Suppose we have a particle moving in a circle on the $xy$ plane. Then the angular momentum operator will be just $L_z = ε_{3jk} x_jp_k$ and $L^2 = L_z ^2$. Then if $m$ are the eigenvalues of $L_z$ we'll have $m^2$ as the eigenvalues of $L^2$ so we'll have 1 quantum number, instead of $l$ and $m_l$ as in the 3D case. However we can't even define creation and annihilation operators as usual ($L_{\pm}= L_x \pm i L_y $), since $L_x$, $L_y$ aren't even defined, so how are we to find the eigenvalues of $L_z$?

  • 2
    $\begingroup$ Algebraic arguments are rather unlikely to help you get the spectrum of $L_z$, as you don't have an algebra (i.e. a closed multiplication structure in the space of operators) because you only have one operator to multiply. $\endgroup$ Commented Jun 20, 2018 at 19:46
  • $\begingroup$ Is there anything we can do instead? $\endgroup$
    – Arbiter
    Commented Jun 20, 2018 at 20:15
  • $\begingroup$ Solve the wave equation on a circle (classically). What are the eigenfunctions/values? If it's in a circle and not on a circle, it will be a bit more interesting. $\endgroup$
    – JEB
    Commented Jun 20, 2018 at 21:13
  • $\begingroup$ If you are trying to quantize this system, the uncertainty principle forbids $L^2 = L_z^2$. If that equality is true, then $L_x$ and $L_y$ are known to be zero (0) with no uncertainty. That's not allowed in quantum systems. $\endgroup$
    – Bill N
    Commented Jun 21, 2018 at 14:09
  • $\begingroup$ I thought of that and that's why I said that $L_x$ and $L_y$ are not defined in a two dimensional world, not that they are zero, is that wrong? Of course in the physical world this problem cannot occur- a particle cannot move strictly on a circle due to the uncertainty principle. $\endgroup$
    – Arbiter
    Commented Jun 21, 2018 at 14:17

1 Answer 1


Let us start from the fact that $L^2(\mathbb R^2, dxdy)$ is isomorphic to $L^2(\mathbb R_+, rdr)\otimes L^2(\mathbb S^1, d\theta)$, the unitary identification being the unique linear continuous extension of $$U : L^2(\mathbb R_+, rdr)\otimes L^2(\mathbb S^1, d\theta) \ni u_n(r) \otimes \frac{e^{i m \theta}}{\sqrt{2\pi}}\mapsto u_n(r)\frac{e^{i m \theta}}{\sqrt{2\pi}} \in L^2(\mathbb R^2, dxdy) \quad n \in \mathbb N\:, m \in \mathbb Z$$ with $dxdy = rdr d\theta$ and $x= r\cos \theta$, $y = r \sin \theta$ and where $\{u_n\}_{n\in \mathbb N}$ is a Hilbert basis of $L^2(\mathbb R_+, rdr)$.

$U$ does not depend on the choice of this basis.

Restrict $X_j$ and $P_j$ to the Schwartz space $\cal S(\mathbb R^2)$ that is a dense invariant core for them, that is equivalent to saying that they are essentially self adjoint thereon: they are symmetric on $\cal S(\mathbb R^2)$ and their closures are selfadjoint so that restricting to that space preserves the full information on these selfadjoint operators.

It is possible to construct a Hilbert basis of $L^2(\mathbb R^2, dxdy)$ whose elements stay in $\cal S(\mathbb R^2)$ and have the form $u_n(r) \frac{e^{i m \theta}}{\sqrt{2\pi}}$ where $u_n \in C^\infty_0((0,+\infty))$.

Next define on $\cal S(\mathbb R^2)$ the operator $$L := XP_y -YP_x$$ By direct inspection, using the basis written above, one sees that $$U L U^{-1} = -i\hbar I\otimes \frac{d}{d\theta}\:.$$ With this expression, it turns out that the full set of functions $u_n(\cdot) \frac{e^{i m \cdot}}{\sqrt{2\pi}}$ is a Hilbert basis of eigenvectors of $L$ with eigenvalues $\hbar m$.

As a consequence of Nelson's theorem, the symmetric densely defined operator $L$ is therefore

  1. essentially self-adjoint on $\cal S(\mathbb R^2)$,

  2. its spectrum is a pure point-spectrum with eigenvalues $\hbar m \in \mathbb Z$.

Another more indirect approach, exploiting Peter-Weyl's theorem, would concern the theory of strongly-continuous irreducible representations of $SO(2)$ giving rise to the same result.


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