# A Question about Path-Integral Measure

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I want to do the following path integral.

$$\mathcal{Z}=\int\mathcal{D}x e^{iS[\dot{x}]}$$

The action only denpends on $\dot{x}$. For some reason, I want to replace the integral measure $\mathcal{D}x$ by $\mathcal{D}\dot{x}$.

So I have

$$\mathcal{Z}=\int\mathcal{D}\dot{x}\mathrm{Det}\left(\frac{\delta x}{\delta\dot{x}}\right)e^{iS[\dot{x}]}.$$

The variable $x$ is related with $\dot{x}$ via the linear transformation

$$x(t)=\int_{0}^{t}\dot{x}(s)ds,$$

which implies

$$\mathrm{Det}\left(\frac{\delta x}{\delta\dot{x}}\right)\equiv 1.$$

Am I correct in the above derivation?

No.

@arnold neumaier Can you elaborate?

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''which implies'' does not follow. Rewrite the derivative as an operator applied to $x$, and perform the functional differentiation by working out the virtual displacement to get the correct answer.

answered Feb 13 by (13,660 points)

Is $\det(\frac{\delta x}{\delta\dot{x}})=\frac{1}{\det(\frac{d}{dt})}$ the correct answer?

Formally yes, but the result is not well-defined without regularization since the determinant of an operator with continuous spectrum makes no sense. On the other hand, the regularized determinant is a constant, hence can be absorbed in the normalization of the functional integral.

Thank you very much. Would you please look at another question I posted here?

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