We know that for a position variable $x$ and momentum $p$, the uncertainties of the two quantities are bounded by

$$\Delta x \Delta p \gtrsim \hbar$$

Now, this is usually first explained with $x$ being a simple linearly measured position and $p$ being linear momentum. But it should apply to any good coordinate and its conjugate momentum. It should, for instance, apply to angle $\phi$ about the $z$ axis, and angular momentum $L_z$:

$$\Delta \phi \Delta L_z \gtrsim \hbar$$

The thing is, $\Delta \phi$ can never be greater than $2\pi$. I mean, you have to have some value of $\phi$ and $\phi$ only runs from 0 to $2\pi$. Therefore

$$\Delta L_z \gtrsim \hbar/\Delta \phi \geq \hbar/2\pi$$

But, uh-oh! This means it is impossible for $\Delta L_z$ to be zero, and we should never be able to have angular momentum states with definite $L_z$ values.

Of course, it doesn't mean that. But I have never figured out how this is not in contradiction with the Schroedinger eqn. calculations that give us states with definite values of $L_z$. Can anyone help me out?

One answer I anticipate is that $\phi$ is sort of "abstract" in that if you chose your origin at some other point you will get completely different values of $\phi$ and $L_z$, and ipso facto, usual considerations don't apply. I don't think this will work, though. Consider a "quantum bead" sliding around on a rigid circular ring and you get the exact same problem with no ambiguity in $\phi$ or $L_z$. (Well, there will be some limited ambiguity in $\phi$, but still, there won't be in $L_z$.)


The problem here is there is at this time still no "legitimate" self-adjoint phase operator. As you phrase the problem, you assume that $\hat \phi$ and $\hat L_z$ would have the same commutation relations as $\hat x$ and $\hat p$, and in particular given that $\hat L_z\mapsto -i\hbar d/d\phi$ the $\hat \phi$ operator would be multiplication of an arbitrary function $f(\phi)$ by $\phi$, i.e. $$ \hat L_zf(\phi)=-i\hbar \frac{df}{d\phi}\, ,\qquad \hat \phi f(\phi)= \phi f(\phi) $$ Thus far everything is fine except that, when it comes to boundary condition, we must have $f(\phi+2\pi)=f(\phi)$. However, the function $\phi f(\phi)$ does not satisfy this. As a result, the action of a putative $\hat \phi$ as defined above takes a "legal" function $f(\phi)$ that satisfies the boundary conditions to an "illegal" one $\phi f(\phi)$, and make $\hat \phi$ NOT self-adjoint (which means trouble).

The uncertainty relation assumes that the operators involved as self-adjoint. Since there is (thus far) no known definition of $\hat \phi$ that makes it self-adjoint, the quantity $\Delta \phi$ cannot be computed in the usual way and indeed is not necessarily well defined for arbitrary states. In other words, there is no mathematical reason to believe that $\Delta \phi\Delta L_z\ge \hbar /2$.

Indeed an obvious "problem" with your expression is obtained by taking $f(\phi)$ to be an eigenstate of $\hat L_z$. Then clearly $\Delta L_z=0$ so the putative variance $\Delta \phi$ would have to be arbitrarily large, which is impossible given that $\phi$ physically ranges from $0$ to $2\pi$.

The problem of constructing a self-adjoint phase operator is an old one. It has been the subject of several questions on this site, including this one. Finding a good definition of a phase operator remains an open research problem.

Edit: added some clarifications after a query.

  • $\begingroup$ Saw this question years ago and never saw an answer. Didn't realize what I was stepping into! Thanks for the education! $\endgroup$ – bob.sacamento Jun 8 '17 at 17:10
  • $\begingroup$ Question: Is this another way to get at the difficulty of a phase operator: A position operator is straightforward. There's the position. It's eigenfunction is a delta distribution at that position. But phase is different. You look at the value of a ... field at some point and time and there's it's value. But there is no mapping from value to phase. To know the phase, you need to know how it's "vibrating", what the other values are in its neighborhood. And that is completely different from getting a simple position. $\endgroup$ – bob.sacamento Jun 8 '17 at 17:17
  • $\begingroup$ @bob.sacamento: Yes there is some work where one starts with Wigner distributions in phase space (the $Q$-function IIRC) from which one can integrate some degrees of freedom and be left with some phase POVM. It's the most sensible solution I know because it has an operational meaning. I've been searching in the last 15 minutes for papers on this and I can't find them but I'll sort it out and can provide references if interested. $\endgroup$ – ZeroTheHero Jun 8 '17 at 17:53
  • $\begingroup$ Don't go to alot of trouble, but I would be interested if you come across anything. $\endgroup$ – bob.sacamento Jun 8 '17 at 18:23
  • $\begingroup$ @bob.sacamento see section 5.2, 5.3 and 7.3 of this: iopscience.iop.org/article/10.1088/1751-8121/50/32/… $\endgroup$ – ZeroTheHero Jul 19 '17 at 2:32

To complement ZeroTheHero's answer, it is possible to deduce uncertainty relations by using the operators $\cos\phi$ and $\sin\phi$ because those now get the necessary periodicity to match what $L_z$ expect. See section 4 in [1].

Here is a summary of the main results. The commutation relations are (in $\hbar=1$ units),

$$\begin{align} [\sin\phi, L_z] &= i\cos\phi,\\ [\cos\phi, L_z] &= -i\sin\phi, \end{align} $$

leading to the uncertainty relations

$$\begin{align} (\Delta L_z)^2(\Delta\sin\phi)^2 &\ge \frac{1}{4}\langle\cos\phi\rangle^2,\\ (\Delta L_z)^2(\Delta\cos\phi)^2 &\ge \frac{1}{4}\langle\sin\phi\rangle^2. \end{align}$$

But the beauty of this approach is that by choosing a state sufficiently localised about an angle $\phi_0$, by performing a Taylor expansion in $\delta\phi=\phi-\phi_0$, they degenerate to

$$\Delta L_z \sqrt{\langle(\delta\phi)^2\rangle} \ge \frac{1}{2},$$

which is a kind of uncertainty relation with $\Delta\phi = \sqrt{\langle(\delta\phi)^2\rangle}$. An approximate one, though.

[1] P. Carruthers and Michael Martin Nieto, Phase and angle variables in quantum mechanics, Rev. Mod. Phys. 40 (1968), 411–440

  • $\begingroup$ Of course the usual argument: why do you need two operators to specify one phase $\phi$ remains unresolved. See doi.org/10.1016/0003-4916(76)90283-9 (behind paywall) for more discussion on this fascinating topic. $\endgroup$ – ZeroTheHero Jun 8 '17 at 0:09
  • $\begingroup$ I had that interesting paper in my personal library (when it comes to think out of the box Lévy-Leblond is king!). Do you know whether the experimental tests he proposes at the end of the paper could be done? I am talking about equation (4.4), his result, versus (4.3), Carruthers' and Nieto's result, for small a small number of photons $\nu$. $\endgroup$ – user154997 Jun 8 '17 at 8:12
  • $\begingroup$ Yes it's a very nice paper. That people of such caliber cannot find a complete solution points to the intrinsic difficulty of the problem. I do not know of any experimental tests of this. $\endgroup$ – ZeroTheHero Jun 8 '17 at 12:25

I will give two answers, a naive, and a high minded. The naive answer is that $\Delta\phi$ can be greater than $2\pi$. Consider the naive derivation of the angle/angular momentum uncertainty relation from the position/linear momentum one: we simply multiply $\Delta\phi=\Delta x/R$ and $\Delta L=R\,\Delta p$, where $R$ is the radius of rotation. Now the issue becomes clear $\Delta x/R$ can still take arbitrarily large values, it is only the fact that we interpret them modulo $2\pi$ that makes them "bounded". In other words, the "true" angle should be measured in the universal cover of the circle, which is $\mathbb{R}$, and can take arbitrarily large values, for it accounts for the entire history of motion from the "initial" position.

The above explanation is too classical to work in the rigorous sense. To be rigorous we have to replace classical angles and angular momenta with self-adjoint operators, which satisfy the canonical commutation relation $[\hat L_z,\hat\phi]=i\hbar$ (i.e. are "conjugate"). The issues with defining such a pair are discussed in Quantum theory of rotation angles by Barnett and Pegg, from which I quote:

"If we represent an angular momentum operator as $\hat L_z=-i\hbar \frac{\partial}{\partial\phi}$ and the angle operator as multiplication by $P$ , then the commutator (2.4) is satisfied. However, this representation of the angle operator causes problems. If $u(\phi)$ is a periodic wave function, then $\phi u(\phi)$ will not be and is therefore outside the angular momentum state space. Judge and Lewis realized that the eigenvalues of a well-behaved angle operator would have to be restricted to a $2\pi$ interval. Their solution was to modify the angle operator so that it corresponded to multiplication by $\phi$ plus a series of step functions. These step functions sharply change the angle by $2\pi$ at appropriate points. The resulting commutation relation between this operator and $\hat L$, has a $\delta$-function term in addition to the $i\hbar$ term from the commutator (2.4)...

Another approach is to avoid the problem of multivaluedness by not dealing with an Hermitian angle operator at all, but rather only periodic functions of the angle operator. Naturally, this approach does not allow us to investigate the properties of the angle operator itself."

The second paragraph is the rigorous version of the naive answer above. For the Judge-Lewis type operators the uncertainty relation is modified into $\Delta\hat\phi \Delta \hat L_z \gtrsim \frac12\hbar|1-2\pi P|$, where $P$ is the angular probability density at the boundary of the angular range, i.e. at the $\pi$/$-\pi$ stitch, where the $\delta$-functions live. In particular, there can be states that exhibit a discontinuity in the gradient at $\pi$, but the minimum uncertainty states exhibit no such discontinuity, see Uncertainty principle for angular position and angular momentum by Franke-Arnold et al.

  • $\begingroup$ One should add that the approach of Barnett has been controversial. See here: doi.org/10.1016/0003-4916(91)90037-9 (unfortunately behind a paywall). Note that the Robertson uncertainty relation is airtight: it depends only on the triangle inequality and the assumption that the operators are self-adjoint. The only way to beat this inequality is to have operators that are NOT self-adjoint, a delicate point of often not-so-delicate discussions between supporters and opponents of the work of Barnett on this topic. $\endgroup$ – ZeroTheHero Jun 8 '17 at 0:08
  • $\begingroup$ Don't worry about the paywalls. There are no walls on the internet ;) . $\endgroup$ – Vendetta Jun 14 '17 at 19:55

protected by Qmechanic Jun 8 '17 at 3:28

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