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My questions are about worldline path integrals from the book Gauge Fields and Strings of Polyakov. On page 153, chapter 9, he says

Let us begin with the following path integral \begin{align} &\mathscr{H}(x,y)[h(\tau)]=\int_{x}^{y}\mathscr{D}x(\tau)\delta(\overset{\,\centerdot}{x}{}^{2}(\tau)\boldsymbol{-}h(\tau)) \nonumber\\ &=\int\mathscr{D}\lambda(\tau)\exp\left(i\int_{0}^{1}d\tau\lambda(\tau)h(\tau)\right)\int_{x}^{y}\mathscr{D}x(\tau)\exp\left(-i\int_{0}^{1}d\tau\lambda(\tau)\dot{x}^{2}(\tau)\right) \tag{9.8}\label{9.8} \end{align} where $h(\tau)$ is the worldline metric tensor.

The action in (9.8) is invariant under reparametrizations, if we transform: \begin{align} x(\tau)&\rightarrow x(f(\tau)) \nonumber\\ h(\tau)&\rightarrow\left(\frac{df}{d\tau}\right)^{2}h(f(\tau)) \tag{9.9}\label{9.9}\\ \lambda(\tau)&\rightarrow\left(\frac{df}{d\tau}\right)^{-1}\lambda(f(\tau)) \end{align}

Polyakov continued with the following statement.

It is convenient to introduce instead of the worldline vector $\lambda(\tau)$, the worldline scalar Lagrange multiplier $\alpha(\tau)$: \begin{align} \lambda(\tau)&\equiv\alpha(\tau)h(\tau)^{-1/2} \nonumber\\ \alpha(\tau)&\rightarrow\alpha(f(\tau)) \tag{9.11} \end{align} So that: \begin{align} &\mathscr{H}(x,y)[h(\tau)] \nonumber\\ &=\int\mathscr{D}\alpha(\tau)e^{i\int_{0}^{1}d\tau\alpha(\tau)\sqrt{h(\tau)}}\int_{x}^{y}\mathscr{D}x(\tau)\exp\left(-i\int_{0}^{1}d\tau\frac{\alpha(\tau)\dot{x}^{2}(\tau)}{\sqrt{h(\tau)}}\right) \tag{9.12} \end{align}

  1. My first question is about equation (9.12). What Polyakov did there is boldly replace the integral measure $\mathscr{D}\lambda$ by $\mathscr{D}\alpha$. Didn't he miss the Jacobian factor?

$$\mathscr{D}\lambda=\mathscr{D}\alpha\det\left(\frac{\delta\lambda}{\delta\alpha}\right)$$

My second question is the following.

He introduced another parameter $t$, called proper time, defined as

\begin{align} t\equiv\int_{0}^{\tau}\sqrt{h(s)}ds;\quad T\equiv t(1) \tag{9.13} \end{align} and so \begin{align} &\mathscr{H}(x,y)[h(\tau)]\equiv\mathscr{H}(x,y;T) \nonumber\\ &=\int\mathscr{D}\alpha\exp i\int_{0}^{T}\alpha(t)dt\int_{x}^{y}\mathscr{D}x\exp-i\int_{0}^{T}\alpha(t)\dot{x}^{2}(t)dt \tag{9.14} \end{align}

  1. Can anybody tell me how he derived the equation (9.14) via using the "proper time" parameter $t$?

New Edition: As pointed out by @Qmechanics in his answer, equation (9.12) has a missing Jacobian factor. The correct path integral should be

$$\int\mathscr{D}\alpha(\tau)(h(\tau))^{-1/2}e^{i\int_{0}^{1}d\tau\alpha(\tau)\sqrt{h(\tau)}}\int\mathscr{D}x(\tau)e^{-i\int_{0}^{1}\alpha(\tau)\frac{\dot{x}^{2}(\tau)}{\sqrt{h(\tau)}}d\tau}.$$

On page 152, Polyakov started with the two point function

\begin{align} &G(x,y)=\int\frac{\mathscr{D}x(\tau)}{\mathrm{Vol}\mathfrak{Diff}}\exp\left(-m\int_{0}^{1}\sqrt{\dot{x}^{2}(\tau)}d\tau\right) \tag{9.6} \\ &=\int\frac{\mathscr{D}h(\tau)}{\mathrm{Vol}\mathfrak{Diff}}\exp\left(-m\int_{0}^{1}\sqrt{h(\tau)}d\tau\right)\int\mathscr{D}x(\tau)\delta(\dot{x}^{2}(\tau)-h(\tau)) \tag{9.7} \end{align}

Then, using the above corrected result, one has \begin{align} &G(x,y)=\int\frac{\mathscr{D}h(\tau)}{\mathrm{Vol}\mathfrak{Diff}}\exp\left(-m\int_{0}^{1}\sqrt{h(\tau)}d\tau\right)\cdot \nonumber\\ &\cdot\int\mathscr{D}\alpha(\tau)(h(\tau))^{-1/2}e^{i\int_{0}^{1}d\tau\alpha(\tau)\sqrt{h(\tau)}}\int\mathscr{D}x(\tau)e^{-i\int_{0}^{1}\alpha(\tau)\frac{\dot{x}^{2}(\tau)}{\sqrt{h(\tau)}}d\tau}. \end{align}

Therefore, one has

$$G=\int\frac{\mathscr{D}h(\tau)}{\mathrm{Vol}\mathfrak{Diff}}(h(\tau))^{-1/2}e^{-m\int_{0}^{1}\sqrt{h(\tau)}d\tau}\int\mathscr{D}\alpha(\tau)e^{i\int_{0}^{1}d\tau\alpha(\tau)\sqrt{h(\tau)}}\int\mathscr{D}x(\tau)e^{-i\int_{0}^{1}\alpha(\tau)\frac{\dot{x}^{2}(\tau)}{\sqrt{h(\tau)}}d\tau}$$

The problem is that the jacobian factor $h^{-1/2}$ already breaks the reparametrization invariance of the effective action. How does the above path integral make any sense?

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  1. Yes.

  2. It is perhaps helpful to realize that Polyakov is performing a worldline (WL) reparametrization (9.9) such that the metric component $h\equiv 1$ is identically equal to one in the new coordinate $t$.

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  • $\begingroup$ Thank you very much for your answer. I have one more question about the path integral. I edited my post. Would you please have a look at my new edition? $\endgroup$ – The Last Knight of Silk Road Feb 15 at 15:16

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