Eigen value of total angular momentum

[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?

 

[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!

 

[Idoubt]

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.

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

 

[dauto]

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

\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 J_x 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.

 

[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?

 

[Idoubt]

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

 

[dauto]

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

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

 

[Idoubt]

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

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