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Angular Mumentum: Question for my exam tomorrow

  1. Jul 25, 2011 #1
    Hello!
    My exam is tomorrow... Help will be much appreciated...

    Say l=0 is given. My hemiltonian is [itex] \bar {H}=\frac {1} {\hbar} \Omega (\vec {n} \cdot \vec {L})^2 [/itex]

    And we use "base states" that are [itex] L_z [/itex] eigenstates, |1 m>
    n is a general direction unit vector.

    Now, I am looking for all the possible energies I can measure.

    What I thought was the answer is - sincs [itex] L^2=l(l+1)\hbar^2 |lm> [/itex]

    There is only one option- since l=1. But the solution shows a different thing:
    They refer to the possible m's: since l=1, m=-1,0,1

    So they actually use [itex] L_z^2 [/itex] instead of [itex] L^2 [/itex]
    Why is that true?

    In the next section of the question, the vector [itex] \vec {n}= \frac {1} {\sqrt 2} (1,1,0) [/itex] is given.
    Now, we put the system(for t=0) in such a way so that [itex] |\psi>= |1,-1> [/itex] and we are looking for the wave function for any time t.

    So now is what I don't understand: in the first section, they used [itex] L_z^2 [/itex] in the hemiltonian, instead of [itex] L^2 [/itex] .
    But now, using the given n, in the hemiltonian, since n does not have a z component, the Hemiltinian should be zero!

    What an I not getting right?
    Thank you so much!!
     
  2. jcsd
  3. Jul 25, 2011 #2

    Fredrik

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    In the first part of the problem, I would guess that they're just choosing the coordinate system so that z axis coincides with [itex]\vec n[/itex]. Then [itex]\vec n\cdot L=L_z[/itex].

    In the second part, with the given [itex]\vec n[/itex], your Hamiltonian is a constant (I assume that [itex]\Omega[/itex] is a number) times [itex]L_x^2+L_y^2=\vec L^2-L_z^2[/itex].

    Edit: BruceW is right. I got the second part wrong. The Hamiltonian is a constant times [itex](L_x+L_y)^2=L_x^2+L_y^2+L_xL_y+L_yL_x[/itex], i.e. what mathfeel said.
     
    Last edited: Jul 25, 2011
  4. Jul 25, 2011 #3
    Given the choice of coordinate direction is arbitrary, we can pick [itex]\hat{n} = \hat{z}[/itex]. In this case the Hamiltonian reduces to:
    [itex]H = \frac{\Omega}{\hbar} L_z^2[/itex]

    Now in this case, the problem specifically define [itex]\hat{n}[/itex], so H is now:
    [itex]H = \frac{\Omega}{2\hbar} (L_x + L_y)^2 = \frac{\Omega}{2\hbar} (L_x^2 + L_y^2 + L_x L_y + L_y L_x)[/itex]
    You are right, there is no more [itex]L_z[/itex] operator, but that simply says that [itex]|l,m\rangle[/itex] might no longer be the eigenstates of this Hamiltonian, it does not mean, for example that [itex]H | 1,1\rangle = 0[/itex].

    And the problem asks you to find [itex]\psi(t) = e^{itH/\hbar} |1,-1\rangle[/itex].
     
    Last edited: Jul 25, 2011
  5. Jul 25, 2011 #4
    Thank you! but if one can choose the any direction, why is it wrong to look at it generally, assuming [itex] L^2= l(l+1)\hbar ^2 [/itex] Why choose if I can loot at the general case?

    In the second part:
    I still have [itex] \hat {H} =\frac {\Omega} {\hbar} (\vec {n} \vec {L} )^2 [/itex]
    So, I still have L- then why can it be that |lm> will no longer be eigenstates of H ?

    Thank you!
     
  6. Jul 25, 2011 #5

    BruceW

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    Fredrik has got the second part of the question right. The hamiltonian still commutes with [itex] L^2 [/itex] and [itex] {L_z}^2 [/itex], and since the system starts off in an eigenstate of both of these, the system will stay in the same eigenstate (since it is in an energy eigenstate). (So the state doesn't change with time).

    EDIT: I got this totally wrong.
     
    Last edited: Jul 25, 2011
  7. Jul 25, 2011 #6

    BruceW

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    Woops, no. Me and Fredrik got the second part wrong. mathfeel got it right.
     
  8. Jul 25, 2011 #7

    BruceW

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    The Hamiltonian is the component of angular momentum in the direction of n, not the total angular momentum.
     
  9. Jul 25, 2011 #8

    BruceW

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    n is now given, so the hamiltonian is represented by a component of angular momentum that is not in the z-direction. Therefore, the hamiltonian no longer commutes with the z-component. Therefore, |lm> cannot be an eigenstate of H.
     
  10. Jul 25, 2011 #9
    Thank you! I understand now...

    But why can't I assume (in the second part) that |l m> are eigenstates of the Hamiltonian?
     
  11. Jul 25, 2011 #10
    Ok... thank you!
     
  12. Jul 25, 2011 #11
    Oh boy, and you have an exam tomorrow?

    [itex]L_z^2[/itex] and [itex]L^2[/itex] are two different operator regardless of what you choose for [itex]n[/itex]. The Hamiltonian in this coordinate is [itex]H \propto L_z^2[/itex] does NOT mean a states has ZERO [itex]L_x[/itex] or [itex]L_y[/itex]. In fact, since:
    [itex]L_x^2 + L_y^2 |l,m\rangle = L^2 - L_z^2 |l,m\rangle = \left( l(l+1) - m^2 \right) \hbar^2 |l,m\rangle[/itex]
    is always positive regardless where you point you z axis: even if you point the angular momentum vector completely in the z-direction, there is still some magnitude in the x and y direction. This is unlike classical mechanics.
     
  13. Jul 25, 2011 #12
    Yeah, my exam is tomorrow...

    The problem is, I am having a really hard time struggling with the English language's logic... My logic is taken from another language... Hebrew. despite the fact I understand every word alone- it is very hard to understand the whole thing. That is why, I guess, it seems to you I don't understand a word I'm talking about...
    Thank you for your help...
     
  14. Jul 25, 2011 #13
    mathfeel- I know these things. I just didn't understand why it is wrong to use just L^2.

    Thank you.
     
  15. Jul 25, 2011 #14

    Fredrik

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    Because the operator in the Hamiltonian isn't [itex]\vec L^2[/itex], it's [itex](\vec n\cdot\vec L)^2[/itex].

    If you really meant "why is it wrong to use another coordinate system?", the answer is that it's not. You can use any coordinate system you want to. It's just smarter to choose the one that simplifies the calculations.
     
  16. Jul 25, 2011 #15
    Yes... that was the question... and I understood it, with your help... Thank you again!
     
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