# I What is the angle in the action-angle uncertainty principle?

1. Jul 6, 2016

There are lots of basic explanations on the Internet of the two most famous uncertainty principles, that of momentum-position and Energy-time, but I do not find any basic explanation of the action-angle uncertainty principle. I do not even know what angle (or what operator) is being referred to here. Could someone give me an experiment which corresponds to this principle? (That is, preparing a large number of particles in identical quantum states, measure (um, calculate) the action of one half, measure the angle (?) of the other half, get the corresponding statistical distributions, and multiply the respective standard deviations.) Thanks.

2. Jul 6, 2016

### vanhees71

It is very hard to define angles as observables in QT. Before you haven't precisely given a definition, it is also not clear how to formulate an uncertainty relation.

To understand the problem, let's discuss the most simple case of a particle confined to a circle. In the wave-mechanics formulation it should be described by wave functions $\psi(\phi)$ living in $\mathrm{L}^2([0,2 \pi))$, i.e., the square-integrable functions on the interval $[0,2 \pi)$. Since $\phi$ is the angle describing where the particle sits on the circle, these functions must be $2 \pi$ periodic (let's forget for a moment that there could be an additional phase to keep things simple).

Now assume that $\hat{\phi}$ is an observable, i.e., a self-adjoint operator on this Hilbert space. Since the wave-mechanics formulation is in position space, i.e., in $\phi$ space one should have
$$\hat{\phi} \psi(\phi)=\phi \psi(\phi).$$
However, for $\psi$ being $2\pi$ periodic the right-hand side is not a $2 \pi$-periodic function anymore, i.e., $\hat{\phi}$ leads out of the Hilbert space we just want to use to describe the particle on the circle, which means this $\hat{\phi}$ is not a self-adjoint operator. So you cannot use the angle as an observable to begin with, and thus it's hard to define an uncertainty relation for it.

3. Jul 6, 2016

Thank you very much, vanhees71. The following sentence from http://plato.stanford.edu/entries/qt-uncertainty/ led me to this question:
"[Heisenberg] went on to consider other experiments, designed to measure other physical quantities and obtained analogous relations for time and energy:
δtδE~h and action J and angle w δJδw~h ..." Your pointing out that w is not Hermitian is enlightening, given that the proof of the uncertainty principles that I know depends on both operators being Hermitian. Come to think of it, I don't think J is either, so I am not sure one could design an experiment to get a distribution out of J either. So I am not sure what Heisenberg had in mind (the Stanford Encyclopedia article did not elaborate).

4. Jul 6, 2016

### kith

The action-angle coordinates stem from classical Hamiltonian mechanics and seem to have played an important role in the old quantum theory. QM texts usually don't touch on this but there probably are papers which link them to the modern formalism.

I find it interesting that there's quite a number of handwavy uncertainty relations involving quantities which cannot be associated with self-adjoint operators easily. Besides angles, there are also time (energy-time uncertainty relation) and phase (number-phase uncertainty relation) which don't correspond to well-defined self-adjoint operators.

Last edited: Jul 6, 2016
5. Jul 7, 2016

### vanhees71

Forget the handwavy relations. There's the Heisenberg-Robertson uncertainty relation
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|,$$
where the average is taken with respect to any (pure or mixed) state.

An exception is the time-energy uncertainty relation, which has to be treated separately since time is not an observable in quantum theory but a parameter. See Landau-Lifshitz vol. 4 for a very illuminating discussion of this in the context of relativistic QT.

I'm not an expert on questions concerning angle and phase variables. One has to carefully check how corresponding uncertainty relations are defined. In the latter case they should be defined by (gauge independent, i.e., observable) photon correlation functions. I guess one can find discussions in textbooks on quantum optics like Mandel & Wolf.

6. Jul 7, 2016

### A. Neumaier

The right way to quantize angles $\phi$ is to quantize $s=\sin\phi$ and $c=\cos\phi$ (which are the things actually measured, from which one computes the angle). $s$ and $c$ together with the conjugate angular momentum $J$ have good though noncanonical commutation rules closing to a Lie algebra for the group $E(2)=ISO(2)$ (rather than a Heisenberg algebra for canonically conjugate pairs of observables), and the resulting operators satisfy $s^2+c^2=1$ as in the classical case. (This is loosely analogous to the 3-dimensional case where Euler angles also cannot be quantized but angular momentum can, again non-canonically in terms of the Lie algebra for the group $SO(3)$.)

Since $s$ and $c$ commute, they can be measured simultaneously to arbitrary accuracy (by measuring one and and the quadrant of the angle, and computing the other from it), but there is an uncertainty relation between $J$ and $s$ or $c$ (derived in the usual way) since these do not commute.

Everything you might want to know about this situation (and probably much more) is in a paper by Kastrup at http://arxiv.org/abs/quant-ph/0510234

Last edited: Jul 8, 2016
7. Jul 7, 2016

### Truecrimson

Not only that, but any eigenstate of the angular momentum operator will have $\Delta J =0$. In general, since bound operators can't have a constant commutator, the RHS of the uncertainty relation may depend on the state.