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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.

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I cannot totally agree with @dmckee.

First it is totally wrong to write something such as: $$\hat{P} = -i\hbar\partial /\partial x.$$ The correct way is 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 $\langle x|\hat{P}|\phi\rangle = -i\hbar\frac{\partial}{\partial x}\langle x|\phi\rangle$

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    $\begingroup$ Of course you are right about the expression of an operator being dependent on the representation chosen, but when I'm introducing this idea it is to students who are having their first encounter with QM. For the moment the SE in a position representation is quantum mechanics to them. For the rest I don't see a meaningful distinction between saying I'm going to assume "the form of the momentum operator is this" and saying "the commutator $[x,p]$ is that". $\endgroup$ – dmckee Nov 4 '16 at 1:37
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    $\begingroup$ Thanks! So the answer to my question would be something like this: First, define momentum in QM. The definition is 1., and from it you get 2., which is, "the P operator generates the translation operator". Second, use a Postulate, [x,p] = iℏ, to show that the above condition demands that ⟨x|P^|ϕ⟩=−iℏ∂⟨x|ϕ⟩/∂x. So in the end, there IS a postulate regarding the form of p. $\endgroup$ – Juan Perez Nov 4 '16 at 2:37
  • $\begingroup$ @dmckee you are right! $\endgroup$ – HanaKaze Nov 4 '16 at 3:09
  • $\begingroup$ I'm sorry @HanaKaze, but I don't think it's logical to say "Now let me derive it for you" and then go on to make a number of assumptions about the framework of quantum mechanics. Classical mechanics does not say $[x,p]=i\hbar$, and in fact you mentioned this in your answer, so nothing is "derived". These are all motivations, and beautiful ones to say the least. Read my answer below. $\endgroup$ – Arturo don Juan Nov 5 '16 at 0:24
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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.

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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.


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.

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    $\begingroup$ There is a derivation, using Noether's theorem and the quasi-classical limit, c.f. Landau QM sec. 15: books.google.co.uk/… $\endgroup$ – bolbteppa Nov 4 '16 at 0:46
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    $\begingroup$ @bolbteppa The thing is that the quasi-classical limit is a procedure justified because it works (i.e. a postulate) rather than something that arises inexorably from the mathematics. We have to found QM on some set of assumptions. The OP is asking if the meaning of $\hat{p}$ in the position representation is one of them and you say "no I can use a different assumption". Which is true but dodges the issue because the two assumption are equivalent to one another. $\endgroup$ – dmckee Nov 4 '16 at 1:52
  • $\begingroup$ As I said before, I understand why the form of p makes sense. But in the end, it seems you have to give it as a postulate, directly or indirectly. If something cannot be derived, it has to be postulated. Thats how I see things, at least. $\endgroup$ – Juan Perez Nov 4 '16 at 2:42
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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

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


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,

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

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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 $p$.

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  • $\begingroup$ Another reply suggesting there is a postulate specifically for the form of p (with respect to x). Thanks. Guess the textbooks I read weren't that thorough. $\endgroup$ – Juan Perez Nov 4 '16 at 2:45
  • $\begingroup$ @JuanPerez No, why would you say that? It's a representation. The content of the theory are states and operarors where the norm gives probability. A representation is just a way to represent these things. It doesn't change the physical content! It is also not necessary, it's just convenient. $\endgroup$ – OkThen Nov 4 '16 at 2:56
  • $\begingroup$ You just said that you take the abstract algebra [x,p] = iℏ as a postulate. What I meant to say is that that postulate specifically gives the relationship between x and p, and that you couldn't obtain a working expression of the p operator (which, as you say does depend on what representation you choose) without it. $\endgroup$ – Juan Perez Nov 5 '16 at 0:15
  • $\begingroup$ Unfortunately, I don't think I made myself clear. There is no need for an additional postulate to obtain P = d/dx. Anything that satisfies [x,P]=i is valid. And you can write these representations much in the same way you obtain the Pauli Matrices just from the angular momentum algebra. All of these people answers show you that exact same thing. $\endgroup$ – OkThen Nov 5 '16 at 2:13
  • $\begingroup$ Sorry to keep replying, but I'm a bit confused. I understand now that P = -iℏd/dx is not a postulate. But [x,p] = iℏ is one, right? (If it's not a postulate, but a choice, I could choose something else, like [x,p] = 0, or [x,p] = 78, and get equivalent results?) $\endgroup$ – Juan Perez Nov 5 '16 at 3:08
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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.

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  • $\begingroup$ It is stated that "none of quantum mechanics is "derived" from any preceding theories". This is not correct. Quantum mechanics is founded in classical field theory. $\endgroup$ – my2cts Jun 1 '18 at 23:52
  • $\begingroup$ Could you define "founded"? $\endgroup$ – Arturo don Juan Jun 2 '18 at 0:07

protected by Qmechanic Nov 4 '16 at 7:26

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