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How to find S(+) 3 x 3 matrix?

  1. Dec 14, 2008 #1
    For spin 1, the states are |1>, |-1> and |0>

    These are written as:

    |1> = column matrix[1 0 0]

    |0> = column matrix [0 1 0]

    |-1> = column matrix [0 0 1]

    I need to find the 3 x 3 matrices for S(+) and S(-) which operate on these kets to give the correct answers eg.

    S(-)|1> = sqrt(2)*h(bar)*|0>

    I had tried getting linear eqns from the following:

    [a b c] * [1] = h(bar)*sqrt(2)* [0]
    [d e f] [0] [1]
    [g h i] [0] [0]


    but i just get

    a + 0 + 0 = 0
    d + 0 + 0 = 1* h(bar) * sqrt(2)
    g + 0 + 0 = 0

    This doesn't look remotely useful... I know the lowering operator is:

    [ 0 0 0 ]
    [sqrt(2) 0 0 ]
    [ 0 sqrt(2) 0 ]

    But how do I get there?

    Cheers for any help

    Philip
     
  2. jcsd
  3. Dec 14, 2008 #2
    it took out my carefully placed spacing in the eqns, hope this is still legible....
     
  4. Dec 14, 2008 #3

    dextercioby

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    Apply the operator S_{-} both on |1>, |0> and |-1> and see what you get.
     
  5. Dec 14, 2008 #4
    But i'm meant to be working out S_{1}..

    I have checked the answer given and it works, by trying S_{-}|1> i get the right thing but i don't understand how you would work out S_{-} if you weren't given the answer?
     
  6. Dec 14, 2008 #5
    Your column matrices for the spin states only have one nonzero term. Let the lowering operator matrix have the general form:

    [tex]
    \textbf{S}_{-} = \begin{bmatrix} A & B & C \\ D & E & F \\ G & H & I \end{bmatrix}
    [/tex]

    Now,

    [tex] \textbf{S}_{-}\left|1\right> = [/tex] ?

    What is the question mark equal to in terms of the generalized matrix elements? Do the same for the other two spin-vectors, and it should be evident how you can construct the matrix.
     
  7. Dec 14, 2008 #6
    I think you get a 1 x 3 matrix with h(bar)*sqrt(2) in front of it, but only because I know those answers from a previous question I did.

    [tex]
    \textbf{S}_{-} = \begin{bmatrix} A \\ D \\ G \end{bmatrix}
    [/tex]


    The others, would be:

    [tex]
    \textbf{S}_{-} = \begin{bmatrix} B \\ E \\ H \end{bmatrix}
    [/tex]

    [tex]
    \textbf{S}_{-} = \begin{bmatrix} C \\ F \\ I \end{bmatrix}
    [/tex]

    Is this what you meant by a generalised matrix?
     
  8. Dec 14, 2008 #7
    argh, obviously ignore the S(-) i was using your code cause I don't know how to do it myself. The S(-)s are meant to be ?s
     
  9. Dec 14, 2008 #8
    ohhhhhhhhhhhhhhhhhhhhh, i just got it cheers.

    once you put the column vectors i just posted equal to the states you are expecting ie. B= 1 and H = 1

    cheers
     
  10. Dec 14, 2008 #9
    Exactly right. And what exactly are these generalized 1x3 matrices? What is the purpose of the lowering operator?

    EDIT: I think your coefficient values B=1 and H=1 may be off. You know what the matrix is supposed to be, so you can verify.

    Actually, to be more clear,

    [tex]

    \textbf{S}_{-}\left|1\right> = \begin{bmatrix} A \\ D \\ G \end{bmatrix} =\sqrt2\hbar \left|0\right>

    [/tex]
     
    Last edited: Dec 14, 2008
  11. Dec 14, 2008 #10
    sorry was being lazy by 1 i meant [tex] sqrt(2)\hbar [\tex]

    they are the eigenvalues?
     
  12. Dec 14, 2008 #11
    Be careful using the word "eigenvalue" here, since the lowering operator does not return an eigenvector (the same vector that it was acting on). I almost made the same mistake when thinking about how to explain it to you. Now, if you have to construct the Pauli Spin matrices, then these operators do return eigenvectors with the appropriate eigenvalue.
     
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