Dirac equation as Hamiltonian system

Question

Let us consider Dirac equation $$(i\gamma^\mu\partial_\mu -m)\psi =0$$ as a classical field equation. Is it possible to introduce Poisson bracket on the space of spinors $\psi$ in such a way that Dirac equation becomes Hamiltonian equation $$\dot{\psi}=\{H_D, \psi\}?$$

Of course, such bracket would be graded (super bracket), but if it exists this would explain on classical level why $\frac{1}{2}$-spinors correspond to fermions.

Answer 1

In my opinion, one cannot rigorously [*] define the bracket. Suppose that you use the Dirac field equation for arriving to the ordinary Lagrangian density

$$ \mathcal{L} = c \bar{\psi} \left( i\hbar \gamma^\mu \frac{\partial}{\partial x^\mu} \right) \psi $$

This is a function of the spinor components $\psi_i$ and their adjoints $\bar{\psi_i}$. The problem begins when you try to obtain the conjugate momentum for the adjoints (the dot denotes time derivative)

$$\bar{\pi_i} = \frac{\partial \mathcal{L}}{\partial \dot{\bar{\psi_i}}} = 0$$

which implies that not all the canonical variables are independent and not true 'phase-space' structure exists.

You could try to formally define Poisson brackets in the usual fashion,

$$ \{ A, B \} \equiv \sum_i \frac{\partial A}{\partial \psi_i} \frac{\partial B}{\partial \pi_i} - \frac{\partial A}{\partial \pi_i} \frac{\partial B}{\partial \psi_i} + \sum_j \frac{\partial A}{\partial \bar{\psi}_j} \frac{\partial B}{\partial \bar{\pi}_j} - \frac{\partial A}{\partial \bar{\pi}_j} \frac{\partial B}{\partial \bar{\psi}_j} $$

but note that this is only formally valid, because the variables are not all independent. The equations of motion would be written somewhat as

$$ \dot{A} \approx \{ A, \mathcal{H} \} $$

using Dirac weak equality sign, because this is a constrained dynamics. The Hamiltonian density is obtained from

$$ \mathcal{H} \approx \sum_i \pi_i \dot{\psi}_i + \sum_j \bar{\pi}_j \dot{\bar{\psi}}_j - \mathcal{L} $$

Notice that all of this is a quantum treatment. There is not classical spinor theory.

[*] I suppose that all depends on what are you trying to do.

Answer 2

Now, I don't know what the word rigorous means, but here is a straight off the bat naive answer. Given $$ H = \int d^3 x \, \bar{\Psi}(i \gamma_i \partial_i +m)\Psi $$ from $$ \mathcal{L} = i\bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi - \bar{\Psi} m \Psi \quad \text{and}\quad H = \int d^3x \, (\pi \dot{\Psi}-\mathcal{L}) $$ with $(+,-,-,-)$. Let's use the possoin bracket $$ \{A,B\} = \int d^3 x \, (\delta_{\Psi}A \, \delta_{\pi}B-\delta_{\pi}A \, \delta_{\Psi}B) $$ and let's remember that $\pi_{\Psi} = i \bar{\Psi}\gamma_0 \implies \bar{\Psi} = -i \pi \gamma_0$ so that $$ H = \int d^3x \, -i\gamma_0\pi(i\gamma_i\partial_i +m)\Psi $$ consider $$ \delta_{\pi}H = \int d^3x \, \gamma_0 (\delta_{\pi}\pi) \gamma_i \partial_i \Psi + \gamma_0\pi\gamma_i\partial_i(\delta_{\pi}\Psi) - i\gamma_0 (\delta_{\pi}\pi) m\Psi - i\gamma_0 \pi m (\delta_{\pi}\Psi) $$ $$ =\gamma_0 \gamma_i \partial_i \Psi - i\gamma_0 m \Psi $$ Then looking at $$ \{H,\Psi\} = -\gamma_0 \gamma_i \partial_i \Psi + i\gamma_0 m \Psi = \dot{\Psi} $$ God knows if this is right, but this sounds like what you are looking for.

Answer 3

It is possible to construct a Hamiltonian. In fact this is the way how Dirac initially wrote his equation. For that the time and space coordinates have to be treated differently. In the Schrödinger picture, the Hamiltonian generates the time dynamics via ($\hbar =0$) $$i \partial_t \psi = H \psi.$$ We see that we can obtain this structure from the Dirac equation by multiplying it by $\gamma^0$ (and using $\gamma_0^2=1$). With that we obtain the result $$ H = {\bf\alpha} \cdot {\bf p} + \beta m$$ where we introduced the conventional notation $\beta =\gamma^0$, $\alpha^k =\gamma^0 \gamma^k$, and $p_k = -i\partial_k$.

If you want to write this down in terms of a classical field theory, with the field $\psi$ evolving as $$\partial_t \psi({\bf r}) = \{\mathcal{H},\psi ({\bf r})\}, $$ the Hamiltonian is given by $$\mathcal{H} = \int\!d^3r \,\psi^*({\bf r})H \psi({\bf r}).$$

Edit:

I define the Poisson bracket to be $$\{ A, B\} = \int\!d^3r\left[\frac{\delta A}{\delta \psi({\bf r})}\frac{\delta B}{\delta \psi^*({\bf r})} -\frac{\delta B}{\delta \psi({\bf r})}\frac{\delta A}{\delta \psi^*({\bf r})} \right]$$ where the derivatives are functional derivatives and we have assumed (as usual) that $\psi$ and $\psi^*$ are independent variables with the defining relations $$ \frac{ \delta \Psi({\bf r})}{\delta \Psi({\bf r}')} = \frac{ \delta \Psi^*({\bf r})}{\delta \Psi^*({\bf r}')} = \delta^3({\bf r}-{\bf r}'), \quad\frac{ \delta \Psi({\bf r})}{\delta \Psi^*({\bf r}')} = \frac{ \delta \Psi^*({\bf r})}{\delta \Psi({\bf r}')} = 0.$$