# For general 2d complex traceless $\sigma$, the U that diagonalises $\sigma^\dagger \sigma$ leaves $U^\dagger \sigma U$ with zeros on the diagonal?

The general $$2\times2$$ complex traceless matrix can be written in terms of the Pauli matrices $$\sigma = a^i \sigma^i$$. Consider also its conjugate $$\sigma^\dagger=a^{*i}\sigma^i$$. I was interested in diagonalising the hermitian matrix $$\sigma^\dagger \sigma$$, which I did and found a relatively complicated unitary matrix $$U$$ that did so $$U^\dagger(\sigma^\dagger \sigma)U = D ={\rm diag}(\lambda_-,\lambda_+),$$ where the eigenvalues are $$\lambda_\pm = |a|^2\pm|a^* \times a |$$. I was then interested in the diagonal elements of $$U^\dagger \sigma U$$ and found that they were $$\propto \vec{a}\cdot (a^* \times a)=0$$. My question is, could I have realised that $$U^\dagger \sigma U$$ had only off-diagonal non-zero elements without having to compute $$U$$ itself?

My thoughts: Inserting $$1= U^\dagger U$$ above I realised that $$(U^\dagger \sigma U)^\dagger (U^\dagger \sigma U)=D$$ $$U^\dagger \sigma U$$ is a complex traceless matrix and so was wondering if any such matrix multiplied by its conjugate transpose giving a diagonal matrix with distinct entries must have zero on the diagonal. However upon examining $$\left(\begin{matrix} \alpha & \beta \\ \gamma & -\alpha \end{matrix} \right) \left(\begin{matrix} \alpha^* & \gamma^* \\ \beta^* & -\alpha^* \end{matrix} \right)= \left(\begin{matrix} \lambda_- & 0 \\ 0 & \lambda_+ \end{matrix} \right) \quad \alpha,\beta,\gamma \in \mathbb{C}$$ one has only constraints $$\alpha \gamma^*-\beta \alpha^*=0, \quad |\alpha|^2+|\beta|^2=\lambda_1, \quad |\alpha|^2+|\gamma|^2=\lambda_2.$$ Sure, $$\alpha=0$$ is one solution, but it does not seem to be unique...

Edit: I have provided below a proof of a properly modified version of my statement. However I would still appreciate a more elegant proof!

The answer is no. In particular, it does not hold if $$\sigma$$ is traceless and Hermitian (e.g. for your problem in the case where vector $$a$$ is real). Without loss of generality, scale $$\sigma$$ so that $$\sigma^2 = I$$. If we select a unitary $$U$$ that diagonalizes $$\sigma$$, then we find that $$U^\dagger \sigma U = \pmatrix{1&0\\ 0&-1} \implies U^\dagger \sigma^\dagger \sigma U = [U^\dagger \sigma U]^\dagger U^\dagger \sigma U = \pmatrix{1&0\\ 0&1}.$$ So, $$U$$ diagonalizes $$\sigma^\dagger \sigma$$, but $$U^\dagger \sigma U$$ does not have zeros on the diagonal.
• I agree. However, I did specify (although did not stress) in the body of the question that I was considering the case of distinct eigenvalues. I think I have now solved the problem and shown that with the exception of cases when $a$ is real, the statement is correct. – Rudyard May 22 at 20:05
I can write $$U^\dagger \sigma U = b^i\sigma^i$$ for some $$b^i \in \mathbb{C}$$. Notice, if $$b^3=0$$ then diagonal elements are zero. Now consider $$(b^i\sigma^i)^\dagger b^j\sigma^j=\left( \begin{matrix} |b|^2 +i(b^*\times b)^3 & i(b^*\times b)^1+(b^*\times b)^2 \\ i(b^*\times b)^1-(b^*\times b) & |b|^2 -i(b^*\times b)^3\end{matrix}\right) \overset{!}{=} \left( \begin{matrix} \lambda_- & 0 \\0 & \lambda_+ \end{matrix} \right)$$ The off-diagonal constraints imply $$(b^*\times b)^1 = b^{2*}b^3 - b^{3*}b^2 = 0$$ $$(b^*\times b)^2 = b^{3*}b^3 - b^{3*}b^2 = 0$$ which imply $$b^3 \left( 1- \frac{b^{2*}b^1}{b^2 b^{1*}}\right)=0$$ $$\implies {\rm either} \; b^3=0 \;({\text{ diagonal elements are zero}}) \quad {\rm or} \; (b^* \times b)^3 =0 \;$$ but in the latter case we would have $$\lambda_\pm = |b|^2$$ equal (which happens when initial vector $$a^i$$ is real).
So in conclusion, if $$a^i$$ is not a completely real vector, then $$U$$ that diagonalises $$(a^{i*}\sigma^i a^j \sigma^j)$$ leaves $$U^\dagger a^i\sigma^i U$$ with zeros on the diagonal