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?

  • $\begingroup$ Are you sure there are no constraints in the system? $\endgroup$ Dec 3, 2021 at 12:07
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    $\begingroup$ With one Hamiltonian, the solution is interpreted as a wavefunction and with the other it's interpreted as a classical field. So there is no reason for them to be the same. $\endgroup$ Dec 3, 2021 at 12:24

1 Answer 1


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.

  • $\begingroup$ Thank you, Frederic! How could I see that special case explicitly? What should I impose? $\endgroup$
    – Pablo T.
    Dec 4, 2021 at 15:34
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    $\begingroup$ In order to see it, one has to develop the field operators and get H expressed with creation and annihilation operators. Then you let act this operator on a 1-electron state and you get the energy of 1 particle x 1-electron state. On the other hand you plug the 1-particle solution of the Dirac equation u(p) into the first equation, again you get the energy of the particle x 1-particle solution. Both provide the same result. But the H built with field operators can also act on multi-particle states which makes it more general. $\endgroup$ Dec 7, 2021 at 14:03

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