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!