It's a very basic question, where does the relation $$\hat{P}\psi(x) = -i\hbar \partial_x \psi(x)$$ for any square integrable $\psi(x)$ come into existence? Some texts I found states that the above relation comes as a consequence of momentum being defined as generator of translation. But what is the basis of this definition? If momentum were defined to be generator of other form of symmetry, then it wouldn't have had the form as it does now.

In some other text, it's the other way around. Namely the action of momentum on a wavefunction is defined to be $$\hat{P}\psi(x) = -i\hbar \partial_x \psi(x)$$ and thence it leads to momentum being the generator of translation.

Which one is the correct one? How was such action of momentum on wavefunction historically developed?

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    $\begingroup$ More on momentum operator in QM. $\endgroup$
    – Qmechanic
    May 24, 2016 at 8:05
  • $\begingroup$ Some of the answers there brought up the commutator between x and p. But as knzhou explained below, this commutator was also actually postulated out of nowhere. I have reservation for this, I would like to know why they postulated that commutator. $\endgroup$
    – nougako
    May 24, 2016 at 8:27
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    $\begingroup$ @nougako I think out of nowhere is a bit harsh. I believe schrodinger was initially motivated to form his wave equation by de broglie's particle wave ideas, hence he considered plane waves and postulated that the ideas generalise. Heisenberg on the other hand was thinking of matrices, I think because some of the equations he was working with looked like matrix multiplication. If you have matrices, and you want to check if they commute, you look at commutators, and $i\hbar$ was consistent for $x$ and $p$. -Add Sources $\endgroup$
    – snulty
    May 24, 2016 at 12:06

5 Answers 5


Historically, you probably want to start with the de Broglie relations (i.e. $p = \hbar k$), which are just a wild guess. This immediately pops out the form of $p$ as an operator if the wavefunction is a plane wave.

Mathematically, $p$ should be defined as the generator of translations (or equivalently the conserved quantity corresponding to translational invariance), from which we derive its action on wavefunctions as $-i\hbar \partial_x$. You can do it the other way (which is logistically easier for some textbooks) but that's awkward.

Physically, it doesn't matter. You ask, "what if momentum were defined to be the generator of some other symmetry?", but this is missing the point, because then it would represent a different physical quantity. The only important thing is that momentum is the amount of "oomph" a particle has when it hits something, and you can derive that from any of the three options above.

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    $\begingroup$ (I think a similar misconception is behind your asking for a "historical development". You might think it's like math, where people accumulate lemmas and eventually prove a big theorem. That's backwards: QM wasn't historically developed. Schrodinger postulated his equation out of nowhere. The development was filled in afterward.) $\endgroup$
    – knzhou
    May 24, 2016 at 8:05
  • $\begingroup$ I am still not getting how I should connect your first explanation about the de Broglie hypothesis and the rest of your explanation. As to my own view of your answer, it seems like the first paragraph suffices to explain the history. I imagine de Broglie hypothesis was first tested, and turns out to be a success. This implies that momentum operator acting on a plane wave is given by the derivative of that plane wave. Then people made an induction, what if we generalize this relation to any wavefunction and test it. It turns out to be successful as well so people use this relation up till now. $\endgroup$
    – nougako
    May 24, 2016 at 8:09
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    $\begingroup$ @nougako Another concurrent historical development was Heisenberg's matrix mechanics, where they postulated $[x, p] = i \hbar$ out of nowhere. (This is 100% equivalent to $p = -i\hbar \partial_x$ by the Stone-von Neumann theorem.) $\endgroup$
    – knzhou
    May 24, 2016 at 8:15
  • $\begingroup$ @nougako But the point of my answer is that there really isn't logic going on here: the content is just postulated. It's inspired guesses. $\endgroup$
    – knzhou
    May 24, 2016 at 8:17
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    $\begingroup$ @nougako You got me to dig around; have a look at this! It was to fit atomic spectra. Some motivation came from the 'old quantum theory'. Some parts are surprisingly prescient. But lots of it is redundant, or just logically in the wrong place. $\endgroup$
    – knzhou
    May 24, 2016 at 8:36

Momentum is the generator of spacial translations, even in classical physics. Anyway, you can find a derivation here or in Sakurai's book Modern Quantum Mechanics. They are more or less the same and go like this:

The translation operator is the operator $T( a)$ such that

$$T( a) \mid x \rangle = \mid x+a\rangle$$

From the definition it follows that the adjoint of $T$ performs a backwards translation:

$$T^\dagger(a) \mid x \rangle = \mid x-a\rangle$$

Of course, we must require that if we translate and then translate back the state is unchanged:

$$T^\dagger(a) T(a) \mid x \rangle = \mid x \rangle$$

From which it follows that $T$ must be unitary: $T^\dagger=T^{-1}$

Any unitary operator can be written in the form

$$T(a) =e^{-iKa}$$

with $K$ hermitian. Now you will find that the eigenstates of $K$ in the position basis are plane waves:

$$\langle x \mid k \rangle = \psi_k(x) \sim e^{ikx}$$

Now (and this is the crucial passage), the De Broglie hypothesis comes into play:

$$p = \hbar k$$

so that


And with some math (the passages are in the paper I linked) you can show that

$$P \psi(x) = \langle x \mid P \mid x \rangle = - i \hbar \frac{\partial \psi}{\partial x}$$

The De Broglie hypotesis is not strictly necessary. For example Sakurai observes that for an infinitesimal translation you have

$$T(dx) = 1-i K dx$$

and that in classical mechanics the generating function of the infinitesimal translation

$$x'=x+dx$$ $$p'=p$$


$$F(x,p')=x p'+ p dx$$

where $xp'$ is the generating function of the identity transformation. From the similarity between $F(x,p')$ and $T(dx)$ he then speculates that $K$ is related to momentum, and since $K \ dx$ must be dimensionless we must have

$$K=\frac{P}{\text{constant with dimensions of an action}}$$

It turns out from experiments that our constant is exactly $\hbar$.

  • $\begingroup$ You basically starts from defining the momentum operator as a translation generator in QM. My original question is why is this so? Why, in QM, one can define momentum as a translation generator as it is in classical mechanics? What I would like to know from you is, did they just take a correspondence between QM and CM in defining momentum as such? $\endgroup$
    – nougako
    May 24, 2016 at 11:17
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    $\begingroup$ @nougako I'm pretty sure this is the whole process of quantizing that Dirac suggests. Take a classical system and replace position and momentum with operators satisfying canonical commutation relations. Even the hamiltonian operator comes from the classical idea of energy, single particle eg $E=p^2/2m+U\to \hat{H}=\hat{P}^2/2m+\hat{U}$ $\endgroup$
    – snulty
    May 24, 2016 at 11:22
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    $\begingroup$ Wait, wait: no one is defining $P$ as the generator of space translations. We just define a translation operator $T$, from which a generator $K$ comes out. Then we find out that the eigenfunctions of $K$ in the position basis are plane waves and so discover that its eigenvalues are wavenumbers. Ten we use the De Broglie hypothesis to link $K$ to $P$. $\endgroup$
    – valerio
    May 24, 2016 at 12:16
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    $\begingroup$ Also, @snulty is right: the process of quantization is just this, replacing classical functions with operators. The symmetries and their generators stays exactly the same (for example angular momentum $\to$ rotational symmetry) but they aren't functions anymore but operators in QM. QM was built from the hamiltonian formulation of CM after all. Actually, I would say that if you learn very well hamiltonian CM the step to QM is almost trivial. $\endgroup$
    – valerio
    May 24, 2016 at 12:19
  • $\begingroup$ The deBroglie hypothesis is a red herring. Momentum is the generator of translations because "momentum", by its very definition, is the Noether charge of translation symmetry, and Noether charges generate their symmetries in the Hamiltonian formulation (already purely classically!), and it is the Hamiltonian formulation that we quantize. $\endgroup$
    – ACuriousMind
    May 24, 2016 at 16:15

Momentum and position are conjugate variables in classical mechanucs, which means they satisfy the Poisson bracket relationship. When quantum mechanics was invented the Poison bracket relation was replaced by the operator commutation relationship which results in the relation under consideration.


That $\hat{P} = - i \hbar \partial_x$ generates translations comes from a straight-forward computation: if $\psi$ is continuously differentiable, and $\Psi$ as well as its derivative are square integrable, then you can prove that \begin{align} i \frac{\mathrm{d}}{\mathrm{d} y} \bigl ( \psi(x - y) \bigr ) \big \vert_{y = 0} = - i \partial_x \psi(x) \end{align} holds, and you write $\mathrm{e}^{- i y \cdot \hat{P}} \psi(x) = \psi(x - y)$.

The best physical motivation in my opinion why $\hat{P}$ should be called “momentum” (operator) is via a semiclassical limit using standard techniques. You can use Wigner-Weyl calculus to show that if the potentials vary slowly compared to the wavelength of your wave function, then \begin{align} \hat{P}(t) = \widehat{p(t)} + \mathrm{error} \end{align} holds true, i. e. the Heisenberg observable $\hat{P}(t)$ associated to momentum is approximately equal to the quantization of the classically evolved momentum $p(t)$. You can make similar arguments for position, angular momentum and other observables. Making this precise is quite difficult.

A simplified, but in my opinion excellent explanation can be found in Ehrenfest's 1927 paper. Unfortunately, most quantum mechanics text books I have seen do a very bad job explaining this point (perhaps because they can't read Ehrenfest's paper, it is written in German), though.


Ab initio the momentum operators can be constructed using de Broglie Plane waves

In one dimension, using the plane wave solution of the Schrodinger equation,the wave function

Psi = exp. i (kx -wt) ,

if one takes the partial derivative w.r. to x of the wave function

delta/delta x (Psi) = ik. Psi

and using de-Broglie relation p = hbar . k we get

delta/delta x (Psi) = i p/hbar . Psi

The above relation suggests the operator equivalence of momentum:

p-operator = -ihbar. Delta/deltax

so the momentum value p is a scalar factor, the momentum of the particle and the value that is measured, is the eigenvalue of the momentum operator.

As the partial derivative is a linear operator the momentum operator is also linear, (one can think of momentum as generator of translational symmetry)

and because any wave function can be expressed as a superposition of other possible states

when this momentum operator acts on the entire superimposed wave, it furnishes the momentum eigenvalues for each plane wave component.

  • $\begingroup$ Using the deBroglie hypothesis is not an "ab initio derivation". Modern quantum mechanics starts from the canonical commutation relations, and $p=\hbar k$ is a derived statement. $\endgroup$
    – ACuriousMind
    May 24, 2016 at 16:18
  • $\begingroup$ well, i was trying to think in terms of historical perspective! $\endgroup$
    – drvrm
    May 24, 2016 at 17:30

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