Derivative of square root of unitary matrix

Question

Given a (special) unitary matrix $U(x) \in SU(2)$ which is a function of $x\in\mathbb{R}$, along with its unitary square root $u(x) = \sqrt{U(x)}$ such that $u(x)^2 = U(x)$ and $u(x)^\dagger u(x) = \mathbb{1}_{2\times2}$, what is the derivative $$\frac{d}{dx}u(x)=\frac{d}{dx}\sqrt{U(x)}?$$

Can it be expressed purely in terms of products and sums of $u(x)$, $u(x)^\dagger$, $U(x)$, $U(x)^\dagger$, $\frac{d}{dx}U(x)$ and $\frac{d}{dx}U(x)^\dagger$? Naively, it seems like it should be something like (suppressing arguments) $$\frac{du}{dx} = \frac{1}{2}\left(\frac{1}{2} u^\dagger \frac{dU}{dx} + \frac{1}{2}\frac{dU}{dx}u^\dagger \right),$$ following from a "symmetrized" form of the regular chain rule $\tfrac{d}{dx}\left[f(x)\right]^{1/2}=\frac{1}{2} \left[f(x)\right]^{-1/2}f'(x)$ with the identification of $u\leftrightarrow \left[f(x)\right]^{1/2}$ and $u^\dagger = u^{-1} \leftrightarrow \left[ f(x)\right]^{-1/2}$.

Note: Here $U^\dagger$ is the conjugate transpose.

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Edit: I am actually interested in the quantity $$ u^\dagger \frac{du}{dx} + u \frac{du^\dagger}{dx}, $$ so even if there is not a way to write $\tfrac{du}{dx}$ simply, an expression for this would be sufficient.

Answer 1

$ \def\R#1{{\mathbb R}^{#1}} \def\o{{\tt1}} \def\mbrace#1{\left\lbrace\begin{array}{r}#1\end{array}\right\rbrace} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{tr}\LR{#1}} \def\qiq{\quad\implies\quad} $Given an $\R{4}$ unit vector $$p=\mbrace{a\\b\\c\\d},\qquad\quad \|p\|^2=\o$$ Construct an $SU(2)$ matrix (corresponding to a unit quaternion) $$\eqalign{ U &= \m{a+ib & c+id\\-c+id & a-ib} \\ U^\dagger U&=I \\ \det(U) &= (a^2+b^2)+(c^2+d^2) = \o \\ \trace{U} &= 2a \\ \trace{I+U} &= \LR{2+2a} \\ }$$ The square root of a quaternion has a known formula $${\sqrt U} = \frac{\qquad I+U}{\sqrt{\trace{I+U}}} \qiq u = \LR{2+2a}^{-1/2}\LR{I+U}$$ Differentiation yields $$\eqalign{ \dot u &= \LR{2+2a}^{-1/2}\dot U \;\:-\; \LR{2+2a}^{-3/2}\LR{I+U}\dot a \\ \dot u^\dagger &= \LR{2+2a}^{-1/2}\dot U^\dagger \;-\; \LR{2+2a}^{-3/2}\LR{I+U^\dagger}\dot a \\ }$$ where a dot is used to denote the derivative with respect to $x$.

The variable $a$ can be replaced by the trace and $U$ can be expanded in terms of Pauli matrices (if those are your preferences).