Cfrgtkky

I'd like to evaluate generic expressions of the following form:

$$frac{d}{da}expleft[aX + bYright]$$

where $a,b$ are scalars and $X,Y$ are arbitrary complex matrices. Replacing the exponential with its defining series gives us an infinite series of terms which are each symmetrized by the derivative:

$$sum_{n=0}^{infty}frac{1}{n!}frac{d}{da}left(aX+bYright)^{n}$$

For say $n=3$ this gives a term of the form

$$X(aX+bY)^{2}+(aX+bY)X(aX+bY)+(aX+bY)^{2}X$$

I don't see a clean way to resum such a series into a nice analytic form, but I thought maybe there was a clean result for this up to commutators. Any insights or literature suggestions are very welcome.

In general, the derivative of the exponential map is given by
$$
frac{d}{dt}e^{Z(t)} = e^{Z}frac{1 - e^{-mathrm{ad}_{Z}}}{mathrm{ad}_{Z}}frac{dZ(t)}{dt}
$$
Thus, for your case of $Z(a) = aX + bY$, we have
$$
begin{align*}
frac{d}{da}e^{Z} &= e^{Z}frac{1 - e^{-mathrm{ad}_{Z}}}{mathrm{ad}_{Z}}(X)
\ & = e^Z sum_{k=0}^infty frac{(-1)^k}{(k+1)!} mathrm{ad}_Z^k(X)
end{align*}
$$
Where $mathrm{ad}_Z^k(X)$ denotes $[overbrace{Z,[Z,cdots,[Z}^{k text{ times}},X]cdots]]$.

From the other answer based on Hall's text, we also have
$$
left. frac{d}{da} e^Z right|_{a = 0} = e^{bY}left{
X - frac 1{2!}b[Y,X] + frac 1{3!}b^2[Y,[Y,X]] - cdots
right}
$$

The number $b$ is really irrelevant to your question.

For all $X,Yin M_n(mathbb{C})$, we have

$$
frac{d}{dt}e^{X+tY}big|_{t=0}=e^Xleft{
Y-frac{[X,Y]}{2!}+frac{[X,[X,Y]]}{3!}-cdots
right}.
$$

More generally, if $X(t)$ is a smooth matrix-valued function, then
$$
frac{d}{dt}e^{X(t)}=e^{X(t)}left{
frac{I-e^{-operatorname{ad}_{X(t)}}}{operatorname{ad}_{X(t)}}bigg(frac{dX}{dt}bigg)
right}
$$

See Theorem 5.4 (Derivative of Exponential) and its proof in Brian Hall's Lie Groups, Lie Algebras, and Representations.

If $X$ and $Y$ commute, then
$$
frac{d}{da}e^{aX+bY}=frac{d}{da}e^{aX}e^{bY}=Xe^{aX+bY}.
$$

Let me add a different way of writing the result.
If $X(t)$ is an operator-valued function, then
$$ frac{mathrm d}{mathrm dt} mathrm e^{X(t)} = int_0^1 mathrm e^{(1-lambda) X(t)} frac{mathrm dX}{mathrm dt} mathrm e^{lambda X(t)}; mathrm dlambda . tag{*} $$

Proof.
We begin with the identity
$$ mathrm e^{-Lambda X} frac{mathrm d}{mathrm dt} mathrm e^{Lambda X} = int_0^Lambda mathrm e^{-lambda X} frac{mathrm dX}{mathrm dt} mathrm e^{lambda X}; mathrm dlambda . $$
These two expressions are equal because they satisfy the same initial value problem as functions of $Lambda$.
Setting $Lambda = 1$ gives the desired result (*).

Applied to your example:
$$ frac{mathrm d}{mathrm da} mathrm e^{aX + bY} = int_0^1 mathrm e^{(1-lambda)(aX+bY)}, X, mathrm e^{lambda(aX+bY)}; mathrm dlambda . $$