Is the Momentum Operator a Postulate?

Question

I've been studying the postulates of QM and seeing how to derive important ideas from them. One thing that I haven't been able to derive from them, however, is the identity of the momentum operator.

For simplicity, I'm only thinking about no relativistic effects, no spin, no time-dependent potentials, and one spatial dimension. Also I'm assuming the position operator is simply multiplication by $x$, as in, I'm in position space. So the Hamiltonian operator is $ H = -\frac{\hbar^2}{2m}\nabla^2+V$.

I know that the momentum operator is $p = -i\hbar \frac{\partial}{\partial x}$.

But how do I get there from the postulates? I know that it makes sense, as it results in the Ehrenfest Theorem, the De Broglie wavelength hypothesis, the Heisenberg Uncertainty Principle (for $x$ and $p$), the momentum operator being the generator of the translation operator, and possibly many other desirable theorems, and correlations with classical momentum.

But none of these are postulates (at least, not in the various formalisms I encountered), so you can't derive $p = -i\hbar \frac{\partial}{\partial x}$ from them. Rather, they are consequences of it. You need to know the operator beforehand to see that they are correct. Yes, this is just semantics, but that is the core issue for me:

Regardless of how much sense it makes, is the identity $p = -i\hbar \frac{\partial}{\partial x}$ (under the assumptions I made) a Postulate, meaning that you can't derive it from other postulates, or can it in fact be obtained from them? And in the latter case, could you show me how?

Note: I know that there are many different and equivalent sets of postulates for QM. But in none that I saw did they name it as a postulate nor properly derived it.

Answer 1

First it is a bit scrappy to write something like: $$\hat{P} = -i\hbar\partial /\partial x.$$ It's more rigorous to write: $$\langle x|\hat{P}|\phi\rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\phi\rangle,$$ and it should be interpreted as the momentum operator in spatial representation.

Derivations:

The physical meaning behind momentum is that: 1. It is the conserved quantity corresponding to spatial translation symmetry. 2. Because of 1, the momentum operator (Hermitian) is the generator of the spatial translation operator (unitary).

In terms of equations:

Define the spatial translation operator $D(a)$ s.t. $$C|x+a \rangle = D(a)|x \rangle,$$ and: $$D(a) = e^{-ia\hat{p}/\hbar}$$

I assume you have no problem deriving this.

Please note that this only depends on the quantization condition $[x,p] = i\hbar$, which is one of the postulates of quantum mechanics.

Take an arbitrary state $|\phi\rangle$ and apply $D(a)$ on it: $$D(a)|\phi\rangle = \int D(a)|\phi\rangle |x\rangle \langle x|dx$$ Change of variable, RHS = $$\int C|x\rangle \langle x-a|\phi\rangle dx$$

Take $a\to 0$, plug in to RHS: $$\phi(x-a) = \phi(x) - a\frac{\partial}{\partial x}\phi(x)$$ and to LHS: $$D(a) = 1-ia\hat{p}/\hbar$$

you can recover $\langle x|\hat{P}|\phi\rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\phi\rangle$

Answer 2

There is no derivation, but there is a heuristic argument.

Assume that it is 1926 and Derby has just challenged us to show him the wave equation that goes with de Broglie "waves" (as he did challenge Schrödinger). That means that we are working on a wave equation. The solutions should be of the form (in one dimension) $$ \Psi(x,t) = A e ^{i(kx - \omega t)} \,$$ where $k = 2 \pi / \lambda$ is the wave-number and $\omega = 2 \pi / T$ is the angular frequency.

We also want \begin{align*} p &= h/\lambda = \hbar k\\ E &= h f = \hbar \omega \end{align*} to agree with de Broglie and Plank's ad hoc assumptions that are working.

We could notice (as I presume that Schrödinger did) that the spatial and temporal derivatives that usually appear in a wave equation will give us factors of $k$ and $\omega$ respectively (with some inconvenient factors of $i$ hanging around, but we just have to live with that.). That is, we've just decided to go with \begin{align*} p &\mapsto \frac{\hbar}{i}\frac{\partial}{\partial x} \\ E &\mapsto -\frac{\hbar}{i}\frac{\partial}{\partial t} \\ \end{align*}

From there it is just a matter of saying that for a particle moving in a potential $V$ the total energy (Hamiltonian in many cases) is \begin{align*} E &= T + V \\ &= \frac{p^2}{2m} + V \;, \end{align*} Seeing this as one derivative with respect to time and two derivatives with respect to space and then fixing up the constants, we can arrive at $$ \left[-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)\right] \Psi(x,t) = i\hbar\frac{\partial}{\partial t} \Psi(x,t) \tag{TDSE}$$

I should re-iterate that this is in no way a proof. It's a kind of extended plausibility argument. And one that rather strains the suspension of disbelief except that it works.

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I have a more carefully constructed version of this argument that I give to my modern physics students and variations can be found in many place that predate my version.

Answer 3

The fact that $\hat{P} \to -i \hbar\, \partial / \partial x $ in the position basis is neither derivable nor a postulate, because it's not always true. The canonical commutation relation $[\hat{X}, \hat{P}] = i \hbar \hat{I}$ is generally taken as a postulate, but even if you choose the representation $\hat{X} \to x$ in the position basis, then the CCR allows infinitely many representations of $\hat{P}$ of the form $\hat{P} \to (-i \hbar\, \partial / \partial x) + f(x)$ for any function $f(x)$. The choice of representation corresponds to a gauge choice for the wavefunction, and does not affect any physically observable quantities. See Exercise 7.4.9 on pgs. 213-214 of Shankar for further discussion.

Answer 4

I know that the momentum operator is P = -iℏ ∂/∂x.

To be sure, it is the momentum operator in the position basis. The momentum operator in the momentum basis is $P = p$ in analogy with the position operator in the position basis is $X = x$.

(Borrowing heavily from Brian Hatfield's "Quantum Field Theory of Point Particles and Strings")

The key is to start with the commutation relation

$$[X,P] = i\hbar$$

If $|x\rangle$ denotes a position eigenstate, then

$$X|x\rangle = x|x\rangle$$

and

$$\langle x|X|x'\rangle = x\,\delta(x - x^\prime)$$

which is to say that the operator $X$ is diagonal in the position basis. We seek

$$\langle x|P|x^\prime\rangle$$

Since

$$\left[x, \frac{\partial}{\partial x}\right] = -1$$

it follows that the operator $\frac{\hbar}{i}\frac{\partial}{\partial x}$ serves as representation of $P$ in this basis and thus

$$\langle x|P|x^\prime\rangle = \frac{\hbar}{i}\frac{\partial}{\partial x}\,\delta(x - x^\prime)$$

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If the Hamiltonian operator is

$$H = \frac{P^2}{2m} + V(X)$$

then

$$\langle x|H|x^\prime\rangle = \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right)\,\delta(x - x^\prime)$$

Now, the Schrödinger equation is

$$i\hbar\frac{\partial}{\partial t} \langle x |\psi(t)\rangle = \langle x|H|\psi(t)\rangle$$

Inserting the identity

$$1 = \int\mathrm{d}x^\prime|x^\prime\rangle\langle x^\prime |$$

yields

$$i\hbar\frac{\partial}{\partial t} \langle x |\psi(t)\rangle = i\hbar \frac{\partial}{\partial t}\psi(x,t)= \int \mathrm{d}x^\prime\langle x|H|x^\prime\rangle \langle x^\prime |\psi(t)\rangle = \int \mathrm{d}x^\prime\langle x|H|x^\prime\rangle \psi(x^\prime,t)$$

and finally, using the result from the top of this section,

$$\begin{align} i\hbar\frac{\partial}{\partial t} \psi(x,t) &= \int \mathrm{d}x'\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right)\,\delta(x - x')\psi(x',t)\\ &= \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right)\psi(x,t)\end{align}$$

Answer 5

You have the abstract algebra:

$$ \left[ x , p \right] = i \hbar $$

as a postulate. It either comes from usual Poisson bracket goes to commutator rule, or it's just abstractly set as a definition.

Anyway, you can look for representations of this algebra. The first thing to see is that there are no finite dimensional representations (also known as matrices). A proof by absurd goes by assuming that it is possible and then one should take the trace of the commutation relation. It is obvious that you get $$ 1 = 0 .$$

The second thing you should see: this is an infinitesimal translation. Indeed, consider $a$ an infinitesimal parameter, then

$$ \delta x = [x , a p] = i \hbar a, $$

which is the usual infinitesimal translation one would expect. You are naturally led to think about this $ap$ as the generator of translations.

Unfortunately, the classification of representations of infinite dimensional algebras is a subtle subject. I point you to the Stone-von Neumann theorem.

The best I can do is to motivate the usual representation. And it is actually not that hard, because we only have left the diffeomorphism algebra (remember it should be infinite dimensional), where $x$ and $p$ should act on functions.

Given a function of x, called $\psi (x)$, a translation can be obtained by the Taylor series:

$$ \psi (x + a) = \psi(x) + a \psi'(x) + \frac{a^{2}}{2} \psi^{''}(x) + \cdots = \exp{ \left( a \frac{d}{dx} \right)} \psi(x), $$

and there you have it: $p = -i \hbar \frac{d}{dx}$. The algebra is then realized by the usual vector field algebra:

$$ \mathcal{L}_{p} x = - i \hbar. $$

You see, derivatives always generate translations. Quantum Mechanics tells you to call them momentum.

I leave to you to work out what would have happened if I had chosen to act on functions of $p$.

Answer 6

As a quick extension to the above answers, let me repeat that none of quantum mechanics is "derived" from any preceding theories. Yes, there are many correspondences that are quite striking - canonical quantization, geometric quantization, action waves in Hamilton-Jacobi theory, extending the deBroglie dispersion relation (what @dmckee was talking about), etc. - and many people utilize these to motivate the development of quantum mechanics from the viewpoint of classical physics. But at the end of the day, quantum mechanics is the more fundamental theory, so it is postulated (they call them "postulates of QM" for a reason :).

Another way of seeing it is that recovering quantum mechanics from classical physics is not a well-posed problem. Information is lost when taking the "classical limit" of quantum mechanics, so "deriving" quantum mechanics from classical physics in its strict definition doesn't make sense.

This message is morally identical to what Feynman stresses in this popular video.

Answer 7

The more adequately "fundamental" way is to derive it from the commutator, which basically tells you how the informational tradeoff works between position and momentum, which is the heart of the lesson of quantum mechanics: the Universe contains an information content limit, just as it has an information speed limit. Note that the momentum operator only looks like that with respect to position, so, in a sense, this presumes also that we have defined position as well.

It is an empirical fact, established to obscene levels of confidence, by much repeated trial and failure, that it is impossible to query more information from a system regarding both its position and momentum together than that given by the limit

$$H_x + H_p \ge 1 + \lg(\pi\hbar)$$

expressed via the Shannon entropy (note that the limit depends on what units you are using; technically, the entropy is relative to a scale), or more crudely (not as strongly, i.e. there are cases where the bottom relation holds but not the top, and they are not physically valid cases, e.g. the sum of two suitably far-separated delta functions in both positional and momental space) and typically given as

$$\Delta x \Delta p \ge \frac{1}{2}\hbar$$

.

In the linear-algebra based language that quantum theory provides, that means the operators $\hat{x}$ and $\hat{p}$ must satisfy

$$[\hat{x}, \hat{p}] = i\hbar$$

If you have any two operators satisfying this, then it is possible to show that if you use the eigenstates of one of them as a basis, indexed by $x$ below, then the other must have the form

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

. So you don't even need to try and work out what the basis for $\hat{x}$ or $\hat{p}$ is, just assume it exists, and derive accordingly. If you used a different basis than what someone else used, the maths will still work out in the same way.

Hence the momentum operator itself would not be a postulate directly. Rather, as part of the description of a single particle, we should specify the commutation relation between position and momentum, and that defines both of them at once.