Eigenvectors of rotation matrix

In summary, the conversation discusses finding the corresponding eigenvectors for a given operator with specific eigenvalues. The solution involves letting the matrix operate on a generic vector and solving for the resulting vector. The correct eigenvectors are determined to be \frac{1}{\sqrt{2}} \left[ \begin{array}{c}1 \\i \\\end{array} \right] and \frac{1}{\sqrt{2}} \left[ \begin{array}{c}i \\1 \\\end{array} \right].
  • #1
timon
56
0

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?
 
Last edited:
Physics news on Phys.org
  • #2
timon said:

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

which implies that [itex] x_1 = x_2 = 0 [/itex]. Am i missing something crucial?
No, it does not imply that. Let x1=1 (and worry about normalization later), then what is x2?
 
  • #3
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?

[itex]
\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]
[/itex]
 
  • #4
Are these the correct eigenvectors?

[itex]
\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]
[/itex]
Yes, that's right.

timon said:
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
[tex]x_1 \cos{\theta} + x_2\sin{\theta} = x_1 \cos{\theta} + x_1 i\sin{\theta}[/tex]
which by inspection implies [itex]x_2 = i x_1[/itex], not [itex]x_2 = i x_2[/itex]
 
  • #5
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.
 

FAQ: Eigenvectors of rotation matrix

What are eigenvectors and why are they important in a rotation matrix?

Eigenvectors are special vectors that remain in the same direction when multiplied by a transformation matrix, such as a rotation matrix. They are important in a rotation matrix because they provide a frame of reference for the rotation and can help determine the axis and angle of rotation.

How do you calculate the eigenvectors of a rotation matrix?

To calculate the eigenvectors of a rotation matrix, you first need to find the eigenvalues of the matrix. Then, you can use the eigenvalues to find the corresponding eigenvectors using the formula (A - λI)v = 0, where A is the rotation matrix, λ is the eigenvalue, and v is the eigenvector.

What do the eigenvectors of a rotation matrix represent?

The eigenvectors of a rotation matrix represent the principal axes of rotation. These are the axes around which the rotation occurs, and the length of the eigenvectors corresponds to the amount of rotation around that axis.

Can a rotation matrix have complex eigenvectors?

Yes, a rotation matrix can have complex eigenvectors. However, in most cases, the eigenvectors will be real numbers since rotations typically occur in a 3-dimensional space. Complex eigenvectors are more common in higher dimensional spaces.

How do you interpret the eigenvectors of a rotation matrix in terms of the rotation itself?

The eigenvectors of a rotation matrix can be interpreted as the axis of rotation. The direction of the eigenvector represents the direction of the axis, and the magnitude of the eigenvector represents the amount of rotation around that axis. Additionally, the eigenvectors are orthogonal to each other, meaning they are at right angles, which corresponds to the fact that the rotation matrix is an orthogonal matrix.

Back
Top