This is a fairly basic question which has bothered me for some time and I haven't been able to satisfactorily resolve.

In the ground state of the rigid rotor the energy is zero. That is, there is no zero point energy for this system. This means we definitely need to make sure we aren't violating the uncertainty principle for position and momentum.

So, the first place to look at is the wavefunction of the ground state. The wavefunction of the rigid rotor are the associated Legendre polynomials, and for the ground state (ignoring normalization) this is:$$P_0^0(x)=1.$$The point being, this is just a constant.

So, my first intuition is to say that we aren't violating position-momentum uncertainty because the wavefunction is only a constant so we know nothing about its location. This, however, I think is wrong because we solve this in spherical coordinates and the angle $\phi$ is constrained to be within $[0,2\pi]$. That is, our particle is somewhere on a sphere. Thus, this explanation doesn't work because the particle is somewhere on the sphere so we have some certainty of its position.

My next thought was that the uncertainty principle I'm thinking of is written as follows:$$\Delta p_x\Delta x \ge \frac{\hbar}{2}$$This, however, is wrong because it means that when the ground state energy is zero, the angular momentum is zero (that is, this is a closed spherically symmetric wave).

Thus, the next explanation I thought is that if I solved this in Cartesian coordinates rather than spherical coordinates, I would still get the same answer that the ground state energy is zero, but the x, y, and z position and x, y, and z momenta would all have positive, finite values which do not violate the uncertainty principle.

So, hopefully I've convinced you that this requires a slightly more nuanced answer than one might expect.

Is the last explanation correct in that when solving this in spherical coordinates, we should actually think about some other uncertainty principle than the one depending on Cartesian coordinates written above? Something like $$\Delta L \Delta r \ge \frac{\hbar}{2}$$

So, let me know if my reasoning has gone astray and where the solution lies. I feel a bit embarrassed that this isn't obvious, but I think my last point is where the answer lies.

Specifically, I'd like to see if anyone can address the question of the uncertainty principle in the ground state when this problem is solved in Cartesian Coordinates.

  • $\begingroup$ I think the uncertainty relation you want is between the angle $\phi$ and the angular momentum $l_z = -\mathrm{i}\hbar(\mathrm d/\mathrm d\phi)$. Using the generalised relations here you should find that $\Delta\phi \Delta l_z \geq \hbar/2$ Not sure about Cartesian coordinates though. $\endgroup$ Feb 19, 2017 at 23:21
  • $\begingroup$ Ya, I think you're right. Or maybe the angle $\phi$ and the total angular momentum? $\endgroup$
    – jheindel
    Feb 20, 2017 at 0:03
  • $\begingroup$ Your intuition is correct; because you know the energy is zero you cannot know in what direction the rotor is positioned if uncertainty is to be obeyed. $\endgroup$
    – porphyrin
    Feb 20, 2017 at 11:24
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    $\begingroup$ I have chanced upon a book that discusses angle-angular momentum uncertainty, and addresses the exact contradiction you raised about $\theta$ being restricted to $[0,2\pi)$, but I don't think I am qualified to explain it: Fayngold & Fayngold, Quantum Mechanics and Quantum Information, pp 384-388. It is to do with the domain over which the operators involved are Hermitian. I've uploaded the relevant pages here. $\endgroup$ Mar 3, 2017 at 19:56
  • $\begingroup$ @orthocresol Wow, thanks for posting that here. That turns out to be a much more subtle problem than I expected. Maybe once I get the time to fully understand what's going on there I'll write up the answer. I get the general idea though. That's pretty interesting. $\endgroup$
    – jheindel
    Mar 4, 2017 at 19:46

1 Answer 1


As far as the uncertainty relation between $L_{z} = -i\hbar \partial/\partial\phi$ and $\phi$ is concerned, there is a simple way to understand why we cannot use the commutation relation $[\partial/\partial\phi,\, \phi] = 1$, which leads to the problematic inequality $\Delta L_z \Delta\phi \ge \hbar/2$.

Remember that the variable $\phi$ is defined on a circle, i.e., $\phi = 0$ and $\phi=2\pi$ represent the same point. A function of the form $\phi\psi(\phi)$, where $\psi(\phi)$ is $2\pi$-periodic, has a discontinuous jump by the amount $-2\pi \psi(0)$ at $\phi = 0$ (or $2\pi$). Hence, when applying $\partial/\partial\phi$ to $\phi\psi(\phi)$, one should make sure to differentiate the discontinuity as well, which leads to an additional term: $-2\pi \psi(0)$ times a Dirac delta function. Then, the correct commutation relation should be $[\partial/\partial\phi,\, \phi]\,\psi(\phi) = \psi(\phi) - 2\pi\psi(0)\delta(\phi)$, and we have the modified uncertainty principle $$ \Delta L_z \Delta\phi \ge \frac{1}{2}\big|\langle[L_{z}, \phi]\rangle\big| = \frac{\hbar}{2}\Big|\langle\psi|[\partial/\partial\phi,\, \phi]|\psi\rangle\Big| = \frac{\hbar}{2}[ 1- 2\pi|\psi(0)|^2]. $$ For a normalized angular momentum eigenfunction $\psi_{m}(\phi) = e^{im\phi}/\sqrt{2\pi}$, the right-hand side of the above yieids zero. This is exactly what should be the case when the uncertainties in the angular momentum and the angle are respectively zero and $2\pi$.

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    $\begingroup$ Very well put! This is almost exactly what the reference that orthocresol linked above says, but you put it in a much more understandable way. I see where the problem arises now, and it will be good to keep this in mind as this is an artifact which could arise in lots of different periodic systems like this. $\endgroup$
    – jheindel
    Mar 5, 2017 at 22:00
  • $\begingroup$ @jheindel I'm glad because you seem to like this way of understanding the anomalous term in the uncertainty relation :) Compared to what's found in the standard literature, this approach is less formal (i.e., doesn't involve discussions on domains of operators and all that) but, I believe, is much more intuitive. $\endgroup$
    – higgsss
    Mar 6, 2017 at 11:02

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