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Spin Matrices for Multiple Particles

  1. Jul 16, 2015 #1
    I have two questions, but the second is only worth asking if the answer to the first is yes:
    Are the spin matrices for three particles, with the same spin,
    σ ⊗ II,
    I ⊗ σ ⊗ I
    and
    II ⊗ σ
    for particles 1, 2 and 3 respectively, where σ is the spin matrix for a single one of the particles?

    I know that the spin matrices for two particles are
    σ ⊗ I and
    I ⊗ σ
    for particles 1 and 2 respectively, so I am guessing that the same is the case for when there are more than two particles.
     
  2. jcsd
  3. Jul 16, 2015 #2

    blue_leaf77

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    Yes, it is.
     
  4. Jul 16, 2015 #3
    Nice - thanks for the quick answer ;)

    What I was wanting to ask was whether the following is a way to generalise the spin matrices for each of N particles:

    upload_2015-7-13_18-32-25-png.85924.png

    Where w is the direction, a is the matrix/particle number, N is the number of particles and dim(V) = (2s + 1)^N, where s is the spin of each particle (1/2 for spin 1/2, 1 for spin 1 etc.).
     
  5. Jul 16, 2015 #4

    blue_leaf77

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    You have to check again what the notations used in that expression mean. For instance, does the first ##I_{dim(V)}## means that it is actually a series of (Kronecker) product of ##a-1## individual identity matrices? If yes, then ##dim(V) = 2s+1##, that is, the dimension of the spin space corresponding to a single particle.
    I think that is indeed the case there since the dimensions of the first and second ##I## are denoted identically.
     
  6. Jul 16, 2015 #5
    Ah yes, I made a mistake there - yeh, the subscript of the identity matrices should be just 2s+1, and the a-1 and N-a under the brackets are the number of Kronecker/tensor products to have - for example, for 8 spin 1 particles, the 3rd matrix would be:

    upload_2015-7-17_0-13-13.png

    ...which is a rather large matrix...
     
  7. Jul 19, 2015 #6
    There is a problem though:

    If I add together the three spin matrices (z-direction) for 3 spin 1/2 particles, I get:
    upload_2015-7-19_13-11-35.png

    But surely it should be:
    upload_2015-7-19_13-12-31.png
     
  8. Jul 19, 2015 #7

    blue_leaf77

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    It doesn't really matter as long as you remember, which eigenvector (when you have to calculate it) belongs to which eigenvalue. For example the eigenvector ##(0,0,0,1,0,0,0,0)^T## belongs to eigenvalue ##-\hbar## in the upper matrix, but belongs to ##\hbar## in the lower one.
     
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