Sometimes, I see this called the Dirac Hamiltonian:
$$\hat{H} =\vec{\mathbf{\alpha}}\cdot\hat{\vec{p}} + \mathbf{\beta}m ;$$
but, if we take the Dirac Lagrangian density, obtain the corresponding Hamiltonian density via Legendre transforming and integrate, we get — if I'm not mistaken —
$$\hat{H} = \iiint_{\mathbb{R}^3}\hat{\bar\Psi}(x)\, (-\mathrm{i} \gamma^i \partial_i + m)\hat\Psi(x)\,\mathrm{d}^3{x}.$$
If both expressions are equal, this would mean
$$ \mathbf{\beta}\hat{1} = \iiint_{\mathbb{R}^3} \hat{\bar\Psi}(x)\,\hat\Psi(x)\,\mathrm{d}^3{x},\hspace{1em}\mathbf{\alpha}^i\hat{p}_i=-\mathrm{i}\iiint_{\mathbb{R}^3}\hat{\bar\Psi}(x)\gamma^i\partial_i\hat\Psi(x)\,\mathrm{d}^3x.$$
Could this be? Or am I missing something, and these two Hamiltonians aren't the same object?
Best Answer
The main difference between the first and the second expression for the Hamiltonian of Dirac-equation is that the first is considering the Dirac-equation as an description for one single particle whereas the second assumes that the $\hat{\Psi}$s are field operators (that the little hat on top of them suggests) that act on the Fock space which is a space of multi-particle states ranging from the vacuum state (no particle) over the 1 particle state up to states of arbitrary high number of particles. Fock space is a short name for a space in "occupation number representation". So the second representation of the Hamiltonian is more general and contains some kind of the first expression as a special case.
The modern approach to the Dirac-equation prefers the multi-particle representation, since the viewpoint of the Dirac-equation as a description for one particle is actually compromised by the occurrence of states of negative energy/frequency. This issue actually can be solved by considering the Dirac equation as a description in multi-particle space, i.e. the Fock space.