The Relativistic Point Particle

Before we quantize a string, it is useful to quantize the object that a string generalizes: a relativistic point particle. A particle sweeps out a worldline; a string sweeps out a worldsheet. Many of the structural ideas of string theory are already visible in this one-dimensional model:

• the parameter along the history is a gauge label;
• an auxiliary metric field turns a square-root action into a quadratic action;
• varying that metric imposes a constraint;
• gauge fixing leaves a finite-dimensional modulus, whose integral gives the propagator.

The string version of these four statements is the Polyakov formulation.

Conventions

We use a mostly-plus target-space metric,

$$\eta_{\mu\nu}=\operatorname{diag}(-,+,\ldots,+),$$

and write

$$\dot X^\mu \equiv {dX^\mu\over d\lambda}, \qquad \dot X^2\equiv \eta_{\mu\nu}\dot X^\mu\dot X^\nu.$$

A timelike trajectory has $\dot X^2<0$. We set $\hbar=c=1$.

The square-root worldline action

Let a particle trajectory in $D$-dimensional Minkowski space be described by embedding functions

$$X^\mu(\lambda), \qquad \mu=0,1,\ldots,D-1,$$

where $\lambda$ is an arbitrary parameter along the curve. The invariant line element is

$$ds=\sqrt{-\eta_{\mu\nu}dX^\mu dX^\nu}.$$

The relativistic action is proportional to the proper time swept out by the particle:

$$S_{\rm pp} =-m\int ds =-m\int d\lambda\sqrt{-\dot X^2}.$$

This is the one-dimensional prototype of the Nambu—Goto action. A particle action measures the invariant length of a worldline; a string action will measure the invariant area of a worldsheet.

In a curved target spacetime with metric $G_{\mu\nu}(X)$, the same expression becomes

$$S_{\rm pp} =-m\int d\lambda \sqrt{-G_{\mu\nu}(X)\dot X^\mu\dot X^\nu}.$$

For the rest of this page we mostly work in flat spacetime.

The same geometric worldline can be labeled by many different parameters. Reparametrization invariance says that the labels are gauge; the curve in spacetime is physical.

Reparametrization invariance

The parameter $\lambda$ has no direct physical meaning. Under an orientation-preserving reparametrization

$$\lambda'=f(\lambda), \qquad f'(\lambda)>0,$$

the same curve is described by $X'^\mu(\lambda')=X^\mu(\lambda)$. The velocity transforms as

$${dX^\mu\over d\lambda'} ={d\lambda\over d\lambda'}{dX^\mu\over d\lambda}.$$

Therefore

$$d\lambda'\sqrt{-\left({dX\over d\lambda'}\right)^2} =d\lambda\sqrt{-\left({dX\over d\lambda}\right)^2},$$

so $S_{\rm pp}$ is invariant. This is a gauge symmetry, not an ordinary global symmetry: many functions $X^\mu(\lambda)$ represent the same physical history.

A simple gauge choice for a timelike trajectory is temporal gauge,

$$\lambda=X^0\equiv t.$$

Then

$$S_{\rm pp}=-m\int dt\sqrt{1-\dot{\mathbf X}^{\,2}},$$

where now $\dot{\mathbf X}=d\mathbf X/dt$. The canonical momentum is

$$p_i={m\dot X_i\over \sqrt{1-\dot{\mathbf X}^{\,2}}},$$

and the Hamiltonian is

$$H=\sqrt{\mathbf p^2+m^2}.$$

This is the correct relativistic energy, but the square root is awkward for covariant quantization. We would like a form where Lorentz covariance and the constraint structure are manifest.

The einbein action

Introduce an auxiliary field $e(\lambda)$, called the einbein. The one-dimensional worldline metric may be written as

$$ds_{\rm wl}^2=e(\lambda)^2d\lambda^2.$$

The Polyakov-like particle action is

$$S[X,e] ={1\over2}\int d\lambda \left(e^{-1}\dot X^2-e m^2\right).$$

It is invariant under reparametrizations provided

$$e'(\lambda')d\lambda'=e(\lambda)d\lambda.$$

Thus $e(\lambda)d\lambda$ is the invariant worldline measure.

Varying $X^\mu$ gives, in flat spacetime,

$${d\over d\lambda}\left(e^{-1}\dot X_\mu\right)=0.$$

Varying $e$ gives

$$-e^{-2}\dot X^2-m^2=0,$$

or

$$\dot X^2+e^2m^2=0.$$

For $m>0$ and $e>0$, this equation gives

$$e={\sqrt{-\dot X^2}\over m}.$$

Substituting this solution back into $S[X,e]$ recovers the square-root action:

$$S[X,e_{\rm cl}] =-m\int d\lambda\sqrt{-\dot X^2}.$$

So the einbein does not introduce a new physical degree of freedom. It replaces the square root by a gauge field and a constraint.

The canonical momentum is

$$P_\mu={\partial L\over \partial \dot X^\mu}=e^{-1}\dot X_\mu.$$

The canonical Hamiltonian is

$$H_{\rm can}=P_\mu\dot X^\mu-L ={e\over2}\left(P^2+m^2\right).$$

Since $e$ has no kinetic term, it is a Lagrange multiplier imposing

$$P^2+m^2=0.$$

This is the covariant mass-shell condition. In temporal gauge it becomes

$$-E^2+\mathbf p^2+m^2=0,$$

or

$$E^2=\mathbf p^2+m^2.$$

The einbein $e(\lambda)$ is the worldline metric degree of freedom. Its variation imposes the mass-shell constraint $P^2+m^2=0$.

Why the einbein matters

The square-root action becomes singular in the massless limit, while the einbein action remains meaningful:

$$S_{m=0}[X,e]={1\over2}\int d\lambda\,e^{-1}\dot X^2, \qquad P^2=0.$$

For strings, the analogous auxiliary field is the worldsheet metric $h_{\alpha\beta}$, and its equation of motion gives the Virasoro constraints.

Gauge fixing and the proper-time modulus

The local profile of $e(\lambda)$ is gauge. For a worldline segment with fixed endpoints, one can use reparametrization invariance to set $e$ to a constant. But one quantity cannot be removed: the total invariant proper time

$$\mathcal T=\int_0^1 e(\lambda)d\lambda.$$

This positive number is a modulus of the gauge-fixed worldline geometry.

Worldline diffeomorphisms remove the local shape of $e(\lambda)$, but the total proper time $\mathcal T$ remains. In string perturbation theory, the analogous residual data are moduli of Riemann surfaces.

In constant-einbein gauge, take $0\leq \lambda\leq1$ and

$$e(\lambda)=\mathcal T.$$

The Lorentzian equations of motion are

$$\ddot X^\mu=0,$$

with the constraint

$$\dot X^2+\mathcal T^2m^2=0.$$

The classical trajectory is a straight line. In a curved target spacetime the corresponding statement is that the particle follows a geodesic.

Euclidean continuation and the propagator

After Wick rotation to Euclidean target space, the action becomes

$$S_E[X,e] ={1\over2}\int d\lambda \left(e^{-1}\dot X_E^2+e m^2\right).$$

In the gauge $e=\mathcal T$ on $0\leq\lambda\leq1$,

$$S_E[X;\mathcal T] ={1\over2}\int_0^1 d\lambda \left({\dot X^2\over\mathcal T}+\mathcal T m^2\right).$$

The path integral with fixed endpoints is the heat kernel

$$K(x_f,x_i;\mathcal T) =\int_{X(0)=x_i}^{X(1)=x_f}\mathcal D X\, \exp[-S_E[X;\mathcal T]].$$

Because the action is Gaussian,

$$K(x_f,x_i;\mathcal T) =\int{d^D p\over(2\pi)^D} \exp\left[ip\cdot(x_f-x_i)-{\mathcal T\over2}(p^2+m^2)\right].$$

Equivalently, after performing the Gaussian momentum integral,

$$K(x_f,x_i;\mathcal T) ={1\over(2\pi\mathcal T)^{D/2}} \exp\left[-{(x_f-x_i)^2\over2\mathcal T}-{m^2\mathcal T\over2}\right].$$

The Euclidean scalar propagator is obtained by integrating over the modulus:

$$G_E(x_f,x_i) ={1\over2}\int_0^\infty d\mathcal T\,K(x_f,x_i;\mathcal T).$$

The factor $1/2$ appears because of our normalization of the einbein action. It is equivalent to the Schwinger representation

$${1\over p^2+m^2} ={1\over2}\int_0^\infty d\mathcal T\, \exp\left[-{\mathcal T\over2}(p^2+m^2)\right].$$

Thus

$$G_E(x_f,x_i) =\int{d^D p\over(2\pi)^D} {e^{ip\cdot(x_f-x_i)}\over p^2+m^2}.$$

A particle propagator may be written as a momentum-space denominator, as a Schwinger proper-time integral, or as a sum over Euclidean worldlines with fixed endpoints.

The lesson is important: after gauge fixing, we do not choose one value of the modulus $\mathcal T$. We integrate over it. String perturbation theory follows the same philosophy, but the moduli spaces are those of punctured Riemann surfaces.

The particle-to-string dictionary

The Polyakov string action is nearly forced on us by the particle example.

Relativistic particle	Relativistic string
Worldline coordinate $\lambda$	Worldsheet coordinates $\sigma^\alpha=(\tau,\sigma)$
Embedding $X^\mu(\lambda)$	Embedding $X^\mu(\tau,\sigma)$
Einbein $e(\lambda)$	Worldsheet metric $h_{\alpha\beta}(\tau,\sigma)$
Reparametrization invariance	Two-dimensional diffeomorphism invariance
Constraint $P^2+m^2=0$	Constraints $T_{\alpha\beta}=0$
Proper-time modulus $\mathcal T$	Moduli of Riemann surfaces
Sum over worldlines	Sum over worldsheets

The natural string analogue of the einbein action is

$$S_P =-{1\over4\pi\alpha'} \int d^2\sigma\sqrt{-h}\,h^{\alpha\beta} \partial_\alpha X^\mu\partial_\beta X_\mu.$$

The next page explains how this Polyakov action is related to the geometric area action of a string.

Exercises

Exercise 1: Reparametrization invariance of the square-root action

Show explicitly that

$$S_{\rm pp}=-m\int d\lambda\sqrt{-\dot X^2}$$

is invariant under an orientation-preserving reparametrization $\lambda'=f(\lambda)$.

Solution

Under $\lambda'=f(\lambda)$ with $f'(\lambda)>0$,

$${dX^\mu\over d\lambda'} ={d\lambda\over d\lambda'}{dX^\mu\over d\lambda}.$$

Therefore

$$\sqrt{-\left({dX\over d\lambda'}\right)^2} ={d\lambda\over d\lambda'}\sqrt{-\left({dX\over d\lambda}\right)^2}.$$

Multiplying by $d\lambda'$ gives

$$d\lambda'\sqrt{-\left({dX\over d\lambda'}\right)^2} =d\lambda\sqrt{-\left({dX\over d\lambda}\right)^2}.$$

So the integrand is an invariant line element and the action is unchanged.

Exercise 2: Eliminating the einbein

Starting from

$$S[X,e] ={1\over2}\int d\lambda \left(e^{-1}\dot X^2-e m^2\right),$$

vary with respect to $e$ and show that substituting the solution for $e$ gives the square-root action.

Solution

The variation is

$$\delta_e S ={1\over2}\int d\lambda \left(-e^{-2}\dot X^2-m^2\right)\delta e.$$

Thus

$$-e^{-2}\dot X^2-m^2=0,$$

so

$$e^2={-\dot X^2\over m^2}.$$

For $e>0$,

$$e={\sqrt{-\dot X^2}\over m}.$$

Substitute into the action:

$$\begin{aligned} S[X,e_{\rm cl}] &={1\over2}\int d\lambda \left({m\over\sqrt{-\dot X^2}}\dot X^2 -{\sqrt{-\dot X^2}\over m}m^2\right) \\ &={1\over2}\int d\lambda \left(-m\sqrt{-\dot X^2}-m\sqrt{-\dot X^2}\right) \\ &=-m\int d\lambda\sqrt{-\dot X^2}. \end{aligned}$$

Exercise 3: Hamiltonian constraint

Compute the canonical momentum and Hamiltonian for the einbein action, and show that varying $e$ imposes the mass-shell condition.

Solution

The momentum is

$$P_\mu={\partial L\over\partial\dot X^\mu}=e^{-1}\dot X_\mu.$$

Thus $\dot X_\mu=eP_\mu$. The Hamiltonian is

$$\begin{aligned} H &=P_\mu\dot X^\mu-L \\ &=eP^2-{1\over2}\left(eP^2-e m^2\right) \\ &={e\over2}\left(P^2+m^2\right). \end{aligned}$$

Since $e$ is a Lagrange multiplier, varying it gives

$$P^2+m^2=0.$$

With $P^0=E$ and mostly-plus signature, this is

$$-E^2+\mathbf p^2+m^2=0,$$

or $E^2=\mathbf p^2+m^2$.

Exercise 4: Proper-time representation

Verify that, for Euclidean $p^2\geq0$ and $m^2>0$,

$${1\over p^2+m^2} ={1\over2}\int_0^\infty d\mathcal T\, \exp\left[-{\mathcal T\over2}(p^2+m^2)\right].$$

Then explain how this identity leads to the worldline path-integral representation of $G_E(x_f,x_i)$.

Solution

Let $A=p^2+m^2>0$. Then

$${1\over2}\int_0^\infty d\mathcal T\,e^{-\mathcal T A/2} ={1\over2}\cdot {2\over A} ={1\over A}.$$

Therefore

$${1\over p^2+m^2} ={1\over2}\int_0^\infty d\mathcal T\, \exp\left[-{\mathcal T\over2}(p^2+m^2)\right].$$

Fourier transforming gives

$$G_E(x_f,x_i) =\int{d^D p\over(2\pi)^D}{e^{ip\cdot(x_f-x_i)}\over p^2+m^2}.$$

Using the proper-time identity inside the momentum integral gives

$$G_E(x_f,x_i) ={1\over2}\int_0^\infty d\mathcal T \int{d^D p\over(2\pi)^D} \exp\left[ip\cdot(x_f-x_i)-{\mathcal T\over2}(p^2+m^2)\right].$$

The inner integral is the heat kernel for a particle moving from $x_i$ to $x_f$ in fixed proper time $\mathcal T$. The configuration-space worldline path integral is another representation of this same heat kernel.