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Spin matrices for particle of spin 1

  1. Aug 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Construct the spin matrices ([tex]S_{x}[/tex], [tex]S_{y}[/tex], and [tex]S_{z}[/tex]) for a particle of spin 1. Determine the action of [tex]S_{z}[/tex], [tex]S_{+}[/tex], and [tex]S_{-}[/tex] on each of these
    states.

    2. Relevant equations
    [tex]S=\sqrt{1(1+1)}\hbar [/tex]
    m=-s,-s+1,...,s-1,s

    3. The attempt at a solution
    m=-1,0,0,0,0,0,0,0,-1
    [tex]S=\sqrt{2}\hbar [/tex]
    [tex]S_{z}=\hbar \[ \left( \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & -1 \end{array} \right)\] [/tex]

    So I understand where [tex]S_{z}[/tex] comes from... I know what the answers for the matrices are from http://en.wikipedia.org/wiki/Pauli_matrices but I don't know how to go about finding [tex]S_{x}[/tex] and [tex]S_{y}[/tex] even with the commutation laws etc... and I don't really understand what it means by 'the action of' in the second part despite numerous searchs on google....

    Any help is much appreciated, thankyou!
     
  2. jcsd
  3. Aug 15, 2007 #2

    George Jones

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    From where does [itex]S_z[/itex] come?

    Depending on how you found [itex]S_z[/itex], you might be able to use the same technique to find [itex]S_x[/itex] and [itex]S_y[/itex] first, or to find [itex]S_+[/itex] and [itex]S_-[/itex] first.

    Neither do I. What states?

    The eigenstates of [itex]S_z[/itex]? General states?

    Is this question from a course? From a book? If so which course and book?
     
  4. Aug 15, 2007 #3
    Thanks very much for your reply....

    But I'm very confused now... This exact question "Construct the spin matrices ([tex]S_{x}[/tex], [tex]S_{y}[/tex], and [tex]S_{z}[/tex]) for a particle of spin 1. Determine the action of [tex]S_{z}[/tex], [tex]S_{+}[/tex], and [tex]S_{-}[/tex] on each of these states." is from a quantum mechanics course. The relevant equations were equations I thought were relevant... and I thought that the equation listed for m gave the matrix [tex]S_{z}[/tex].

    How would I find eigenstates of [tex]S_{z}[/tex]? And what is the usual way for determining spin matrices? I can't find this info out anywhere on the internet and am getting more and more confused by the second...
     
  5. Aug 15, 2007 #4

    George Jones

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    Have you seen the general theory of angular momentum in your quantum course?

    Stuff like ([itex]\hbar = 1[/itex]):

    [tex]J_z \left| jm \right> = m \left| jm \right>;[/tex]

    [tex]J_+ \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m + 1 \right)} \left| j,m+1 \right>;[/tex]

    [tex]J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.[/tex]

    If so, you can find the matices by considering stuff like

    [tex]\left< jm \right| J_\pm \left| jm \right>[/tex]

    for [itex]j=1[/itex] and [itex]m = -1, 0, 1.[/itex]
     
    Last edited: Aug 15, 2007
  6. Aug 15, 2007 #5

    malawi_glenn

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    George Jones, you mean

    [tex]J_- \left| jm \right> = \sqrt{j \left( j + 1 \right) - m \left(m - 1 \right)} \left| j,m-1 \right>.[/tex]

    ?
     
  7. Aug 15, 2007 #6

    George Jones

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    Yes, thanks. I have corrected my previous post.
     
  8. Oct 6, 2008 #7
    This is an old thread, but people are bound to come back looking for these answers, so here's my 2 cents.
    For a given [tex]S[/tex], [tex]S_z[/tex] comes from demanding [tex]S_z|S,m> = m |S,m>[/tex] and is thus a diagonal (2S+1)x(2S+1) matrix with elements [tex]S, S-1,...,-S[/tex]
    (Note: I assume [tex]|S,S> = (1,0,...,0)[/tex] and [tex]|S,-S> = (0,0,...,1)[/tex] )
    The [tex]S_x[/tex] and [tex]S_y[/tex] matrices - and consequently [tex]S_+[/tex] and [tex]S_-[/tex] - come from rotations of Sz about the y and z axes, [tex]S_x = U_y(\pi/2).S_z.U^{\dagger}_y(\pi/2) \text{ and } S_y = U_z(\pi/2).U_y(\pi/2).S_z.\left(U_z(\pi/2).U_y(\pi/2)\right)^{\dagger}[/tex],
    where the Uy matrix is comprised by the elements ([tex]m, m' = S, S-1,..., -S[/tex] and [tex]\theta\in [0,\pi][/tex]):

    [tex]U^{y}_{m,m'}(\theta) = \sqrt{(S-m)! (S+m)! (S-m')! (S+m')!}\sum_{x=Max(0,m'-m)}^{Min(S+m',S-m)}\frac{(-1)^x cos(-\theta/2)^{2 S + m' - m - 2 x} sin(-\theta/2)^{2 x + m - m'}}{(S + m' - x)! (S - m - x)! x! (x + m - m')!}[/tex]

    and Uz is diagonal with ([tex]\phi\in [0,2 \pi][/tex]):

    [tex]U^z_{k,k}(\phi) = e^{i (k-1) \phi}, \text{ } k = 1,2,...,2 S+1[/tex]

    There is some ambiguity here however, since another Uz is quoted:

    [tex]U^z_{m,m}(\phi) = e^{-i m \phi}, \text{ } m = S,...,-S[/tex]

    In the former case, [tex]k[/tex] refer to matrix index (e.g. [tex]U_{1,1}[/tex] is the upper left element of the matrix), while in the latter case [tex]m[/tex] refer to magnetic number indexing (e.g. [tex]U_{S,S}[/tex] is the upper left element of the matrix). I use the former formula and it gives the expected results.
     
  9. Dec 4, 2008 #8

    turin

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    How do you get the rotation matrix about the y-axis when you don't know what the Sy generator is?
     
  10. Sep 1, 2009 #9
    Hello,
    I've to construct Ix and Iy for I=1.

    so, I can construct lowering and raising operator but how do you construct cartesian operator from this equation ?

    there are no definition for Ix and Iy by their action on eigenstate vector like Iz, I+ and I-....

    How can I do that easily ?
     
  11. Sep 1, 2009 #10
    By using the spinor representation.
    In essence you are using combinations of spin-1/2 to represent the behaviour of arbitrarily large spins. This way you can generate operators and wavefunctions of large spins starting from the known spin-1/2 matrices.

    This was shown originaly by Majorana in 1932.
    I have retrieved the info from W.Thompson's Angular Momentum book.
     
  12. Sep 1, 2009 #11

    kuruman

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    If you know what the 3x3 ladder operators look like, start with the definitions

    [tex]S_{+}=S_{x}+iS_{y}[/tex]

    [tex]S_{-}=S_{x}-iS_{y}[/tex]

    and write the cartesian matrices as linear combinations of the ladder operators.

    [tex]S_{x} = ...[/tex]

    [tex]S_{y} = ...[/tex]
     
  13. Sep 1, 2009 #12
    but this relation is available only for pauli matrix (2x2) and so for spin 1/2....

    please help me !!!! or give the matrix representation of Ix and Iy for I=1 if you know it...
     
  14. Sep 1, 2009 #13

    kuruman

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    The relations are good for any dimensionality.
     
  15. Sep 1, 2009 #14
    The equations I have posted solve your problem for any spin, integer or half-integer.
    Anyway, for spin = 1:

    [tex]
    S_x = \left(
    \begin{array}{ccc}
    0 & \frac{1}{\sqrt{2}} & 0 \\
    \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
    0 & \frac{1}{\sqrt{2}} & 0
    \end{array}
    \right)
    [/tex]

    [tex]
    S_y = \left(
    \begin{array}{ccc}
    0 & -\frac{i}{\sqrt{2}} & 0 \\
    \frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\
    0 & \frac{i}{\sqrt{2}} & 0
    \end{array}
    \right)
    [/tex]

    [tex]
    S_z = \left(
    \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & -1
    \end{array}
    \right)
    [/tex]
     
  16. Sep 1, 2009 #15
    so, why kaltsoplyn talk about a really hard way to determine Ix and Iy if it is so easy to construct its by this relations ?
     
  17. Sep 1, 2009 #16
    Synthesizing high order spins starting from spin-1/2's is hard, although the procedure is straightforward.

    The formulas I give are general that's why they look so complex. However, it's not very hard to produce any such matrix if you plug these formulas in a symbolic mathematical software package like, say, Mathematica or Maple.
     
  18. Sep 1, 2009 #17
    I don't undersand why you are this square root ? It is in reality 1/2 and not [tex] \frac{1}{\sqrt{2}}[/tex]
     
  19. Sep 1, 2009 #18
    [tex]
    S_x = \frac{1}{2} \left(\left(
    \begin{array}{ccc}
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    0 & 0 & 0
    \end{array}
    \right) + \left(
    \begin{array}{ccc}
    0 & 0 & 0 \\
    1 & 0 & 0 \\
    0 & 1 & 0
    \end{array}
    \right)\right)
    [/tex]

    [tex]
    S_y = \frac{i}{2} \left(\left(
    \begin{array}{ccc}
    0 & 0 & 0 \\
    1 & 0 & 0 \\
    0 & 1 & 0
    \end{array}
    \right) - \left(
    \begin{array}{ccc}
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    0 & 0 & 0
    \end{array}
    \right)\right)
    [/tex]

    this is what I find with your relations.
     
  20. Sep 1, 2009 #19

    Strange, you should get the square root and I don't see anything wrong in my eqs.
    Besides your results should check against:

    [tex]
    [S_i,S_j]=i \hbar\epsilon_{i,j,k} S_k
    [/tex]
     
  21. Sep 1, 2009 #20

    kuruman

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    [tex]
    S_+ \left| s m \right> = \sqrt{s \left( s + 1 \right) - m \left(m + 1 \right)} \left| s,m+1 \right>;
    [/tex]

    [tex]
    S_+ \left| 10 \right> = \sqrt{1* \left( 1 + 1 \right) - 0* \left(0 + 1 \right)} \left| 1,1 \right> = \sqrt{2}\left| 1,1 \right>;
    [/tex]

    That's where the square root comes from. When you multiply by 1/2, you get it in the denominator.
     
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