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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?

Valac
  • 2,893

1 Answers1

2

  1. For the corresponding problem with discretized time, the Jacobian determinant of the coordinate transformation $$(x^0,x^1,\ldots, x^N)\qquad\longrightarrow \qquad (x^0,v^{1/2},\ldots, v^{N-1/2}), $$ where $$v^{j+1/2}~:=~\frac{x^{j+1}-x^j}{\Delta t} ,$$ would be $\det=(\Delta t)^{-N}$, not unity.

  2. For continuum time, the velocity is $$v(t)~=~\frac{dx(t)}{dt}~=~\int \!dt^{\prime} x(t^{\prime}) \frac{d}{dt}\delta(t\!-\!t^{\prime}). $$ Whether the functional determinant $${\rm Det}\frac{\delta v(t)}{\delta x(t^{\prime})} $$ is unity (or not) depends on regularization scheme and boundary conditions. However, see also this related Phys.SE post.

Qmechanic
  • 201,751