Eigenvector of 2-by-2 unitary matrix in spherical coordinates

Question

I was reading basic quantum mechanics, and stumbled upon this math problem.

$$\begin{pmatrix} \cos\theta & e^{-i\phi} \sin \theta \\ e^{i\phi}\sin\theta & -\cos\theta\end{pmatrix}$$

Here I need to find the eigen values and eigen vectors corresponding to this matrix. I have successfully found out the Eigen Value of this vector, which is +1, and -1.

When I am trying to find the eigen vector, I am getting lost. I have been to this position (by choosing eigen value = +1) -->

$$\begin{pmatrix} \cos\theta-1 & e^{-i\phi} \sin \theta\\ e^{i\phi}\sin\theta & -\cos\theta-1\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$$ The answer is $\begin{pmatrix} \cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2}\end{pmatrix}$.

Whatever solution I try, I am not reaching to this solution. Can anyone of you please at least show me the right direction?

My background is NOT hardcore maths, and I am a software engineer, studying quantum mechanics and quantum computing by passion. Sorry if this question sounds silly. But any help is appreciated. I need to solve it to move to the forward chapters. Thank you.

Answer 1

A good starting point here are the double-angle identities from trigonometry: $$\cos\theta=1-2\sin^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}-1,\quad \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.$$ Inserting these into the second matrix of interest and simplifying a little gives

\begin{align} \begin{pmatrix} \cos\theta-1 & e^{-i\phi} \sin \theta\\ e^{i\phi}\sin\theta & -\cos\theta-1\end{pmatrix} &=\begin{pmatrix} (1-2\sin^2\frac{\theta}{2})-1 & e^{-i\phi} \cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\ e^{i\phi}\cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} & -(2\cos^2\frac{\theta}{2}-1)-1\end{pmatrix}\\ &=\begin{pmatrix} -2\sin^2\frac{\theta}{2} & 2e^{-i\phi} \sin\frac{\theta}{2}\cos\frac{\theta}{2}\\ 2e^{i\phi}\sin\frac{\theta}{2}\cos\frac{\theta}{2} & -2\cos^2\frac{\theta}{2}\end{pmatrix}\\ &=2\sin\frac{\theta}{2} \cos\frac{\theta}{2} \begin{pmatrix} -\sin\frac{\theta}{2}/\cos\frac{\theta}{2} & e^{-i\phi} \\ e^{i\phi} & -\cos\frac{\theta}{2}/\sin\frac{\theta}{2}\end{pmatrix}\\ \end{align} The prefactor is generally non-zero, so $(x_1,x_2)$ must be annihilated by the remaining matrix. Typically one then deduces $x_1,x_2$ "by inspection", i.e., by looking at this matrix and reading off the answer. But to be more explicit, we need $$\begin{pmatrix} -\sin\frac{\theta}{2}/\cos\frac{\theta}{2} & e^{-i\phi}\\ e^{i\phi} & -\cos\frac{\theta}{2}/\sin\frac{\theta}{2}\end{pmatrix} \begin{pmatrix} x_1\\ x_2\end{pmatrix} = \begin{pmatrix} x_2 e^{-i \phi} -x_1 \sin\frac{\theta}{2}/\cos\frac{\theta}{2}\\ x_1 e^{i\phi} -x_2 \cos\frac{\theta}{2}/\sin\frac{\theta}{2} \end{pmatrix}$$ to vanish. This amounts to $x_2/x_1=e^{i\phi}\sin\frac{\theta}{2}/\cos\frac{\theta}{2}$; the given eigenvector satisfies this, and thus is a genuine solution. It's worth noting, though, that this isn't the only solution: any multiple of $(x_1,x_2)$ also solves this. The most obvious motivation for this choice of solution is that it's normalized: $|x_1|^2+|x_2|^2=1$. Beyond that, it's just a fairly convenient option: $x_1$ will always be real, and the vector is explicitly $2\pi$-periodic in $\phi$.

Answer 2

Eignevalues of $A$ are roots of the characteristic equation $$ \det(xI-A)=0 $$ in your case $$ (x-\cos\theta)(x+\cos\theta)-(-e^{i\phi}\sin\theta)(-e^{-i\phi}\sin\theta)=x^2-1 $$ so the eigenvalues are $\pm1$.

Once you have an eigenvalue $\lambda$, you can find solutions to $Ax=\lambda x$ or $(A-\lambda I)x=0$ (as you say you're trying). This is simple linear algebra with which you should be familiar (e.g. row reduction/Gaussian elimination). Note that in this circumstance there is a whole subspace of solutions (i.e. eigenvectors are not unique).

In your case, there are two 1-dimensional eigenspaces (i.e. the eigenvector for each $\pm1$ eigenvalue is unique up to scalar multiplication).

In your example for $\lambda=1$, you can take something orthogonal to, say, the second row of $\lambda I-A$ (which will be orthogonal to the first row as well), say $$ (1+\cos\theta, \ e^{i\phi}\sin\theta). $$ Some trigonometric identities will get this to the answer you gave (as noted in another answer), say $$ 1+\cos\theta=2\cos^2(\theta/2), \quad \sin\theta=2\cos(\theta/2)\sin(\theta/2). $$ These can be derived from the Pythagorean $$ \sin^2\theta+\cos^2\theta=1 $$ and sum $$ \cos(a+b)=\cos a\cos b-\sin a\sin b, \quad \sin(a+b)=\sin a\cos b+\cos a\sin b $$ identities.

Answer 3

A nice way to write an eigenvector of $A=\begin{pmatrix}a&b\\ c&d \end{pmatrix}$ corresponding to the eigenvalue $\lambda$ is $v=\begin{pmatrix}b\\ \lambda-a\end{pmatrix}$. Using this result for your matrix, the eigenvector must be $v=\begin{pmatrix}e^{-i\phi}\sin\theta\\ 1-\cos\theta\end{pmatrix}$.

A good possibility is that the your eigenvector is scalar multiple of $\begin{pmatrix} \cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2}\end{pmatrix}$ which is actually the case as:
If $e^{-i\phi}\sin\theta=k\cos\theta/2$
and $1-\cos\theta=ke^{i\phi}\sin\theta/2$
then on dividing you get an identity $\frac{1-\cos\theta}{\sin\theta}=\tan\theta/2$.