# A particle on a ring has quantised energy levels - or does it?

The free particle on a ring is a toy example in QM, with the wavefunction satisfying $$-\frac {\hbar^2}{2mr^2}\psi_{\phi\phi}=E\psi$$ and the cyclic BCs $$\psi(\phi+2\pi)=\psi(\phi)$$. This problem is easily solved to give $$\psi_m=e^{\pm im\phi}$$ for $$m=0,1,2,\dots$$

However, it occured to me that the BCs are too restrictive for the usual quantum mechanical axioms. As only the probability amplitude, $$|\psi|^2$$, is directly observable, we should strictly only require that the modulus stays the same after a $$2\pi$$ rotation - i.e., that $$\psi(\phi+2\pi)=e^{i\alpha}\psi(\phi)$$. This obviously gives rise to a multivalued function in general, but the probability amplitude is still well defined.

However, any function $$e^{ik\phi}$$ is now a solution to the equation, where $$k$$ can be any real number, so we do not get quantisation.

Does this discussion relate to the rotation group and spin? And what is the justification for imposing the restrictive cyclic BCs, even though we strictly shouldn't? I believe that it is somehow resolved by relativistic QM.

• This (PDF link) might be of interest. Dec 6, 2022 at 13:43
• Using $f_m(\phi)=e^{im \phi}$ does satisfy $f_m(\phi+2\pi)=f_m(\phi)$ if $m\in \mathbb{Z}$ so what's the problem? You $\psi$ does not satisfy this... Dec 6, 2022 at 16:49
• @ZeroTheHero I think the question is why the wave function must be periodic? Dec 6, 2022 at 17:05
• Possibly related Dec 6, 2022 at 18:22
• This is the same as the "Bloch theorem" but for "zero" crystal potential, see e.g. physics.stackexchange.com/a/320147/226902, physics.stackexchange.com/q/577224/226902 Dec 7, 2022 at 13:38

Note that when checking to see whether the Hamiltonian operator is self-adjoint, we obtain

$$\langle \psi,H\rho\rangle = \int_0^{2\pi} \overline{\psi(\phi)}\left(-\frac{1}{2}\rho''(\phi)\right)\mathrm d\phi = -\frac{1}{2}\left[\overline{\psi(\phi)} \rho'(\phi) - \overline{\psi'(\phi)} \rho(\phi)\right]\bigg|_0^{2\pi} + \langle H\psi,\rho\rangle$$

Self-adjointness requires that the boundary term in brackets vanishes. Periodic boundary conditions $$\psi(2\pi)=\psi(0), \psi'(2\pi) = \psi'(0)$$ on the domain of $$H$$ certainly satisfy this requirement, but they are not the most general; one can check that $$\psi(2\pi) = e^{i\alpha}\psi(0), \psi'(2\pi) = e^{i\alpha} \psi'(0)$$ also does the trick.

However, this only works if the phase $$\alpha$$ is fixed. The requirement that $$|\psi(2\pi)|^2=|\psi(0)|^2$$ is insufficient to guarantee this, because it would be satisfied for $$\psi(2\pi)=e^{i\alpha}\psi(0)$$ and $$\rho(2\pi) = e^{i\beta}\rho(0)$$ for arbitrary $$\alpha$$ and $$\beta$$, in which case the boundary term would become $$-\frac{1}{2}\big(e^{i(\beta-\alpha)}-1\big)\left[\overline{\psi(0)}\rho'(0) -\overline{\psi'(0)}\rho(0)\right] \neq 0$$

As a result, the proper generalization of periodic boundary conditions is quasiperiodic boundary conditions $$\psi(2\pi) = e^{i\alpha} \psi(0)$$, where $$\alpha$$ is fixed. Satisfying this requirement still leads to quantization, with $$\psi(\phi) \propto e^{ik\phi}, \qquad k\in \mathbb Z + \frac{\alpha}{2\pi}$$

This generalization does come with some technically subtle strings attached, however. The primary issue is that the wavefunction can no longer be strictly considered to be an element of $$L^2(\mathrm S^1)$$, since it is no longer single-valued.

There are a number of ways to handle this, but the most straightforward is to recall how to do quantum mechanics in the presence of a magnetic field with vector potential $$\vec A$$. In general, we have that

$$H \psi = \frac{1}{2}\left(-i\nabla + \vec A\right)^2 \psi$$ Furthermore, a gauge transformation is implemented via $$\vec A \mapsto \vec A - \nabla \chi, \qquad \psi \mapsto e^{i\chi} \psi$$

If we begin from the Hamiltonian $$H = \frac{1}{2}{\left(-i\frac{d}{d\phi} + \frac{\alpha}{2\pi}\right)}^2$$ (which corresponds to a particle moving on a ring which is threaded by a magnetic flux $$\alpha$$) and impose the familiar periodic boundary conditions on its domain, then we find solutions of the form $$e^{ikx}, k\in \mathbb Z$$ while the spectrum of the Hamiltonian is given by $$\frac{1}{2}\left(\mathbb Z + \frac{\alpha}{2\pi}\right)^2$$. If we now implement a gauge transformation generated by $$\chi(\phi) = \alpha \phi/2\pi$$, the Hamiltonian becomes $$-\frac{1}{2}\frac{d^2}{d\phi^2}$$ while the eigenfunctions take the form $$e^{i(k+\alpha/2\pi)\phi}$$.

From this perspective, the multi-valued wavefunctions which arise in the presence of quasiperiodic boundary conditions may be interpreted as the result of applying a singular gauge transformation $$\chi(\phi) = \alpha\phi/2\pi$$ (which fails to be continuous on $$\mathrm S^1$$) to ordinary single-valued ones. Such a transformation works perfectly well locally - e.g. on the neighborhoods $$(0,2\pi)\subset \mathrm S^1$$ or $$(-\pi,\pi)\subset \mathrm S^1$$ - but cannot be smooth extended globally.

In turn, the existence of a constant function $$\alpha/2\pi$$ which cannot be expressed as the derivative of another function is an indicator that the underlying space has non-trivial topological properties. In more technical language, it is a reflection of the fact that the first de Rham cohomology group of the circle $$H^1_{dR}(\mathrm S^1)$$ is non-zero.

• I am curious: can we make some sense of this using OP's "Riemann surface"-alike method? That is to say, suppose we define the "true" wave function as living on domain $\mathbb{R}$ understood as an infinite helix, that is then suitably projected down onto $S^1$ to produce an "apparent" wave function or at least pdf on the ring. In this setup, how can we understand the quantization/boundary conditions you refer to and how can we work the solution from there? Dec 6, 2022 at 20:43
• Note that twisted boundary conditions can be thought of as turning on a (flat) background field for the $U(1)$ shift symmetry of this problem. It is indeed a well-defined, useful operation. Dec 7, 2022 at 21:10
• When we perform the gauge transformation to gauge away $\alpha/2\pi$, what is now the spectrum of the Hamiltonian? I've tried calculating it and I find that the magnetic flux is now not affected by the quasi-periodic BCs, so that the energy spectrum is that of a free particle on a ring, $E\propto \mathbf{Z}^2$. Also, what is $k$ after we gauge transform? I.e. what is quantised in the wavefunction $e^{i(k+\alpha/2\pi)}$? Oct 21, 2023 at 16:49
• @TheQuantumMan The (gauge-invariant) spectrum of the Hamiltonian is $\frac{1}{2}\left(\mathbb Z + \frac{\alpha}{2\pi}\right)^2$. The only question is whether $\alpha/2\pi$ appears explicitly in the Hamiltonian (in which case the boundary conditions on the energy eigenfunctions are periodic) or whether it appears implicitly as a twisted boundary condition on the energy eigenfunctions. $k$ is an element of $\mathbb Z/2\pi$ in either case. Oct 21, 2023 at 23:58

For the particle on the ring, you have two (or, in fact, infinitely many) ways to label the very same point: Indeed, $$\phi$$ and $$\phi +2\pi$$ label the same point and thus all wave functions should satisfy $$\psi(\phi)=\psi(\phi+2\pi)$$.

Consider for instance the interval $$[0,2\pi] \subset \mathbb R$$. If you identify both endpoints with each other $$0 \sim 2\pi$$ you obtain, topologically, the ring or $$1$$-torus $$S^1$$. It turns out that the set of functions on $$S^1$$ is in one-to-one correspondence to the set of $$2\pi$$-periodic functions and hence you can construct your function (Hilbert) space from the set of all $$2\pi$$-periodic functions. Put differently, by construction all your wavefunctions are $$2\pi$$-periodic. See for example this excellent answer by @J. Murray.

As explained in the first reference, this is different to the situation of e.g. a periodic crystal with lattice constant $$L$$, where we indeed "only" have that $$|\psi(x+L)|^2=|\psi(x)|^2$$. In this case, $$x$$ and $$x+L$$ are "physically equivalent", but they are not the same point.

References:

1. Zwiebach, Barton. Mastering Quantum Mechanics: Essentials, Theory, and Applications. MIT Press, 2022. Section 5.4

This boils down to choosing an appropriate model for the particle on a ring. There might be many models, however experimental physics only will tell you which model is deemed ,,physical" However, there may be different experimental setups which are considered "particles on a ring", which have differing experimental outcomes and are modelled by different theoretical setups. This way the meaning of a "quantum mechanical particle on a ring" can become multi-valued and different non-equivalent theoretical models may be admissible.

This said, what does a theoretical particle on a ring look like in the language of quantum mechanics? This boils down to two ingredients:

• A Hilbert space $$\mathfrak{H}$$ which is interpreted as the collection of all wave functions,
• and a self-adjoint operator $$H: \text{dom}(H) \to \mathfrak{H}$$ defined on a dense subset $$\text{dom}(H) \subseteq \mathfrak{H}$$.

The spectrum $$\text{spec}(H)$$ is then interpreted as the energy values of $$H$$ (which can be determined in experiment!).

The "setup with periodic boundary condition" is then as follows:

• We set $$\mathfrak{H} := \text{L}^2(\mathbb{S}^1)$$ and interpret $$\psi \in \text{L}^2(\mathbb{S}^1)$$ as a function $$\psi : \mathbb{S}^1 \to \mathbb{C}$$, where $$\int_U \left| \psi \right|^2 \text{d} \sigma$$ is the probability of finding the particle in state $$\psi$$ in the region $$U \subseteq \mathbb{S}^1$$. Here $$\sigma$$ denotes the standard measure on $$\mathbb{S}^1$$.

• Then the Hamiltonian will be $$H := \frac{P^2}{2m}$$, where $$P \psi = -\text{i} \hbar \psi'$$ taken on a suitable dense subset such that $$H$$ becomes self-adjoint.

Note that this is very natural: we want $$\psi$$ to be a function on the sphere $$\mathbb{S}^1$$, since we want $$\left| \psi \right|²$$ to be the probability density of the position of the particle. In this model it does not make sense to ask whether $$\psi (x + 2\pi) = \text{e}^{\text{i}\alpha}\psi(x)$$, since $$x \in \mathbb{S}^1$$, not $$x \in \mathbb{R}$$. In other words: there is no sense in asking whether the particle is in some region $$U \subseteq \mathbb{R}$$, since the particle moves on the sphere.

Usually when one does calculations, $$\psi$$ is interpreted as a function $$\tilde{\psi}$$, now defined on $$\mathbb{R}$$, by setting

$$\tilde{\psi} (\theta + 2 \pi n) := \psi(\text{e}^{\text{i} \theta})$$

for all $$\theta \in [0, 2\pi]$$ and $$n \in \mathbb{N}$$. Note that every $$x \in \mathbb{R}$$ can be written as $$\theta_x + 2 \pi n_x$$ for suitable $$\theta_x \in [0, 2\pi]$$ and $$n_x \in \mathbb{N}$$, and the value of $$\tilde{\psi}(\theta_x + 2\pi n_x$$ does not depend on the choice of $$\theta_x$$ and $$n_x$$. Thus $$\tilde{\psi}$$ is a well-defined function on $$\mathbb{R}$$ which now by definition satisfies for every $$k \in \mathbb{N}$$

$$\tilde{\psi}(x+2\pi k) = \tilde{\psi}(x).$$

So the so called "boundary condition" is in fact just a way of making calculations easier, since now we can apply the usual calculus on $$\mathbb{R}$$.

If one wants $$\tilde{\psi}$$ to satisfy different boundary conditions, for example

$$\tilde{\psi}(x + 2 \pi) = \text{e}^{\text{i}\alpha} \tilde{\psi} (x),$$

one needs a different choice of theoretical model. On the sphere, it does not make sense to say that $$\psi$$ has a different value after going round once (functions are single-valued after all!). In particular one needs a different choice of $$\mathfrak{H}$$ and a corresponding interpretation of the wave function: what does the quantity $$\int_U \left| \psi \right|^2 \text{d} \mu$$ represent? What, for example, if $$\psi : [0, 4 \pi] \to \mathbb{C}$$ (with the requirement $$\psi (x + 2\pi) = \text{e}^{\text{i} \alpha} \psi(x)$$ for $$x \in [0, 2\pi]$$) does the quantity

$$\int_{\left( \pi, 3 \pi \right)} \left|\psi(x)\right|^2 \text{d}x = \int_{\left( \pi, 2 \pi\right)} \left|\psi(x)\right|^2 \text{d}x + \int_{\left( 0, \pi\right)} \left|\psi(x)\right|^2 \text{d}x = \int_{\left( 0, 2\pi\right)} \left|\psi(x)\right|^2 \text{d}x$$

represent?

It cannot be the probability of finding the particle in $$(0, 2\pi)$$, since that should be $$1$$, and for general $$\psi$$ only

$$1 = \int_{\left( 0, 4\pi\right)} \left|\psi(x)\right|^2 \text{d}x \geq \int_{\left( 0, 2\pi\right)} \left|\psi(x)\right|^2 \text{d}x$$

holds.

If $$\psi$$ is not a function on the sphere anymore, the interpretation of $$\psi$$ begin the probability density of the position might not make sense anymore – so do we still model a particle on a ring?

• Why are the two cases physically different? At the end of the day, both these choices are suitable when one checks the self-adjointness of the Hamiltonian. Oct 21, 2023 at 16:42

The free particle on a ring is a toy example in QM, with the wavefunction satisfying $$-\frac {\hbar^2}{2mr^2}\psi_{\phi\phi}=E\psi$$ and the cyclic BCs $$\psi(\phi+2\pi)=\psi(\phi)$$. This problem is easily solved to give $$\psi_m=e^{\pm im\phi}$$ for $$m=0,1,2,\dots$$

However, it occurred to me that the BCs are too restrictive for the usual quantum mechanical axioms.

No, they are not "too restrictive" for the "usual" axioms.

We usually apply our boundary conditions to the wave function. For example, in the Hydrogen atom, we say that the wave function behaves like $$r^\ell$$ as $$r \to 0$$ and we require that the wave function be singly valued in space (which gives us the usual condition on $$m$$ as being an integer between $$-\ell$$ and $$\ell$$).

As only the probability amplitude, $$|\psi|^2$$, is directly observable, we should strictly only require...

You have posited this "strict" interpretation as an axiom of your question, but without justification. Applying boundary conditions on the wavefunction is perfectly fine. For example, we apply them to solve for hydrogenic orbitals, we apply them to solve the particle-in-a-box problem, we apply them to solve for Bloch waves. Etc.

For more information, also see this website, in particular the section titled "The Wavefunction Postulate" and "Constraints on Wavefunction."

As only the probability amplitude, |ψ|2, is directly observable

How is that observable? There's no way to measure the probability amplitude for a single particle. The restriction isn't based on the condition being "observable", it's based on logic: if two sets of coordinates describe the same point, then those two points much have the same properties (because a point has the same properties as itself). Your argument makes as much sense as saying "The Schrödinger equation says $${\displaystyle i\hbar {\frac {\partial }{\partial t}}|\psi (t)\rangle ={\hat {H}}|\psi (t)\rangle }$$, but $${\hat {H}}|\psi (t)$$ isn't observable, only its amplitude is, so $$i\hbar {\frac {\partial }{\partial t}}|\psi (t)$$ can be equal to $${\hat {H}}|\psi (t)$$ times any unit term". A basic foundation of QM is that its equations continue to hold even when we can't directly observe whether they are holding true at a particular instant/location. The universe can't violate QM just because no one is looking.

$$\psi^2$$ is most certainly not observable: Observables are the eigenvalues of hermitian operators. For a state $$\psi$$ to be physical, the expectation values for any observable under this state needs to be physical, ie for instance $$\langle\psi|\hat{H}|\psi\rangle<\infty$$. The problem is that for the proposed $$\psi$$ (which is discontinous), the energy would have to be infinite.

• The answer given by J. Murray above would seem to imply that $\langle \psi | H | \psi \rangle$ is in fact finite for the "twisted" boundary conditions $\psi(2 \pi) = e^{i \alpha} \psi(0)$. Dec 7, 2022 at 12:41