# Given Hamiltonian . Find eigenvalues and eigenfunctions

1. Nov 15, 2014

### tasos

1. The problem statement, all variables and given/known data
We have the hamiltonian $$H = al^2 +b(l_x +l_y +l_z)$$
where a,b are constants.
and we must find the allowed energies and eigenfunctions of the system.
2. Relevant equations

3. The attempt at a solution

I tried to complete the square on the given hamiltonian and the result is:
$$H = a\mathcal{L} ^2 +\frac {3}{4} \frac {b^2}{a}$$

Where $$\mathcal{L} ^2$$ here is the new operator "of angular momentum" with components :
$$\mathcal{L} ^2=(\mathcal{L} _x +\mathcal{L} _y +\mathcal{L} _z)$$
$$\mathcal{L} _x=(l_x + \frac {b}{2a}), \mathcal{L} _x=(l_x + \frac {b}{2a}), \mathcal{L} _x=(l_x + \frac {b}{2a})$$

I calculated all the commutators of $$(\mathcal{L}^2_x),(\mathcal{L}_x),(\mathcal{L}_y),(\mathcal{L}_z),(\mathcal{L}_+),(\mathcal{L}_-)$$
and i found the same results from angular momentum theory.

So i assumed that the eigenvalues here are $$ħl(l+1)+ \frac {3}{4} \frac {b^2}{a}$$
from the eigenvalues equation $$Hf = λf$$

and since we have the same theory for "Big L" of angular momentum. We have the same eigenvalues
for $$(\mathcal{L}^2 , \mathcal{L}_z)≡ (ħl(l+1), ħm$$
And about the eigenfunctions we have the spherical harmonics $$Y_l^m$$

Is this corrrect or i lost on the way???

Last edited: Nov 15, 2014
2. Nov 16, 2014

### vela

Staff Emeritus
You might want to test out your idea for the $l=1$ case (referring to $\hat{L}^2$ in the original Hamiltonian). It should be much work to write down the 3x3 matrix representing $\hat{H}$ explicitly and diagonalize it. Then you can see if you get the results you expect.

3. Nov 16, 2014

### tasos

new calculations showed that $$[\mathcal{L}^2,\mathcal{L}_+]\neq0$$ does not commute, so i dont see how this proccess can help.

Vela i dont understand.

Whats the method you are saying i must follow.

4. Nov 16, 2014

### vela

Staff Emeritus
I didn't suggest a method to solve the problem. I said you should try a relatively concrete example to see if your results panned out since you were wondering if your logic was valid. If you were to look at the case $l=1$, you can use the 3x3 matrices that represent $\hat{L}^2, \hat{L}_x, \hat{L}_y, \hat{L}_z$ in the usual basis and calculate the matrix representing $\hat{H}$ explicitly. Then you could simply find the eigenvalues of that matrix and compare them to what you think they should have been based on your other calculation. If they didn't match, you know you made a mistake somewhere.

I haven't thought about the problem, much less worked it out, so take this with a grain of salt. I noticed we can express the Hamiltonian as
$$\hat{H} = a\hat{L}^2 + b\vec{n}\cdot\vec{L}$$ where $\vec{n} = (1, 1, 1).$ The Hamiltonian appears to pick out $\vec{n}$ as a preferred direction, so perhaps reorienting the coordinate system to point along that direction would be fruitful.

5. Nov 17, 2014

### tasos

if we use the Hamitlonian operator on the YLM basis we get
$$\hat{H}|Y_l^m> = \dfrac{b}{a}(1-i)l_+ + \dfrac{b}{2}(1+i)l_- + al_z(l_z + \dfrac{b}{a} )|Y_l^m> =$$
$$c_-\dfrac{b}{a}(1-i) Y_l^(m+1) +c_+\dfrac{b}{a}(1+i) Y_l^(m-1) + (a\hbar m^2 + b\hbar m)Y_l^m$$

So i sould demand that the $$Y_l^(m-1), Y_l^(m+1)$$ must be zero so that the hamiltonian operator gives us exact energies and the exact eigenvalues of angular momentum.

6. Nov 17, 2014

### tasos

Also i've seen a lot of exercises with angular momentum that we use$$\hat{n}l$$
But in general case with no given value of l i cant see how i can work this out.

Last edited: Nov 17, 2014