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Trace of the spin matrix of spin-1

  1. Feb 20, 2014 #1
    Spin-1 matrix Sx, Sy, Sz are traceless 3*3 matrix, and have the property ##[S_i, S_j] = i\epsilon_{ijk}S_k##, and we know that ##Tr(S_i^2) = 1^2+0^2+(-1)^2=2##.
    All of the above are independent of representation, of course, the trace of a matrix is representation-independent.

    so, if we want to know ##Tr(S_xS_z)=?##, we can use ##Tr(S_yS_z^2)=Tr[(S_zS_y+iS_x)S_z]=Tr(S_yS_z^2)+iTr(S_xS_z)##, thus ##Tr(S_xS_z)=0##.

    And my question is, what about ##Tr(S_xS_yS_xS_y)=?## Using the similar method mentioned above.

    Regards!
     
  2. jcsd
  3. Feb 21, 2014 #2

    samalkhaiat

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    Science Advisor

    What is this? Why should the sum (of squares of the eigenvales of [itex]S_{ z }[/itex]) gives the trace of [itex]S_{ i }^{ 2 }[/itex]? For spin one matrices, you have
    [tex]
    \sum_{ 1 }^{ 3 } S_{ i }^{ 2 } = \left( \ \begin{array} {rrr} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end {array} \right) .
    [/tex]
    So, the trace of this is 6 not 2.

    No, who told you this? The trace depends on the dimension of the representation. In any representation [itex]\rho[/itex], the Casmir matrix is given by
    [tex]
    \sum_{ i = 1 }^{ 3 } J_{ i }^{ 2 } = j ( j + 1 ) \mbox{ I }_{ \rho ( j ) \times \rho ( j )} ,
    [/tex]
    where [itex]\rho ( j ) = 2 j + 1[/itex] is the dimension of the representation and I is the identity matrix in that representation. So, clearly
    [tex]
    \mbox{ Tr } \left( \sum_{ i = 1 }^{ 3 } J_{ i }^{ 2 } \right) = j ( j + 1 ) ( 2 j + 1 ) .
    [/tex]
    So, for spin-1/2 the trace is [itex]( 1/2 ) ( 3/2 ) ( 2 ) = ( 3/2 )[/itex]. And, for spin one you get [itex]( 1 ) ( 2 ) ( 3 ) = 6[/itex].

    Why do you need to do all this? The spin-1 matrices are simple enough to determine the trace of products of any number of them. However, you can use the identity
    [tex]
    ( S_{ x } S_{ y } )^{ 2 } = S_{ x }^{ 2 } S_{ y }^{ 2 } - i S_{ x } S_{ z } S_{ y }
    [/tex]
    Now, take the trace and use the fact that
    [tex]\mbox{ Tr } ( S_{ x }^{ 2 } S_{ y }^{ 2 } ) = 1 , \ \mbox{ and } \ \ \mbox{ Tr } ( S_{ x } S_{ z } S_{ y } ) = - i
    [/tex]

    Sam
     
  4. Feb 21, 2014 #3

    Maybe my notations could cause ambiguity, for ##Tr(S_i^2)##, I didn't mean to sum from 1 to 3, ##i## stands for ##x## or ##y## or ##z##. And we can go to the representation which makes ##S_i## diagonal, and the diagonal elements are the eigenvalues, i.e. 0, 1, -1, so I got ##Tr(S_i^2)=2##.

    I know that we can choose a specific representation(usually Sz diagonal), and do the matrix product explicitly, then get the traces. But I want to do it in another way, using the properties of spin operators like the commutation relations. I think this is a good homework. ^_^
    I had proved that ##Tr(S_xS_zS_y)=-i##, but got some trouble for ##Tr(S_x^2S_y^2)=1##. Is there any chance you could possibly show me the details of ##Tr(S_x^2S_y^2)=1##?

    Thanks a lot!
     
  5. Feb 22, 2014 #4

    Avodyne

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    Science Advisor

    [tex]S_x^2+S_y^2+S_z^2=2I[/tex][tex]S_x^2+S_y^2=2I-S_z^2[/tex][tex](S_x^2+S_y^2)^2=(2I-S_z^2)^2[/tex][tex]S_x^4+S_y^4+S_x^2S_y^2+S_y^2S_x^2=4I-4S_z^2+S_z^4[/tex]Take the trace of both sides, and use
    [tex]{\rm Tr}\,S_i^2={\rm Tr}\,S_i^4=2[/tex][tex]{\rm Tr}\,I=3[/tex]
     
  6. Feb 22, 2014 #5
    Wow! That's amazing, thank you so much!
     
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