# Eigen value of total angular momentum

1. Nov 8, 2013

### Idoubt

I'm trying to understand why the eigen value of the total angular momentum $L^{2}$ is $\hbar ^2 l(l+1)$. The proofs I have seen go like this. Using the ladder operators $L_{\pm} = L_x \pm iL_y$ we can see and the $|l, m \rangle$ state with maximum value of m (eigen value of $L_z$ )

$\langle l,m_{max} | L^2 | l, m_{max} \rangle = \langle l, m_{max} | L_{-}L_+ + L_z^2 + \hbar L^z |l,m_{max}\rangle$
= $l^2 \hbar ^2 + \hbar ^2 l = \hbar ^2 l(l + 1)$

Because $L_+$ acting on the state with highest $m$ value annihilates it. So far so good, as long as I can prove that $L_+ |l , m_{max} \rangle = 0$ , I'll be happy.

Now all proof's I have seen for the eigen values of $L_+$ use the same identity again but now say

$\langle l, m | L_-L_+| l , m \rangle = \langle l, m | L^2 - L_z^2 - \hbar L_z | l , m \rangle = \hbar ^2 \left( l(l+1) - m(m+1) \right)$

and claim that when $l = m$ this is zero. So you are now using the fact that $L^2 |l , m \rangle = \hbar ^2 l(l+1)$ which was what I wanted to prove in the first place. Is there an independent proof that avoids this cycle?

2. Nov 8, 2013

### vanhees71

There is no circle here. You just have to argue a bit differently. It's clear that there are common eigenvectors of $L^2$ and $L_z$ with
$$L^2 |\alpha,m \rangle=\alpha |\alpha m \rangle, \quad L_z |\alpha,m \rangle=m |\alpha,m \rangle.$$
I've set $\hbar=1$ to save a bit of typing :-).

Then you define the ladder operators and prove that there's a maximum value for $L_z$, called $l=m_{\text{max}}$. Using the ladder operator $L_-$ and the fact that there is also a minimal value $m_{\text{min}}$, you get $m_{\text{min}}=-l$ and that $l \in \mathbb{N}_0/2$.

Now you use
$$L^2=L_- L_+ +L_z^2+L_z,$$
which follows from the definition of the ladder operators (no properties of the eigenvectors are used). Then you take the expectation value wrt. $|\alpha,l \rangle$. You find, because of $L_+|\alpha,l \rangle=0$
$$\langle \alpha,l | L^2 | \alpha,l \rangle=\alpha=\langle \alpha,l|L_z^2+L_z|\alpha,l\rangle=l^2+l=l(l+1),$$
where I've only used the eigenvalue equation for $L_z$. As you see, at the end you get the eigenvalue for $L^2$
$$\alpha=l(l+1)$$
without assuming this relation before. So there is no circle in the argument!

3. Nov 8, 2013

### Idoubt

My question is how do you prove that there is a maximum or minimum value for $m$ and also how does this imply that $L_+ | l, m_{max} \rangle = 0$ ? Why not some other state ? And could you explain what is $\mathbb{N}_0/2$

4. Nov 8, 2013

### dauto

$$\langle\alpha,m|J^2|\alpha,m\rangle=\langle\alpha,m|J_x^2+J_y^2+J_z^2| \alpha,m\rangle$$

but we also have

$$\langle\alpha,m|J_x^2| \alpha,m\rangle=\langle\alpha,m|J_x^\dagger J_x| \alpha,m\rangle$$

Because Jx is Hermitian. That shows that $\langle\alpha,m|J_x^2| \alpha,m\rangle$ is positive because it is the magnitude (squared) of a vector. Samething is true about $\langle\alpha,m|J_y^2| \alpha,m\rangle$

Putting that back into the original equation we have

$$\langle\alpha,m|J^2|\alpha,m\rangle \ge \langle\alpha,m|J_z^2| \alpha,m\rangle$$

and from that you obtain

$\alpha \ge m^2.$ Clearly, for fixed alpha, m is limited in scope. Given that the rising operator lifts the eigenvalue m by a unit, eventually that inequality will be violated unless at some point the operator gives a vanishing eigenvector as a result.

5. Nov 11, 2013

### Idoubt

yes, I see that there must be a upper limit on m for a fixed α. But is it completely rigorous to say that since the equality would be violated $L_+| \alpha , m_{max} \rangle$ is zero? I was reading this proof in Griffith and he mentions that strictly we can only conclude that the function is not normalizable. I am not fully comfortable with the whole argument, is just me or is there some mathematical 'iffyness' here?

6. Nov 11, 2013

### Idoubt

Also except for the state $| 0 , 0 \rangle$ $\alpha = m$ does not seem to be allowed, why is that?

7. Nov 11, 2013

### dauto

The solutions are $\alpha = l(l+1)$, and $|m| \le l$. How could they be equal unless they are both zero?

Last edited: Nov 11, 2013
8. Nov 12, 2013

### Idoubt

Oh I see. It again follows from the fact that $\alpha = l(l+1)$ which follows from $L_{+} | \alpha, m_{max} \rangle = 0$