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Traceless Hermitian Matrices

  1. May 19, 2015 #1
    Hello,
    Here's a textbook question and my solution, please check if it is correct, I'm slightly doubtful about the second part.

    Consider Hermitian matrices [itex]M_1, M_2, M_3,\ and\ M_4[/itex] that obey:
    [itex]M_i M_j+M_j M_i = 2 \delta_{ij} I \hspace{10mm} for\ i,\ j\ = 1,\ ... ,4 [/itex]

    (1) Show that the eigenvalues of [itex]M_i[/itex] are [itex]±1[/itex].
    Solution: When [itex]i = j, \hspace{20mm} M_i M_i = I[/itex].
    Since [itex]M_i[/itex] are Hermitian, this equation implies that they are unitary. The eigenvalues of a unitary matrix are complex numbers of unit modulus, since it's also Hermitian, they have to be real. So, the eigenvalues have to be [itex]±1[/itex]. [itex]\hspace{20mm} Q.E.D.[/itex]

    (2) Show that [itex]M_i[/itex] are traceless.
    Solution: When [itex]i ≠ j, \\
    M_i M_j = -M_j M_i\\
    \Rightarrow M_i M_j M_i = -M_j M_i M_i\\
    \Rightarrow M_i M_j M_i = -M_j I\\
    \Rightarrow Tr(M_i M_j M_i) = -Tr(M_j) \\
    \Rightarrow Tr(M_j M_i M_i) = -Tr(M_j)\\
    \Rightarrow Tr(M_j) = -Tr(M_j)\\
    \Rightarrow Tr(M_j) = 0. \hspace{20mm} Q.E.D.[/itex]

    (3) Show that [itex]M_i[/itex] cannot be odd-dimensional matrices.
    Solution: For some eigenbasis [itex]U, D = U^†M_iU, D[/itex] is a diagonal matrix with [itex] Tr(D) = Tr(M_i)[/itex] and the diagonal elements of [itex]D[/itex] are the eigenvalues, [itex]±1[/itex]. Also, [itex] Tr(D) = Tr(M_i) = 0. [/itex] An odd dimension cannot result in a traceless matrix, hence by argument of parity, [itex]M_i[/itex] are even dimensional matrices. [itex]\hspace{20mm}Q.E.D.[/itex]

    Thanks!
     
  2. jcsd
  3. May 19, 2015 #2

    wabbit

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    Yes, I think the three proofs are fine. Nothing suspicious about (2) in particular.
     
  4. May 19, 2015 #3
    Thanks for verifying, I'm still not very confident. :)
     
  5. May 19, 2015 #4

    wabbit

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    I can't see why, you've spelled out every step in detail, which part gives you doubt ?
     
  6. May 19, 2015 #5
    I mean, I got it right, but I'm not confident that I can get math right, yet. Hopefully more problems and I'll feel confident about my solutions. Thanks for asking! :)
     
  7. May 19, 2015 #6

    wabbit

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    Ah yes I feel that way about physics all the time : )
     
  8. Oct 28, 2015 #7
    I think there is a mistake on the 2nd question. Between line 4 and 5 you comutate Mi with Mj which you can't do it. You should put a (-) in front so the final result would be 0=0 . There is another way ;) .
     
  9. Feb 4, 2017 #8
    No, it's okay - it's a cyclic permutation, just cycled the "wrong" way (or cycled twice)
     
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