Hamilton formalism for Dirac spinors

Question

Let's have the Dirac free lagrangian: $$ L = \bar {\Psi} (i\gamma^{\mu}\partial_{\mu} - m) \Psi . $$ I can rewrite it as $$ L = i\Psi^{\dagger}\partial_{0}\Psi - H_{d}, \quad H_{d} = \Psi^{\dagger}(-i(\hat {\alpha} \cdot \nabla ) + \beta m)\Psi , \quad \hat {\alpha} = \gamma_{0}\gamma . $$ Here $H_{d}$ is the Hamiltonian density. Is it possible to build Hamilton equations and Poisson brackets for spinors (for example, $\Psi$ may be the canonical coordinate and $\Psi^{\dagger}$ be the canonical impulse)?

Answer 1

The Hamiltonian density for any classical field is given by: \begin{equation} \mathcal{H} = \pi \dot{\phi} - \mathcal{L} \end{equation} where $\pi$ is the canonical momentum density: \begin{equation} \pi(\mathbf{x},t) = \frac{\partial \mathcal{L}}{\partial \dot{\phi}(\mathbf{x},t)} \end{equation} In classical point particle mechanics the Poisson brackets for two functions $f$ and $g$ are defined as: \begin{equation} \left\{f,g\right\}_{PB} = \frac{ \partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} \end{equation} where $q_j$ are the generalized coordinates and $p_j$ are the canonical momenta. Clearly: \begin{equation} \left\{q,p\right\}_{PB} = 1 \tag{1} \end{equation} In field theory, the Poisson bracket for two functionals $f$ and $g$ at equal times is defined as: \begin{equation} \left\{f(t),g(t)\right\}_{PB} = \int \mathrm{d}^3 \mathbf{x} \; \left(\frac{\delta f}{\delta \phi(\mathbf{x},t)} \frac{\delta g}{\delta \pi(\mathbf{x},t)} - \frac{\delta f}{\delta \pi(\mathbf{x},t)} \frac{\delta g}{\delta \phi(\mathbf{x},t)} \right) \end{equation} Now, using the rules of functional differentiation, it is easy to see that: \begin{equation} \left\{\phi(\mathbf{x},t),\pi(\mathbf{y},t)\right\}_{PB} = \delta^3(x-y) \end{equation} which is the classical field version of equation $(1)$.

Furthermore, according to Dirac's quantization rule, we can go back and forth between classical point particle mechanics and quantum mechanics via the following recipe: \begin{equation} \begin{array}{ccc} \text{classical mechanics} & \leftrightarrow & \text{quantum mechanics} \\ \displaystyle \left\{A,B\right\}_{PB} & \leftrightarrow & \displaystyle \frac{1}{i\hbar}\left[A,B \right] \end{array} \end{equation} provided the quantities we are considering exist in the classical world (for instance, quantum mechanical spin does not have a classical equivalent and so the rule does not work). To go back and forth between between classical field theory and the operatorial formulation of quantum field theory, we use the rule: \begin{equation} \begin{array}{ccc} \text{classical field theory} & \leftrightarrow & \text{quantum field theory} \\ \displaystyle \left\{A,B\right\}_{PB} & \leftrightarrow & \displaystyle \frac{1}{i\hbar}\left[A,B \right]_\mp \end{array} \end{equation} where the subscript $-$ means the normal commutator and is relevant for bosonic fields, and the $+$ subscript implies the anti-commutator which is relevant for fermionic fields (such as the Dirac field).

Answer 2

The classical Lagrangian for the free electron field is, $$ L=\int d^{3}x(i\psi^{\dagger}\frac{\partial \psi}{\partial t}+i\psi^{\dagger}\alpha_{r}\frac{\partial \psi}{\partial x^{r}}-m\psi^{\dagger}\beta \psi) \ . $$ The q's are $q^{i}(t)\rightarrow q^{(a,x)}\rightarrow \psi^{a}(t,x)$ and so the velocities are $\dot{q}^{i}(t)\rightarrow \frac{\partial \psi^{a}(t,x)}{\partial t}$. Varying the velocities, $$ \delta L= \int d^{3}x i\psi^{\dagger}\delta(\frac{\partial \psi}{\partial t}) $$ shows that the conjugate momenta - the $p_{i}(t)$ - are $i\psi^{\dagger}$. The classical Hamiltonian is, $$ H=\int d^{3}x(-i\psi^{\dagger}\alpha_{r}\frac{\partial \psi}{\partial x^{r}}+m\psi^{\dagger}\beta\psi) $$ The PBs corresponding to $[q^{i},p_{j}]_{PB}=\delta^{i}_{j}$ are, $$ [\psi^{a}(t,x),i\psi^{\dagger}_{b}(t,y)]_{PB}=\delta^{a}_{b}\delta(x-y) $$ Applying Dirac's quantization rule to the above PB gives the commutator, $$ [\psi^{a}(t,x),\psi^{\dagger}_{b}(t,y)]_{-}=\delta^{a}_{b}\delta(x-y) $$ and the $\psi$ are now operators. However, if one quantizes the classical Hamiltonian using this commutator there is no ground state, but one can quantize the classical Hamiltonian by using the anticommutator, $$ [\psi^{a}(t,x),\psi^{\dagger}_{b}(t,y)]_{+}=\delta^{a}_{b}\delta(x-y) $$ It appears that quantum dynamics of the electron field still goes with commutators. In the classical theory, $$ \frac{\partial \psi}{\partial t}=[\psi,H]_{PB} $$ and applying Dirac's quantization rule gives, $$ \frac{\partial \psi}{\partial t}=[\psi,H]_{-} $$ where $\psi$ and $H$ are operators. Everything looks consistent because writing down the quantum EoM directly from the quantum Hamiltonian (quantized using the anticommutator), $$ \frac{\partial \psi}{\partial t}=-i[\psi,H]_{-} $$ shows that the last two equations are identical. Note that the $H$ in the last equation is the quantized Hamiltonian which has been changed from the classical Hamiltonian by the use of the anticommutator, whilst the H in the penultimate equation is the classical Hamiltonian with all the fields simply promoted to operators. This is my understanding of the notion of the classical electron field, which comes from reading section 11 of Dirac's "Lectures on Quantum Field Theory."