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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.

I think these equations are relevant:

$$\begin{align}\Omega &= \begin{bmatrix}\cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{bmatrix} \\ \Omega \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} &= \begin{bmatrix} x_1 \cos{\theta} + x_2 \sin{\theta} \\ -x_1 \sin{\theta} + x_2 \cos{\theta} \end{bmatrix} \\ e^{i\theta} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} &= \begin{bmatrix} x_1 \cos{\theta} + x_1 i\sin{\theta} \\ x_2 \cos{\theta} + x_2 i\sin{\theta} \end{bmatrix}\end{align}$$

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?

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Who says $x_1$ and $x_2$ have to be real? –  DJBunk Jun 26 '12 at 17:52
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Also, I am not sure about both your $x_1$ and $x_2$ constraints are correct. –  DJBunk Jun 26 '12 at 18:28
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The first equations is $ix_1=x_2$. –  MBN Jun 26 '12 at 20:16

2 Answers 2

No, I think you're just making a math mistake. Take a close look at the origin of the condition $x_2 = ix_2$ and I believe you'll find it.

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When you know that the eigenvalues are $e^{\pm i\theta}$, there's an easier way to write down the eigenvectors.

Start with the vector (1,x) where x is unknown. Apply your matrix to it. You get back the vector $(\cos(\theta)+x\sin(\theta),zzz)$ where $zzz$ is stuff we don't care about.

The point is that this new vector needs to be $e^{\pm i\theta}$ times the old one. That means we have: $$1\times e^{\pm i\theta} = \cos(\theta) + x\sin(\theta).$$ But the left hand side is $\cos(\theta)\pm i\sin(\theta)$. Therefore, $x=\pm i$.


Another decent way to do this problem:

Let's rewrite that rotation matrix into its Pauli spin components: $$\Omega = \begin{bmatrix}\cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{bmatrix} = \cos(\theta)\begin{bmatrix}1&0\\0&1\end{bmatrix} +i\sin(\theta)\begin{bmatrix}0&-i\\+i&0\end{bmatrix}\\ =\cos(\theta)1 + i\sin(\theta)\sigma_y,$$ where "1" is the unit matrix and $\sigma_y$ is the Pauli spin matrix.

Well you know that any vector is an eigenvector of "1". So what you really want are the eigenvectors of $i\sin(\theta)\sigma_y$. But these have to be the same as the eigenvectors of $\sigma_y$ (assuming $\sin(\theta)$ isn't zero).

Now to get the eigenvectors of $\sigma_y$, note that: $$(\sigma_y)^2 = 1.$$ Therefore: $$(1\pm \sigma_y)\;\;\sigma_y = \sigma_y \pm (\sigma_y)^2 = \sigma_y\pm 1 = \pm(1\pm \sigma_y),$$ and therefore $1\pm \sigma_y$ is an eigenvector of $\sigma_y$ with eigenvalue $\pm 1$.

Writing this out in matrix form we have the eigenvectors: $$1\pm\sigma_y = \begin{bmatrix}1&\mp i\\\pm i&1\end{bmatrix},$$ and so two choices of the eigenvectors are the columns of the above matrix, i.e.: $$\begin{bmatrix}1\\\pm i\end{bmatrix},\;\;\;\textrm{and}\;\;\; \begin{bmatrix}\mp i\\1\end{bmatrix}.$$ Note that these two solutions are the same, except for an arbitrary phase; if you multiply the first solution by $\mp i$ you get the second one. Your choice.

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