There is one thing that always troubled me in quantum mechanics, how do you justify the expression of the energy and momentum operators, namely $\hat{E} = i\hbar\frac{\partial}{\partial t}$ and $\mathbf{\hat{p}} = -i \hbar \mathbf{\nabla}$?

The only justifications I've seen were always the same: take a plane wave function with a certain frequency / wave vector, inject Planck's / De-Broglie's relation to convert to energy / momentum, time differentiate / take the gradient, and associate what comes out.

But a plane wave is only one particular wavefunction (not even physical), how can you infer anything general about wavefunctions starting from properties of particular solutions?

Is there any more fundamental justification?


This was closed and tagged duplicate but the associated question doesn't answer the question. Although it has a very similar title, it is actually asking about some sort of force operator. (Only a link is given in an answer, but it refers to the demonstration I gave.) Please check this before closing ! :(

I also checked other related questions but none cover the subject as it has started to be here.


5 Answers 5


First, an important note - $i\hbar\frac{\partial}{\partial t}$ is not the energy operator. It is not an operator at all. Remember that an operator acts on elements of the Hilbert space, such as $L^2(\mathbb R)$; time derivatives do not act on these functions. See my answer here.

The most fundamental justification I know has to do with symmetry groups. To be clear, this is not the historical approach taken by Schrödinger, et al in the 1920's. As pioneers into a new frontier of physics, their work was guided by intuition, heuristic arguments, and presumably no shortage of trial and error. However, we have the luxury of seeing the subject through a modern lens; doing so reveals a beautiful structure which underlies both classical and quantum theory, so that's the approach I will take here.

The non-relativistic spacetime symmetry group consists of translations, rotations, and Galilean boosts; those are the operations under which Newtonian physics is invariant. In the Hamiltonian formulation of mechanics, these symmetry transformations are manifested as flows through phase space, which are generated by observables. Briefly, one starts with an observable $F$ which is a smooth function of your $x$'s and $p$'s. As an example, I've plotted the Hamiltonian function for the standard pendulum below.

enter image description here

Each smooth function $F(x,p)$ induces a Hamiltonian vector field $\mathbf X_F$ given by

$$\mathbf X_F = \pmatrix{\frac{\partial F}{\partial p}\\-\frac{\partial F}{\partial x}}$$

For the Hamiltonian plotted above, that looks like this: enter image description here

From here, we can define a flow by "connecting" the arrows of the vector field, creating streamlines:

enter image description here

The flow $\Phi_F$ generated by $F$ is the map which takes phase space points and pushes them along their respective streamlines, as shown here. Note: this is an animated GIF, and you may need to click on it.

enter image description here

The rate of change of a quantity $G$ along the flow $\Phi_F$ generated by the quantity $F$ is given by

$$\frac{dG}{d\lambda} = \big\{G,F\big\}$$ where $\{\bullet,\bullet\}$ is the Poisson bracket.

This reveals the underlying structure of classical mechanics. The flow generated by the momentum function causes a constant shift in the corresponding position; $\{x,p\} = 1$, so following the flow a distance $\lambda$ simply causes $x\rightarrow x+\lambda$. Thus, we say that momentum is the generator of spatial translations. In the same way, the Hamiltonian is the generator of time translations - the flow generated by the Hamiltonian is simply identified with time evolution.

The Poisson bracket allows us to define an algebra of observables. Note that $$\{x,p\} = 1$$ $$\{x,x\}=\{p,p\}=0$$ $$\{L_i,L_j\}=\epsilon_{ijk} L_k$$ where the $L$'s are the angular momentum observables. This structure is preserved when we move to quantum mechanics.

The canonical formulation of quantum mechanics is ostensibly completely different than Hamiltonian mechanics. In the latter, a state corresponds to a point in phase space; in the former, a state is (more or less) an element of some underlying Hilbert space, such as $L^2(\mathbb R)$. Observables in Hamiltonian mechanics correspond to smooth functions on phase space, while observables in quantum mechanics correspond to self-adjoint operators.

However, a substantial amount of structure is retained. Rather than flows, symmetry transformations in quantum mechanics are represented by unitary operators. Just as flows are generated by smooth functions, unitary transformations are generated by self-adjoint operators.

As stated above, momentum is the generator of spatial translations; the flow generated by $p$ serves to shift $x$ by some constant amount. We might guess, then, that the unitary operator generated by $\hat p$ corresponds to the same thing in quantum mechanics.

Explicitly, to go from a self-adjoint operator to the corresponding unitary operator, we exponentiate; it therefore follows that we would expect the (unitary) translation operator to take the form$^\dagger$

$$T_\lambda = e^{i\lambda\hat p}$$

and the transformed position operator would be $$\hat x \rightarrow e^{i\lambda \hat p}\hat x e^{-i\lambda \hat p}$$ For an infinitesimal translation, this can be expanded to yield $$\hat x \rightarrow \hat x - i\lambda [\hat x,\hat p]$$ Compare this to what you get when you follow the momentum flow for an infinitesimal distance $\lambda$ in classical mechanics: $$x \rightarrow x + \lambda \{x,p\}$$

If we want $\hat x \rightarrow \hat x + \lambda$, we must identify $$\frac{[\hat x,\hat p]}{i} = 1$$

In the position space representation where $\hat x \psi(x) = x\psi(x)$, then this implies that $\hat p\psi(x) = -i\psi'(x)$.

Note also that with this identification, we see that the Poisson bracket is "deformed"$^{\dagger\dagger}$ into the quantum mechanical commutator bracket:

$$\{x,p\}=1 \iff\frac{[x,p]}{i} = 1$$

We can repeat the procedure for the other observables, noting each time that the structure from Hamiltonian mechanics is preserved. It's not the same as classical physics, but it certainly rhymes.

$^{\dagger}$ I'm leaving out factors of $\hbar$ because they get in the way of the structure I'm trying to illustrate, but you're free to replace of the $\hat p$'s with $\hat p/\hbar$ if you'd like.

$^{\dagger\dagger}$ For more information on the translation of Poisson brackets to commutator brackets, you might google the phrase deformation quantization.

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    $\begingroup$ This is a great answer and I love it because it connect the more modern machinery of vector fields to quantization but it's a posteriori justification. Schrödinger (see onlinelibrary.wiley.com/doi/10.1002/andp.19263840404) didn't know or cared about vector fields or groups, and certainly didn't know about commutation relations, but explicitly used Hamilton's partial differential equation "Die üblische Form der letzteren knüpft an die Hamiltonsche partielle Differentialgleischung". $\endgroup$ Jun 19, 2020 at 13:56
  • $\begingroup$ @ZeroTheHero Thank you for the compliment! Yes, perhaps I should make it more clear that this is not the historical route that was taken. $\endgroup$
    – J. Murray
    Jun 19, 2020 at 14:06
  • $\begingroup$ In the end, Schrödinger made a kind of lucky guess (a "quantum leap" if you permit the pun) that is mysterious. Methinks this vector field business is of course more foundational to the understanding the dynamics of quasi-probability distributions, which is equivalent to the original Schrödinger approach but depends on it in the sense that the original Wigner function is written in terms of bilinears in the wavefunctions. $\endgroup$ Jun 19, 2020 at 14:16
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    $\begingroup$ Landau said it best in L&L's Quantum mechanics: "Thus quantum mechanics occupies a very unusual place among physical theories: it contains classical mechanics as a limiting case, yet at the same time requires this limiting case for its own formulation”. That's the classic chicken-and-egg problem. $\endgroup$ Jun 19, 2020 at 14:18

I will restrict my answer to the 1-dimensional momentum operator, which is enough to understand what is going on.

The momentum operator you have written has the following form in 1D:

$$ \hat{p}=-i\hbar\frac{d}{dx}. $$

This is not a general expression for the momentum operator. It is the momentum operator written in a particular representation, the position representation. As an example of another representation, you can consider the momentum representation, and in that representation the momentum operator is simply:

$$ \hat{p}=p, $$

it acts on the momentum wave function by multiplying it by the momentum $p$. Therefore, your question really is: why does the momentum operator look like it does in the position representation?

To understand what an operator or a state looks like in a particular representation, you need to project the operator or ket on that representation. The position representation is made up of the eigenstates of the position operator, $\hat{x}|x\rangle=x|x\rangle$. Therefore, to understand what the momentum operator looks like in the position basis, you need to calculate:

$$ \langle x|\hat{p}|\psi\rangle. $$

There are various ways of evaluating this expression. One I really like involves the translation operator $\hat{T}(\alpha)$, the operator that translates a position eigenket by an amount $\alpha$, $\hat{T}(\alpha)|x\rangle=|x+\alpha\rangle$. This operator is given by $\hat{T}(\alpha)=e^{-i\alpha\hat{p}/\hbar}$. For an infinitesimal translation $-\epsilon$, we get:

$$ \langle x|\hat{T}(-\epsilon)|\psi\rangle=\langle x+\epsilon|\psi\rangle=\psi(x+\epsilon), $$

where I used the action of the translation operator on bras, $\langle x|\hat{T}(\alpha)=\langle x-\alpha|$. Taylor expanding the translation operator for an infinitesimal translation, I can also write the following:

$$ \langle x|\hat{T}(-\epsilon)|\psi\rangle=\langle x|\left(1+\frac{i\epsilon}{\hbar}\hat{p}+\cdots\right)|\psi\rangle=\psi(x)+\frac{i\epsilon}{\hbar}\langle x|\hat{p}|\psi\rangle+\cdots, $$

This expression has the term $\langle x|\hat{p}|\psi\rangle$ we need. We can therefore equate this second expression for $\langle x|\hat{T}(-\epsilon)|\psi\rangle$ to the first one above, and isolate $\langle x|\hat{p}|\psi\rangle$ to obtain:

$$ \langle x|\hat{p}|\psi\rangle=-i\hbar\lim_{\epsilon\to 0}\left(\frac{\psi(x+\epsilon)-\psi(x)}{\epsilon}\right)=-i\hbar\frac{d\psi}{dx}. $$

In the last equality I used the definition of derivative as a limit. This is your result: for an arbitrary state $|\psi\rangle$, the momentum operator in the position representation acts by calculating the derivative of the wave function (which is the position representation of the state).

If you want further details, I recently went through this here.

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    $\begingroup$ But the translation operator is really the Taylor expansion operator, so by expressing it in terms of the momentum operator, you have already assumed the answer... $\endgroup$
    – Godzilla
    Jun 18, 2020 at 20:16
  • $\begingroup$ Not really: the translation operator is indeed defined through the momentum operator, but this makes no reference to what that operator looks like in the position representation (which is what the question was). $\endgroup$
    – ProfM
    Jun 18, 2020 at 20:33
  • $\begingroup$ And to clarify: the one assumption that goes behind the explanation is the commutator between position and momentum. $\endgroup$
    – ProfM
    Jun 18, 2020 at 20:52
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    $\begingroup$ This was precisely my point. The following three are mathematically equivalent: 1. the canonical commutation relation; 2. the Fourier transform relationship between the position and momentum wavefunctions; 3. the form of the momentum operator in the position representation. You can derive any of these 3 from one of them and that's what you have done. None of them is more fundamental than the other and as far as I understood, the OP wanted something more fundamental than that. $\endgroup$
    – Godzilla
    Jun 18, 2020 at 21:52
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    $\begingroup$ @user655870, you are correct, I take the canonical commutation relation as my starting point, assuming it without proof. An alternative to this is to define the translation operator from scratch, and then if you insist that it obeys some "reasonable" properties (unitarity, combination of two translations, negative translation, etc.), then you can arrive at the canonical commutation relation. A textbook that takes this approach is "Modern Quantum Mechanics" by Sakurai. $\endgroup$
    – ProfM
    Jun 20, 2020 at 20:33

Another way to look at it (but not necessarily more "fundamental") is to see that the translation operator of a localised particle $\left|x\right>$ can be expressed as a Taylor series operator:

$$\left|x-a\right> = e^{a\frac{d}{dx}}\left|x\right>$$

Since translations have to leave the momentum invariant, we need to require that the translation operator and the momentum operator commute:

$$[\hat{p}_x, e^{a\frac{d}{dx}}] = 0$$

The simplest nontrivial operator that satisfies this is $\hat{p}_x \propto \frac{d}{dx}$. Also see here for a related explanation as to why higher-derivative theories are not common in QM and QFT.

Moreover, we require that $\hat{p}_x$ has appropriate units, so it must be proportional to the experimentally measured constant $\hbar$. Finally we want $\hat{p}_x$ to be Hermitian, since we want it to correspond to a real physical observable. Since the operator $\frac{d}{dx}$ is anti-Hermitian, we have to introduce a complex factor $-i$, which means that the simplest Hermitian operator with appropriate units that commutes with the translation operator and is not just a c-number (as in classical mechanics) is:

$$\hat{p}_x = -i\hbar\frac{d}{dx}$$

  • $\begingroup$ Interesting reply thanks! Here the only assumption is the commutation relation right? Which you set to zero as an extension of classical mechanics or there there is already some quantum touch here? $\endgroup$
    – user655870
    Jun 20, 2020 at 23:33
  • $\begingroup$ Two comments: - What if we exploit le freedom and take another possible expression? For example a second order derivative wouldn't be that heretic would it? - I thought, once you found the operator to be proportional to the derivative, then the plane wave case can actually be used to find the value of the constant, what do you think? $\endgroup$
    – user655870
    Jun 20, 2020 at 23:44
  • $\begingroup$ Some of the assumptions here stem from the other QM postulates, such as that an observable has to be Hermitian and that observables are expressible as operators which act on the ket state. If you assume that, then the commutator has to be zero. But you are right, there are other possible solutions to that equation, such as the momentum being a c-number (in which case you recover classical mechanics) and it having higher-order terms (which as far as I understand violates locality but I could be wrong, someone more knowledgeable should clarify that). So that's why this isn't more fundamental. $\endgroup$
    – Godzilla
    Jun 21, 2020 at 5:59
  • $\begingroup$ But at the end of the day, as other users suggested, the real unsatisfying answer is "it's true because it's true". Classical mechanics isn't more intuitive than QM, we are just used to it - QM just shows that two quantities that emerge completely uncorrelated are actually dependent on each other (spacetime and 4-momentum). As a matter of fact, you can have similar behaviour in thermodynamics, where you work with pairs of conjugate variables all the time (though comparatively different to the QM case). So the idea of conjugacy is not even strictly quantum - it's just a feature of our universe. $\endgroup$
    – Godzilla
    Jun 21, 2020 at 6:08

The justification is heuristic.

Start with the plane wave: $$ \Psi(x,t)=e^{i(kx-\omega t)} $$ The momentum $p=\hbar k$ “is recovered” by taking $-i\hbar \frac{\partial\Psi(x,t)}{\partial x}$ and the energy “is recovered” by taking $i\hbar\frac{\partial\Psi(x,t)}{\partial t}$.

Thus, the energy relation for a free particle, described by a plane wave, is $$ E=\frac{p^2}{2m}\qquad \Rightarrow\qquad i\hbar\frac{\partial }{\partial t}\Psi=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)\,. $$ and this is extended to hold when one includes potential energy, although of course $\Psi(x,t)$ will no longer be a plane wave.

The root idea of representing momentum and energy by a derivative can be traced back to the Hamilton-Jacobi formulation of mechanics, where $p$ can be replaced by a derivative w/r to position, i.e. $$ -\frac{\partial S}{\partial t}=H(x,\frac{\partial S}{\partial x},t) $$ with $p=\partial S/\partial x$ and $H=-\partial S/\partial t$.

  • $\begingroup$ What troubles me is precisely this "extension" from plane to non plane wave. Is there any simple justification for this to hold for arbitrary waves? $\endgroup$
    – user655870
    Jun 20, 2020 at 23:49
  • $\begingroup$ No and yes. If you imagine that you have a system where you allow the potential $V(x)$ to go to $0$ smoothly, then you can imagine that for very small potential the “heuristic” prescription should hold. In the end, people believe in this not because it’s well justified but because it’s been validated by now 75 years of correct predictions. It doesn’t mean it’s right of course, but currently there’s no reason to believe it’s not right and any other recipe would have to explain the same huge amount of data. $\endgroup$ Jun 21, 2020 at 0:20
  • $\begingroup$ see also physics.stackexchange.com/q/345859/36194 $\endgroup$ Jun 21, 2020 at 0:23

The only justifications I've seen were always the same: ... [De Brogile]...[Planck]...[Dispersion]

But thats it. Thats the ultimate reason. There isn't any deeper explanation. You can mathematically revolve around in a lot of ways but this doesn't explain it on a deeper level whatsoever.

So let's look at your actual question here:

But a plane wave is only one particular wavefunction (not even physical), how can you infer anything general about wavefunctions starting from properties of particular solutions?

Your Confusion is solved by Linearity. If you have a linear equation you can superimpose solutions and get another solution. Introducing energy and momentum for a plane wave will hence automatically introduce those for any wavefunction because you can expand it into plane waves.

  • $\begingroup$ I thought of linearity but each component doesn't have necessarily the same energy / momentum. For example if a wave function is a superposition of two plane waves with different momenta, differentiating with respect to position gives you a different momentum in factor of each exponential, so it's no longer possible to identify terms, is it? $\endgroup$
    – user655870
    Jun 20, 2020 at 23:27
  • $\begingroup$ exactly. You get a distribution of momenta. That is the fourier transform of your wavefunction. you can calculate the expectation value (mean value) or various types of fluctuation (higher "momenta" of the distribution (not to be confused with physical momentum (k-space)). Energy is the same game, just fourier transform (== decomposition into complex oscillations/plane waves) with respect to t instead of x. $\endgroup$
    – hagebutte
    Jun 21, 2020 at 9:55

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