How to find the eigenstates of a general $2\times 2$ Hermitian matrix?

Question

Given a measurement operator in the general Hemitian form $$ M = \begin{pmatrix} z_1 & x+iy \\ x-iy & z_2\end{pmatrix}, $$ where $x,y,z_1,z_2 \in \mathbb{R}$, show that the eigenvalues are $$ m_{1,2} = \frac{z_1 + z_2}{2} \pm \frac{1}{2}\Big[ (z_1-z_2)^2 + 4(x^2+y^2)\Big]^{1/2} $$ with eigenvectors $$ |m_1\rangle = \begin{pmatrix}e^{i\varphi}\cos(\theta/2) \\ \sin(\theta/2)\end{pmatrix} $$ and $$ |m_2\rangle = \begin{pmatrix}-e^{i\varphi}\sin(\theta/2) \\ \cos(\theta/2)\end{pmatrix}, $$ where $\varphi = \tan^{-1}(y/x)$ and $\theta = \cos^{-1}([z_1-z_2]/[m_1-m_2])$.

I have found the eigenvalues but I am not able to find eigenstates in that form after hours of work.

Answer 1

1. You can find this from the general expressions for eigenvalues and eigenvectors of an arbitrary $2\times2$ matrix. Let $$A\equiv\begin{pmatrix}a&b\\c&d\end{pmatrix}.$$ You can readily verify that its eigenvalues can be written as $$\newcommand{tr}{\operatorname{tr}}\lambda_\pm = \frac{a+d}{2} \pm \sqrt{\left(\frac{a-d}{2}\right)^2 + bc} = \frac{\tr(A)}{2} \pm \sqrt{\left(\frac{\tr(A)}{2}\right)^2 - \det(A)^2}.$$ You can check the consistency of these with the expressions you have for the case of Hermitian matrices.

2. You can also verify that, given any eigenvalue $\lambda$, you can write the corresponding (not normalised) eigenvector as $$v_\lambda = \begin{pmatrix}b \\ \lambda - a\end{pmatrix} = \begin{pmatrix}\lambda-d \\ c\end{pmatrix},$$ where the two expressions coincide precisely when $\lambda$ is an eigenvalue.

3. So, let us specialise to the Hermitian case, where $a,d\in\mathbb R$ and $c=\bar b$. Using the first form given above for the eigenvector, we immediately get $$v_\pm = \begin{pmatrix} b\\ -\Delta \pm\sqrt{\Delta^2+|b|^2}\end{pmatrix} = \begin{pmatrix} b\\ -\Delta \pm S\end{pmatrix},$$ where I defined $\Delta\equiv(a-d)/2$ and $S\equiv \sqrt{\Delta^2+|b|^2}$. To see the similarity with your expression, we need to normalise this. Observe then that $$\|v_\pm\|^2 = 2\left(|b|^2 +\Delta^2 \mp \Delta S\right) = \pm 2S\left(-\Delta \pm S\right).$$ If we were to just look at the corresponding expression for $|\pm\rangle\equiv v_\pm/\|v_\pm\|$, things would get ugly pretty fast. But don't really need to. Instead, we make the following observations:

   1. The only phase in our expression for $v_\pm/\|v_\pm\|$ comes from $b$, so that's easy to take into account: we just have $b=e^{i\varphi}|b|$.
   2. Generally speaking, if $x^2+y^2=1$, then we can write $x=\cos\alpha$ and $y=\sin\alpha$ for $\alpha=\arctan(y/x)=\arccos(x)$. So to get the $\theta$ for our eigenvectors, as an expression with an $\arccos$, it is enough to look at its first component. You also want a final expression containing half the angle, and thus we have $\theta/2=\arccos(b/\|v_\pm\|)$.
   3. We thus reduced the problem to finding a nice expression for $$\theta = 2\arccos\left(\frac{|b|}{\|v_\pm\|}\right). %= \arccos\left(\frac{a-d}{\lambda_+-\lambda_-}\right) %= \arccos\left(\frac{a-d}{2S}\right).$$ Remembering that $\cos(2\arccos(A))=2A^2-1$, we get $$2\arccos\left(\frac{|b|}{\|v_\pm\|}\right) = \arccos\left(2\frac{|b|^2}{\|v_\pm\|^2}-1\right) = \arccos\left(\frac{2|b|^2-\|v_\pm\|^2}{\|v_\pm\|^2}\right).\tag A$$ We are pretty much there now. Observe that $$2|b|^2-\|v_\pm\|^2 = -2(\Delta^2 \mp \Delta S) = 2\Delta(-\Delta\pm S),$$ and thus (A) becomes $$\theta = \arccos\left(\frac{2\Delta(-\Delta\pm S)}{\pm2S (-\Delta\pm S)}\right) = \arccos\left(\pm\frac{\Delta}{S}\right).$$ But also, $\lambda_+-\lambda_-=2S$, and thus you get the final expression you were looking for.
   4. To be closer to the way the problem was stated: you can think of the above procedure as applied only to the eigenvalue $\lambda_+$, and thus obtaining your expression for $|m_1\rangle$. Once you have that, the expression for $|m_2\rangle$ follows immediately, because given any vector $(a,b)$ you can write its orthogonal as $(\bar b,-\bar a)$. You then just need to remember that you can always multiply by a global phase factor without problems.