When is a *-derivation inner (in matrix- or C*-algebra)? Let $A$ be a C*-algebra (actually, the finite dimensional case of $A=Mat(n\times n,\mathbb{C})$ is my main interest at the moment) and $\delta\colon A\to A$ a *-derivation, i.e. a linear map with $\delta(ab)=\delta(a)b+a\delta(b)$ and $\delta(a^*)=\delta(a)^*$. Clearly, the commutator with a self-adjoint element is an example, that is $\delta(a)= i[h,a] = i(ha-ah)$.
My question is: What are the conditions on $\delta$ such that such an $h$ exists (and how can I obtain it)?
A bit of background: This is like a non-commutative version of Noether's theorem: If $A$ arises from the algebra of functions on a symplectic space then derivations are vector fields and Hamiltonian ones are those that arise from taking Poisson brackets with the "Noether charge". There is a differential condition (making sure the Lie derivative of the symplectic structure vanishes) plus a topological one making sure the closed 1-form obtained from contracting the vector field with the symplectic form is in fact exact.
The case of 2x2 matrices, I can figure out myself, but that makes use of Pauli matrices and the fact that traceless 2x2 matrices can be represented by 3-vectors with the vector product (i.e. this might be very special due to low dimensionality).
 A: Derivations on a von Neumann algebra are always inner (even if $\delta\ne\delta^*$).
In the concrete case of a $^*$-derivation in $A=M_n(\mathbb C)$, one can show that $\alpha_t=\exp(t\,\delta)$ is a one-parameter group of automorphisms. Every automorphism of $A$ is inner, so $\alpha_t=u_t\cdot u_t^*$ for some one-parameter unitary group (proof at the end). By Stone's Theorem, $u_t=e^{ith}$ for some selfadjoint $h\in A$.
So
$$
\alpha_t(x)=e^{ith}xe^{-ith},\ \ \ t\in\mathbb R,\ \ x\in A. 
$$
Then
\begin{align}
\delta(x)&=\lim_{t\to0}\frac1t\,(e^{t\delta }-\text{Id})(x)=\lim_{t\to0}\frac1t\,(e^{t\delta(x)}-x)\\ \ \\
&=\lim_{t\to0}\frac1t\,(e^{ith}xe^{-ith}-x)=\lim_{t\to0}\frac1t\,(e^{ith}x-xe^{ith})e^{-ith}\\ \ \\
&=ihx-ixh=i[h,x].
\end{align}

Edit: it is not obvious at first sight why $\{u_t\}$ is continuous. What one has automatically is that $\alpha_t=u_t\cdot u_t^*$ but there is a priori no connection between the different $u_t$. But because $\alpha_t$ is continuous, $$u_{t+h}Au_{t+h}^*-u_tAu_t^*\to0$$ for all $A$. This we may write as $$w_hA-Aw_h\to0,$$ with $w_h=u_t^*u_{t+h}$. Since the unitary group is compact, the net $\{w_h\}$ has a convergent subsequence. If $w_{h_j}\to w$, then $w$ is a unitary and $wA-Aw=0$ for all $A$, so $w=I$; in other words, the only cluster point of $\{w_h\}$ is $I$. Now suppose that $\{w_h\}$ is not convergent; then we can find $\varepsilon>0$ and a subnet $\{w_{h_j}\}$ such that $\|w_{h_j}-I\|>\varepsilon$ for all $j$; this is a contradiction, since by the above this subset would have a subnet converging to $I$. So $w_h\to I$, which is the same as $u_{t+h}\to u_t$.
A: Every derivation $\,\delta\colon A\to A\,$ where $A=M_n(\mathbb{C})\,$ is inner. You may take the following track towards
$\,\delta = [h,\,\cdot\:]\,$ with $\,h\in A$,
originally proposed by I. Kaplansky (to my best knowledge).
Let $\,p\in A\,$ be a rank-one projection.
By the following argument one may assume WLOG that $\delta(p)=0\,$:
From $\,\delta(p) = \delta(p^2) = p\delta(p) + \delta(p)p\,$ one concludes that $\,p\delta(p)p = 0$. With $y:=[p, \delta(p)]$ one has
$$[p,y] = p\big(p\delta(p)-\delta(p)p\big)
 -\big(p\delta(p)-\delta(p)p\big)p = \delta(p)$$
and after defining $\,\delta'(a):=\delta(a)-[a,y]\,$ one reaches $\delta'(p)=0$.
Now $\delta(ap)=\delta(a)p\;\forall\, a\,$ shows that
$$ap\longmapsto\delta(a)p$$
is a well-defined linear map on the left-ideal $\,Ap$. Because of $Ap\cong\mathbb{C}^n\,$ there's an $h\in A$ satisfying $hap=\delta(a)p$, and this $h$ brings us to the closing line:
$$haxp\:=\:\delta(ax)p \:=\: \delta(a)xp+a\delta(x)p
 \:=\: \delta(a)xp+ahxp\qquad\forall\, x\in A \\[2.7ex]
\implies\;\delta(a)\,=\,[h,a]$$
