# Convert Hamiltonian to matrix in weird basis

1. Dec 1, 2014

### emeriska

Hi guys,

I'm having a hard time with that one from Cohen-Tannoudji, $F_{VI}$ # 6. I'm translating from french so sorry if some sentence are weird or doesn't use the right words.

1. The problem statement, all variables and given/known data

We consider a system of angular momentum l = 1; A basis from it sub-space of states is constituted by these 3 eigenvectors of $L_z : |1>, |0>, |-1>$ of eigenvalues $h, 0, -h$ respectively.

This system, that has a electric quadrupolar moment, is submerged in an electric field gradient, so that the hamiltonian is:
$H = \frac{w_0}{h}(L_u^2-L_v^2)$

Where $L_u$ and $L_v$ are the components of $L$ on the 2 directions $O_u$ and $O_v$ of the plan $xOz$, at 45 degree from $Ox$ and $Oz$; $w_0$ is a real constant.

2. Question
a) Write down the matrix representing $H$ in the basis ${ |1>, |0>, |-1>}$. What are the stationary states of the system and their energy? (These states will be named $|E_1> |E_2> |E_3>$ in crescent order of energy)

3. The attempt at a solution
I'm really stuck here. I don't know how to deal with the whole $L_u, L_v$ thing.

I assumed that $L_u = \frac{L_x + L_z}{\sqrt{2}}$ but I'm really not sure if that's the right assumption...

Also I'm having a hard time with the whole "Matrix" thing...not knowing how to convert in matrix :S

Thanks a lot for helping out!

2. Dec 2, 2014

### stevendaryl

Staff Emeritus
You're not really stuck, yet. You have a correct expression for $L_u$. Get a similar one for $L_v$, and figure out the Hamiltonian and its effect on the basis states.

3. Dec 2, 2014

### emeriska

So I figured $L_v = \frac{L_z - L_x}{\sqrt{2}}$

But I don't see how to convert that into a matrix. We use Griffiths and I feel like my book is not really helping me on that :S

4. Dec 2, 2014

### stevendaryl

Staff Emeritus
A matrix is just a way to organize a system of equations. There is nothing fundamental about it. If you compute the Hamiltonian $H$ in terms of $L_x, L_y, L_z$, then you will find that:

$H | -1 \rangle = \alpha_1 | -1 \rangle + \beta_1 | 0 \rangle + \gamma_1 | +1 \rangle$
$H | 0 \rangle = \alpha_2 | -1 \rangle + \beta_2 | 0 \rangle + \gamma_2 | +1 \rangle$
$H | +1 \rangle = \alpha_3 | -1 \rangle + \beta_3 | 0 \rangle + \gamma_3 | +1 \rangle$

You can compute the values $\alpha_i, \beta_i, \gamma_i$. Then you can think of these as a matrix equation as follows:

Let $\left( \begin{array}& u \\ v \\ w \end{array} \right)$ represent the state $u | -1 \rangle + v | 0 \rangle + w | +1 \rangle$. Then the action of $H$ on such a matrix is given by:

$H \left( \begin{array}& u \\ v \\ w \end{array} \right) = \left( \begin{array}& \alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3 \end{array} \right) \left( \begin{array}& u \\ v \\ w \end{array} \right)$

As I said, matrices is just a way to organize the equations.

5. Dec 2, 2014

### emeriska

I read this section of my book at least 100 times and I still have no idea how to start this problem. If I start with what you just gave me, how can I compute the values $\alpha , \beta , \gamma$ ? Of maybe you know somewhere else where that is explained?

6. Dec 2, 2014

### stevendaryl

Staff Emeritus
Well, your Hamiltonian is a combination of factors of $L_x, L_y, L_z$. Do you know how $L_x$ acts on $|+1\rangle$ and on $|0\rangle$ and $|+1\rangle$? What about $L_y$? What about $L_z$?

Have you learned about raising and lowering operators for angular momentum?

7. Dec 4, 2014

### emeriska

Yes I do know about raising and lowering operators, that I guess I will have to find in terms of $L_u, L_v$ right?

But I'm stuck at the very basic. I don't know where to start to write $H$ in a basis. I can't find any example on the web that is even remotely similar to this question.

Thanks for you help by the way, really appreciated.

8. Dec 4, 2014

### stevendaryl

Staff Emeritus
Yes, you know everything you need to know about $H$. You know how to write it in terms of $L_u$ and $L_v$, and you know how to write $L_u$ and $L_v$ in terms of $L_x$ and $L_z$. And you know how to write $L_x$ in terms of $L_+$ and $L_-$. Right? So put it all together to get $H$ in terms of $L_+$ and $L_-$.

The equations that you need:

1. $L_u = \frac{1}{\sqrt{2}} (L_z + L_x)$
2. $L_v = \frac{1}{\sqrt{2}} (L_z - L_x)$
3. $L_x = \frac{1}{2} (L_+ + L_-)$
4. $H = \frac{w_0}{\hbar}((L_u)^2 - (L_v)^2)$
So before you do anything else, write down $H$ in terms of $L_z, L_+$ and $L_-$. Post the answer.

9. Dec 4, 2014

### emeriska

If my calculus are right I should have $H = \frac{w_0}{\hbar} L_z(L_++L_-)$

10. Dec 4, 2014

### stevendaryl

Staff Emeritus
That's not what I got. I got $H = \frac{w_0}{\hbar} (L_x L_z + L_z L_x) = \frac{w_0}{2\hbar} (L_+ L_z + L_- L_z + L_z L_+ + L_z L_-)$. I may have made a mistake, also.

But next, compute what the effect of $H$ is on the three states $|-1\rangle, |0\rangle, |+1\rangle$

Use the following rules:

$L_+ |m\rangle = \sqrt{l(l+1) - m(m+1)} |m+1\rangle$
$L_- |m\rangle = \sqrt{l(l+1) - m(m-1)} |m-1\rangle$
$L_z |m\rangle = m |m\rangle$

where $m=-1, 0, +1$ and where $l=1$

So compute $H|-1\rangle$, $H|0\rangle$, $H|+1\rangle$

11. Dec 4, 2014

### emeriska

You were right on the equation of $H$
So, I have

$L_+ |-1\rangle = \sqrt{2} |0\rangle$
$L_+ |0\rangle = \sqrt{2} |1\rangle$
$L_+ |1\rangle = 0 |2\rangle$

$L_- |-1\rangle = 0 |-2\rangle$
$L_- |0\rangle = \sqrt{2} |-1\rangle$
$L_- |1\rangle = \sqrt{2} |0\rangle$

$L_z |-1\rangle = - |-1\rangle$
$L_z |0\rangle = 0|0\rangle$
$L_z |1\rangle = |1\rangle$

So that $H|-1\rangle$, $H|0\rangle$, $H|+1\rangle$ would look like this:

$H|-1\rangle = \frac{w_0}{2\hbar}(\sqrt{2}|0\rangle(-)|-1\rangle+(-)|-1\rangle\sqrt{2}|0\rangle)$

And etc for the other $H$ right?

And then I use the matrix convertion stuff you gave me a couple posts back?

How do I deal with a multiplication of $|m\rangle$ states?

Last edited: Dec 4, 2014
12. Dec 4, 2014

### emeriska

I just edited that last post because you were right for the equation of $H$

Last edited: Dec 4, 2014
13. Dec 4, 2014

### stevendaryl

Staff Emeritus
Right, except 0 times anything is still 0.

I don't know what you're doing there. $H$ has 4 terms:
$\frac{w_0}{2\hbar} L_+ L_z$
$\frac{w_0}{2\hbar} L_- L_z$
$\frac{w_0}{2\hbar} L_z L_+$
$\frac{w_0}{2\hbar} L_z L_-$

What is $L_+ L_z |-1\rangle$?
What is $L_- L_z |-1\rangle$?
What is $L_z L_+ |-1\rangle$?
What is $L_z L_-|-1\rangle$?

(Note, there is only 1 of those 4 that is nonzero)

Add them all together, multiply by $\frac{w_0}{2\hbar}$, and that gives you $H |-1\rangle$

14. Dec 4, 2014

### stevendaryl

Staff Emeritus
That doesn't happen.

Let me just work out one example: $L_z L_+ |-1\rangle = L_z (L_+ |-1\rangle) = L_z( \sqrt{2} |0\rangle) = \sqrt{2} (L_z |0\rangle) = \sqrt{2} \cdot 0 \cdot |0\rangle = 0$

15. Dec 4, 2014

### emeriska

OH! I see! Rhat's not what I was doing at all. Thanks a lot, I'll see where this leads me

16. Dec 4, 2014

### emeriska

$H \Psi = \left( \begin{array}& 0 & \sqrt{2} & 0 \\ -\sqrt{2}& 0 & \sqrt{2} \\ 0 & \sqrt{2} & 0 \end{array} \right) \Psi$

Thanks a lot!! I guess I only have to find the stationary states from this matrix?

The stationary states would be the line vectors of the matrix is that right? So something like $\frac{w_0}{2\hbar}\sqrt{2}|-1>$ for the first one...

And I guess to find their energy I have to write down $H\Psi=E\Psi$ with H being my matrix and $\Psi$ the vector? Not quite sure about that either...

17. Dec 4, 2014

### stevendaryl

Staff Emeritus
I think it's something like that (you left off the $w_0$ and $\hbar$, but that's easily fixable).

Now your problem has been reduced to solving an eigenvalue-eigenvector problem involving matrices. Have you studied that?

18. Dec 4, 2014

### emeriska

Well I tried something but never got to the energies...anyway, my deadline was tomorrow so I should have the solutions by the weekend.

I certainly won't have 100% but if I have anything more than 0 it will be because of you! Thank you very much for your time, much appreciated! You helped me a lot!