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

$\Omega = \begin{vmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{vmatrix}$

$\Lambda =$$\begin{vmatrix} 2 & 1 & 1\\ 1 & 0 & -1\\ 1 & -1 & 2 \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 $\Omega\ and\ \Lambda$:

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

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

Next computer the eigenvectors:

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

$\Lambda | \omega = 2 >$= $\begin{vmatrix}1\\1\\-1\end{vmatrix}$

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

$\Lambda | \omega = 3 >$= $\begin{vmatrix}1\\0\\1\end{vmatrix}$

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

$\Lambda | \omega = -1 >$= $\begin{vmatrix}1\\-2\\-1\end{vmatrix}$

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

$\Omega | \omega = 0 >$= $\begin{vmatrix}1\\0\\-1\end{vmatrix}$ twice

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

$\Omega | \omega = 2 >$= $\begin{vmatrix}1\\0\\1\end{vmatrix}$

In summary the eigenvectors are:

$\begin{vmatrix}1\\1\\-1\end{vmatrix}$ , $\begin{vmatrix}1\\0\\1\end{vmatrix}$ , $\begin{vmatrix}1\\-2\\-1\end{vmatrix}$ , $\begin{vmatrix}1\\0\\-1\end{vmatrix}$ , $\begin{vmatrix}1\\0\\1\end{vmatrix}$

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

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

$$\begin{vmatrix}1\\1\\-1\end{vmatrix}\ ,\ \begin{vmatrix}1\\0\\1\end{vmatrix}\ , \ \begin{vmatrix}1\\-2\\-1\end{vmatrix}$$

Then $$U = \begin{vmatrix}1&1&1\\1&0&-2\\-1&1&-1\end{vmatrix}\ U^\dagger = \begin{vmatrix}1&1&-1\\1&0&1\\1&-2&-1\end{vmatrix}$$

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.