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

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.

• Welcome to QCSE! Please do not use images for text and formulas. For text, use text. For formulas, use mathjax. Note that you can edit your post to improve it. Dec 7, 2021 at 22:55
• While it's not clear from the wording whether this is an acceptable approach, as a fallback position, you could always just verify that the eigenvectors satisfy the equation $\hat M|m_i\rangle=m_i|m_i\rangle$. Dec 8, 2021 at 7:44
• Also, have you tried, just to get started, rewriting $x\pm iy$ inside $\hat M$ as $e^{\pm i\phi}\sqrt{x^2+y^2}$? Dec 8, 2021 at 7:47
• related on math: math.stackexchange.com/q/4103294/173147
– glS
Dec 8, 2021 at 10:38

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.