Disclaimer: this is a technical question about regularization of functional determinants which comes from a person with (relatively) strong background in QFT, string theory and path integrals, who wishes to understand this subject even better.

Most of my questions regarding this topic have been left unanswered. However, I remain an optimist :)

I am trying to follow the derivation from the book by Polyakov called "Gauge fields and strings". He derives the relativistic Klein-Gordon propagator in the Euclidean first-quantized theory spanned by the Polyakov path integral.

The functional integral is taken over the intrinsic geometries (the quotient space of all worldline metrics $h(\tau)$ over diffeomorphisms $f(\tau)$; the measure on the quotient space is denoted $Dh/Df$) and over the embeddings of the worldline into target Euclidean spacetime (which are encoded by embedding functions $X^{\mu}(\tau)$). The propagator renders

$$ \tilde{G}(p) = \int \frac{Dh}{Df} \int DX \exp\left\{ -\frac{1}{2} \intop_{0}^{1} d\tau \sqrt{h(\tau)} \left[ h^{-1} \dot{X}^2(\tau)-\mu^2 \right] \right\}\cdot e^{ i\, p_{\mu} \,\cdot\, (X(1) - X(0))^{\mu} } = \frac{Z_R}{p^2 + \mu_R^2}. $$

In the first part of the derivation one acquires the quotient space measure

$$ \int \frac{Dh}{Df} = \int \frac{dT}{\sqrt{T}} \cdot \sqrt{\det{\Delta}}, $$


$$ T[h] = \intop_0^1 d\tau \sqrt{h(\tau)} $$

is the length of the worldline (the only invariant of 1-dimensional Riemannian geometry) and the 1-dimensional Laplacian acts on functions $\omega: [0, T] \rightarrow [0, T]$ as

$$ \left( \Delta \omega \right) (t) = - \frac{d^2 \omega(t)}{dt^2}. $$

This part I have successfully reproduced.

The second part is to take the $DX$ path integral which is gaussian:

$$ \int DX \dots = \frac{(2\pi N) ^{ND/2}}{(T)^{ND/2}} \cdot \exp \left\{ \frac{T}{2} \left( p^2 + \mu^2 \right) \right\}, $$

where $N$ is a number of steps in the discretization ($N = T / \Delta t$) and $D$ is the dimensionality of spacetime. This part I have reproduced as well.

The final part would be to take the (numeric, 1-dimensional) integral

$$ \tilde{G} (p) = \int \frac{dT}{\sqrt{T}} \cdot \sqrt{\det \Delta} \cdot \frac{(2\pi N) ^{ND/2}}{(T)^{ND/2}} \cdot \exp \left\{ \frac{T}{2} \left( p^2 + \mu^2 \right) \right\}. $$

Since $\det \Delta$ diverges we have to regularize it.

I have tried to write the product of eigenvalues of $\Delta$ with a cutoff $\epsilon = N^{-1}$:

$$ \det \Delta = \prod_{n=1}^{\infty} \left( \frac{\pi n}{T} \right)^2 = \lim_{N \rightarrow \infty} \pi^n T^{-N} \cdot \Gamma(N+1) = \pi ^{\frac{1}{\epsilon }} T^{-1/\epsilon } \exp \left(\frac{-\log (\epsilon )-1}{\epsilon }+\frac{1}{2} \log \left(\frac{2 \pi }{\epsilon }\right)+O\left(\epsilon ^1\right)\right) = u(\epsilon) \cdot \exp \left\{\epsilon ^{-1} \log T \right\} $$

where $u(\epsilon)$ is some $T$-independent diverging factor. The problem with this expression is that it gives incorrect propagator.

Polyakov, instead, uses a regularizer which I don't understand:

$$ \log \det \Delta = - \intop_{\epsilon^2}^{\infty} \frac{d\tau}{\tau} \sum_n e^{-\tau \lambda_n}, $$

where $\lambda_n = (\pi n / T)^2$ are nonzero eigenvalues of $\Delta$. This regularization gives a correct form of $\tilde{G}(p)$ (with the overall normalization factor $Z$ and the mass $\mu$ renormalized), which at my present level of understanding I consider a miracle.

I have given some thought to this question, and the most promising explanation of the failure of my regularization and success of Polyakov's regularization (other than the fact that he is so much smarter than me, of course) is that I might use different $\epsilon = N^{-1}$ in the regularizations of the determinant and the $DX$ path integral which would give me an incorrect overall result.

I have the following two questions:

  1. Is it correct to assume that there lies the issue?

  2. Even if it is, I still don't understand how to show that his regularization uses the correct definition of $N$ (which coincides with the number of slices used in the $DX$ integration) and mine doesn't.


1 Answer 1


OK, I have tried to solve this one ever since I posted the question. Since three people have starred it, I will post what I have found.

Probably the best way to regularize the determinant is to use the Riemann zeta-function regularization.

We start by writing the product of eigenvalues (with dependence on $T$ which we want to acquire) as an exponent:

$$ \prod \lambda_n(T) = \exp \left\{ \sum \, \ln \lambda_n(T) \right\}. $$

Now we want to compute the (diverging) sum of the logarithms. The important observation here is that we only want to compute it up to some (presumably diverging) constant summand:

$$ \sum \, \ln \lambda_n(T) \rightarrow \sum \, \ln \lambda_n(T) + C, $$

since $C$ contributes only to the renormalization of the overall normalization constant:

$$ Z \rightarrow Z \cdot e^C. $$

We now use the identity

$$ \ln x = - \lim _{s \rightarrow 0} \frac{d}{ds}\, x^{-s}: $$

$$ \sum \, \ln \lambda_n = \sum \, \ln \frac{\pi n}{T} = - \sum \, \lim _{s \rightarrow 0} \frac{d}{ds}\, \left( \frac{\pi n}{T} \right)^{-s} = - \lim _{s \rightarrow 0} \frac{d}{ds}\, \sum \, \left( \frac{\pi n}{T} \right)^{-s} = $$

$$ = - \lim _{s \rightarrow 0} \frac{d}{ds} \left( \frac{T}{\pi} \right)^s \left\{ \zeta(s) + c \right\} = C + \left( \frac{1}{2} - c \right) \ln T $$

where on the last line I used the fact that for any plausible regularization the sum $\sum n^{-s}$ is equal to the Riemann zeta function $\zeta(s)$ plus some diverging constant $c$.

This means that

$$ \sqrt{\det \Delta} = Z\cdot T^{\frac{1}{2} - c}. $$

Now comes the part which I don't understand: somehow we set $c = 0$. In this case $\sqrt{\det \Delta} \sim \sqrt{T}$ which is the correct behavior and it gives the correct propagator.

I would generally expect the $c$ divergence to cancel with the same $T^a$ divergence from the Gaussian $DX$ integral. But it seems like this could be done in a lot of ways, rendering a finite $T^a$ expression with any $a$ we want. This is an example of how a divergence corresponds to the loss of information in the theory.

So we need a kind of renormalization condition which would fix $c = 0$. I am still trying to find it.

P.S. sadly, I don't believe in the zeta-function magic. So answers like "just plug the zeta function instead of the diverging sum and forget about the diverging constant $c$" don't satisfy me. In all cases we've done so in the past this constant didn't matter. In this case it seems like it does.


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