The zeroth component of a particle's four-momentum is the energy divided by the speed of light. For a free particle of mass $m$, that is

$$p_0 = \frac{E_{p}}{c} = \sqrt{\vec{\bf p}^2 + m^2c^2}.$$

Now, the Dirac equation for a free electron can be put in the hamiltonian form

$$ \hat{H} \psi = i \hbar \frac{\partial \psi}{\partial t} $$


$$ \hat{H} = -ic\hbar \mathbf{\alpha}\cdot\mathbf{\nabla} + \beta m c^2 .$$

My question is: if the Hamiltonian is still to be interpreted as the energy, shouldn't there be a correspondence (equality, I would expect) between said hamiltonian and the particle energy given by the relativistic dispersion relation?

More specifically, if I promote the momenta in the relativistic dispertion relation to operators, shouldn't the resulting operator $\hat{p}_0$ be equal to $\hat{H}/c$?

Edit: my reasoning to believe they are different is this article: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.89.052101 in which they define the $\hat{p_0}$ in the same way I did (eq. 5 in the article) and they call $\hat{H}_0$ the Dirac free particle hamiltonian, and later, both these operators appear in the same equation as if they are different things. For example, in eq. 8 they give the energy subspace projectors as:

$$ \hat{\Lambda}^{\pm} = \frac{1}{2}\left(1 \pm \frac{\hat{H}_0}{c \hat{p}_0}\right) $$

Edit 2: arvix version of said article: https://arxiv.org/abs/1403.0550

  • 5
    $\begingroup$ $p^0$ is the Hamiltonian. $\endgroup$
    – Prahar
    Commented Jan 27, 2021 at 22:58
  • 1
    $\begingroup$ Prahar is correct, but more generally see volume 1 of Weinberg's quantum theory of fields for a systematic development. $\endgroup$ Commented Jan 27, 2021 at 23:29
  • $\begingroup$ @RichardMyers Thank you for the answers, but... in this article they seem to use $\hat{p}_0$ and $\hat{H}_0$ with different meaning. I will edit the question and include this extra info. $\endgroup$ Commented Jan 28, 2021 at 15:26
  • $\begingroup$ The article is paywalled, consider linking to an arxiv version. $\endgroup$
    – Javier
    Commented Jan 28, 2021 at 15:37
  • $\begingroup$ Edited the question to include the arxiv link. $\endgroup$ Commented Jan 28, 2021 at 16:29

1 Answer 1


The way I understand the notation* in the linked article is as follows: $\hat p_0$ is an operator that acts as $\sqrt{\hat p^2 + m^2c^2}$ on each component of the spinor and should maybe be written as $\hat p_0 = \sqrt{\hat p^2 + m^2c^2}\, \mathbb 1_4$. $\hat H_0$ is the standard Dirac Hamiltonian (for $\vec A = 0$ and $\Phi = 0$) which, of course, mixes the spinor components. The two operators $\hat H_0$ and $c\hat p_0$ are thus indeed different.

Note however that the square of both operators is the same: $$ (c\hat p_0)^2 = \hat H_0^2 = \hat p^2 c^2 + m^2 c^4 . $$ When the classical expression $E = \sqrt{p^2 c^2 + m^2 c^4}$ is "quantized", it is not a priori clear whether the resulting equation should be $$ \mathrm i\hbar \partial_t |\psi\rangle = c\hat p_0 |\psi\rangle , \quad \mathrm i\hbar \partial_t |\psi\rangle = \hat H_0 |\psi\rangle , \quad -\hbar^2 \partial_t^2 |\psi\rangle = (\hat p^2 c^2 + m^2 c^4) |\psi\rangle $$ or possibly something else entirely. (Note that the second one is the Dirac equation and the third one is the Klein-Gordon equation.) The Dirac and K-G equations are well known; I refer to this question for a discussion about the shortcomings of the first version.

*I do not know whether this notation is standard.


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