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Invariance of the determinant under spin rotations

  1. Sep 20, 2015 #1

    ShayanJ

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    1. The problem statement, all variables and given/known data
    Show that the determinant of a ##2 \times 2 ## matrix ## \vec\sigma \cdot \vec a ## is invariant under ## \vec \sigma\cdot \vec a \rightarrow \vec \sigma\cdot \vec a' \equiv \exp(\frac{i\vec \sigma \cdot \hat n \phi}{2})\vec \sigma\cdot \vec a \exp(\frac{-i\vec \sigma \cdot \hat n \phi}{2}) ##.

    2. Relevant equations
    ## \sigma_1=\left( \begin{array}{cc} 0 \ \ \ \ \ \ \ \ 1 \\ 1 \ \ \ \ \ \ \ \ 0 \end{array} \right) ##
    ## \sigma_2=\left( \begin{array}{cc} 0 \ \ \ \ -i \\ i \ \ \ \ \ \ 0 \end{array} \right) ##
    ## \sigma_3=\left( \begin{array}{cc} 1 \ \ \ \ \ \ \ \ 0 \\ 0 \ \ \ \ -1 \end{array} \right) ##

    ##\exp(\frac{\pm i\vec \sigma \cdot \hat n \phi}{2})=\cos{(\frac \phi 2)}\pm i(\vec\sigma \cdot \hat n)\sin{(\frac \phi 2)} ##

    3. The attempt at a solution

    I tried to do it by matrix multiplication but the calculations turned out to be so much tedious that I thought there should be a better way. If this is the only way, just tell me. Otherwise just point to the right direction. Any other hint is welcome too.
    Thanks
     
  2. jcsd
  3. Sep 21, 2015 #2

    fzero

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    You should be able to use the commutation relation for the Pauli matrices ##[\sigma_i , \sigma_j ] = 2 i \epsilon_{ijk} \sigma_k## (sum over ##k## assumed) rather than actually write out explicit matrix products.
     
  4. Sep 21, 2015 #3

    ShayanJ

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    I used the properties of Pauli matrices and got the following:

    ## \cos^2 \frac \phi 2 \ (\vec \sigma \cdot \vec a)-i\sin \phi \left[(\vec a \times \hat n)\cdot \vec \sigma \right]+\sin^2\frac \phi 2 \left[(\vec \sigma \cdot \hat n)(\vec a \cdot \hat n )+i \vec a \cdot \vec \sigma- i(\hat n \cdot \vec \sigma)(\vec a \cdot \hat n)\right] ##

    But I don't know how to proceed!
     
  5. Sep 21, 2015 #4

    fzero

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    I think that in the ##\sin^2(\phi/2)## term, the 1st and 3rd terms should have the same coefficient (to give an overall ##2(\hat n \cdot \vec \sigma)(\vec a \cdot \hat n)##) and there shouldn't be an ##i## in the 2nd term. If you want to recheck the term, the identity ##\sigma_i \sigma_j = \delta_{ij} I + i \epsilon_ijk}\sigma_k## should be useful (though you might already be aware of it).

    So you will have an expression ##\vec{A} \cdot \vec{\sigma}## and you should show that ##\det ( \vec{A} \cdot \vec{\sigma}) = - \vec{A}\cdot \vec{A}##, by explicit muliplication or otherwise. Then apply this to your particular expression.
     
  6. Sep 21, 2015 #5

    ShayanJ

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    I corrected some mistakes and I got ## \vec A \cdot \vec \sigma ## where ## \vec A=\cos^2 \frac \phi 2 \vec a +\sin \phi (\vec a \times \hat n)+\sin^2 \frac \phi 2 \left[2(\vec a \cdot \hat n)\hat n-\vec a \right] ##.
    Now my problem is, ##\vec A \cdot \vec A ## is not a straightforward expression and I can't write it as a function of only ##\vec a \cdot \vec a ##.
     
  7. Sep 22, 2015 #6

    strangerep

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    Can't you just use general properties of determinants?
    I.e., ##\det(AB) = \det A \; \det B## and ##\det A^{-1} = 1/\det A##, etc ?

    Am I missing something here??
     
  8. Sep 22, 2015 #7

    ShayanJ

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    thanks strangerep. The determinants of those exponentials are 1 and so things are very simple.
    I have this bad habit of always trying the hardest way first!
     
  9. Sep 22, 2015 #8

    strangerep

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    Actually, you don't even need to know their value... :oldwink:
     
  10. Sep 22, 2015 #9

    ShayanJ

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    I don't understand!
     
  11. Sep 22, 2015 #10

    strangerep

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    Well, I already noted these properties of determinants:
    ##\det(AB) = \det A \; \det B## and ##\det\left(A^{-1}\right) = 1/\det A##, so...
    $$ \det\left(ABA^{-1}\right) ~=~ .... ~?$$

    [Edit: you answered my post before I'd finished. Work out the above...]
     
  12. Sep 22, 2015 #11

    ShayanJ

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    Yes, but those are useful in proving the invariance only when you know that the determinant of those exponentials are one. But you said I don't need to know the value of the determinants. This is what I don't understand.
     
  13. Sep 22, 2015 #12

    strangerep

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    Re-read my post #10...
     
  14. Sep 22, 2015 #13

    ShayanJ

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    I get it now, thanks.
     
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