# I No r dependence in L operator?

1. Feb 20, 2017

### DoobleD

In classical mechanics, angular momentum, L = r x p, depends of r. For a given momentum p, the bigger r is, the bigger is the angular momentum. Event in spherical coordinates, r still appears in the classical angular momentum.

However, the angular momentum operator in QM has no r dependence, it's only a function of angles phi and theta. While I can follow the derivation of the operator, I find it surprising that the r dependency vanishes. How can we explain this physically ?

2. Feb 20, 2017

### BvU

3. Feb 20, 2017

### DoobleD

Thank you. Is there a specific part adressing my question ? This seem to be quite complete course on QM angular momentum, but after a quick search all I have found regarding r, is around the end when it says they haven't adressed r yet. But they refer to the case of the hydrogen atom I think, which has a hamiltonian with a r dependent potential.

4. Feb 20, 2017

### BvU

First page

5. Feb 20, 2017

### DoobleD

After reading it I don't see where is it ? It shows how we derive the QM version of angular momentum from the classical picture. This doesn't bother me. And even if the maths of this derivation also works out for the angular momentum operator, I just wonder, on a conceptual level, why the r dependence vanishes.

6. Feb 20, 2017

### Staff: Mentor

The $x$, $y$, and $z$ that appear in the first page are the components of $\vec{r}$. So it's there.

7. Feb 20, 2017

### DoobleD

Right, I should have been more precise : why r vanishes in the spherical coordinate version of L ?

8. Feb 20, 2017

### BvU

Math, math and math. The transition from cartesian to spherical makes it look like a magic disappearance, I concede.

http://quantummechanics.ucsd.edu/ph130a/130_notes/node216.html

My hunch is the dependence sits in the wavefunction itself, but I quit trying to form such mental pictures a long time ago.
I do feel with you, so let's ask @Nugatory whether this is only seemingly counter-intuitive, or whether there's some other insight we overlook ...

Last edited: Feb 20, 2017
9. Feb 20, 2017

### DoobleD

Exactly. Good to know I am not alone. :D I came accross the same link BTW searching an answer on Google.

10. Feb 20, 2017

### DoobleD

What adds to the weirdness is that angular momentum in classical mechanics, even if expressed in spherical coordinates, does have a r dependency :

Source (page 11).

11. Feb 20, 2017

### blue_leaf77

Since it has been confirmed by many experiments that such formulation of angular momentum operator does work, I think I will just accept it as granted. Besides that, angular momentum is defined as the generator of rotation in space in quantum mechanics, so I think it makes sense that it should not depend on r.

12. Feb 20, 2017

### Staff: Mentor

The quantum operator has no $r$ dependence, but that does not mean the actual measured values of angular momentum have no $r$ dependence. The latter is what you would expect to be analogous to the classical angular momentum.

13. Feb 20, 2017

### DoobleD

Right ! So, as @BvU suggested, the r dependence would be in the wavefunction ? But...The eigenstates of the angular momentum operator are spherical harmonics, which are not functions of r, but of theta and phi only.

14. Feb 20, 2017

### Staff: Mentor

No, they are spherical harmonics combined with a function of $r$ only.

15. Feb 20, 2017

### DoobleD

Do you mean if the particle is in a r-dependent potential ? The courses I read present the eigenstates of L as being solely spherical harmonics. After a quick search on Google, same thing.

16. Feb 20, 2017

### BvU

central potential, e.g. hydrogen atom allows separation into radial and angular eigenfunctions. L is 'on the angular side'

17. Feb 20, 2017

### DoobleD

Yes I have seen that separation in the case of the hydrogen atom. But I'm trying to consider L just for itself, outside any specific potential considerations. Maybe this doesn't make sense then ?

18. Feb 20, 2017

### DoobleD

Well, I suppose I am considering the case of a free particle then. And it is true that even in classical mechanics, the angular momentum of a free particle is a constant, so not really dependent of r. Mmh, is this a viable answer ? I'm not sure at all. :D

19. Feb 20, 2017

### Staff: Mentor

No, it doesn't. Any quantum system will have some Hamiltonian, and therefore some potential. That determines the energy eigenstates, and it's meaningless to talk about the action of any operator, including L, without some specification of the states it's acting on.

20. Feb 20, 2017

### DoobleD

What about the free particle case ?

21. Feb 20, 2017

### BvU

Take a free particle wave function and see what $L$ does ...

22. Feb 20, 2017

### Staff: Mentor

Then the Hamiltonian is $p^2/2m$, with $V=0$.

23. Feb 21, 2017

### vanhees71

Of course the orbital angular-momentum operators,
$$\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}$$
make sense for themselves. They obey the commutation relation of the Lie algebra so(3) of the rotation group SO(3),
$$[\hat{L}_j,\hat{L}_k]=\mathrm{i} \epsilon_{jkl} \hat{L}_l,$$
where I've used units with $\hbar=1$ to make things simpler.

Now you can show that there's a common eigenbasis for $\hat{\vec{L}}^2$ and $\hat{L}_z$ with eigenvalues $l(l+1)$ and $m \in \{-l,-l+1,\ldots,l-1,l \}$, where $l \in \{0,1,2,\ldots \}$.

In position representation, using spherical coordinates, the eigenfunctions are the spherical harmonics $Y_{lm}(\vartheta,\varphi)$. Since $\hat{\vec{L}}$ commutes with $\hat{r}=|\hat{\vec{x}}|$ you can multiply it with any function of $r$. So any wave function of a single spin-0 particle can be expanded in spherical harmonics
$$\psi(\vec{x})=\sum_{l=0}^{\infty} \sum_{m=-l}^{l} R_{lm}(r) Y_{lm}(\vartheta,\varphi).$$

24. Feb 21, 2017

### DoobleD

Yes all of that is essentially what a textbook presents. But that doesn't tell why r doesn't appear in L operator. If there is such an answer in the first place of course, maybe there is nothing more here other than "the maths works out that way, and the model matches experiments", I don't know.

In the last equation you wrote, the function of r is "artifically" added in the mix so that we get a generic state for any potential. Or it is added by necessity if we solve the hamiltonian of the hydorgen atom. But it doesn't come from a requirement of the L operator itself. The L operator in itself doesn't care about r, neither do its eigenstates. That just seems weird to me, when compared to classical angular momentum. The classical angular momentum does have a r dependency, without the need for any particular potential or further assumption. It is part of its definition. The QM L operator in itself (and its eigenstates) is only concerned with angles.

EDIT: I should have written L operator in spherical coordinates, every time I wrote L.

Seems a very reasonable thing to do. :D I'll try that as soon as I can.

25. Feb 21, 2017

### vanhees71

The angular momentum operator is the generator for rotations. The length of the vector is not affected by rotations (which is the definition of rotations in the first place). So it's intuitive that the angular-momentum operator does not depend on $r$. Also the expansion of the wave function in terms of spherical harmonics given in my previous posting is not "artificial" but very natural thinking in terms of generalized Fourier transformations. Here we expand in terms of angular-momentum eigenfunctions. The reason why this is a clever expansion is that the Laplace operator separates in spherical coordinates. This you find indeed in any good textbook on quantum theory.