Simutanious diagonalization of 2 matrices

[Gary Roach]

Homework Statement

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.

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

Homework Equations

\Omega = \begin{vmatrix}<br /> 1 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 1<br /> \end{vmatrix}

\Lambda =<br /> \begin{vmatrix}<br /> 2 &amp; 1 &amp; 1\\<br /> 1 &amp; 0 &amp; -1\\<br /> 1 &amp; -1 &amp; 2<br /> \end{vmatrix}

[\Omega , \Lambda] = 0

The Attempt at a Solution

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

\Omega= \begin{vmatrix}<br /> 1-\omega &amp; 0 &amp; 1 \\<br /> 0 &amp; -\omega &amp; 0 \\<br /> 1 &amp; 0 &amp; 1-\omega<br /> \end{vmatrix} <br /> \ \Rightarrow (1-\omega)^2 (-\omega) + \omega) = 0 \ \Rightarrow \omega = 0, 0,<br /> 2

\Lambda=\begin{vmatrix}<br /> 2-\omega &amp; 1 &amp; 1 \\<br /> 1 &amp; -\omega &amp; -1 \\<br /> 1 &amp; -1 &amp; 2-\omega<br /> \end{vmatrix} <br /> \ \Rightarrow (2-\omega)(\omega^2 -2\omega - 1) -2(2-\omega) = 0 \Rightarrow<br /> \omega = 2, 3, -1

Next computer the eigenvectors:

\Lambda | \omega = 2 &gt;= \begin{vmatrix}<br /> 0 &amp; 1 &amp; 1 \\<br /> 1 &amp; -2 &amp; -1 \\<br /> 1 &amp; -1 &amp; 0<br /> \end{vmatrix} = 0 \Rightarrow x_2 + x_3 = 0\ ;\ x_1 - x_2 = 0 \Rightarrow<br /> x_1= 1, X_2=1, x_3=-1

\Lambda | \omega = 2 &gt;= \begin{vmatrix}1\\1\\-1\end{vmatrix}

\Lambda | \omega = 3 &gt;= \begin{vmatrix}<br /> -1 &amp; 1 &amp; 1 \\<br /> 1 &amp; -3 &amp; -1 \\<br /> 1 &amp; -1 &amp;-1<br /> \end{vmatrix} = 0 \Rightarrow -x_1 + x_2 + x_3 = 0\ ;<br /> x_1 -3x_2 - x_3 = 0\ ;\ x_1 - x_2 - x_3 = 0 \Rightarrow x_1=1, x_2=0, x_3=1

\Lambda | \omega = 3 &gt;= \begin{vmatrix}1\\0\\1\end{vmatrix}

\Lambda | \omega = -1 &gt;= \begin{vmatrix}<br /> 3 &amp; 1 &amp; 1 \\<br /> 1 &amp; 1 &amp; -1 \\<br /> 1 &amp; -1 &amp;3<br /> \end{vmatrix} = 0 \Rightarrow 3x_1 + x_2 + x_3 = 0\ ;<br /> x_1 + x_2 - x_3 = 0\ ;\ x_1 - x_2 + 3x_3 = 0 \Rightarrow x_1=1, x_2=-2, x_3=-1

\Lambda | \omega = -1 &gt;= \begin{vmatrix}1\\-2\\-1\end{vmatrix}

And from \Omega
\Omega | \omega = 0 &gt; = \begin{vmatrix}<br /> 1 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0 \\<br /> 1 &amp; 0 &amp; 1<br /> \end{vmatrix} = 0 \Rightarrow x_1 + x_3 = 0\ ;\ x_2=0 <br /> \Rightarrow x_1=1, x_2=-0, x_3=-1

\Omega | \omega = 0 &gt;= \begin{vmatrix}1\\0\\-1\end{vmatrix} twice

\Omega | \omega = 2 &gt; = \begin{vmatrix}<br /> -1 &amp; 0 &amp; 1 \\<br /> 0 &amp; -2 &amp; 0 \\<br /> 1 &amp; 0 &amp; -1<br /> \end{vmatrix} $ = 0 \Rightarrow -x_1 + x_3 = 0\ ;\ x_2=0&lt;br /&gt; \Rightarrow x_1=1, x_2=-0, x_3=1<br /> <br /> \Omega | \omega = 2 &amp;gt;= \begin{vmatrix}1\\0\\1\end{vmatrix}<br /> <br /> In summary the eigenvectors are:<br /> <br /> \begin{vmatrix}1\\1\\-1\end{vmatrix} , &lt;br /&gt; \begin{vmatrix}1\\0\\1\end{vmatrix} , &lt;br /&gt; \begin{vmatrix}1\\-2\\-1\end{vmatrix} , &lt;br /&gt; \begin{vmatrix}1\\0\\-1\end{vmatrix} , &lt;br /&gt; \begin{vmatrix}1\\0\\1\end{vmatrix} <br /> <br /> And here is the problem. There is no way to build a unitary matrix (say<br /> \textbf{U}) from these vectors. All possible combinations lead to non-Hemitian<br /> matrices. With out a unitary matrix the diagonalization process -<br /> \textbf{U} \varLambda \textbf{U}^\dagger= diagonalized matrix - can not be<br /> completed.<br /> <br /> Where have I gone wrong?

 

[vela]

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

The first thing to note is that \vert \omega=2 \rangle and \vert \lambda=3 \rangle are the same vector, namely (1, 0, 1)^T. That means {\vert \omega=0_1 \rangle, \vert \omega=0_2 \rangle} and {\vert \lambda=2 \rangle, \vert \lambda=-1 \rangle} 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.

 

[Dick]

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.

 

[Gary Roach]

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.

 

[Gary Roach]

I think I need to back up on this problem. Per Shankar:

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

So let's use Lambda from above since it is Hermitian. And we have the eigenvectors

\begin{vmatrix}1\\1\\-1\end{vmatrix}\ ,\ \begin{vmatrix}1\\0\\1\end{vmatrix}\ , \ <br /> \begin{vmatrix}1\\-2\\-1\end{vmatrix}

Then U = \begin{vmatrix}1&amp;1&amp;1\\1&amp;0&amp;-2\\-1&amp;1&amp;-1\end{vmatrix}\ <br /> U^\dagger = \begin{vmatrix}1&amp;1&amp;-1\\1&amp;0&amp;1\\1&amp;-2&amp;-1\end{vmatrix}

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

This isn't. What have I done wrong.

 

[vela]

The columns of U are the normalized eigenvectors.

 

[Gary Roach]

Finallly. The normalization of the eigenvectors of \Lambda fixed the problem.
U^\dagger*U now = I, U^\dagger*\Lambda * U= diagonal with eigenvalues in the diagonal. The same with \Omega.

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

Gary R. and thank you all for the help

 

[Dick]

Finallly. The normalization of the eigenvectors of \Lambda fixed the problem.
U^\dagger*U now = I, U^\dagger*\Lambda * U= diagonal with eigenvalues in the diagonal. The same with \Omega.

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

Gary R. and thank you all for the help

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