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Matrix representation of spin

  1. Jan 26, 2013 #1

    Please check the attachment and explain why are they arranged (the matrix entries) the way they are?
    I mean, what is the rule for building up a matrix?
    In other words, why did not we start with <-1/2, -1/2 l S^2 l -1/2, -1/2 > and placed it as being the first entry?


    Attached Files:

  2. jcsd
  3. Jan 26, 2013 #2
    it is represented as
    Sxx Sxy
    Syx Syy
    Sx will denote |++> and Sy as |+->
  4. Jan 26, 2013 #3
    I see thanks for your reply, but my question was why? What's the rule for matrix representation? Why isn't it
  5. Jan 26, 2013 #4


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    For each linear transformation ##A:U\to V##, and each pair of ordered bases (one for U, one for V), there's a matrix [A] that corresponds to A in the following way: Let the number on row i, column j of [A] be ##(Au_j)_i##. Here ##u_j## is the jth member of the given ordered basis for U, and ##(Au_j)_i## is the ith component of the vector ##Au_j##, in the given ordered basis for V.

    If you change the order of the basis vectors, you will also change the order of the rows or columns of the matrix. Suppose e.g. that you swap the 3rd and the 5th member of the ordered basis for U. This swaps the 3rd and the 5th column of [A].

    If U=V, then it's convenient to choose the two ordered bases to be the same. Then we can talk about the matrix of A with respect to one ordered basis, instead of two. You seem to be dealing with a vector space that's spanned by the basis
    $$\left\{\left|\frac 1 2,\frac 1 2\right\rangle,\left|\frac 1 2,-\frac 1 2\right\rangle\right\}.$$ Since there are exactly two ordered bases that consist of these two vectors, there are exactly two ways to arrange the rows and columns of the matrix. You can choose to use the other one, but then you have to make the corresponding change in the 2×1 matrices that represent members of the vector space.
    Last edited: Jan 26, 2013
  6. Jan 26, 2013 #5
    Oh, okay thanks!!
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