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Convert Hamiltonian to matrix in weird basis

  1. Dec 1, 2014 #1
    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. jcsd
  3. Dec 2, 2014 #2

    stevendaryl

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    You're not really stuck, yet. You have a correct expression for [itex]L_u[/itex]. Get a similar one for [itex]L_v[/itex], and figure out the Hamiltonian and its effect on the basis states.
     
  4. Dec 2, 2014 #3
    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
     
  5. Dec 2, 2014 #4

    stevendaryl

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

    [itex]H | -1 \rangle = \alpha_1 | -1 \rangle + \beta_1 | 0 \rangle + \gamma_1 | +1 \rangle[/itex]
    [itex]H | 0 \rangle = \alpha_2 | -1 \rangle + \beta_2 | 0 \rangle + \gamma_2 | +1 \rangle[/itex]
    [itex]H | +1 \rangle = \alpha_3 | -1 \rangle + \beta_3 | 0 \rangle + \gamma_3 | +1 \rangle[/itex]

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

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

    [itex]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)[/itex]

    As I said, matrices is just a way to organize the equations.
     
  6. Dec 2, 2014 #5
    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?
     
  7. Dec 2, 2014 #6

    stevendaryl

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

    Have you learned about raising and lowering operators for angular momentum?
     
  8. Dec 4, 2014 #7
    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.
     
  9. Dec 4, 2014 #8

    stevendaryl

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

    The equations that you need:

    1. [itex]L_u = \frac{1}{\sqrt{2}} (L_z + L_x)[/itex]
    2. [itex]L_v = \frac{1}{\sqrt{2}} (L_z - L_x)[/itex]
    3. [itex]L_x = \frac{1}{2} (L_+ + L_-)[/itex]
    4. [itex]H = \frac{w_0}{\hbar}((L_u)^2 - (L_v)^2)[/itex]
    So before you do anything else, write down [itex]H[/itex] in terms of [itex]L_z, L_+[/itex] and [itex]L_-[/itex]. Post the answer.
     
  10. Dec 4, 2014 #9
    If my calculus are right I should have ##H = \frac{w_0}{\hbar} L_z(L_++L_-)##
     
  11. Dec 4, 2014 #10

    stevendaryl

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    That's not what I got. I got [itex]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_-)[/itex]. I may have made a mistake, also.

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

    Use the following rules:

    [itex]L_+ |m\rangle = \sqrt{l(l+1) - m(m+1)} |m+1\rangle[/itex]
    [itex]L_- |m\rangle = \sqrt{l(l+1) - m(m-1)} |m-1\rangle[/itex]
    [itex]L_z |m\rangle = m |m\rangle[/itex]

    where [itex]m=-1, 0, +1[/itex] and where [itex]l=1[/itex]

    So compute [itex]H|-1\rangle[/itex], [itex]H|0\rangle[/itex], [itex]H|+1\rangle[/itex]
     
  12. Dec 4, 2014 #11
    You were right on the equation of ##H##
    So, I have

    [itex]L_+ |-1\rangle = \sqrt{2} |0\rangle[/itex]
    [itex]L_+ |0\rangle = \sqrt{2} |1\rangle[/itex]
    [itex]L_+ |1\rangle = 0 |2\rangle[/itex]

    [itex]L_- |-1\rangle = 0 |-2\rangle[/itex]
    [itex]L_- |0\rangle = \sqrt{2} |-1\rangle[/itex]
    [itex]L_- |1\rangle = \sqrt{2} |0\rangle[/itex]

    [itex]L_z |-1\rangle = - |-1\rangle[/itex]
    [itex]L_z |0\rangle = 0|0\rangle[/itex]
    [itex]L_z |1\rangle = |1\rangle[/itex]


    So that [itex]H|-1\rangle[/itex], [itex]H|0\rangle[/itex], [itex]H|+1\rangle[/itex] 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
  13. Dec 4, 2014 #12
    I just edited that last post because you were right for the equation of ##H##
     
    Last edited: Dec 4, 2014
  14. Dec 4, 2014 #13

    stevendaryl

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    Right, except 0 times anything is still 0.

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

    What is [itex]L_+ L_z |-1\rangle[/itex]?
    What is [itex]L_- L_z |-1\rangle[/itex]?
    What is [itex]L_z L_+ |-1\rangle[/itex]?
    What is [itex]L_z L_-|-1\rangle[/itex]?

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

    Add them all together, multiply by [itex]\frac{w_0}{2\hbar}[/itex], and that gives you [itex]H |-1\rangle[/itex]
     
  15. Dec 4, 2014 #14

    stevendaryl

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    That doesn't happen.

    Let me just work out one example: [itex]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[/itex]
     
  16. Dec 4, 2014 #15
    OH! I see! Rhat's not what I was doing at all. Thanks a lot, I'll see where this leads me
     
  17. Dec 4, 2014 #16
    [itex]H \Psi = \left( \begin{array}& 0 & \sqrt{2} & 0 \\ -\sqrt{2}& 0 & \sqrt{2} \\ 0 & \sqrt{2} & 0 \end{array} \right) \Psi[/itex]

    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...
     
  18. Dec 4, 2014 #17

    stevendaryl

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

    Now your problem has been reduced to solving an eigenvalue-eigenvector problem involving matrices. Have you studied that?
     
  19. Dec 4, 2014 #18
    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!
     
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