# Does the existence of a ladder operator imply that the eigenvalues are discrete?

1. Oct 17, 2009

### Unkraut

Hi!
I don't know much about QM. I'm reading lecture notes at the moment. Angular momentum is discussed. The ladder operators for the angular-momentum z-component are defined, it is shown that <L_z>^2 <= <L^2>, so the z component of angular momentum is bounded by the absolute value of angular momentum. And then, I don't know why, it is stated that "evidently" the angular momentum eigenvalues are discrete. Why is that so? I see somehow that this is the case when I solve the Schrödinger equation in spherical coordinates by seperation. But this did not happen in the text yet. Can this be seen purely algebraically? Is it true, maybe, that the eigenvalue spectrum is discrete if it is bounded?

2. Oct 18, 2009

### Unkraut

Maybe my question was a bit unclear. My problem is the following:
We have an operator $$L_+$$ and an operator $$L_-$$ such that for a simultaneous eigenvector $$\psi$$ of $$L^2$$ and $$L_z$$ with eigenvalues $$\lambda$$ and $$\mu$$ correspondingly we have: $$L_zL_+\psi=(\mu+\hbar)L_+\psi$$ and $$L_zL_-\psi=(\mu-\hbar)L_-\psi$$. That means that each of $$L_+\psi$$ and $$L_-\psi$$ is either also an eigenvector of $$L_z$$ or 0. We also know that the eigenvalue of $$L_z^2$$ cannot exceed that of $$L^2$$ and so $$L_z$$ is bounded from above and below. Of course, from that follows that there exist natural numbers j and k such that $$L_+^j\psi=0$$ and $$L_-^k\psi=0$$. But how does it follow now that the eigenvalues of $$L_z$$ are integer multiples of $$\hbar$$, i.e. that $$\mu$$ is an integer multiple of $$\hbar$$ in the first place? I only see that there exist operators which jump from eigenstate to eigenstate, increasing or decreasing the eigenvalue by $$\hbar$$, but I don't see why there can be no eigenstates in between.

3. Oct 27, 2009

### conway

This was a good question that deserved an answer. Sorry that nobody was able to contribute.

I kind of think you're right about the sloppy logic whereby the existence of ladder operators supposedly shows the states to be discrete.

4. Oct 27, 2009

### strangerep

answer to Unkraut's question, but then realized it needed quite a long
answer (i.e., a derivation of the angular momentum spectrum in QM),
and I didn't have enough time to do it properly.

The short inadequate answer is that the discreteness of the angular momentum
eigenvalues follows rigorously from nothing more than the so(3) Lie algebra
and the assumption that the representation is unitary (ie in Hilbert space).

The full derivation can be found in these textbooks:

Greiner, "Quantum Mechanics - Symmetries",

Ballentine, "Quantum Mechanics - A Modern Development",

(and probably others - those are just the ones where I studied
this stuff).

Unkraut, if you're still having trouble with this issue, post another message
here, and I'll try to write something more comprehensive.

5. Oct 27, 2009

### Bob_for_short

Lz is something like ∂/∂φ, right? Its eigenfunctions are exp(cφ), OK? The periodicity conditions make c to be integer.

6. Oct 29, 2009

### conway

Would it be sloppy to say that the dicreteness follows from the existence of a ladder operator? That was the original question.

7. Oct 29, 2009

### humanino

A criterion for the discreteness of the spectrum of a self-adjoint positive definite operator is that the inverse of its square root be compact.

8. Oct 29, 2009

### Bob_for_short

That's cool!

9. Oct 29, 2009

### Avodyne

There must be a maximum e'val of Lz, call it k. But Lx, -Ly, and -Lz have the same algebra as Lx, Ly, and Lz, so the minimum e'val of Lz must be -k. Now act j times with L- on the state with Lz e'val k; we get a state with Lz e'val k-j. This must eventually equal -k. So there is an integer j which obeys k-j=-k; hence k must be an integer or half integer.

10. Oct 29, 2009

### Bob_for_short

I like it better.

11. Oct 29, 2009

### Fredrik

Staff Emeritus
You also need the fact that the spectrum is bounded.

Suppose that there exists an operator that raises the eigenvalue of any eigenstate by 1. Now the spectrum can't be bounded from above. So suppose instead that the raising operator raises the eigenvalue of every eigenstate by 1, except for the eigenstate with the maximum eigenvalue, which is taken to 0. Now the spectrum can be bounded from above if and only if every eigenvalue is equal to the maximum eigenvalue minus an integer.

Last edited: Oct 29, 2009