# Finding eigenvectors of an observable [closed]

I am trying to find the eigenvectors of the operator: $$S_{u}=\frac{\hbar}{2}\left( \begin{matrix} \cos{\theta} & e^{-i\phi}\sin{\theta} \\ e^{i\phi}\sin{\theta} & -\cos{\theta} \end{matrix} \right)$$

I have already found that the eigenvalues are $+\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$, and that the eigenvectors are something like:

$$|+\rangle_u=\left( \begin{array}{c} 1 \\ \frac{1-\cos{\theta}}{\sin{\theta}}e^{i\phi} \end{array} \right) \quad\text{and}\quad |-\rangle_u=\left( \begin{array}{c} 1 \\ -\frac{1+\cos{\theta}}{\sin{\theta}}e^{i\phi} \end{array} \right)$$

But how do I write these in a more elegant way (in terms of half angles $\phi/2$ and $\theta/2$)?

## closed as off-topic by ZeroTheHero, Jon Custer, stafusa, Cosmas Zachos, tpg2114♦Aug 5 at 2:57

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Your goal is a little bit excessive: you want to rephrase things in terms of the half-angle $\theta/2$, but $\phi$ should be left as-is. With that in mind, then, you simply see $\sin(\theta)$ and $\cos(\theta)$ as double-angle trigonometrics applied to $\theta/2$, and use the standard trigonometric identities with this in mind: $$\sin(\theta) = 2\sin(\theta/2)\cos(\theta/2),$$ and $$\sin^2(x)=\frac12-\frac12\cos(2x)$$ with $x=\theta/2$, rephrased as $$1-\cos(\theta)=2\sin^2(\theta/2)$$ (and similarly $1+\cos(\theta) = 2\cos^2(\theta/2)$). You then put these into your formulas to get $$\frac{1+\cos(\theta)}{\sin(\theta)} = \frac{2\cos^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \frac{\cos(\theta/2)}{\sin(\theta/2)}$$ and similarly $$\frac{1-\cos(\theta)}{\sin(\theta)} = \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \frac{\sin(\theta/2)}{\cos(\theta/2)}.$$
The only thing left to do now is to multiply by the denominators, and that leaves your states in the canonical form, $$|+\rangle_u=\left( \begin{array}{c} {\cos(\theta/2)} \\ {\sin(\theta/2)}e^{i\phi} \end{array} \right) \quad\text{and}\quad |-\rangle_u=\left( \begin{array}{c} {\sin(\theta/2)} \\ -{\cos(\theta/2)}e^{i\phi} \end{array} \right),$$ though it can help to rephrase the second state as $$|-\rangle_u=\left( \begin{array}{c} {\sin(\theta/2)} \\ -{\cos(\theta/2)}e^{i\phi} \end{array} \right) =\left( \begin{array}{c} {\cos\left(\frac{\pi-\theta}{2}\right)} \\ \sin\left(\frac{\pi-\theta}{2}\right)e^{i(\phi+\pi)} \end{array} \right),$$ where the transformation $\theta\mapsto\pi-\theta$, $\phi\mapsto\phi+\pi$ makes it explicit that $|-\rangle_u$ has the same form as $|+\rangle_u$ under an inversion in the Bloch sphere.
Normalizing the kets and using trigonometric identities : $\cos\theta=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$ and $\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$. You can write the eigenkets as :
$$|+\rangle_{u}=\begin{pmatrix}e^{-i\frac{\phi}{2}}\cos\frac{\theta}{2} \\e^{i\frac{\phi}{2}}\sin\frac{\theta}{2}\end{pmatrix}$$
$$|-\rangle_{u}=\begin{pmatrix}e^{-i\frac{\phi}{2}}\sin\frac{\theta}{2} \\-e^{i\frac{\phi}{2}}\cos\frac{\theta}{2}\end{pmatrix}.$$