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Pauli matrices

  1. Oct 5, 2013 #1
    1. Consider the 2x2 matrix [tex]\sigma^{\mu}=(1,\sigma_{i})[/tex] where [tex]\sigma^{\mu}=(1,\sigma)
    [/tex] where 1 is the identity matrix and [tex]\sigma_{i}[/tex] the pauli matrices. Show with a direct calcuation that [tex]detX=x^{\mu}x_{\mu}[/tex]








    3. I'm not sure how to attempt this at all...
     
  2. jcsd
  3. Oct 5, 2013 #2
    I think you're missing a part out, where we define how the big X is related to the little x's
    As a hint, how would you normally go about finding the determinant of a 2x2 matrix?
     
  4. Oct 6, 2013 #3
    Well usually ab - cd assuming a, b are elements in the first row..

    However not sure what to do next...
     
  5. Oct 6, 2013 #4

    vanhees71

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    Write down [itex]X[/itex] explicitly and then calculate its determinant by just doing the algebraic manipulations. It's nothing deep in this calculation, but its result is very important for the description of spin-1/2-particles in relativistic quantum theory!
     
  6. Oct 6, 2013 #5
    Well I don't know what x is therefore I don't know how to write X, and I'm not sure how sigma would change to x
     
  7. Oct 6, 2013 #6

    George Jones

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    Isn't [itex]X[/itex] given as [itex]X = x^\mu \sigma_\mu[/itex]?
     
  8. Oct 6, 2013 #7
    Yes...
     
  9. Oct 6, 2013 #8

    George Jones

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    For an [itex]n \times n[/itex] matrix [itex]A[/itex] and a number [itex]a[/itex], what is [itex]det \left(aA\right)[/itex] in terms of [itex]detA[/itex]?
     
  10. Oct 6, 2013 #9
    [tex]det(cA)=c^{n}(detA)
    [/tex]
     
  11. Oct 6, 2013 #10

    George Jones

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    So [itex]det \left(x^\mu \sigma_\mu\right) = [/itex] ?
     
  12. Oct 6, 2013 #11
    Well,you need to write it as the product of four vectors,use the explicit form of σ matrices,and calculate the determinant of resulting 2×2 matrix.Got it.you will find it's use in showing the relation SO(3,1)=SL(2,C)/Z2.
     
  13. Oct 6, 2013 #12
    Well from George's previous post, I had realised the conclusion beforehand, however I'm not sure how that will result in the answer I need.. but I guess if I follow andrien it should fall out?
     
  14. Oct 6, 2013 #13

    George Jones

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    Each [itex]x^\mu[/itex] is a number and each [itex]\sigma_\mu[/itex] is a number. As an intermediate step, apply
    to [itex]det \left(x^\mu \sigma_\mu\right)[/itex]. What do you get?
     
  15. Oct 6, 2013 #14

    vanhees71

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    No! [itex]\sigma_{\mu}[/itex] are complex [itex]2 \times 2[/itex] matrices, namely
    [tex]\sigma_0=\mathbb{1}[/tex] and [itex]\sigma_{j}[/itex] ([itex]j \in \{1,2,3 \}[/itex]) the three Pauli matrices. This means that in standard representation of the Pauli matrices with [itex]\sigma_3[/itex] diagonalized you have
    [tex]X=x^{\mu} \sigma_{\mu}=x^0 \mathbb{1} + \vec{x} \cdot \vec{\sigma}=\begin{pmatrix}
    x^0+x^3 & x^1-\mathrm{i} x^2 \\
    x^1 + \mathrm{i} x^2 & x^0-x^3
    \end{pmatrix}.
    [/tex]
    Now evaluate its determinant!
     
  16. Oct 6, 2013 #15

    George Jones

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    Oops! The above was a slip of my fingers while typing. Obviously, each [itex]\sigma_\mu[/itex] is a matrix, and this what I meant to type.
     
  17. Oct 6, 2013 #16
    sweeet! I see it now, I had the error of not including x^0 when writing out X which was why I couldn't solve the problem. Thank you!!
     
  18. Oct 6, 2013 #17

    George Jones

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    Now that you see it using the explicit matrix given by vahees71, I will write out the method at which I hinted.

    [tex]det\left( x^\mu \sigma_\mu \right) = \left( x^\mu \right)^2 det \sigma_\mu = \left( x^0 \right)^2 - \left( x^1 \right)^2 - \left( x^2 \right)^2 - \left( x^3 \right)^2 ,[/tex]

    since [itex]det\sigma_0 = 1[/itex] and [itex]det\sigma_i = -1[/itex] for each [itex]i[/itex].

    I think it is important to see the calculation done both ways, i.e., summing before taking the determinant and summing after taking determinants.
     
  19. Oct 6, 2013 #18
    yeah I noticed that the det for sigma(i) was -1, but I didn't know what the determinant for sigma(0) was, so I guess I wasn't able to put everything together...as I wasn't sure how to include sigma(0)
     
  20. Oct 6, 2013 #19

    George Jones

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    But [itex]\sigma_0[/itex] is just the identity matrix ...
     
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