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The Polyakov action of a point-particle is

$$S[X,e]=\frac{1}{2}\int d\tau\left(\frac{\dot{X}^{2}}{e}-m^{2}e\right)$$

with the $(−,+,+,+)$ Minkowski sign convention. How to perform the path-integral

$$Z[X]=\int\mathcal{D}e\,e^{(i/\hbar)S[X,e]}~?$$

Qmechanic
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Xenomorph
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1 Answers1

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That path integral is the relativistic point-particle action $$ Z=\int\mathcal{D}X^\mu\,e^{im\int d\tau\sqrt{\dot{X}^\mu\dot{X}_\mu}}, $$ but it's easier to derive the Polyakov action from it (the reverse of what you ask), as follows. The Lagrangian is $$ L=m\sqrt{\dot{X}^\mu\dot{X}_\mu}, $$ thus the canonical momentum is $$ P_\mu=\frac{\partial L}{\partial\dot{X}^\mu}=\frac{m\dot{X}_\mu}{\sqrt{\dot{X}^\sigma\dot{X}_\sigma}}. $$ The Hamiltonian vanishes identically (which is a consequence of time reparametrization invariance): $$ H=P_\mu\dot{X}^\mu-L=m\frac{\dot{X}_\mu\dot{X}^\mu}{\sqrt{\dot{X}^\sigma\dot{X}_\sigma}}-m\sqrt{\dot{X}^\mu\dot{X}_\mu}=0. $$ However, the theory is not trivial, because the canonical momenta are not independent: they obey the constraint $$ P_\mu P^\mu-m^2=0, $$ which must be imposed on phase space $(X,P)$. You can impose this constraint quantum-mechanically in the phase-space path integral (where both $X$ and $P$ are integrated over), by a Lagrangian multiplier field $e$: $$ \begin{align} Z&=\int\mathcal{D}X^\mu\int\mathcal{D}P_\mu\,\delta[P_\mu P^\mu-m^2]e^{i\int d\tau\left(P_\mu\dot{X}^\mu-H\right)}\\ &=\int\mathcal{D}X^\mu\int\mathcal{D}P_\mu\int\mathcal{D}e\,e^{i\int d\tau\left(P_\mu\dot{X}^\mu-\frac{e}{2}(P_\mu P^\mu-m^2)\right)}\\ &=\int\mathcal{D}X^\mu\int\mathcal{D}e\,e^{iS[X^\mu,e]}. \end{align} $$ Thus your action is $$ Z[X]=\int\mathcal{D}e\,e^{iS[X^\mu,e]}=e^{im\int d\tau\sqrt{\dot{X}^\mu\dot{X}_\mu}}. $$

MathPhysGeek
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