# Commuting momentum and position operators?

I think this is a question about the mathematical axioms of quantum mechanics...

Consider the following operators $\tilde{x}$ and $\tilde{p}$ on Hilbert space $L^2(R)$, defined for fixed $\delta>0$: $$(\tilde{x}\psi)(x) = (n+1/2) \delta\ \psi(x)$$ when $n\delta \leq x <(n+1)\delta$ for some integer $n$; and $$(\tilde{p}\psi)(x) = -i \hbar \frac{d\psi}{dx}(x)$$ when $n\delta < x < (n+1) \delta$, and $(\tilde{p}\psi)(x) = 0$ when $x=n\delta$.

Both these operators are self-adjoint and they commute. Further, $\tilde{p}$ is identical to the usual momentum operator $\hat{p} = -i\hbar \frac{d}{dx}$ on the subspace of $L^2(R)$ on which $\hat{p}$ is defined, and thus has identical spectrum. Finally, by taking $\delta$ small we can make $\tilde{x}$ as close as we like to the usual position operator, $\hat{x}$. Thus naively it looks like $\tilde{x}$ and $\tilde{p}$ represent compatible observables which can be made arbitrarily close to $\hat{x}$ and $\hat{p}$.

I'm pretty sure that I've cheated somewhere, but I can't find anything in the axioms of quantum mechanics that rules out defining observables $\tilde{x}$ and $\tilde{p}$ as above (or at least, the axioms as they are usually presented in under/graduate physics texts). Is there some requirement that observable operators should map $C^\infty$ functions to $C^\infty$ functions? In any case, I'd be grateful if someone could tell me why the above is ruled out.

(As background, I am teaching an undergraduate class on quantum mechanics next semester and I wanted to explain to students that while it's not possible to make exact simultaneous measurement of position and momentum; it is possible to make an approximate simultaneous measurement. In other words, I wanted to find operators which approximate $\hat{x}$ and $\hat{p}$ on 'macroscopic' scales and which commute. This was before I looked in the literature and realised how technical constructing such approximate operators is.)

• This seems like a classic case of limits that do not commute. The fact that $\tilde{x}$ and $\tilde{p}$ commute relies on the fact that for every $x$, there is a small neighbourhood (of size $\sim \delta$) where $\tilde{x}$ acts as multiplication by a constant. Or in the sense of distributions, you should note that even if $\psi$ is smooth, $\tilde{x}\psi$ is discontinuous, but don't have time to write a longer answer. – Vibert Jan 17 '14 at 2:35
• @Vilbert: I agree the $\delta\to 0$ limit is problematic... but even if I just take $\delta$ very small (e.g. less than the Planck scale) it still looks wrong. Also, is there a reason why the fact $\tilde{x}\psi$ discontinuous is a problem? I thought the only requirement on $\tilde{x}$ to represent an observable was that it is self-adjoint. – Mark A Jan 17 '14 at 22:55

Concerning $\tilde{x}$, it is self-adjoint (the proof is easy) if it is defined on its natural domain $D(\tilde x)$. For $\gamma_n(x):= (n+1/2)\delta$ if $n \delta \leq x <(n+1)\delta$ and $n \in \mathbb Z$

$$D(\tilde{x}) = \left\{\psi\in L^2(\mathbb R) \:\left|\:\int_{\mathbb R} |\gamma_n(x)\psi(x)|^2 dx < +\infty \right.\right\}$$

The spectrum is $\sigma(\tilde{x})= \{\delta (n+1/2)\:|\: n \in \mathbb Z\}$ as expected.

The fact that your claimed approximated momentum operator $\tilde{p}$ makes any sense as it stands is questionable. The point is that $\tilde{p}$ defined this way is not self-adjoint but only symmetric, while observables must be self-adjoint in order to exploit the spectral calculus technology.

To obtain a symmetric operator (i.e. Hermitean and densely defined) I think that a possibility is to define its domaine $D(\tilde{p})$ as a space of $C^1$ functions (that vanish in $x_n= \delta n$ for all $n \in \mathbb Z$ with their first derivative for a reason I discuss shortly) also requiring that these functions rapidly vanish with the first derivative for $|x|\to \infty$. It should be possible to relax the smoothness condition using weak derivatives, but the situation does not seem to change remarkably.

Consider the anti-linear operator $(C\psi)(x):= \overline{\psi(-x)}$, the bar denoting the complex conjugation. It is norm preserving and $CC=I$, so it is a conjugation. It seems to me that $C\tilde{p}= \tilde{p}C$ with the domain I introduced above. Therefore, in view of a theorem due to von Neumann, $\tilde{p}$ admits some self-adjoint extension. A careful analysis of the defect indices of $\tilde{p}$ would classify these extensions. Therefore we are not authorized to say that $\tilde{p}$ has the meaning of an observable, we have to choose a self-adjoint extension of it. Unfortunately, the spectrum depends on this choice. It is by no means obvious what the spectrum of a self-adjoint extension of $\tilde{p}$ is.

What I am saying is that since we do not know the spectrum of the claimed approximated momentum we cannot say how (if) this claimed approximation works as soon as $\delta \to 0$.

It is possible to prove that if the domain is chosen as $C_0^\infty(\mathbb R)$ or ${\cal S}(\mathbb R)$ then there is a unique self-adjoint extension and coincides with the standard one. But we cannot make this choice in view of the form of $\tilde{x}$.

In fact, when computing $[\tilde{x}, \tilde{p}]=0$ we are assuming that the domain of $\tilde{p}$ is invariant under the action of $\tilde{x}$. This indeed happens if $D(\tilde{p})$ is made of the $C^1$ functions rapidly vanishing as $|x|\to \infty$ also vanishing at each $\delta n$ with their first derivative. The last condition eliminates the discontinuities introduced by the action of $\tilde{x}$ giving rise to a function in the same domain.

We conclude that $[\tilde{x}, \tilde{p}]=0$ holds on a domain which does not fix a unique self-adjoint extension of $\tilde{p}$ (and its spectrum consequently) as it would be if this domain were $C_0^\infty(\mathbb R)$ or ${\cal S}(\mathbb R)$.

An alternative definition of $\tilde{p}$ is obtained re-defining its domain $D(\tilde{p})$ by including the (vanishing sufficiently fast at infinity) functions $C^1$ in each open interval $(n\delta, (n+1)\delta)$ that are separately periodic in each such interval. The values these functions attain in the set of points $\delta n$ is irrelevant as this set has zero measure and $L^2$ does not care of zero measure set: We can safely assume that $\tilde{p}\psi$ vanishes thereon by definition as already assumed by the OP. This only concerns mathematics while, physically speaking, this assumption has devastating implications.

With this definition $\tilde{p}$ become essentially self-adjoint, i.e. it has a unique self adjoint extension. The spectrum should be $\sigma(\tilde{p}_{ext}) = \{ \frac{2\pi m}{\delta} \hbar\:|\: m \in \mathbb Z\}$, since in each said interval we find the standard momentum operator on a segment with periodic boundary conditions (to be completely sure I should check some conditions but I am reasonably confident)

With this definition you find $[\tilde{x}, \tilde{p}]=0$ on $D(\tilde{p})$.

The problem with this construction is that $\tilde{p}$ does not generate space displacements of the wavefunctions, $(e^{-ix_0\tilde{p}}\psi)(x)= \psi(x- x_0)$. And it does not seem that the situation improves as soon as $\delta \to 0$. Instead $\tilde{p}$ generates "periodic" displacements in each interval $(n\delta, (n+1)\delta)$ (as it were a circle). For $\delta \to 0$ these periodic displacements become denser and denser but they have nothing to do with geometric translations along $x$, as the ones generated by the true momentum operator.

• That looks right, thanks. I suspect that the (unique?) self-adjoint extension of $\tilde{p}$ is $\hat{p}$ as they already agree on the subspace of $C^1$ functions (i.e. their results only differ on a set of measure zero). When we consider $\tilde{p}\tilde{x}\psi$ in the commutator the function $\tilde{x}\psi$ is not $C^1$ so we have to take the correct self-adjoint extension, which is $\hat{p}$ and not $\tilde{p}$, (and which does therefore not commute). – Mark A Jan 17 '14 at 23:07
• Note though that the requirement that the $C^1$ functions vanish at $x=n\delta$ is unnecessary, as the right-hand boundary term of one section is cancelled by the left-hand boundary term of the next in the calculation of $\langle f | \tilde{p} g \rangle$ for any continuous functions $f$ and $g$. – Mark A Jan 17 '14 at 23:11
• You are right, boundary terms cancel each other. Regarding the fact that the self-adjoint extension is unique, it happens provided you replace $C^1(\mathbb R)$ for ${\cal S}(\mathbb R)$ or $C^\infty_0(\mathbb R)$ that are cores for the momentum operator. In this way you find the standard momentum operator as the unique self-adjoint extension. – Valter Moretti Jan 18 '14 at 9:08
• The point is that, these spaces are not invariant under the action of your approximated operator $\tilde{x}$, so the commutation relations hold on a subspace which does not fix the right self-adjoint extension of $\tilde{p}$. I am editing my answer to introduce these observations. – Valter Moretti Jan 18 '14 at 9:14
• Actually when computing $[\tilde{x},\tilde{p}]$ it is assumed that the domain of $\tilde{p}$ is invariant under the action of $\tilde{x}$ (and vice versa), with the choice of $C^1$ (or $C^\infty$ or $\cal S$) functions this cannot hold. The simplest way I see is to assume that the functions in $D(\tilde{p})$ vanish at $n\delta$ with their first derivative. So your $\tilde{p}$ does not uniquely extend to the standard $p$. Or, at least it is by no means obvious. – Valter Moretti Jan 18 '14 at 10:01

Is there some requirement that observable operators should map C∞ functions to C∞ functions? In any case, I'd be grateful if someone could tell me why the above is ruled out.

By defining the position operator in the way you've indicated, you would introduce $artificial$ lattice of preferred values of coordinates in space $x_n = (n+1/2)\delta$ with (artificial) spacing $\delta$ and which would be stationary in arbitrary inertial frame. As you say, the new position and momentum operator violate canonical commutation relation, which is viewed as one of the basic features of operators in quantum theory.

By restricting the position operator to discrete eigenvalues $x_n$, one is restricting the set of possible positions to a discrete set. It is not clear why the other values of $x$ in between should be forbidden. The original Born interpretation allows any value of $x$. Most naturally, possible configurations form a continuous set (imagine what would happen to your description if the frame of reference was shifted along $x$ by $\delta x <\delta$).

• Thanks for your answer. I may have misunderstood but it looks to me like treating $\tilde{x}$ as an observable does not violate the Born interpretation. The Eigenvalues are $x_n$, as you say, and the corresponding Eigenspace is the space of functions with support in the interval $[n\delta,(n+1)\delta)$. Result of a measurement returns $x_n$ and projects the original wavefunction into one of these intervals... conservation of probability looks okay to me. – Mark A Jan 17 '14 at 22:59
• I've edited my answer. – Ján Lalinský Jan 18 '14 at 23:42
• I agree the fixed lattice of $x$-values is artificial. (It's worth noting though that the only other solution I've seen to the problem of constructing approximate, commuting momentum and position operators also introduces an artificial lattice. That was by Von Neumann, who argued you could start with a lattice of coherent states and apply Gram-Schmitt orthogonalization.) – Mark A Jan 20 '14 at 1:41