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I am trying to make sense of the following functional integral in the continuous limit: $$ G({\bf x},{\bf y})=\lim_{N \to \infty}\int \prod_{k=1}^{N-1} d^2 {\bf z}_k \prod_{n=1}^N dp_n \exp\Bigg\{i\sum_{n=1}^N \Bigg[\frac{\langle p \rangle N}{2 T}(z^{(1)}_{n}-z^{(1)}_{n-1})^2+p_n (z^{(2)}_{n}-z^{(2)}_{n-1})+\frac{T p_n}{N \langle p \rangle} V\Big({\bf z}_n,\frac{\sum_{k=1}^n p_n}{N \langle p \rangle}T\Big) \Bigg] \Bigg\}, $$ where ${\bf z}=(z^{(1)},z^{(2)})$, ${\bf z}_0={\bf y}$, ${\bf z}_N={\bf x}$ and $\langle p \rangle= \sum_{n=1}^N\frac{p_n}{N}$.

I am tempted to define the measure as $$ \tau_n \equiv \frac{\sum_{k=1}^n p_n}{N \langle p \rangle}T, \qquad \Delta\tau_n \equiv \tau_{n}-\tau_{n-1}=\frac{ T p_n}{N \langle p \rangle}, \qquad \tau_0=0, \tau_N=T, p_n>0 $$ so that $$ G({\bf x},{\bf y})=\lim_{N \to \infty}\int \prod_{k=1}^{N-1} d^2 {\bf z}_k \prod_{n=1}^N dp_n \exp\Bigg\{i\sum_{n=1}^N \Delta\tau_n \Bigg[ \frac{p_n}{2} \frac{(z^{(1)}_{n}-z^{(1)}_{n-1})^2}{\Delta\tau_n^2}+ p_n \frac{(z^{(2)}_{n}-z^{(2)}_{n-1})}{\Delta \tau_n}+V({\bf z}_n, \tau_n) \Bigg] \Bigg\}, $$

Here come my question:

The measure, $\Delta \tau_n$, is not constant and depends on the discretization point $n$, and in some sense it reminds me the gravitational measure: $\sqrt{-g(\tau)} d\tau$. My heuristic argument is that since $p_n>0$, we can assume that in the limit $N \to \infty$ $\Delta \tau_n$ is going to be arbitrary small and positive and therefore we can write $$ \frac{({\bf z}_{n}-{\bf z}_{n-1})}{\Delta \tau_n} \to \frac{d {\bf z}}{d \tau} \equiv \dot{{\bf z}}, \qquad \sum_{n=1}^N \Delta \tau_n \to\int_0^T d\tau, $$ and therefore $$ G({\bf x},{\bf y}) = \int_{{\bf z}(0)={\bf y}}^{{\bf z}(T)={\bf x}} \mathcal{D}{\bf z}(\tau) \mathcal{D}{p}(\tau) \exp\Bigg\{i \int_0^T d\tau \Bigg[ \frac{p(\tau)}{2} (\dot{z}^{(1)})^2+ p(\tau) \dot{z}^{(2)}+V({\bf z}, \tau) \Bigg] \Bigg\}. $$

Is this correct? Can I make sense of derivatives and Riemann integrals when the measure $\Delta \tau_n$ is not constant but arbitrary small and positive? Or do I have to choose a parameter that gives a constant measure such as the usual choice $$ \tau_n \equiv \frac{n}{N}T, \qquad \Delta \tau_n = \frac{T}{N}. $$

Qmechanic
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aruera
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    Path derivative? – Qmechanic Jul 22 '22 at 19:26
  • @Qmechanic By path derivative I mean that given a path, $x^\mu(\tau)$, its derivative is $\partial_{\tau} x^\mu(\tau)$. Sorry, I'm not used to this formalism and therefore to the typical terminology. – aruera Jul 22 '22 at 23:26

2 Answers2

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The $\tau$ integral is a 1 dimensional Riemann integral. The subtlety you are asking about has nothing in particular to do with path integrals. The discretization need not be constant, and you will still obtain an integral in the limit. However, in one dimension, using a non-uniform discretization can be absorbed into a reparameterization of $\tau$. One way to allow for arbitrary parameterizations is to introduce an "einbein field" $e$ and write the action as \begin{equation} \int d \tau e L\left(\frac{1}{e}\frac{d x}{d \tau}, x\right) \end{equation} Then $e$ transforms in a way such that $d\tau e$ is invariant. See, eg, Section 1.1 of Tong's String Theory notes https://www.damtp.cam.ac.uk/user/tong/string.html.

This is my original answer, which addresses the path integral measure, rather than the measure of the action.

Three examples of non-trivial path integral measures come from:

  • The Fadeev-Popov determinant arises in non-Abelian gauge theory. This determinant factor arises from a change of coordinates in field space associated with fixing a gauge condition. Mathematically, it arises as a Jacobian factor.
  • When quantizing a theory with a non-linearly realized symmetry, such a non-linear sigma model, to properly quantize the theory you do need to include a Jacobian measure factor in the path integral. Some examples sources include this stack exchange answer and Path Integration and the Functional Measure by Unz (this link opens a pdf) (especially Section 4).
  • Quantum anomolies can be interpreted as the non-existence of an appropriate path integral consistent with a symmetry transformation after renormalization. This is known as the Fujikawa method.
Andrew
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  • Andrew, although I am sympathetic to your examples, and I had also ready a nice answer about non-trivial path-integral measures, I don't think that's what the OP is asking about. They are asking about the measure in the (finite-dimensional) integral in the exponent. – ɪdɪət strəʊlə Jul 22 '22 at 22:36
  • @ɪdɪətstrəʊlə Oh, I see, you are right. Sometimes I read the title and then just skim the body of the question. It didn't cross my mind someone would use path integrals to motivate a question about a 1D Riemann integral! I'll add something to the beginning of the answer but I'll leave the path integral stuff there in case it's interesting for anyone, based on the title. – Andrew Jul 22 '22 at 23:30
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  1. Measure factor in the action $S$:

    • The scalar action $S=\int d^nx~{\cal L}$ is an integral over the Lagrangian density ${\cal L}$. The Lagrangian density ${\cal L}=e L$ typically consists of a density $e$ times a scalar $L$. Often $e=\sqrt{|\det(g_{\mu\nu})|}$, where $g_{\mu\nu}$ is components of the worldvolume metric. (In OP's case $n=1$, i.e. a worldline.) This can be non-trivial factor, cf. OP's title question.

  2. Measure factor in the path integral $Z$:

    • The canonical measure factor $$\rho~=~ {\rm Pf}(\omega_{IJ})$$ in the Hamiltonian phase space path integral is given by the (super) Pfaffian, where $\omega_{IJ}$ are components of the symplectic 2-form.

    • In canonical/Darboux coordinates $\rho=1$ is trivial.

    • If we go to the corresponding Lagrangian path integral by integrating out momenta, the measure factor typically becomes a non-trivial functional determinant. This becomes more pronounced in a curved spacetime background.

    • Feynman's fudge factor can be viewed as a non-trivial path integral measure factor, cf. e.g. this Phys.SE post.

Qmechanic
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