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timon
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Homework Statement
This question is from Principles of Quantum Mechanics by R. Shankar.
Given the operator (matrix) [itex] \Omega [/itex] with eigenvalues [itex] e^{i\theta}[/itex] and [itex] e^{-i\theta} [/itex], I am told to find the corresponding eigenvectors.
Homework Equations
[itex]
\Omega = \left[ \begin{array}{cc}
\cos{\theta} & \sin{\theta} \\
-\sin{\theta} & \cos{\theta} \\
\end{array} \right]
[/itex]
[itex]
\Omega \left[ \begin{array}{c}
x_1 \\
x_2 \\
\end{array} \right] = \left[ \begin{array}{c}
x_1 \cos{\theta} + x_2 \sin{\theta} \\
-x_1 \sin{\theta} + x_2 \cos{\theta} \\
\end{array}
\right]
[/itex]
[itex]
e^{i\theta} \left[ \begin{array}{c}
x_1 \\
x_2 \\
\end{array} \right] = \left[ \begin{array}{c}
x_1 \cos{\theta} + x_1 i\sin{\theta} \\
x_2 \cos{\theta} + x_2 i\sin{\theta} \\
\end{array}\right]
[/itex]
The Attempt at a Solution
I let the matrix operate on the generic vector [itex] (x_1, x_2)^T [/itex] and demand that the resulting vector is equal to [itex] (e^{i\theta}x_1, e^{i\theta}x_2)^T [/itex]. From this i get the condition that [itex] x_2 = ix_2 [/itex]and [itex] x_1 = -ix_2 [/itex], which implies that [itex] x_1 = x_2 = 0 [/itex]. Am i missing something crucial?
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