-1
$\begingroup$

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 $\theta/2$)?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ZeroTheHero, Jon Custer, stafusa, Cosmas Zachos, tpg2114
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 :

$$|+\rangle_{u}=\begin{pmatrix}e^{-i\frac{\phi}{2}}\cos\frac{\theta}{2} \\e^{i\frac{\phi}{2}}\sin\frac{\theta}{2}\end{pmatrix}$$

and

$$|-\rangle_{u}=\begin{pmatrix}e^{-i\frac{\phi}{2}}\sin\frac{\theta}{2} \\-e^{i\frac{\phi}{2}}\cos\frac{\theta}{2}\end{pmatrix}.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.