1
$\begingroup$

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 $L_z$?

$\endgroup$
  • 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. $\endgroup$ – Emilio Pisanty Jun 20 '18 at 19:46
  • $\begingroup$ Is there anything we can do instead? $\endgroup$ – Dimitris Jun 20 '18 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 Jun 20 '18 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 Jun 21 '18 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. $\endgroup$ – Dimitris Jun 21 '18 at 14:17
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.