# Four-momentum operator in relativistic QM

From quantum mechanics we know the energy operator is $$\hat{E} = i\partial_0,$$and the momentum operator is $$\hat{p}_i = -i\partial_i.$$ These operators are naturally covariant as they are formed from gradients. I also know the covariant four-momentum vector is $$p_\mu = (p_0,p_1,p_2,p_3).$$ In order to promote the four-momentum to an operator we just substitute in the operators to get

$$\hat{p}_\mu = i(\partial_0,-\partial_1,-\partial_2,-\partial_3).$$

This is wrong, I want $\hat{p}_\mu = i\partial_\mu$. I have negatives on the spatial components. Where am I going wrong?

• In $p_\mu$ you should have minus signs in the spatial part. – Javier Apr 18 '17 at 21:49
• Yes but that comes from $p_\mu = \eta _{\mu \nu} p^\nu$ so I would have $p_\mu = (p^0,-p^1,-p^2,-p^3)$. I can't substitute the covariant momentum operator $-i \partial_i$ in for the contravariant $p^i$ can I? – Matt0410 Apr 18 '17 at 21:52
• Recall that $-i\vec{\nabla}$ is the vector part of the four momentum $p^i$ - not the last three components $p_i$ of the dual one form. – BRT Apr 18 '17 at 22:06
• But surely as it is formed of gradients, the operator $-i\partial_i$ is naturally covariant and cannot be substituted into the contravariant momentum $p^\mu$? – Matt0410 Apr 18 '17 at 22:12
• Try it the other way around - set $p_\mu=i\partial_\mu=i(\partial_t,\vec{\nabla})$. Then $p^\mu = i(\partial_t,-\vec{\nabla})$. – BRT Apr 18 '17 at 22:19

First off, it should be clear that the only possible equations, modulo a sign on the $i$, are

$$\hat{P}_\mu = i \partial_\mu \quad \text{or} \quad \hat{P}^\mu = i \partial^\mu$$

(these are equivalent) because of the indices.

Now, the momentum and the derivative are naturally contravariant and covariant respectively:

$$p^\mu = (p^0, p^1, p^2, p^3) = (E, \mathbf{p})$$

$$\partial_\mu = (\partial_0, \partial_1, \partial_2, \partial_3)$$

so there will necessarily be a sign difference somewhere. For example, from $\hat{P}^\mu = i \partial^\mu$ we get $\hat{P}^0 = i \partial^0 = i \partial_0$ and $\hat{P}^i = i \partial^i = -i \partial_i$, which are correct. If we take $\hat{P}_\mu = i \partial_\mu$ then we get $P_0 = P^0 = i \partial_0$ and $P_i = - P^i = i \partial_i$, which is also correct. You're probably getting confused because when you write $p_\mu = (p_0, p_1, p_2, p_3)$, $p_1$, $p_2$ and $p_3$ are not the physical (i.e., contravariant) components of momentum; they differ by a sign. What we usually consider the components of three-momentum are the $p^i$.

• So in Dirac equations, we use $p_\mu=i\partial_\mu$ – Wang Yun Dec 3 '18 at 13:24