Heisenberg Representation of Dirac Equation Quantization I am wondering exactly how to apply the following method to the Dirac equation (and even electromagnetism if it is easy to type up). It is a method of deriving the momentum-space Hamiltonian without using the explicit Hamiltonian, and is a lot easier than working with quadratic terms. Could somebody be so kind as to help me by showing me how to do the same thing in one or both of those cases as the following?
For a scalar field, we have in general that
$$\hat{\Psi}(x') = e^{i(a_{\mu}\hat{P}^{\mu} + \frac{1}{2} \varepsilon_{\mu \nu}J^{\mu \nu})} \hat{\Psi}(x) e^{-i(a_{\mu}\hat{P}^{\mu} + \frac{1}{2} \varepsilon_{\mu \nu}J^{\mu \nu})}$$
holds, so translating the operator 
$$\hat{\Psi}(\vec{x},t) =  \hat{\Psi}^+(\vec{x},t) + \hat{\Psi}^-(\vec{x},t) = \int d^3 \tilde{\mathrm{k}} \ \hat{a}(\vec{k},t)e^{i\vec{k} \cdot \vec{x}} + \int d^3 \tilde{\mathrm{k}} \ \hat{b}^+(\vec{k},t)e^{-i\vec{k}\cdot \vec{x}}$$ 
in time we see 
$$\hat{\Psi}(\vec{x},t + dt) = e^{i\hat{H}dt} \hat{\Psi} e^{-i\hat{H}dt}$$
is it's unitarily equivalent time displacement. But you know that
$$\hat{\Psi}(\vec{x},t + dt) = (1 + i\hat{H}dt)\hat{\Psi}(1 - i\hat{H}dt) = \hat{\Psi}(\vec{x},t) - i[\hat{\Psi},\hat{H}]dt$$
produces
$$\frac{\partial \hat{\Psi}}{\partial t}dt = - i[\hat{\Psi},\hat{H}]dt$$
which, when expanded, looks like
$$\frac{\partial \hat{\Psi}^+}{\partial t}dt + \frac{\partial \hat{\Psi}^-}{\partial t}dt =  - i[\hat{\Psi}^+,\hat{H}]dt - i[\hat{\Psi}^-,\hat{H}]dt$$
so that
$$\frac{\partial \hat{\Psi}^+}{\partial t} = - i[\hat{\Psi}^+,\hat{H}]$$
$$ \frac{\partial \hat{\Psi}^-}{\partial t} = - i[\hat{\Psi}^-,\hat{H}]$$
Fourier expanded
$$\int d^3 \tilde{\mathrm{k}} (- i \omega ) \ \hat{a}(\vec{k},t)e^{i\vec{k} \cdot \vec{x}}  = \int d^3 \tilde{\mathrm{k}} (-i) [\hat{a}(\vec{k},t),\hat{H}]e^{i\vec{k} \cdot \vec{x}}$$
$$ \int d^3 \tilde{\mathrm{k}} \ (i \omega ) \hat{b}^+(\vec{k},t)e^{-i\vec{k}\cdot \vec{x}} = \int d^3 \tilde{\mathrm{k}} (- i)[\hat{b}^+(\vec{k},t),\hat{H}]e^{-i\vec{k}\cdot \vec{x}}$$
means the equations
$$  \omega  \ \hat{a}(\vec{k},t)= [\hat{a}(\vec{k},t),\hat{H}]$$
$$  \omega  \hat{b}^+(\vec{k},t) =  - [\hat{b}^+(\vec{k},t),\hat{H}]$$
should hold.
These equations almost immediately imply, simply by observation, that choosing
$H = \int d^3 \tilde{\mathrm{k}}' \omega'  \hat{a}^+(\vec{k}',t) \hat{a}(\vec{k}',t)$
in the first case, and
$H = \int d^3 \tilde{\mathrm{k}}' \omega'  \hat{b}(\vec{k}',t) \hat{b}^+(\vec{k}',t)$
in the second case, will produce the right answer, because of the commutation relations
$$ [\hat{a},\hat{a}'^+\hat{a}'] = \delta^3(\vec{k}-\vec{k}')\hat{a}$$
$$ [\hat{a}^+,\hat{a}'^+\hat{a}'] = - \delta^3(\vec{k}-\vec{k}')\hat{a}$$
Therefore 
$$\hat{\mathrm{H}} = \int d^3 \tilde{\mathrm{k}} ' \omega (\hat{a}^+(\vec{k}',t)\hat{a}(\vec{k}',t) + \hat{b}(\vec{k}',t)\hat{b}^+(\vec{k}',t))$$
This same method gives us the momentum, angular momentum and charge very easy, I hope it applies to the Dirac equation, electromagnetism and even supersymmetry to avoid those big calculations. Hope to hear from you.
 A: Start with the Dirac field operators, 
$$
\hat{\psi}_D(\vec{x},t) =  \hat{\psi}_D^+(\vec{x},t) + \hat{\psi}_D^-(\vec{x},t) =  \sum_{s=1}^2 {\int d^3 \tilde{\mathrm{k}} \; \hat{b}_s(\vec{k},t)u_s(\vec{k})e^{i\vec{k} \cdot \vec{x}} + \int d^3 \tilde{\mathrm{k}} \ \hat{c}_s^\dagger(\vec{k},t)v_s(\vec{k})e^{-i\vec{k}\cdot \vec{x}}}
$$ 
where now $u_s(\vec{k})$, $v_s(\vec{k})$ are 4-component spinors and operators $\hat{b}_s(\vec{k},t)$, $\hat{c}_s(\vec{k},t)$ satisfy the anticommutation relations
$$
\{\hat{b}_r(\vec{k}',t), \hat{b}^\dagger_s(\vec{k},t) \} = \{\hat{c}_r(\vec{k}',t), \hat{c}^\dagger_s(\vec{k},t) \} = (2\pi)^3\delta_{rs}\delta(\vec{k}'-\vec{k})\\
\{\hat{b}_r(\vec{k}',t), \hat{b}_s(\vec{k},t) \} = \{\hat{c}_r(\vec{k}',t), \hat{c}_s(\vec{k},t) \} = 0
$$
Applying your time-translation procedure to the positive and negative frequency parts of field as before, yields again
$$
\frac{\partial \hat{\psi}_D^+}{\partial t} = - i[\hat{\psi}_D^+,\hat{H}], \;\;\; \frac{\partial \hat{\psi}_D^-}{\partial t} = - i[\hat{\psi}_D^-,\hat{H}]
$$
and after identifying terms in the Fourier expansion,
$$  
\omega  \hat{b}_s(\vec{k},t)u_s(\vec{k})= [\hat{b}_s(\vec{k},t)u_s(\vec{k}),\hat{H}] \\
\omega  \hat{c}_s^\dagger(\vec{k},t)v_s(\vec{k}) =  - [\hat{c}_s^\dagger(\vec{k},t)v_s(\vec{k}),\hat{H}]
$$
Now assume that the Hamiltonian is only bilinear in the $\hat{b}_r(\vec{k},t)u_r(\vec{k})$ and $\hat{c}_r(\vec{k},t)v_r(\vec{k})$. Straightforward algebra will show that terms of the form $\left(\hat{b}^\dagger_r(\vec{k},t)u^\dagger_r(\vec{k})\right) \left(\hat{c}_{r'}(\vec{k},t)v_{r'}(\vec{k})\right)$ and $\left(\hat{c}^\dagger_r(\vec{k},t)u^\dagger_r(\vec{k})\right) \left(\hat{b}_{r'}(\vec{k},t)v_{r'}(\vec{k})\right)$ cannot contribute if the above commutation relations are to hold. So $H$ can have only terms of the form $\left(\hat{b}^\dagger_r(\vec{k},t)u^\dagger_r(\vec{k})\right) \left(\hat{b}_{r'}(\vec{k},t)u_{r'}(\vec{k})\right)$ and $\left(\hat{c}^\dagger_r(\vec{k},t)v^\dagger_r(\vec{k})\right) \left(\hat{c}_{r'}(\vec{k},t)v_{r'}(\vec{k})\right)$. But the spinor closure relations 
$$
u^\dagger_r(\vec{k})u_{r'}(\vec{k}) = 2 \omega \delta_{rs}, \;\;\; v^\dagger_r(\vec{k})v_{r'}(\vec{k}) = 2 \omega\delta_{rr'}
$$
further reduce acceptable terms to $2\omega \hat{b}^\dagger_r(\vec{k},t)\hat{b}_r(\vec{k},t)$ and $2 \omega \hat{c}^\dagger_r(\vec{k},t)\hat{c}_r(\vec{k},t)$ and eventually leave the Hamiltonian as
$$ 
H = \sum_{r=1}^2{\int d^3 \tilde{\mathrm{k}}' \omega'  \left[ \hat{b}^\dagger_r(\vec{k}',t) \hat{b}_r(\vec{k}',t) + \hat{c}^\dagger_r(\vec{k}',t) \hat{c}_r(\vec{k}',t)\right]}
$$
Indeed, now we have
$$
[\hat{b}_s(\vec{k},t)u_s(\vec{k}),\hat{H}]  \equiv  \sum_{r=1}^2{\int d^3 \tilde{\mathrm{k}}' \omega'  \left[\hat{b}_s(\vec{k},t)u_s(\vec{k}), \hat{b}^\dagger_r(\vec{k}',t) \hat{b}_r(\vec{k}',t) + \hat{c}^\dagger_r(\vec{k}',t) \hat{c}_r(\vec{k}',t)\right]} = \\
= \sum_{r=1}^2{\int d^3 \tilde{\mathrm{k}}' \omega'  \left[\hat{b}_s(\vec{k},t), \hat{b}^\dagger_r(\vec{k}',t) \hat{b}_r(\vec{k}',t)\right]u_s(\vec{k})} + \sum_{r=1}^2{\int d^3 \tilde{\mathrm{k}}' \omega'  \left[\hat{b}_s(\vec{k},t), \hat{c}^\dagger_r(\vec{k}',t) \hat{c}_r(\vec{k}',t)\right]u_s(\vec{k})}
$$
and using the anticommutation relations for the $\hat{b}_s(\vec{k},t)$, $\hat{c}_s(\vec{k},t)$,
$$
\left[\hat{b}_s(\vec{k},t), \hat{b}^\dagger_r(\vec{k}',t) \hat{b}_r(\vec{k}',t)\right] = \hat{b}_s(\vec{k},t)\hat{b}^\dagger_r(\vec{k}',t) \hat{b}_r(\vec{k}',t) - \hat{b}^\dagger_r(\vec{k}',t) \hat{b}_r(\vec{k}',t) \hat{b}_s(\vec{k},t) = \\
= (2\pi)^3\delta_{rs}\delta(\vec{k}'-\vec{k})\hat{b}_r(\vec{k}',t) - \hat{b}^\dagger_r(\vec{k}',t)\hat{b}_s(\vec{k},t) \hat{b}_r(\vec{k}',t) + \hat{b}^\dagger_r(\vec{k}',t)\hat{b}_s(\vec{k},t) \hat{b}_r(\vec{k}',t) = \\
= (2\pi)^3\delta_{rs}\delta(\vec{k}'-\vec{k})\hat{b}_r(\vec{k}',t) 
$$
$$
\left[\hat{b}_s(\vec{k},t), \hat{c}^\dagger_r(\vec{k}',t) \hat{c}_r(\vec{k}',t)\right] = \hat{b}_s(\vec{k},t)\hat{c}^\dagger_r(\vec{k}',t) \hat{c}_r(\vec{k}',t) - \hat{c}^\dagger_r(\vec{k}',t) \hat{c}_r(\vec{k}',t) \hat{b}_s(\vec{k},t) = \\
= - \hat{c}^\dagger_r(\vec{k}',t)\hat{b}_s(\vec{k},t) \hat{c}_r(\vec{k}',t) + \hat{c}^\dagger_r(\vec{k}',t)\hat{b}_s(\vec{k},t) \hat{c}_r(\vec{k}',t) = 0
$$
Substituting gives, as expected,
$$
\sum_{r=1}^2{\int d^3 \tilde{\mathrm{k}}' \omega'  \left[\hat{b}_s(\vec{k},t), \hat{b}^\dagger_r(\vec{k}',t) \hat{b}_r(\vec{k}',t)\right]u_s(\vec{k})} = \sum_{r=1}^2{\int d^3 \tilde{\mathrm{k}}' \omega' (2\pi)^3\delta_{rs}\delta(\vec{k}'-\vec{k})\hat{b}_r(\vec{k}',t) u_s(\vec{k})}  = \omega  \hat{b}_s(\vec{k},t)u_s(\vec{k})
$$
And similarly for $[\hat{c}^\dagger_s(\vec{k},t)v^\dagger_s(\vec{k}),\hat{H}]$.
