Homework Help: Eigenvectors of rotation matrix

1. Jul 13, 2011

timon

1. The problem statement, all variables and given/known data
This question is from Principles of Quantum Mechanics by R. Shankar.

Given the operator (matrix) $\Omega$ with eigenvalues $e^{i\theta}$ and $e^{-i\theta}$, I am told to find the corresponding eigenvectors.

2. Relevant equations

$\Omega = \left[ \begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \\ \end{array} \right]$
$\Omega \left[ \begin{array}{c} x_1 \\ x_2 \\ \end{array} \right] = \left[ \begin{array}{c} x_1 \cos{\theta} + x_2 \sin{\theta} \\ -x_1 \sin{\theta} + x_2 \cos{\theta} \\ \end{array} \right]$
$e^{i\theta} \left[ \begin{array}{c} x_1 \\ x_2 \\ \end{array} \right] = \left[ \begin{array}{c} x_1 \cos{\theta} + x_1 i\sin{\theta} \\ x_2 \cos{\theta} + x_2 i\sin{\theta} \\ \end{array}\right]$

3. The attempt at a solution

I let the matrix operate on the generic vector $(x_1, x_2)^T$ and demand that the resulting vector is equal to $(e^{i\theta}x_1, e^{i\theta}x_2)^T$. From this i get the condition that $x_2 = ix_2$and $x_1 = -ix_2$, which implies that $x_1 = x_2 = 0$. Am i missing something crucial?

Last edited: Jul 13, 2011
2. Jul 13, 2011

Redbelly98

Staff Emeritus
Typo? I think you mean $x_2 = ix_1$ here. In which case these two equations are really the same equation.

No, it does not imply that. Let x1=1 (and worry about normalization later), then what is x2?

3. Jul 13, 2011

timon

Thanks for the quick reply. That actually wasn't a typo. I managed to switch the indices the same way in three separate calculations, getting the same erroneous result each time. Are these the correct eigenvectors?

$\frac{1}{\sqrt{2}} \left[ \begin{array}{c} 1 \\ i \\ \end{array} \right] , \frac{1}{\sqrt{2}} \left[ \begin{array}{c} i \\ 1 \\ \end{array} \right]$

4. Jul 13, 2011

Redbelly98

Staff Emeritus
Yes, that's right.

That's weird. From your equations in Post #1, you get
$$x_1 \cos{\theta} + x_2\sin{\theta} = x_1 \cos{\theta} + x_1 i\sin{\theta}$$
which by inspection implies $x_2 = i x_1$, not $x_2 = i x_2$

5. Jul 13, 2011

timon

It is indeed quite amazing that I managed to screw the same trivial manipulation up three times in a row, always in the same manner. Thank you for the help.