# Computing a Gaussian path integral with a zero-mode constraint

I have the following partition function: $$$$Z=\int_{a(0)=a(1)} \mathcal{D}a\,\delta\left(\int_0^1 d\tau \,a -\bar{\mu}\right)\exp\left(-\frac{1}{g^2}\int_0^1d\tau\, a^2\right)$$$$ where $$\bar{\mu}$$ is a constant. How can I explicitly compute this partition function? In particular, I do not know how to deal with that Dirac delta containing an integral. Of course if the Dirac delta wasn't there, the calculation would be very easy, since it would reduce to the usual computation of a propagator.

The final result should be $$$$Z=\frac{1}{g}\exp\left(-\frac{\bar{\mu}^2}{g^2}\right).$$$$

Edit: I understood Prahar's method. I would like to get to the same answer using zeta function regularization, as proposed by ɪdɪət strəʊlə. So, how do we apply the zeta function regularization here?

• Something to try: introduce a new field $\xi$, and use use $\delta(F[a]) = \int \mathcal{D} \xi e^{i \xi F[a]}$ to re-express the delta function. Switch the order of the $a$ and $\xi$ integrals and complete the square to do the (Gaussian) integral on $a$. Finally, do the $\xi$ integral, which should also be Gaussian. Feb 18 at 11:35
• @Andrew one comment about this approach is that the delta function is not a functional delta function but rather a normal one (since $a$ is integrated and $\mu$ is a constant). One could however introduce a variable (not a filed) $\xi$ so that $\delta(f(a))=\int \mathrm{d}\xi \exp(i \xi f(a))$. This would, of course, work, nevertheless. Feb 18 at 17:20
• @ɪdɪətstrəʊlə Good point! Thank you for the comment. Feb 18 at 17:57

$$a(\tau)$$ is a function on $$[0,1]$$ so you can expand it in a Fourier series $$a(\tau) = c \tau + a_0 + \sum_{n=1}^\infty a_n \cos( 2\pi n \tau) + \sum_{n=1}^\infty b_n \sin( 2\pi n \tau)$$ In your question, you have not specified any boundary conditions on $$a(\tau)$$ so I don't know if it's periodic or not. The aperiodicity of $$a(\tau)$$ is captured by the first term above. Given your final answer, I think it's supposed to be periodic so I'm going to set $$c=0$$.

Using this, we find $$\int_0^1 d\tau a^2 = a^2_0 + \sum_{n=1}^\infty a_n^2 + \sum_{n=1}^\infty b^2_n , \qquad \int_0^1 d\tau a = a_0$$ The path integral measure is $$\int {\cal D} a = {\cal N} \int_{-\infty}^\infty da_0\prod_{n=1}^N\int_{-\infty}^\infty da_n db_n$$ $${\cal N}$$ is an overall normalization which we will fix by renormalizing the path integral. Note that we have also introduced a UV cut-off $$N$$. The $$N \to \infty$$ limit will be taken after renormalizing.

Putting it all back into the path integral, we have \begin{align} Z &= \int {\cal D} a \delta \left( \int_0^1 d\tau a - {\bar \mu} \right) \exp \left( - \frac{1}{g^2} \int_0^1 d\tau a^2 \right) \\ &= {\cal N} \int_{-\infty}^\infty da_0\prod_{n=1}^N \int_{-\infty}^\infty da_n db_n \delta ( a_0 - {\bar \mu}) \exp \left( - \frac{a_0^2}{g^2} - \frac{1}{g^2} \sum_{n=1}^N a_n^2 - \frac{1}{g^2} \sum_{n=1}^N b^2_n ] \right) \end{align} The integral over $$a_0$$ localizes due to the Delta function. The remaining integrals are simple Gaussians. Therefore, \begin{align} Z &= {\cal N} ( g^2 \pi )^N \exp \left( - \frac{{\bar \mu}^2}{g^2} \right) \end{align} We can now set $${\cal N} = \alpha ( g^2 \pi )^{-N}$$ where $$\alpha$$ is an arbitrary finite constant and then take the $$N \to \infty$$ limit so we get \begin{align} Z &= \alpha \exp \left( - \frac{{\bar \mu}^2}{g^2} \right) . \end{align} Now, without any extra information, the normalization $$\alpha$$ of $$Z$$ cannot be fixed. However, OP seems to asking this question in the context of the paper 2112.03793 (this was clarified to me in the comments by @ɪdɪətstrəʊlə) where the authors impose the normalization condition $$\frac{1}{\sqrt{\pi}}\int d {\bar \mu} Z = 1 \quad \implies \quad \alpha = \frac{1}{g}.$$ In summary, $$Z = \frac{1}{g} \exp \left( - \frac{{\bar \mu}^2}{g^2} \right)$$

• How can one see that $c\tau$ accounts for the aperiodicity? And why can you set $\mathcal{N}$ as you do in the end? Feb 18 at 13:53
• The normalization of the path integral is arbitrary. Physical objects are defined by dividing out by the normalization anyway. Also $a(1)-a(0) = c$ so $c$ captures the aperiodicity of $a(\tau)$. Feb 18 at 13:56
• If they are arbitrary constants I would agree, but here the coupling is not on the same footing as let us say $\pi$, see @ɪdɪət strəʊlə answer. Concerning the aperiodicity, I believe the restriction tells us to integrate only over periodic functions, so it is not necessary to consider that specific term, moreover you could add any non-periodic function so there is an ambiguity I think Feb 18 at 14:01
• @Prahar I still disagree, this partition function comes from integrating out other degrees of freedom and the normalisation is already chosen in the initial path integral so you cannot redefine it. The OP's question hides that fact, but see arxiv.org/abs/2112.03793 for more details. Nevertheless, you cannot choose the normalisation to be coupling-constant dependent. I could choose it to be $\mathcal{N}=(g^2\pi)^{-N}\exp(\frac{\bar{\mu}^2}{{g^2}})$ and I would completely ruin the computation then. Feb 18 at 14:02
• @ɪdɪətstrəʊlə edited. Feb 18 at 17:43

While @Prahar's answer is correct when it comes to the exponential behaviour, i.e. the zero-mode, I disagree with their way of obtaining the prefactor. The $$\frac{1}{g}$$ behaviour is crucial and not a matter of normalisation. In particular, they disregard that in such computations one must zeta-function regularise the resulting determinants$$^*$$.

Before delving into the non-zero-mode sector let me outline the effect of the zero-mode for the sake of completeness. If we decompose $$a$$ into $$a= a_0 + a'$$, where $$a_0$$ is a constant (the zero-mode) and $$a'$$ are non-constant, non-winding, periodic functions (and thus $$\int_0^1 \mathrm{d}\tau \;a' = 0$$), it is clear that $$\int_0^1\mathrm{d}\tau a = a_0$$ and $$\mathrm{D}a=\mathrm{d}a_0\; \mathrm{D}a'$$. Then the delta-function will only affect the zero mode, and hence $$Z=\int \mathrm{d}a_0\ \delta\left(a_0-\bar{\mu}\right)\ \mathrm{e}^{-{a_0^2}/{g^2}}\ Z_\text{non-zero-mode}[g] = \exp\!\left(-\frac{\bar{\mu}^2}{g^2}\right)\ Z_\text{non-zero-mode}[g].$$

Now, the non-zero-mode sector is Gaussian and free of constraints and reads simply \begin{aligned} Z_\text{non-zero-mode}[g] &= \int \mathrm{D}a' \exp\!\left(-\frac{1}{g^2}\int \mathrm{d}\tau \ a^2\right) = \\ &= \left[\det'\!\left(\frac{1}{g^2}\mathbb{I}\right)\right]^{-1/2} = \\ &= \left(\prod_{n\neq 0} \frac{1}{g^2}\right)^{-1/2} = \prod_{n=1}^\infty g^2 \overset{\zeta}{=} \frac{1}{g}.\end{aligned} where $$\mathbb{I}$$ is the identity operator acting on the space of periodic functions and $$\det'$$ excludes the zero mode, which is treated separately above. The last equality is understood in zeta-function regularisation$$^\dagger$$

In total we have \begin{aligned}Z[g,\bar{\mu}] &=\int\mathrm{D}a\ \delta\left(\int_0^1\mathrm{d}\tau a - \bar{\mu}\right)\ \exp\!\left(-\frac{1}{g^2}\int\mathrm{d}\tau a^2\right)=\\ &=Z_\text{non-zero-mode}[g,\bar{\mu}]\ Z_\text{zero-mode}[g,\bar{\mu}] = \\ &= \frac{1}{g}\ \exp\!\left(-\frac{\bar{\mu}^2}{g^2}\right), \end{aligned} getting the desired answer.

$$^*$$ As indeed noted by @Prahar in a comment, one can use a different regularisation method and should get the same $$g^{-1}$$ behaviour. One cannot however simply set the normalisation of the partition function to get the desired result, as the normalisation is already fixed in this case.

$$^\dagger$$ Remember that $$\prod_{n=1}^\infty \text{constant} \overset{\zeta}{=} \exp(\log(\text{constant})\ \zeta(0))=\frac{1}{\sqrt{\text{constant}}}$$.

• Can you elaborate on how the zero-mode comes about? Feb 18 at 13:58
• I disagree with this answer. No one said that we should use zeta function regularization and nothing else. At the end of the day, the normalization depends on how we renormalize the path integral. It becomes physical if we can argue that a particular way of renormalizing is relevant for our calculation, but since that information was not provided to us by OP, we cannot conclude that. Feb 18 at 14:01
• @ohneVal and Ruben I have updated my answer in response to your comments Feb 18 at 16:53
• Thank you very much! Feb 18 at 16:59
• split it into $\prod_{n>0}$ and $\prod_{n<0}$ and note that they're the same, so it is $\left(\prod_{n>0}\right)^2$. This kills the $1/2$. The exponent $-1$ can simply move inside the product, based on the properties of products. Feb 18 at 17:30