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QM: Angular Momentum Matrices (Rotating Molecule)

  1. Mar 1, 2010 #1
    1. The problem statement, all variables and given/known data

    For [tex]l=1[/tex] the angular momentum components can be represented by the matrices:

    [tex]
    \hat{L_{x}} = \hbar \left[ \begin{array}{ccc} 0 & \sqrt{\frac{1}{2}} & 0 \\ \sqrt{\frac{1}{2}} & 0 & \sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & 0 \end{array} \right]
    [/tex]

    [tex]
    \hat{L_{y}} = \hbar \left[ \begin{array}{ccc} 0 & -i\sqrt{\frac{1}{2}} & 0 \\ i\sqrt{\frac{1}{2}} & 0 & -i\sqrt{\frac{1}{2}} \\ 0 & i\sqrt{\frac{1}{2}} & 0 \end{array} \right]
    [/tex]

    [tex]
    \hat{L_{z}} = \hbar \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right]
    [/tex]

    Q(a). Confirm that the matrices fulfill the commutation relations of angular momentum.

    Q(b) Calculate the matrix which represents the Hamiltonian:

    [tex]\hat{H} = \frac{1}{2I}\hat^{L}^{2} + \alpha \hat{L}_{z}[/tex]

    of a rotating molecule, where [itex]I[/itex] and [itex]\alpha[/itex] are constants and:

    [tex]\hat{L}^{2} = \hat{L}_{x}^{2} + \hat{L}_{y}^{2} + \hat{L}_{z}^{2}[/tex]

    Q(c) Calculate the energy levels of the molecule.

    2. Relevant equations

    Commutation Relations of angular momentum:

    [tex] \hat{L_{x}},\hat{L_{y}} = i\hbar \hat{L_{z}}[/tex]

    [tex] \hat{L_{y}},\hat{L_{z}} = i\hbar \hat{L_{x}}[/tex]

    [tex] \hat{L_{z}},\hat{L_{x}} = i\hbar \hat{L_{y}}[/tex]

    Commutator Definition:

    [tex]\hat{A},\hat{B} = \hat{A}\hat{B} - \hat{B}\hat{A}[/tex]

    Rest as relevant within the question statement (and subsequent answers)

    3. The attempt at a solution

    I have seen examples using Pauli matrices that are 2x2 but I don’t know how to go about this using these 3x3 matrices, i.e. how to adapt to these matrices from the standard Pauli ones.

    A bit of help and advice to get me going in the right direction would be great, then I think I should hopefully be OK. :wink:
     
  2. jcsd
  3. Mar 1, 2010 #2

    kuruman

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    Can you multiply two 3x3 matrices together? If yes, then take the matrix product LxLy, subtract from it the matrix product LyLx and see if result is equal to i*(hbar)*Lz.

    Repeat as necessary for the rest of the commutation relations.
     
  4. Mar 1, 2010 #3
    .. it was more help with the latter questions, I know now how to do the first part to show the commutation relations.
     
  5. Mar 1, 2010 #4

    kuruman

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    This is vague and not very helpful. Can you do part (b) and write the Hamiltonian as a 3x3 matrix? If not, exactly what can you not do in part (b)? How about part (c)? What troubles you there?
     
  6. Mar 2, 2010 #5
    Sorry, I do see that obviously my last post wasn't very helpful or informative of problems.

    Right, so:

    For the first question (proving the commutation relations):

    I have managed to do this for the first two cases without any problems, but I just can't get it to work for the last case, so I must be overlooking something but I can't see what even after checking calculations. This is what I have:

    [tex]\left[\hat{L_{z}},\hat{L_{x}}\right] = -\hbar \left[ \begin{array}{ccc} 0 & -\sqrt{\frac{1}{2}} & 0 \\ \sqrt{\frac{1}{2}} & 0 & -\sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & 0 \end{array} \right][/tex]

    So I don't see where the [itex]i[/itex] terms come from!?!

    For the second question (with the hamiltonian):

    I have that:

    [tex]\left(\hat{L_{x}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} 0 & {\frac{1}{2}} & 0 \\ \frac{1}{2}} & 0 & \frac{1}{2}} \\ 0 & {\frac{1}{2}} & 0 \end{array} \right][/tex]

    [tex]\left(\hat{L_{y}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} 0 & \frac{1}{2}} & 0 \\ -\frac{1}{2}} & 0 & \frac{1}{2}} \\ 0 & -\frac{1}{2}} & 0 \end{array} \right][/tex]

    [tex]\left(\hat{L_{z}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right][/tex]

    So then:

    [tex]\hat{L}^{2} = \hat{L}_{x}^{2} + \hat{L}_{y}^{2} + \hat{L}_{z}^{2} = 3\hbar^{2}\left(\left[ \begin{array}{ccc} 0 & {\frac{1}{2}} & 0 \\ \frac{1}{2}} & 0 & \frac{1}{2}} \\ 0 & {\frac{1}{2}} & 0 \end{array} \right] + \left[ \begin{array}{ccc} 0 & \frac{1}{2}} & 0 \\ -\frac{1}{2}} & 0 & \frac{1}{2}} \\ 0 & -\frac{1}{2}} & 0 \end{array} \right] + \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]\right) = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right][/tex]

    Which means:

    [tex]\hat{H} = \frac{1}{2I}\hat^{L}^{2} + \alpha \hat{L_{z}} = \frac{1}{2I}\left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right] + \alpha \hbar \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right][/tex]

    .. and then I assume I need to add the matrices together, and combine the constants, which gives:

    [tex]\hat{H} = \left(\frac{1}{2I} + \alpha \hbar\right) \left[ \begin{array}{ccc} 2 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right][/tex]

    .. is this correct?! Not sure if need to do anything else.

    For the thirds question (energy levels):

    .. I don't know how to do this.
     
  7. Mar 2, 2010 #6

    kuruman

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    The matrix on the right is related to Ly. How?

    The rest of your work shows that you are confused about how the square of a matrix is to be calculated. For example, if you want to find Lx2, you don't just square every single matrix element and call that the square of the matrix. That works only if the matrix is diagonal. e.g. with Lz. To do it right, you have to take the 3x3 matrix product LxLx.
     
  8. Mar 2, 2010 #7
    .. oh wait, I do actually see where the ' i ''s come from, since the two matrices are just different by a multiplication of i which has been factored outside of the matrix, if that makes sense. My question therefore should rather be, what's going on with the other negative sign that shouldn't be there?!

    Ok. I've looked up how to take the matrix product but I don't really understand how to do it still.. any tips?
     
  9. Mar 2, 2010 #8

    kuruman

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    Calculate i*hbar*Ly. This is what you want on the right side of the equation. How does it compare with what you already got?

    How did you calculate the matrix products such as LxLy that you needed for the commutators? Do the same thing except that you need to replace Ly with Lx to calculate LxLx and likewise for the other squares.
     
  10. Mar 2, 2010 #9
    i.

    It differs in that what I got has a factor of [itex]\hbar[/itex] opposed to the [itex]\hbar^{2}[/itex] that I need.. oh wait, no, my correct calculation should be:

    [tex]\left[\hat{L_{z}},\hat{L_{x}}\right] = -\hbar^{2} \left[ \begin{array}{ccc} 0 & -\sqrt{\frac{1}{2}} & 0 \\ \sqrt{\frac{1}{2}} & 0 & -\sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & 0 \end{array} \right][/tex]

    .. so I think it's all good now, just have to double check it through though. :wink:

    ii.

    Oh right OK I'll give that another go now then at calculating those..

    [tex]\left(\hat{L_{x}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} {\frac{1}{2}} & 0 & {\frac{1}{2}} \\ 0 & 1 & 0 \\ {\frac{1}{2}} & 0 & {\frac{1}{2}} \end{array} \right][/tex]

    [tex]\left(\hat{L_{y}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} {\frac{1}{2}} & 0 & -{\frac{1}{2}} \\ 0 & 1 & 0 \\ -{\frac{1}{2}} & 0 & {\frac{1}{2}} \end{array} \right][/tex]

    and obviously from previously:

    [tex]\left(\hat{L_{z}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right][/tex]

    So calculating the Hamiltonian expression again:

    [tex]\hat{L}^{2} = \hat{L}_{x}^{2} + \hat{L}_{y}^{2} + \hat{L}_{z}^{2} =

    3\hbar^{2}\left(\left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right]\right) =

    \frac{3\hbar^{2}}{2}\left(\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]\right)[/tex]

    and that matrix is an identity matrix, so can denote it just as I, therefore:

    [tex]\hat{L}^{2} = \frac{3\hbar^{2}I}{2}[/tex]

    hence:

    [tex]\hat{H} = \frac{1}{2I} \hat^{L^{2}} + \alpha \hat{L_{z}} = \frac{1}{2I}\left(\frac{3 \hbar^{2}I}{2}\right) + \alpha \hat{L_{z}} = \frac{3\hbar^{2}}{4} + \alpha \hbar \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right][/tex]

    .. better?!
     
  11. Mar 2, 2010 #10

    kuruman

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    Better, but in need of improvement.

    Where did that factor of 3 come from? If you first put the (hbar)2 inside each matrix for Lx2, Ly2 and Lx2, then add the three matrices, what do you get?

    Also, when you finally construct the Hamiltonian, it should be a 3x3 matrix because it is the sum of two 3x3 matrices. What you have written down is a constant added to a 3x3 matrix. L2 that is part of the Hamiltonian is a 3x3 matrix the product of a constant times the 3x3 identity matrix.

    When the times comes for you to assemble the Hamiltonian, I suggest that you bring all constants inside the matrices, then add everything together.
     
  12. Mar 2, 2010 #11
    [tex]\left(\hat{L_{x}}\right)^{2} = \left[ \begin{array}{ccc} {\frac{\hbar^{2}}{2}} & 0 & {\frac{\hbar^{2}}{2}} \\ 0 & \hbar^{2} & 0 \\ {\frac{\hbar^{2}}{2}} & 0 & {\frac{\hbar^{2}}{2}} \end{array} \right][/tex]

    [tex]\left(\hat{L_{y}}\right)^{2} = \left[ \begin{array}{ccc} {\frac{\hbar^{2}}{2}} & 0 & -{\frac{\hbar^{2}}{2}} \\ 0 & \hbar^{2} & 0 \\ -{\frac{\hbar^{2}}{2}} & 0 & {\frac{\hbar^{2}}{2}} \end{array} \right][/tex]

    [tex]\left(\hat{L_{z}}\right)^{2} = \left[ \begin{array}{ccc} \hbar^{2} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \hbar^{2} \end{array} \right][/tex]

    Hence:

    [tex]\left(\hat{L}\right)^{2} = \left[ \begin{array}{ccc} 2\hbar^{2} & 0 & 0\\ 0 & 2\hbar^{2} & 0 \\ 0 & 0 & 2\hbar^{2} \end{array} \right] [/tex]

    Therefore:

    [tex]\frac{1}{2I}\left(\hat{L}\right)^{2} = \left[ \begin{array}{ccc} \frac{\hbar^{2}}{I} & 0 & 0\\ 0 & \frac{\hbar^{2}}{I} & 0 \\ 0 & 0 & \frac{\hbar^{2}}{I} \end{array} \right][/tex]

    Also:

    [tex]\alpha \hat{L_{z}} = \alpha \hbar \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right] = \left[ \begin{array}{ccc} \alpha \hbar & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -\alpha \hbar \end{array} \right][/tex]

    Then the result:

    [tex]\hat{H} =\left[ \begin{array}{ccc} \left(\frac{\hbar^{2}}{I} + \alpha \hbar\right) & 0 & 0 \\ 0 & \left(\frac{\hbar^{2}}{I}\right) & 0 \\ 0 & 0 & \left(\frac{\hbar^{2}}{I} - \alpha \hbar \right)\end{array} \right][/tex]
     
  13. Mar 2, 2010 #12

    kuruman

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    Bravo. You have successfully completed the first two parts. :approve:

    Now for part (c). Note that the matrix representing the Hamiltonian is diagonal. What does that mean?
     
  14. Mar 2, 2010 #13
    .. finally got there on 3rd attempt! :approve:

    Um, well are the energies encoded within the Hamiltonian matrix, i.e:

    [tex]E_{1} = \left(\frac{\hbar^{2}}{I} + \alpha \hbar\right) , E_{2} = \left(\frac{\hbar^{2}}{I}\right) , E_{3} = \left(\frac{\hbar^{2}}{I} - \alpha \hbar \right)[/tex]

    or something like that?!?
     
  15. Mar 2, 2010 #14

    kuruman

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    Correct. Since the matrix is already diagonal, you can pick out the energies as you have done.

    For future reference: If the matrix is not diagonal, then you will have to diagonalize it first, in which case the resulting eigenvalues will be the energies that you are seeking.
     
  16. Mar 2, 2010 #15
    Brill! and thanks for letting me know that.
     
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