1
$\begingroup$

In general, the eigenvalues of the components of position $\vec{r}$ and momentum $\vec{p}$ are not quantized. Certainly, not quantized for a free particle. Is there a physical explanation of how is it possible that the eigenvalues of $\vec{L}^2$ and $L_z$ are always quantized where $\vec{L}=-i\hbar\vec{r}\times\vec{\nabla}$?

$\endgroup$

2 Answers 2

2
$\begingroup$

I think you mean to say that the spectrum (or eigenvalues) of the quantum operators $p$ and $x$ are continuous (rather than saying they're not quantised) for the free particle. For other systems with non trivial boundary conditions it is in fact the case that the spectrum of the momentum operator can be discrete.

Back to your question, it is precisely this sort of thing that leads to discrete spectra of the angular momentum operator. There are two ways of seeing it.

  1. Do the quantization in position space, solving the differential equation for eigenfunctions of $Lz$ and $L^2$. The eigenvalues of the former, $m$ must be quantised so that the functions be single valued when rotating coordinates by $2 \pi$ (or in the case of spin 1/2, its square is single valued which allows half integer m).
  2. Do the quantization with the algebra of the Lie group, $so(3) ~ su(2)$. Asking for a finite dimensional irreducible representation of the group leads to a finite number of eigenstates of the operators with values of m integer separated.
$\endgroup$
2
$\begingroup$

The angular momentum of a free particle is quantized.

The argument that $L_z$'s eigenvalues are quantized is simple and independent of the potential the particle is in. You simply note that $L_z$ is proportional to $d/d\theta$, so the eigenfunctions must have the form $e^{i n \theta}$, where $n$ is an integer to ensure they are single-valued. Then the eigenvalues are proportional to $n$ and hence quantized.

You might be under the impression that if we took a state with definite $\mathbf{r}$ and definite $\mathbf{p}$, then the state would have definite value $\mathbf{L} = \mathbf{r} \times \mathbf{p}$, which can be arbitrary. The problem is that no such states exist, because of the Heisenberg uncertainty principle. The more definite you choose $\mathbf{r}$ to be, the less definite $\mathbf{p}$ must be. If you do the math in detail, you will find that the expectation value $\langle L_z \rangle$ can take on continuous values, but the actually measured values can't.

$\endgroup$
4
  • $\begingroup$ I asked here a question physics.stackexchange.com/questions/350404/… Mark H reply was that since $L$ changes in discrete steps, the fan speeds up in discrete steps. But the speed/momentum is not quantized. For a free particle, if $L$ changes discretely, how do I interpret that? $\endgroup$ Mar 6, 2019 at 16:01
  • $\begingroup$ @mithusengupta123 That answer is simply wrong. It only has that many upvotes because of the HNQ effect. $\endgroup$
    – knzhou
    Mar 6, 2019 at 16:02
  • $\begingroup$ @lux It is strange that for $\vec{L}^2$ and $L_z$ to take discrete values, it is not at all necessary that both $\vec{r}$ and $\vec{p}$ take discrete values! $\endgroup$ Mar 8, 2019 at 14:48
  • $\begingroup$ Well if you do the calculation you see that the important point, that leads to the angular momentum operators providing a rep of so(3), is that x and p do not commute... This is essentially the origin of the discrete spectrum. $\endgroup$
    – lux
    Mar 9, 2019 at 17:23

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.