Answer: Find Eigenvalues of QM Angular Momentum Operator L2, Lz

[Niles]

Homework Statement

Hi all.

A particle is in the state $\left| {ml} \right\rangle$, and I need to find the eigenvalues for the operator L² and L_z.

If we start by looking at L², then I know that the eigenfunctions of this operator are the spherical harmonics, and I know how to write L² in spherical coordinates. But finding the eigenenergies this way seems to be a long task.

What is the easiest way of doing this?

Regards,
Niles.

 

[Ben Niehoff]

It's much, much simpler than you're making it. The state ket $|l,m \rangle$ is already written in terms of the eigenvalues you need.

 

[Niles]

For L² I know from the WWW that the eigenvalues are $\hbar^2 l(l+1)$, but how can I see this from the state ket?

 

[malawi_glenn]

write L^2 in terms of ladder operators L+ L- and Lz, and do a sandwhich with the bra <lm|

 

[Tian WJ]

Hello, Niles! I wish I could do some help.

You are infact using Dirac notation (DN). All the possible states of a certain quantum system form a Hilbert space (HS). Vectors in HS (generally complex rather than real) are denoted via the bra $|>$, and if one wants to represent a special state, one will enrich the formulism of the bra. For example, we use $|\phi>$ to represent the state of by the wavefunction $\phi$. As to eigen states, the corresponding eigenvalue or quantum number are generally put into the bra $|>$. Hence, $|x'>$ means the eigen state of the coordinate $x$(with eigenvalue x'), $|p'>$ means the eigen state of momemtum(with eigenvalue p'), $|En>$ or $|n>$ means the eigen state of energy(with eigenvalue En), and
$ |m,L> $ , as you put above, means the co-eigenstate of $(L_z, L^2)$ , with eigen values $ m\hbar $ and $ L(L+1)\hbar^2$ respectively.

And please also notice that, in Dirac notations above, we simply employ an abstract state vector, and refer to no concrete quantum mechanical representation.

Hence, indeed you know the answer if you know the meaning of Dirac notation.

 

[Niles]

... $ |m,L> $ , as you put above, means the co-eigenstate of $(L_z, L^2)$ , with eigen values $ m\hbar $ and $ L(L+1)\hbar^2$ respectively.

But my mission is to prove that the ket |m,l> has the eigenvalues (m \hbar, l(l+1)\hbar), respectively.

I'll try writing L² with ladder-operators and L_z.

 

[Niles]

write L^2 in terms of ladder operators L+ L- and Lz, and do a sandwhich with the bra <lm|

This is what I have so far:

L² = L₊L_-+L_z²-\hbar L_z.

This gives us:

$$M = \left\langle {lm} \right|L_ + L_ - \left| {lm} \right\rangle + \left\langle {lm} \right|L_z^2 \left| {lm} \right\rangle - \hbar \left\langle {lm} \right|L_z \left| {lm} \right\rangle = 0 + \hbar ^2 m^2 - \hbar ^2 m = \hbar ^2 (m^2 - m)$$

where M is the eigenvalue of L². Am I on the right track?

 

[Tian WJ]

I see, I will have a try to solve your problem.

 

[malawi_glenn]

I see, I will have a try to solve your problem.

recall that the forum rules forbids you to post full solutions to problems here.

 

[Niles]

recall that the forum rules forbids you to post full solutions to problems here.

How is it that the integer l even comes into play when finding the eigenvalues of L²? In my approach (which I am fairly certain is correct), only m shows up.

 

[malawi_glenn]

hint: write L+L- = L^2 - Lz^2 + \hbar Lz

and act on the ket with m_max.