim! The reason why I'm interested in this is that I'm considering an effective field theory, where I'm first integrating out a field, and then I'm left with another path integral. So absorbing the determinant of the propagator in the normalization is not something I can do in principle $\endgroup$
    – user22710
    Nov 28, 2015 at 17:26
  • $\begingroup$ Ok, I've found this physics.ucsd.edu/~mcgreevy/s13/215C-lectures.pdf . Here, in section 2.3.1 he does the computation and just adds an extra factor of $k^2$ in the integral, since the logarithm would otherwise have a dimensionful argument. In this way the integral is well defined, and converges if one takes enough derivatives. I'm not sure about why we're allowed to just add an extra $k^2$ in the logarithm, though $\endgroup$
    – user22710
    Nov 28, 2015 at 17:41

3 Answers 3


You can simply compute the integral using your preferred regularization method (cut-off, dimensional, Pauli-Villars...), and if all goes well (which is not guaranteed), the divergences will not depend on your parameter and they will eventually disappear when you compute physical stuff. If this does not happen, maybe your theory is simply ill-defined.

And as for adding an extra factor of $k^2$ as mentioned in the comments, we know that adding the trace of $\ln k^2$ would add a (for physical applications) irrelevant constant (which in dimensional regularization is furthermore zero), such that you can add it whenever you feel like it.


As far as I know, the easiest way to do the integration by hand is for integer dimensions (i.e. Mathematica gives you the integral in terms of hypergeometric functions, but that's not really helpful). You can derive the integral (with a hard cut-off $\Lambda$) with respect to $m^2$, perform the integral, expand in $m/\Lambda$, and then integrate back. It might be helpful to subtract a constant $\int_k \ln k^2$.

In the case $d=3$, you should get $\frac{\Lambda}{2\pi^2}m^2-\frac{m^3}{6\pi}$. Note that the first term is not (necessarily) problematic, and can have a very physical interpretation. For example, in the context of $N$ relativistic bosons in the limit $N\to\infty$ (a well known model in condensed matter), this term correspond to the renormalization of the parameter driving the system through the phase transition between the ordered and disordered phase.


Is It Zero? the product of $\frac{1}{k^{2} + m^{2}}$ for $k = -\infty$ to $\infty$, is zero if $m$ is not zero. $\log \left( 0 \right)$ diverges as expected.
If $m = 0$, then we get the product of $\frac{1}{k^{2}}$, which from $1$ to $\infty$, is zero, and from $0$ to $\infty$, is likely undefined (to be resolved).
From $- \infty$, to $\infty$ one asks what is the "average" (I think it's called the geometric mean) of $\frac{1}{k^{2}}$ in terms of multiplying. If it is not $1$, then your answer is $0$ or $\infty$. Looking at it is another question.
What if the geometric mean of $k$? the integral of $\log \left( k \right)$, is $\log \left( k \right) \cdot \left( k - 1 \right)$, which dividing by the very large bounds ($\log \left( 0 \right)$ to $\log \left( k \right)$), get us $k - 1$. If that is correct, massive objects make it become $0$. Massless objects also make it zero. Your approach doesn't include zero in the domain, which needs to be examined.

I have theories of a crystal icy space where the determinant of the connection times a vector, thus the lie algebra has zero determinant. But this is even crazier. Consider that possibility that the determinant may be zero, because your process has a bias against it. If it is zero, your process obviously diverges as expected. I am not sure if the converse is true. It depends on stuff like Borel estimation and so on.

Currently (I am not exactly sure), but an example to convert the Fourier series to make it more manageable is the kernel $Dirac' \left( x_{3}-x_{2} \right) \cdot x_{2} \cdot \arctan \left( x_{1} \cdot x_{2}^{2} \right) \cdot e^{-x_{1}^{2}/x_{2}^{2}}$; Using something the arctangent to work with the bell curve kernel, may allow the Fourier series that has a bad end behavior to eventually become an exponential series with no infinities. One might also put it from $0$ to infinity, and do $e^{-x_{1}^{2}/x_{2}} \cdot \arctan \left(x_{1} \cdot x_{2} \right) \cdot Dirac' \left( x_{3} - x_{2} \right)$, or something similar. Using $\arctan$ 'kills' the bell curve, and the bell curve 'kills' bad [end] behavior from what should be a bunch of cosine waves.

Consider those options. Otherwise your result will diverge, and we won't know why.


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