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Simutanious diagonalization of 2 matrices

  1. Jun 22, 2011 #1
    1. The problem statement, all variables and given/known data

    From Principles of Quantum Mechanics, 2nd edition by R Shankar, problem
    1.8.10:

    By considering the commutator, show that the following Hermitian matrices may be
    simultaneously diagonalized. Find the eigenvectors common to both and verify
    that under a unitary transformation to this basis, both matrices are
    diagonalized.

    [itex]\textbf{Theorem 13:} [/itex] If [itex] \Omega\ and\ \Lambda [/itex] are two commuting Hermitian
    operators, there exists (at least) a basis of common eigenvectors that diagonalizes them
    both.

    2. Relevant equations
    [itex] \Omega = \begin{vmatrix}
    1 & 0 & 1 \\
    0 & 0 & 0 \\
    1 & 0 & 1
    \end{vmatrix}[/itex]

    [itex] \Lambda = [/itex][itex]
    \begin{vmatrix}
    2 & 1 & 1\\
    1 & 0 & -1\\
    1 & -1 & 2
    \end{vmatrix}[/itex]

    [[itex] \Omega [/itex] , [itex] \Lambda[/itex]] = 0

    3. The attempt at a solution

    The two matraces definitely meet the requirements of Theorem 13. Next computer
    the eigenvalues of [itex] \Omega\ and\ \Lambda [/itex]:

    [itex] \Omega [/itex]= [itex] \begin{vmatrix}
    1-\omega & 0 & 1 \\
    0 & -\omega & 0 \\
    1 & 0 & 1-\omega
    \end{vmatrix}
    \ \Rightarrow (1-\omega)^2 (-\omega) + \omega) = 0 \ \Rightarrow \omega = 0, 0,
    2[/itex]

    [itex] \Lambda [/itex]=[itex] \begin{vmatrix}
    2-\omega & 1 & 1 \\
    1 & -\omega & -1 \\
    1 & -1 & 2-\omega
    \end{vmatrix}
    \ \Rightarrow (2-\omega)(\omega^2 -2\omega - 1) -2(2-\omega) = 0 \Rightarrow
    \omega = 2, 3, -1 [/itex]

    Next computer the eigenvectors:

    [itex]\Lambda | \omega = 2 > [/itex]= [itex] \begin{vmatrix}
    0 & 1 & 1 \\
    1 & -2 & -1 \\
    1 & -1 & 0
    \end{vmatrix} [/itex] = 0 [itex] \Rightarrow x_2 + x_3 = 0\ ;\ x_1 - x_2 = 0 \Rightarrow
    x_1= 1, X_2=1, x_3=-1[/itex]

    [itex] \Lambda | \omega = 2 > [/itex]= [itex] \begin{vmatrix}1\\1\\-1\end{vmatrix} [/itex]

    [itex] \Lambda | \omega = 3 > [/itex]= [itex] \begin{vmatrix}
    -1 & 1 & 1 \\
    1 & -3 & -1 \\
    1 & -1 &-1
    \end{vmatrix} [/itex] = 0 [itex] \Rightarrow -x_1 + x_2 + x_3 = 0\ ;
    x_1 -3x_2 - x_3 = 0\ ;\ x_1 - x_2 - x_3 = 0 \Rightarrow x_1=1, x_2=0, x_3=1 [/itex]

    [itex] \Lambda | \omega = 3 > [/itex]= [itex] \begin{vmatrix}1\\0\\1\end{vmatrix} [/itex]

    [itex] \Lambda | \omega = -1 > [/itex]= [itex]\begin{vmatrix}
    3 & 1 & 1 \\
    1 & 1 & -1 \\
    1 & -1 &3
    \end{vmatrix} [/itex] = 0 [itex] \Rightarrow 3x_1 + x_2 + x_3 = 0\ ;
    x_1 + x_2 - x_3 = 0\ ;\ x_1 - x_2 + 3x_3 = 0 \Rightarrow x_1=1, x_2=-2, x_3=-1 [/itex]

    [itex] \Lambda | \omega = -1 > [/itex]= [itex] \begin{vmatrix}1\\-2\\-1\end{vmatrix} [/itex]

    And from [itex] \Omega [/itex]
    [itex] \Omega | \omega = 0 > [/itex] = [itex] \begin{vmatrix}
    1 & 0 & 1 \\
    0 & 0 & 0 \\
    1 & 0 & 1
    \end{vmatrix} [/itex] = 0 [itex] \Rightarrow x_1 + x_3 = 0\ ;\ x_2=0
    \Rightarrow x_1=1, x_2=-0, x_3=-1 [/itex]

    [itex] \Omega | \omega = 0 > [/itex]= [itex] \begin{vmatrix}1\\0\\-1\end{vmatrix} [/itex] twice

    [itex] \Omega | \omega = 2 > [/itex] = [itex] \begin{vmatrix}
    -1 & 0 & 1 \\
    0 & -2 & 0 \\
    1 & 0 & -1
    \end{vmatrix} $ = 0 [itex] \Rightarrow -x_1 + x_3 = 0\ ;\ x_2=0
    \Rightarrow x_1=1, x_2=-0, x_3=1 [/itex]

    [itex] \Omega | \omega = 2 > [/itex]= [itex] \begin{vmatrix}1\\0\\1\end{vmatrix} [/itex]

    In summary the eigenvectors are:

    [itex] \begin{vmatrix}1\\1\\-1\end{vmatrix} [/itex] , [itex]
    \begin{vmatrix}1\\0\\1\end{vmatrix} [/itex] , [itex]
    \begin{vmatrix}1\\-2\\-1\end{vmatrix} [/itex] , [itex]
    \begin{vmatrix}1\\0\\-1\end{vmatrix} [/itex] , [itex]
    \begin{vmatrix}1\\0\\1\end{vmatrix} [/itex]

    And here is the problem. There is no way to build a unitary matrix (say
    [itex]\textbf{U}[/itex]) from these vectors. All possible combinations lead to non-Hemitian
    matrices. With out a unitary matrix the diagonalization process -
    [itex] \textbf{U} \varLambda \textbf{U}^\dagger [/itex]= diagonalized matrix - can not be
    completed.

    Where have I gone wrong???
     
    Last edited: Jun 22, 2011
  2. jcsd
  3. Jun 22, 2011 #2

    vela

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    For the first matrix, Ω, let's call the three eigenvectors [itex]\vert \omega=2 \rangle[/itex], [itex]\vert \omega=0_1 \rangle[/itex], and [itex]\vert \omega=0_2 \rangle[/itex]. Similarly, for the second matrix, Λ, you have the eigenvectors [itex]\vert \lambda=3 \rangle[/itex], [itex]\vert \lambda=2 \rangle[/itex], and [itex]\vert \lambda=-1 \rangle[/itex].

    The first thing to note is that [itex]\vert \omega=2 \rangle[/itex] and [itex]\vert \lambda=3 \rangle[/itex] are the same vector, namely (1, 0, 1)T. That means {[itex]\vert \omega=0_1 \rangle[/itex], [itex]\vert \omega=0_2 \rangle[/itex]} and {[itex]\vert \lambda=2 \rangle[/itex], [itex]\vert \lambda=-1 \rangle[/itex]} span the same subspace. The second thing is that because Ω's vectors are degenerate, any linear combination of those two vectors is also an eigenvector with eigenvalue 0.
     
  4. Jun 22, 2011 #3

    Dick

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    What you missed is that in getting the zero eigenvectors for Ω, x2=0 is not required. x2 can be anything. Saying "[1,0,-1] twice" makes no sense. So a basis for zero eigenvectors is [1,0,-1] and [0,1,0]. Then, as vela points out, the other eigenvectors for Λ lie in that subspace.
     
  5. Jun 23, 2011 #4
    I'm still thinking about your replies (reading like crazy). I haven't disappeared. My Linear Algebra seems to be a bit rusty. I really appreciate the help.

    Gary R.
     
  6. Jun 25, 2011 #5
    I think I need to back up on this problem. Per Shankar:

    If [itex] \Lambda [/itex] is a Hermitian matrix, there exists a unitary matrix U (built out of the eigenvectors of [itex] \Lambda [/itex] such that [itex]U^\dagger \Lambda U [/itex] is diagonalized.

    So lets use Lambda from above since it is Hermitian. And we have the eigenvectors

    [tex] \begin{vmatrix}1\\1\\-1\end{vmatrix}\ ,\ \begin{vmatrix}1\\0\\1\end{vmatrix}\ , \
    \begin{vmatrix}1\\-2\\-1\end{vmatrix} [/tex]

    Then [tex] U = \begin{vmatrix}1&1&1\\1&0&-2\\-1&1&-1\end{vmatrix}\
    U^\dagger = \begin{vmatrix}1&1&-1\\1&0&1\\1&-2&-1\end{vmatrix}[/tex]

    An there is the problem. Per Shankar, U is unitary and [itex]U^\dagger U = I [/itex]

    This isn't. What have I done wrong.
     
  7. Jun 25, 2011 #6

    vela

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    The columns of U are the normalized eigenvectors.
     
  8. Jun 26, 2011 #7
    Finallly. The normalization of the eigenvectors of [itex]\Lambda[/itex] fixed the problem.
    [itex]U^\dagger*U now = I, U^\dagger*\Lambda * U [/itex]= diagonal with eigenvalues in the diagonal. The same with [itex]\Omega[/itex].

    Question: Does this work only because the two matrices share a common eigenvector?

    Gary R. and thank you all for the help
     
  9. Jun 26, 2011 #8

    Dick

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    Yes, you can only simultaneously diagonalize if the matrices have a common basis of eigenvectors. Being Hermitian and commuting guarantees this.
     
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