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Dot product of two pauli matrices

  1. Apr 21, 2009 #1

    KFC

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    In some text, I read something like this

    [tex]\vec{S}_i\cdot\vec{S}_j[/tex]

    where [tex]\vec{S}_i[/tex] and [tex]\vec{S}_j[/tex] are "vectors" with each components be the pauli matrices [tex]S_x, S_y, S_z[/tex] individularly. My question is: if all components of this kind of vector are a 3x3 matrix, so how do you carry out the dot product? So the dot product will give another matrix instead of a number?
     
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  3. Apr 21, 2009 #2

    Matterwave

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    "Vectors" are not matrices. Matrices are tensors of rank 2, while vectors are tensors of rank 1. I don't see how you can have "vectors" which are Pauli-Matrices.

    Matrix multiplication (2 matrices multiplied together, not a matrix and a vector) has specific rules and will result in another matrix with the same number of columns and rows.
     
  4. Apr 21, 2009 #3

    KFC

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    In the text of linear space, it read: matrices can be a special kind of "vectors" (that's why we add qutoation mark). In abstract linear vector space, anything satisfying the RULES of linear space can form a vector. And in many QM text, the author like to form a "vector" with three pauli matrices Sx, Sy, Sz. What I am really confuse is how to carry out the dot product b/w such "vectors"?
     
  5. Apr 21, 2009 #4

    Matterwave

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    Hmmm, I don't know if the dot product is the same as matrix multiplication (I'm not sure how the inner product is defined in a vector space that includes matrices), but if it is then:

    http://en.wikipedia.org/wiki/Matrix_multiplication

    That explains it pretty well.

    Oh, I was thinking of only square matrices before, if the matrix is not square then the resulting matrix won't have the same dimensions as before. Sorry, it's been a while since I've really had to do Matrix multiplication. :P
     
  6. Apr 22, 2009 #5
    A logical way to define a dot product when using pauli matrixes as basis vectors would be to use the anticommutator

    [tex]
    a \cdot b = \frac{1}{2} \{ a, b \} = \frac{1}{2} (a b + b a )
    [/tex]

    EDIT: latex in PF doesn't appear to be working right now. That was:
    a \cdot b = \frac{1}{2} \{ a, b \} = \frac{1}{2} (a b + b a )


    Such a dot product won't be in the span of the pauli matrixes themselves, but will be your typical vector dot product multiplied by the identity matrix. If you wanted a dot product that produced a true scalar instead of scalar times identity, then you should be able to couple the above with a trace operation:

    [tex]
    a \cdot b = \frac{1}{4} TR \{ a, b \} = \frac{1}{4} TR (a b + b a )
    [/tex]

    EDIT: latex in PF doesn't appear to be working right now. That was:
    a \cdot b = \frac{1}{4} TR \{ a, b \} = \frac{1}{4} TR (a b + b a )

    You can similarily and naturally define the cross product or wedge product using the commutator. Some playing with these ideas can be found here:

    http://sites.google.com/site/peeterjoot/math/pauli_matrix.pdf

    ( as written this assumes some clifford algebra background ).
     
  7. Apr 22, 2009 #6

    KFC

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    Thanks a lot. It helps.
     
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