Eigenvectors of rotation matrix

[timon]

Homework Statement

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.

Homework Equations

<br /> \Omega = \left[ \begin{array}{cc}<br /> \cos{\theta} &amp; \sin{\theta} \\<br /> -\sin{\theta} &amp; \cos{\theta} \\<br /> \end{array} \right]<br />
<br /> \Omega \left[ \begin{array}{c}<br /> x_1 \\<br /> x_2 \\<br /> \end{array} \right] = \left[ \begin{array}{c}<br /> x_1 \cos{\theta} + x_2 \sin{\theta} \\<br /> -x_1 \sin{\theta} + x_2 \cos{\theta} \\<br /> \end{array}<br /> \right] <br />
<br /> e^{i\theta} \left[ \begin{array}{c}<br /> x_1 \\<br /> x_2 \\<br /> \end{array} \right] = \left[ \begin{array}{c}<br /> x_1 \cos{\theta} + x_1 i\sin{\theta} \\<br /> x_2 \cos{\theta} + x_2 i\sin{\theta} \\<br /> \end{array}\right]<br />

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_2and x_1 = -ix_2, which implies that x_1 = x_2 = 0. Am i missing something crucial?

 

[Redbelly98]

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_2and x_1 = -ix_2,

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

which implies that x_1 = x_2 = 0. Am i missing something crucial?

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

 

[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?

<br /> \frac{1}{\sqrt{2}} \left[ \begin{array}{c}<br /> 1 \\<br /> i \\<br /> \end{array} \right] ,<br /> \frac{1}{\sqrt{2}} \left[ \begin{array}{c}<br /> i \\<br /> 1 \\<br /> \end{array} \right] <br />

 

[Redbelly98]

Are these the correct eigenvectors?

<br /> \frac{1}{\sqrt{2}} \left[ \begin{array}{c}<br /> 1 \\<br /> i \\<br /> \end{array} \right] ,<br /> \frac{1}{\sqrt{2}} \left[ \begin{array}{c}<br /> i \\<br /> 1 \\<br /> \end{array} \right] <br />

Yes, that's right.

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.

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

 

[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.