# Homework Help: Energies of a two particle system

1. Jan 18, 2013

### Sekonda

Hey,

This question is on determining the energies of a two particle system given the Hamiltonian, I believe it to be simple enough but would like you guys to check it and fill in any gaps in my reasoning

So I believe the eigenvalues of J^2 and J^2(z) are given by:

$$\hat{J}^{2}:j(j+1)\: ,\: \hat{J}^{2}_{z}:m(m+1)\Leftrightarrow \hbar=1$$

('z' subscript same as '3')

and so the energy of state 1,1 is :

$$2(\alpha+\beta)$$

The second part state that j=3, therefore m=3,2,1,0,-1,-2,-3
and so we just pop these into our eigenvalue equations above to attain the energies :

$$|3,3> : 12\alpha+12\beta\: ,\: |3,2>:12\alpha+6\beta$$

etc.

Is this right?

Thanks for any comment/help!
SK

Last edited: Jan 18, 2013
2. Jan 18, 2013

### G01

There is a mistake in your work. What is the eigenvalue of the Jz operator?

3. Jan 18, 2013

### Sekonda

The eigenvalue of the Jz operator is 'm', so does that mean the eigenvalue of the Jz^2 operator is m(m+1)?

Oh and 'z' is the same as '3' for the subscripts!

4. Jan 18, 2013

### G01

No. If $J_z|j,m>=m|j,m>$ then what is

$$J_z^2|j,m>=J_z(J_z|j,m>)=?$$

5. Jan 18, 2013

### Sekonda

Oh, m^2?

6. Jan 18, 2013

### G01

Yep.

7. Jan 18, 2013

### Sekonda

Oh ok, well I just assumed it was the same as the J^2 eigenvalue j(j+1), does the J^2 operator raise the state by 1 then?

8. Jan 18, 2013

### G01

You are confused about the J^2 operator. It is a total angular momentum operator (squared), not a ladder operator. The eigenvalues for the J^2 operator are correct as you have them: $J^2 |j,m>=j(j+1)|j,m>$

If you don't understand why the J^2 operator has a different eigenvalues than the J_z operator, you should review the derivation of these eigenvalue equations in your textbook or with your instructor.

9. Jan 18, 2013

### Sekonda

It's not that I don't understand why they're different but more why the J^2 is equal to j(j+1), I'm sure we've 'shown' it somewhat before but these things are easily forgotten by myself.

Thanks anyway G01 for being patient with me and helping!
SK

10. Jan 18, 2013

### G01

You're welcome!