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How to construct angular momentum matrix

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data
    We just started matrix representation of operators. For the life of me I can't figure out how to construct any of the matrices. For example, the L+ matrix for l=1 (which gives m=-1,0,1). I plug in the 3 values of m into the L+ equation and get sqrt(2), sqrt(2), and 0. Then I have no idea what the next step is to get the matrix.. how do I know where in the matrix to put these values?


    2. Relevant equations
    (It's in the photo.)


    3. The attempt at a solution
    https://docs.google.com/file/d/0B0ByXWZYouOBUTExbmZ4d2lkemM/edit?usp=sharing
    attachment.php?attachmentid=57888&stc=1&d=1366046261.jpg

    I have the L+ values for when m=-1, 0, and 1. I just don't know how I'm supposed to know where to put them in a matrix. At the bottom left of the photo, my friend told me that's the next step, to have the -1,0,1 on the top for the rows, and on the left side for the columns, then I put them together like, for example, the first spot would be <-1|L+|-1> but how do I calculate that, what does <-1|L+|-1> mean? I tried plugging in m=-1 and then subtracting/adding and all kinds of experimental stuff but nothing gave me what I wanted.

    I've spent way too long on this, it's getting frustrating, there must be something obvious that just isn't clicking. I would really, really, really, really appreciate it if someone could help me out. Thanks!
     

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    Last edited by a moderator: Apr 15, 2013
  2. jcsd
  3. Apr 15, 2013 #2

    DrClaude

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    Staff: Mentor

    That is correct. Each element in the matrix corresponds to a pair ##m,m'##.

    $$
    \langle l'=1, m'=-1 | L_+ | l=1, m=-1 \rangle
    $$
    which you can calculate using the formula you have. You understand what the ##\delta_{m', m\pm1}## means?

    You have the formula. Can you be more specific as to where you think you go wrong with it?
     
  4. Apr 15, 2013 #3
    Hey, thanks for responding! Ok, here's my hang-up:
    I did the same thing you said, plug in m' = -1 and m=-1 into the L+ equation. That gives me:
    L+ = sqrt[l(l+1)-m'(m+1)] = sqrt[1(1+1)-(-1)(-1+1)] = sqrt(2-0) = sqrt(2).
    So this tells me that the very first spot of the matrix, the m'=-1, m=-1 spot, should be sqrt(2). But it's not, it's 0. That is where I'm hung up.

    Also, which one is m' and which one is m? Obviously for the first one they're both -1 but then, for example, the second column of the first row... my 2 m numbers are 0 and -1, which one is m' and which one is m?

    Thanks!
     
  5. Apr 15, 2013 #4

    DrClaude

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    You forgot the Kronecker delta ##\delta_{m',m\pm1}##.

    Does it matter?

    At one point, it becomes a question of convention. But if you use ##\langle l,m' | L_\pm | l,m \rangle##, that is, with the ##m## in the ket and the ##m'## in the bra, then when you write a column vector with the values ##(m=-1, m=0, m=1)^T##, then the columns of the matrix correspond to ##m## and the rows to the ##m'##. (Think of what happens when you do matrix-vector multiplication. Each row multiplies the column vector, so the elements of one row correspond to the values of ##m## in the column vector.)
     
  6. Apr 15, 2013 #5

    vela

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    The usual convention is to order the basis vectors such that ##\lvert 1 \rangle## is the first basis vector and ##\lvert -1 \rangle## is the last. You have them in the opposite order.

    You're also being a bit sloppy with your notation. You shouldn't write ##\hat{L}_+ = \sqrt{l(l+1)-m'(m+1)}##. That doesn't really make sense. You should say
    $$\hat{L}_+ \lvert l, m \rangle = \sqrt{l(l+1)-m(m+1)}\ \lvert l, m+1 \rangle,$$ which tells you the effect ##\hat{L}_+## has on an eigenstates of ##\hat{L}_z##. Also, note that it's just ##m## inside the square root, not ##m## and ##m'##.
     
  7. Apr 15, 2013 #6
    Sorry about being sloppy, I just wanted to get it all quickly posted. Ok here's what I'm doing:

    For [itex]\ell = 1[/itex] and m=-1, that means <-1|L+|-1> = <[itex]\ell[/itex][itex]^{'}=1,m^{'}=-1|L_{+}|\ell=1,m=-1> [/itex]

    So what I understand from that is that I plug in m=-1 into the original L+ equation, but I don't understand what the prime is on l and m. What is m'? Anyway, so I plug in m=-1 and get...
    [itex]\sqrt{1(1+1)-(-1)(-1+1)}\delta=\sqrt{2}\delta[/itex]

    I haven't plugged m' (prime) anywhere tho since I don't know what that is. So I have this square root of 2 that should go in the (1,1) spot of the matrix, but the correct answer there is 0.

    And the delta there has a subscript of m', m+1 so for my m=-1 case, it means m',-1+1 = m',0 and I have no idea what that means. What is the meaning of delta here?

    So [itex]L_{+} |l,m> = \sqrt{l(l+1)-m(m+1)} |l,m+1>[/itex]
    [itex]L_{+} |1,-1> = \sqrt{2} |1,0>[/itex] ---> What does that tell me?
     
    Last edited: Apr 15, 2013
  8. Apr 16, 2013 #7

    DrClaude

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    They're just labels. You have two states that are characterized by two quantum numbers, ##l## and ##m##. You could label those two states with ##l_1,m_1## and ##l_2,m_2## or, as it is done here, with ##l,m## and ##l',m'##. The unprimed numbers correspond to the state in the ket, and the primed to the state in the bra.

    You haven't looked up the Kronecker delta, have you? ##\delta_{ij}=1## iff ##i=j##, otherwise, ##\delta_{ij}=0##. In the equation, you have ##\delta_{m',m\pm1}##, so if you put ##m=m'=1##, then ##m' \neq m \pm 1##, so the Kronecker delta is equal to 0.

    Because of the Kronecker delta, there is a strict relation between the possible values of ##m## and ##m'##, such that you can write the equation in terms of ##m## only.

    And the delta there has a subscript of m', m+1 so for my m=-1 case, it means m',-1+1 = m',0 and I have no idea what that means. What is the meaning of delta here?

    That tells you that if you operate on the state ##|1,-1 \rangle## with the operator ##L_{+}##, the state changes to ##|1,0 \rangle## (but that operation does not conserve the norm).

    Putting the ##L_+## matrix together tells you how that operator couples together states with different values of ##m##.
     
  9. Apr 16, 2013 #8
    Ahhh so the delta is the missing piece! I took a quick read over you comment and everything makes great sense, thank you so much! I'm going to go practice constructing more matrices soon and hopefully all will go well then but I'll post back if I'm still hung up.

    Thanks so much to everyone who replied, I really appreciate it!
     
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