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The energy operator is:

$$\hat{E}= i\hbar \frac{\partial}{\partial t}$$

While momentum is:

$$\hat{P}=-i \hbar \frac{\partial}{\partial x}$$

Why is it not like the momentum with $-i$? I suspect this has to do with relativity because of the metric signature $(+,-,-,-)$ though I'm not sure. I think that as long as they're the opposite it works.

Moreover, I think it doesn't matter at all because usually, in nonrelativistic quantum mechanics, momentum is squared, so $-i$ or $i$ doesn't matter at all. Though relativistically, they should be the opposite. Is that true?

Qmechanic
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Habouz
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    The energy energy operator in quantum mechanics is the Hamiltonian. It is not (!) equal to the time derivative, cf. this. – Tobias Fünke Jul 18 '22 at 15:21
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    I would be carful about calling $i \hbar \frac{\partial}{\partial t}$ the energy operator, but I guess that is not important to the question.

    The minus sign in front of energy/time comes from the Minkowski metric. You will find these minus signs all over physics, irrespective of whether you are dealing with Galilean or special/general relativity.

    I guess this is because Galilean relativity is really just the limit $c \to \infty$.

     – leapsheep Jul 18 '22 at 15:40
  • Keep in mind that the definition of momentum you posted is really its action on wavefunctions i.e. the state components in the position representation. The position dependence comes from the position eigenbra $\langle x|\Psi\rangle$. The sign preceding the $i$ for bra and kets is opposite, as bra transform with the adjoint operators. On the other hand, the energy operator (or rather time evolution) acts on kets. This explains your minus sign. – Mr. Feynman Jul 18 '22 at 15:59
  • @leapsheep Hmm but Galilean relativity, at least mathematically, can just as much be thought of as a limit of Euclidean space-time, whose metric doesn't have any minuses. – Ryder Rude Jul 18 '22 at 16:11
  • Possible duplicate: https://physics.stackexchange.com/q/9557/2451 – Qmechanic Jul 18 '22 at 17:57

3 Answers3

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While the commenter is correct in that $i \hbar \partial / \partial t$ should not be regarded as the energy operator, one can nevertheless ask: how is the sign in the Schrodinger equation related to the sign in the definition of the momentum operator?

The answer is that they have to be oppositely signed, if we want particles with positive energy and momentum to move in the $+x$ direction. As a simple example, consider a particle in a 1d box with periodic boundary conditions. The solutions to the Schrodinger equation are simultaneous eigenstates of $H$ and $p$, given by $$ \psi_k(x) = \frac{1}{\sqrt{L}} e^{ikx-i\omega t} $$ The eigenvalues of $p$ and $H$ are $\hbar k$ and $\hbar \omega = \hbar^2 k^2 / 2m$, respectively. Observe that this is a right-moving plane wave, if one were to change the sign of either $p$ or the Schrodinger equation, we would instead obtain a left-moving plane wave.

A couple further comments. First, note that the overall sign in $p$ and the Schrodinger equation together is arbitrary; it’s just a convention. Second, the relative sign between the two operators has nothing to do with relativity, and has no significance in this regard, other than the fact that the wave equation is relativistically invariant.

Zack
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  1. The sign of the momentum operator in the position representation is fixed by demanding that it should fulfill the canonical commutation relation $[x,p] = \mathrm{i}\hbar$ when $x$ is just multiplication by the wavefunction argument, and the canonical commutation relations are in canonical quantization just a consequence of the classical $\{x,p\} = 1$

  2. The sign in the Schrödinger equation, in turn, is likewise fixed by canonical quantization by turning the classical evolution equation $\partial_t f = \{f,H\}$ into the Heisenberg equation of motion $\partial_t f = \frac{1}{\mathrm{i}\hbar}[f,H]$. From there the expression in the Schrödinger picture follows - note that $\mathrm{i}\hbar\partial_t = H$ isn't really a meaningful equation as such, see this question and its answers.

Hence, the sign have nothing to do with relativity, since they follow from the general procedure of canonical quantization and Hamiltonian mechanics without any assumptions about how the generalized positions and momenta we're quantizing behave under Galilean or Lorentzian transformations (and indeed without the assumption that the Hamiltonian is "energy").

ACuriousMind
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  • I think one problem with this is the quantisation axiom : that you have to put a $-\frac{i}{h}$ in front of the commutator. Why not $\frac{i}{h}$? We're not giving any deeper reason why one choice gives forward time evolution while the other choice gives reverse time evolution – Ryder Rude Jul 18 '22 at 16:25
  • I got a deeper reason. In, say, $\frac{dX}{dt}=\frac{-i}{h}[X, \frac{P^2}{2}]$, the $-i$ in the front multiplies with the $+i$ from $[X,P]=i$, to give us what we expect, i.e. $\frac{dX}{dt}=P$ – Ryder Rude Jul 18 '22 at 16:31
  • @RyderRude You can choose to put $\frac{\mathrm{i}}{\hbar}$, but that doesn't change that momentum and the time derivative end up with opposite signs – ACuriousMind Jul 18 '22 at 16:34
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The energy operator is not to be confused with the hamiltonian operator.

Defining the energy operator makes us able to write the schroedinger equation as

$\hat{H} =\hat{E} $

But keep in mind this is an equation not a definition.

The quantum hamiltonian depends on the system and is derived through the "first quantization" ie. Promotion of variables to hermitian operators

Semi-relativistically this sign difference makes us derive the klein-gordon equation

Defining the operator $\hat{P^{\mu} }=(\hat{H}/c, \hat{P_x} , \hat{P_y}, \hat{P_z} )$ and using the coordinate rappresentation we find that (using the (+, -, -, - ) metric) $\hat{P^{\mu} } \hat{P}_{\mu} =-\hbar^2\partial^{\mu}\partial_{\mu}=m^2c^2$ The last relation because of the mass shell equation.

Riccardo Gandolfo
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