# Eigenstates of Orbital Angular Momentum

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1. Feb 9, 2016

### leo.

Recently I've been studying Angular Momentum in Quantum Mechanics and I have a doubt about the eigenstates of orbital angular momentum in the position representation and the relation to the spherical harmonics. First of all, we consider the angular momentum operators $L^2$ and $L_z$. We know that they commute so that we can find a basis of eigenstates commom to those two operators.

Also, we also know from the general theory of Angular Momentum that the eigenvalues of $L^2$ are of the form $l(l+1)\hbar^2$ with integral or half-integral $l$, while fixing $l$ the eigenvalues of $L_z$ are of the form $m\hbar$ with $m\hbar$ being restricted to the numbers $-l,-l+1,\dots,l-1,l$.

In that case the eigenvalue equations for the two operators in the position representation using spherical coordinates are:

$$L^2 \psi(r,\theta,\phi) = l(l+1)\hbar^2 \psi(r,\theta,\phi), \\ L_z\psi(r,\theta,\phi) = m\hbar \psi(r,\theta,\phi).$$

The book then states the following:

I have several doubts regarding that. First of all, in the way the book presents, the spherical harmonics are defined as eigenfunctions of the operators in question. I do know that $r$ appears just as a parameter, but this doesn't mean the solutions do not depend on this variable. The spherical harmonics are functions of $\theta$ and $\phi$ alone, so how can they be eigenfunctions of these operators if they do not depend on the parameter $r$?

And the most important question: how the author concludes that the eigenfunctions are all of the form $\psi_{l,m}(r,\theta,\phi)=f(r)Y^m_l(\theta,\phi)$? I mean, first he defined $Y^m_l$ as the eigenfunctions, later he says that the eigenfunctions have this specific form. How does one following the author's reasoning find out that the eigenfunctions have this form with this exact dependence on the parameter $r$?

Of course we can solve the equations. In that way we indeed get what the spherical harmonics are. But then we see that this is the form of the separable solutions: if we use separation of variables we arrive at this form of eigenfunctions. But why there are no other solutions?

In general I'm quite confused in the way this is being presented and why the eigenstates of orbital angular momentum have that form in the position representation.

2. Feb 9, 2016

### Jilang

You can take the f(r) part and move it to left the operators. Then the Y becomes an eigenstate.

3. Feb 9, 2016

### leo.

I understand that Jilang, but my point here is a little different. We see that:
1. As the author states, we first define $Y^m_l(\theta,\phi)$ so that $L^2Y^m_l(\theta,\phi) = l(l+1)\hbar^2Y^m_l(\theta,\phi)$ and $L_zY^m_l(\theta,\phi)=m\hbar Y^m_l(\theta,\phi)$. Then as presented here, $Y^m_l$ is just a notation for the eigenfunctions which do not depend on $r$ at all.
2. Secondly, the author states that after $Y^m_l$ is so defined, if $\psi_{l,m}$ is any eigenfunction corresponding to the eigenvalues $l(l+1)\hbar^2$ and $m\hbar$ then $\psi_{l,m}$ is necessarily of the form $\psi(r,\theta,\phi)=f(r)Y^m_l(\theta,\phi)$.
Thus, the first assertion is just a matter of notation. The second one is the one which confuses me. Obviously if $\psi(r,\theta,\phi) = f(r)Y^m_l(\theta,\phi)$ by the reason you stated $\psi_{l,m}$ is an eigenfunction of $L^2$ and $L_z$.

On the other hand, the author's assertion is: if $\psi_{l,m}$ is any eigenfunction of $L^2$ and $L_z$ corresponding to the eigenvalues $l(l+1)\hbar^2$ and $m\hbar$ then it is of the form $\psi_{l,m}(r,\theta,\phi)=f(r)Y^m_l(\theta,\phi)$. My question is then: why is that the case? How can one arrive at this result?

4. Feb 9, 2016

### strangerep

@leo.,

FYI, you might get (slightly) better answers if you specify which book you're using.

Does your book give an explicit expression for the $L^2$ operator in spherical polar coordinates? Or if not, are you able to compute one?

I'm reasonably sure that what's going on here is just the standard method of solving the Schrodinger equation by separation of variables, but possible intermixed with a bit of group theory. Do you understand the method of separation of variables for solving DEs, and why it works?

5. Feb 9, 2016

### leo.

@strangerep, I forgot to tell the book I'm using. It's the book "Quantum Mechanics" by Cohen-Tannoudji. He deals with Angular Momentum in chapter 6. His approach is to first deal with general angular momentum operators. Here if $\mathbf{J}$ is an arbitrary angular momentum he finds the spectrum of $J^2$ and $J_z$. Only then he deals with the orbital angular momentum.

The book gives an explicit expression for $L^2$ and $L_z$ and I've also computed it already. Now, I think I do understand the method of separation of variables: in three dimensions using spherical coordinates, for instance, when solving some PDE we suppose the solution is of the form $\psi(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi)$. In general we arrive at Sturm Liouville problems in each of the variables which in turn yields complete set of solutions. That's the idea in general right?

When I studied this method, one example I saw was Laplace's equation $\nabla^2 \psi = 0$ in which the angular part yielded the Spherical Harmonics. As a result the Spherical Harmonics form a complete set of functions on the sphere.

All of this is fine, but the author's approach here seems to be a little different. He already knows the spectrum and instead of constructing the Spherical Harmonics by solving the PDE he simply defines them as the solutions. Later he computes the Spherical Harmonics by using the ladder operators.

In that case I'm not finding out, using the author's reasoning, why the eigenfunctions are of the form $\psi_{l,m}(r,\theta,\phi)=f(r)Y^m_l(\theta,\phi)$. The eigenvalue equations are:

$$-\hbar^2\left(\dfrac{\partial^2}{\partial\theta^2}+\dfrac{1}{\tan \theta} \dfrac{\partial}{\partial \theta}+\dfrac{1}{\sin^2\theta}\dfrac{\partial^2}{\partial\phi^2}\right)\psi(r,\theta,\phi)=l(l+1)\hbar^2\psi(r,\theta,\phi) \\ -i\hbar \dfrac{\partial}{\partial \phi}\psi(r,\theta,\phi)=m\hbar \psi(r,\theta,\phi)$$

I believe that if I use separation of variables here I get the same result, but it doesn't seem to be that what the author is doing. What am I missing here?

6. Feb 9, 2016

### strangerep

Yes.

I'd guess the author is just using separation of variables -- but only using 2 factors at first -- so that the $r$-dependence can be factored out. They're also using knowledge that this concrete $L^2$ is the representation of the abstract $L^2$ on a specific Hilbert space, hence similar features of the angular momentum spectrum can be assumed. (Strictly speaking, one needs also to know how restriction to orbital angular momentum restricts the spectrum.)

I further guess that they're doing it this way (with ladder operators) since that's quicker than solving the angular DE to find the full set of spherical harmonics.

Unfortunately, I don't have the book so I can't check their whole approach.

7. Feb 13, 2016

### leo.

Reading a little bit more I think I understood what is going on based on the following paragraph from the book

Hence I believe what is going on here is the following: the eigenvalue equations are

$$L^2\psi_{l,m}(r,\theta,\phi)=l(l+1)\hbar^2\psi_{l,m}(r,\theta,\phi) \\ L_z\psi_{l,m}(r,\theta,\phi)=m\hbar \psi_{l,m}(r,\theta,\phi)$$

now since $r$ is just a parameter we look for the simplest solutions, which are the ones which do not depend on $r$ at all. We define those solutions to be $Y^m_l(\theta,\phi)$.

On the other hand, the general solutions, which depend on the parameter $r$ satisfies the exact same equations. Now, if we know that the solutions are unique to within a constant factor, $\psi_{l,m}$ must be a constant factor times $Y^m_l$. But the operators do not act on the $r$ coordinate, so that concerning that differential equations the most general constant also depends on $r$. In that case we end up with the general solution $\psi_{l,m}(r,\theta,\phi)=f(r)Y^m_l(\theta,\phi)$.

Now, that the solutions are unique can be proved solving the angular equations with separation of variables as suggested or rather using the ladder operators. Is that the whole idea here? At least for me it seems to make sense.

8. Feb 13, 2016

### strangerep

My only extra remark is that you might be interested to study how the quantum angular momentum spectrum can be derived group-theoretically without even mentioning a differential equation. See Ballentine, ch7, (if you haven't already seen this approach).