- #1
Gary Roach
- 20
- 0
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.
[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.
Homework 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
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: