Convert Hamiltonian to matrix in weird basis

[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. Homework Statement
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)

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 :SThanks a lot for helping out!

 

[stevendaryl]

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. Homework Statement
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)

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 :SThanks a lot for helping out!

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.

 

[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

 

[stevendaryl]

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

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.

 

[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?

 

[stevendaryl]

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?

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?

 

[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.

 

[stevendaryl]

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.

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.

 

[emeriska]

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

 

[stevendaryl]

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

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

 

[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\rangleSo 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?

 

[emeriska]

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

 

[stevendaryl]

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

Right, except 0 times anything is still 0.

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

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

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

 

[stevendaryl]

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

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

 

[emeriska]

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

 

[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...

 

[stevendaryl]

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

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

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...

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

 

[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!