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All 2x2 Hermitian and Unitary Matrices (Check My Proof)

  1. Sep 20, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Find all 2x2 Matrices which are both hermitian and unitary.

    2. Relevant equations
    Conditions for Matrix A:
    [itex]A=A^†[/itex]
    [itex]A^†A=I[/itex]
    I = the identity matrix
    † = hermitian conjugate


    3. The attempt at a solution

    1. We see by the conditions that A^† = A and by the second condition, we see that AA=I. So we write:

    (\begin{pmatrix}
    a & b\\
    c & d
    \end{pmatrix}\begin{pmatrix}
    a & b\\
    c & d
    \end{pmatrix})= \begin{pmatrix}
    1 & 0\\
    0 & 1
    \end{pmatrix}

    Which sets us up with:
    2. Solve
    \begin{pmatrix}
    a^2+bc & ab+bd\\
    ca+dc & cb+d^2
    \end{pmatrix}=\begin{pmatrix}
    1 & 0\\
    0 & 1
    \end{pmatrix}

    3. So our system:
    cb+d^2=1
    a^2+bc=1
    ab+bd=0
    ca+dc=0

    4. In solving, we see that a= +/1, b=+/-1 and c,d = 0.

    So we get 4 total matrices that are diagonal and all possible values +/-1.



    Is this correct? Am I forgetting anything? Any complex elements am I forgetting to check?
     
  2. jcsd
  3. Sep 20, 2015 #2

    fzero

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    The 3rd and 4th equations require ##d=-a## if either ##b## or ##c## is nonzero, so there are more solutions, which include imaginary components .
     
  4. Sep 21, 2015 #3

    RJLiberator

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    Damn, I was afraid of that.

    Well, if d=-a. Then bc=1.

    There seems to be a lot of solutions....

    I'll have to write the general answer down then.

    a=-d, and bc=1
     
  5. Sep 21, 2015 #4

    fzero

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    Since the matrix is Hermitian, ##c =\bar{b}##, and ##a## is real, so you should have a condition ##a^2 + |b|^2 =1##. You can solve this with trig functions.
     
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