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)?
[Physics] Hamilton formalism for Dirac spinors
classical-field-theorydirac-equationhamiltonian-formalismpoisson-bracketsquantum-field-theory
Related Solutions
Given the Dirac Lagrangian density
$$ \tag{1} {\cal L}~=~\overline{\psi}(i\sum_{\mu=0}^3\gamma^{\mu}\partial_{\mu}-m)\psi, \qquad \overline{\psi}~:=~\psi^{\dagger}\gamma^0, \qquad \{\gamma^{\mu},\gamma^{\nu}\}_{+} ~=~2\eta^{\mu\nu}{\bf 1}_{4\times 4}, $$
with Minkowski signature $(+,-,-,-)$, and $\psi$ is a Grassmann-odd Dirac-spinor, the question is How to find the corresponding Hamiltonian formalism?
The Legendre transformation of (1) is singular. The Dirac-Bergmann analysis of the theory (1) leads to constraints, cf. e.g. Ref. 1 or this Phys.SE post. Here we will instead take a shortcut using the Faddeev-Jackiw method.
I) Complex Grassmann-fields. We first identify the Hamiltonian density ${\cal H}$ as (minus) the terms in (1) that don't involve time derivatives:
$$ \tag{2} {\cal L}~=~i\psi^{\dagger}\dot{\psi}-{\cal H}, \qquad {\cal H}~=~ \overline{\psi} (-i\sum_{j=1}^3\gamma^{j}\partial_{j}+m)\psi. $$
The symplectic one-form potential can be transcribed from the kinetic term in (2):
$$ \tag{3} \vartheta(t) ~=~\int\! d^3x~ i\psi^{\dagger}({\bf x},t) ~\mathrm{d}\psi({\bf x},t), $$
where $\mathrm{d}$ denotes the exterior derivative$^1$ on the infinite-dimensional configuration space for the fermion field. The symplectic two-form is then
$$ \omega(t)~=~\mathrm{d}\vartheta(t) ~=~\int\! d^3x~ i\mathrm{d}\psi^{\dagger}({\bf x},t) \wedge \mathrm{d}\psi({\bf x},t) $$ $$ \tag{4} ~=~\int\! d^3x~d^3y~ i\mathrm{d}\psi^{\dagger}({\bf x},t) \wedge \delta^3({\bf x}-{\bf y}) ~\mathrm{d}\psi({\bf y},t). $$
The equal-time super-Poisson/Dirac bracket on fundamental fields is the inverse supermatrix of the supermatrix for the symplectic two-form (4):
$$ \tag{5} \{\psi_{\alpha}({\bf x},t), \psi^{\dagger}_{\beta}({\bf y},t)\}_{PB}~=~ -i \delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})~=~\{\psi^{\dagger}_{\alpha}({\bf x},t), \psi_{\beta}({\bf y},t)\}_{PB}, $$
and other fundamental super-Poisson brackets vanish. Due to the QM correspondence principle, the canonical anticommutation relations (CARs) are the super-Poisson brackets (5) multiplied with $i\hbar$:
$$ \tag{6} \{\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)\}_{+} ~=~ \hbar\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})\hat{\bf 1} ~=~\{\hat{\psi}^{\dagger}_{\alpha}({\bf x},t), \hat{\psi}_{\beta}({\bf y},t)\}_{+}, $$ and other CARs vanish.
I) Real Grassmann-fields. Alternatively, let us decompose the complex Dirac spinor
$$\tag{7}\psi_{\alpha}~\equiv~(\psi^1_{\alpha}+i\psi^2_{\alpha})/\sqrt{2} \quad\text{and}\quad \psi^{\dagger}_{\alpha}~\equiv~(\psi^1_{\alpha}-i\psi^2_{\alpha})/\sqrt{2}, $$
in real and imaginary parts. The Lagrangian density (2) reads up to total derivative terms$^2$
$$ \tag{2'} {\cal L}~=~\frac{i}{2}\left(\psi^{\dagger}\dot{\psi}- \dot{\psi}^{\dagger}\psi\right)-{\cal H} ~=~\frac{i}{2}\sum_{a=1}^2(\psi^a)^T\dot{\psi}^a-{\cal H}.$$
The corresponding symplectic one-form potential is
$$ \tag{3'} \vartheta(t) ~=~\sum_{a=1}^2\int\! d^3x~ \frac{i}{2}\psi^a({\bf x},t)^T ~\mathrm{d}\psi^a({\bf x},t). $$
The symplectic two-form is
$$ \omega(t)~=~\mathrm{d}\vartheta(t) ~=~\sum_{a=1}^2\int\! d^3x~ \frac{i}{2}\mathrm{d}\psi^a({\bf x},t)^T \wedge \mathrm{d}\psi^a({\bf x},t) $$ $$ \tag{4'} ~=~\sum_{a,b=1}^2\int\! d^3x~d^3y~ \frac{i}{2}\mathrm{d}\psi^a({\bf x},t)^T \wedge \delta_{ab}~\delta^3({\bf x}-{\bf y}) ~\mathrm{d}\psi^b({\bf y},t). $$
The equal-time super-Poisson is
$$ \tag{5'} \{\psi^a_{\alpha}({\bf x},t), \psi^b_{\beta}({\bf y},t)\}_{PB}~=~ -i \delta^{ab}~\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y}). $$
The CARs are
$$ \tag{6'} \{\hat{\psi}^a_{\alpha}({\bf x},t), \hat{\psi}^b_{\beta}({\bf y},t)\}_{+} ~=~ \hbar\delta^{ab}~\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})\hat{\bf 1} . $$
References:
- A. Das, Lectures on QFT, (2008); chapter 10.
--
$^1$ In our super-conventions, the exterior derivative $\mathrm{d}$ is Grassmann-even and carries form-degree +1.
$^2$ Note that adding a total time derivative
$$i\psi^{\dagger}\dot{\psi}~\longrightarrow~i\psi^{\dagger}\dot{\psi}+ \frac{d}{dt}(\alpha\psi^{\dagger}\psi)\tag{8} $$
to the kinetic term (2) corresponds to adding an exact term
$$ \vartheta(t)~\longrightarrow~\vartheta(t)+ \mathrm{d} \int\! d^3x~ \alpha\psi^{\dagger}({\bf x},t) \psi({\bf x},t) \tag{9} $$
to the symplectic one-form potential (3), which has no effect on the symplectic 2-form (4).
What follows is what you could describe as naiively applying the Legendre transform for constructing the Hamiltonian from the Lagrangian. Weinberg's "Quantum Theory of Fields" Vol. I chapter 8 goes over the canonical quantization of the electromagnetic field that correctly handles constraint equations.
Rather than finding the $T^{00}$ component of the stress-energy tensor, which often needs to be modified using the equations of motion to bring them into a recognizable form, you can construct the Hamiltonian using the canonical formalism. First, take the Lagrangian (using that instead of the density so that I can be free to move back and forth from mode space - also using Heaviside-Lorentz units with $\hbar=c=1$, and particle physicists' metric signature $(+,-,-,-)$), \begin{align} L & = \int \operatorname{d}^3x \left(\bar{\psi}\left[i\gamma^\mu D_\mu - m\right]\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} \right) \\ & = \int \operatorname{d}^3x \left( \bar{\psi}\left[i\gamma^\mu D_\mu - m\right]\psi + \frac{1}{2} E_iE_i - B_iB_i \right) \\ & = \int \operatorname{d}^3x \left( \vphantom{\frac{1}{2}} \bar{\psi}i\gamma^0\partial_0\psi + \bar\psi i \gamma^i D_i \psi - m\bar{\psi}\psi - e \phi \bar{\psi}\gamma^0\psi \right. \\ & \hphantom{=\int \operatorname{d}^3x}\ \ \left. + \frac{1}{2}\left[-\partial_i\phi-\partial_0A_i\right] \left[-\partial_i\phi-\partial_0A_i\right] - \frac{\epsilon_{ijk}\epsilon_{inm}}{2} \partial_jA_k \partial_nA_m \right). \end{align} The reason for expanding the the covariant inner products is to make explicit the next step: defining the canonically conjugate momenta. \begin{align} \pi_\psi &\equiv \frac{\delta L}{\delta \partial_0 \psi} = \bar{\psi}i\gamma^0 = i\psi^\dagger \\ \Pi_i &\equiv \frac{\delta L}{\delta \partial_0 A_i} = \partial_i\phi + \partial_0 A_i \\ \Pi_\phi &\equiv \frac{\delta L}{\delta \partial_0 A_0} = 0 \end{align} Notice how the momentum canonically conjugate to $\phi=A_0$ doesn't appear in the Lagrangian. This means that, without gauge fixing, $\phi$ is a non-dynamical field that doesn't have any momentum. In classical physics, it plays the roll of a Lagrange multiplier that enforces $\rho - \nabla\cdot\mathbf{E}= e\bar{\psi}\gamma^0\psi + \partial_i \Pi_i =0$, the first of Maxwell's equation.
Now, $H$ is defined as \begin{align} H &= \int \operatorname{d}^3x \left[\pi_\psi \partial_0\psi + \Pi_i \partial_0 A_i\right] - L \\ &= \int \operatorname{d}^3x \left[ \vphantom{\frac{\epsilon_i}{2}} \Pi_i (\Pi_i - \partial_i\phi) - \pi_\psi \gamma^0 \gamma^i D_i\psi -i m\pi_\psi \gamma^0\psi - ie\phi\pi_\psi \psi \right. \\ &\hphantom{= \int \operatorname{d}^3x}\ \ \left.- \frac{1}{2} \Pi_i \Pi_i + \frac{\epsilon_{ijk}\epsilon_{inm}}{2}\partial_jA_k \partial_nA_m \right]. \end{align} Notice that the term containing the time derivative of $\psi$ vanishes entirely because the Dirac equation is first order in time, so all of the needed time derivatives are supplied by the canonical equations of motion.
Now we move the gauge invariant parts of the electromagnetic field into mode space (Fourier transform over space, but not time) to highlight some interesting structure in the Hamiltonian. \begin{align} H & = \int \operatorname{d}^3k \left[ \frac{1}{2}\Pi_i\Pi_i + \frac{k^2}{2} \left(\delta_{ij} - \frac{k_i k_j}{k^2}\right)A_iA_j \right] \\ &\hphantom{=}+\int \operatorname{d}^3x \left[- \pi_\psi \gamma^0 \gamma^i D_i\psi -i m\pi_\psi \gamma^0\psi - ie\phi\pi_\psi \psi - \Pi_i\partial_i\phi \right] \\ &= \int \operatorname{d}^3k \left[ \frac{k_ik_j}{2 k^2}\Pi_i\Pi_j + \frac{1}{2} \left(\delta_{ij} - \frac{k_i k_j}{k^2}\right)\left(\Pi_i\Pi_j + k^2 A_iA_j\right) \right] \\ &\hphantom{=}+\int \operatorname{d}^3x \left[- \pi_\psi \gamma^0 \gamma^i D_i\psi -i m\pi_\psi \gamma^0\psi - ie\phi\pi_\psi \psi + \phi \partial_i \Pi_i\right] \end{align} Note that the electromagnetic field momentum, $\Pi_i$, is gauge invariant (it's just $-E_i$). The quantity $\left(\delta_{ij} - \frac{k_i k_j}{k^2}\right)A_j$ is also gauge invariant since it is the Fourier transform of the solenoidal part of $A_i$; in other words, the purpose of that last line was to collect the solenoidal components of the electromagnetic field (the photons) and move the derivative from $\phi$ to $\Pi_i$ in that term. The longitudinal kinetic term, $\frac{k_i k_j}{2k^2}\Pi_i\Pi_j$, is likewise nicely isolated.
It is also interesting to note that the only terms that aren't manifestly gauge invariant are $$-ie\phi\pi_\psi \psi + \phi \partial_i \Pi_i = e\phi\bar{\psi}\gamma^0\psi - \phi\partial_i E_i,$$ which vanishes by the equations of motion. It strikes me as likely that it is dealing with these terms in a canonical fashion that leads to the requirement for gauge fixing and constraints.
Best Answer
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).