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# Relativistic quantum mechanics

In almost every introduction to quantum field theory, there is some historical review of the topic in which the idea of relativistic quantum mechanics is introduced and is shown to be insufficient. While this is true to some degree, the reasons for this run deeper than the usual introduction will get into, and the notion of a well-defined relativistic quantum mechanics is not reintroduced until the ridiculously far-off point of string theory.

To do this correctly, we'll have to work out the rigorous quantization process of a relativistic point particle, rather than the fairly hand-wavy process usually used in such introductions, using simply the substitutions

\begin{eqnarray} \vec{p} &=& -i\hbar \vec{\nabla}\\ E &=& i\hbar \partial_t \end{eqnarray}

## The relativistic point particle

There are many ways to describe the relativistic point particle, but the two main ones are the Nambu-Goto action and the Polyakov action. For a free particle in Minkowski space, the Nambu-Goto action is

\begin{equation} S[X] = m \int_{\tau_a}^{\tau_b} d\tau \sqrt{\eta_{\mu\nu} \dot{X}^\mu(\tau)\dot{X}^\nu(\tau)} \end{equation}

while the Polyakov action is

\begin{equation} S[X, e] = \int_{\tau_a}^{\tau_b} d\tau \frac{1}{2} \left[ \frac{1}{e} \eta_{\mu\nu} \dot{X}^\mu(\tau)\dot{X}^\nu(\tau) - m^2 e \right] \end{equation}

For now, let's focus on the Nambu-Goto action. It is well-known that the action is invariant under a reparametrization

\begin{equation} \tau \to \tau' = f(\tau) \end{equation}

For some diffeomorphism $f$. For any diffeomorphism leaving the boundaries invariant, we have

\begin{eqnarray} S[X] &=& m \int_{\tau_a}^{\tau_b} d\tau \sqrt{\eta_{\mu\nu} \dot{X}^\mu(\tau)\dot{X}^\nu(\tau)}\\ &=& m \int_{\tau_a}^{\tau_b} d\tau' \frac{d (f^{-1}(\tau')}{d\tau'} \sqrt{\eta_{\mu\nu} \dot{X}^\mu(f^{-1}(\tau'))\dot{X}^\nu(f^{-1}(\tau'))}\\ &=& \end{eqnarray}

Therefore, we have that for any path $X$ minimizing the action, $X \circ f$ will also do so, so that our system is underdetermined. Using the Hessian of our Lagrangian, we can verify that this is true. First computing its derivatives with respect to $\dot{X}$,

\begin{eqnarray} \frac{\partial L}{\partial \dot{X}^\mu} &=& m \frac{\partial \sqrt{\dot{X}^2}}{\partial \dot{X}^2} \frac{\partial}{\partial \dot{X}^\mu} \eta_{\mu\nu} \dot{X}^\mu(\tau)\dot{X}^\nu(\tau)\\ &=& - m \frac{1}{\sqrt{\dot{X}^2}} \dot{X}_\mu(\tau) \end{eqnarray} \begin{eqnarray} \frac{\partial L}{\partial \dot{X}^\mu \partial \dot{X}^\nu} &=& \end{eqnarray}

...

Therefore, we're going to have to use a constrained approach to our Hamiltonian, as well as constrained quantization. This is where the delicate nature of the quantization comes in.

Using a more cautious approach to the Legendre transform, let's consider now the equivalent theory

\begin{equation} S[X, V, P] = \int_{\tau_a}^{\tau_b} d\tau \left[ \sqrt{\eta_{\mu\nu} V^\mu(\tau) V^\nu(\tau)} + P_\mu (\dot{X} - V) \right] \end{equation}

This can easily be checked to have the equations of motion

\begin{eqnarray} \dot{X} &=& V\\ \frac{\partial L(V)}{\partial V^\mu} &=& P_\mu\\ \dot{P}_\mu &=& 0 \end{eqnarray}

Performing all the appropriate substitutions, this is equivalent to the Euler-Lagrange equations. Our momentum here is

\begin{eqnarray} P_\mu &=& \frac{\partial L(V)}{\partial V^\mu}\\ &=& \end{eqnarray}

We have seen previously that due to the vanishing determinant of our Hessian, we cannot invert our momentum to get $\dot{X}$ as a function of $P$.

...

To fix our gauge freedom, we can add some terms to fix our constraint. In our case, the proper time gauge will be an additional term

\begin{equation} H_\lambda[X, P, \lambda] = \lambda(\tau) \left[ P_\mu P^\mu - m^2 \right] \end{equation}

Our total action then becomes

\begin{equation} H[X, P, \lambda] = \end{equation}

...

### BRST quantization

A common way to perform the quantization of a gauge system is the BRST quantization. For this, first we have to introduce some additional variables,

Last updated : 2019-10-21 17:27:42