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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!

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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.

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  • $\begingroup$ 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. $\endgroup$ – Rudyard May 22 at 20:05
  • $\begingroup$ @rudyard sorry I missed that $\endgroup$ – Ben Grossmann May 22 at 22:32
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I think I have solved this:

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

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