# On the usage of derivative in operator theory

In quantum mechanics, we work with linear operators on Hilbert spaces $$\mathscr{H}$$. Suppose I have two bounded ones, defined on the same space $$A, B: \mathscr{H}\to\mathscr{H}$$.

It seems to me there is an ambiguity on the way to deal with the derivative.

On one way, the operator $$AB$$ is usually interpreted as the composition $$(A\circ B)f:=A(B(f))$$ for every test function $$f$$ on $$\mathscr{H}$$. If so, the derivative $$D$$ operator on $$AB$$ should act as follows $$D[AB]f = D[(A\circ B)f]= D[A(Bf)]D(B(f))$$ On the other way, the following right examples treat $$AB$$ as if it were literally a product of operators instead of a composition. In other words, the preferred way to compute the derivative is the Leibniz rule $$D[AB]= AD[B]+BD[A]$$

$$1^{\rm{st}}$$ example, by E. Pisanty:

The exponential of an operator $$\hat A(t)$$ does not obey the differential equation $$\frac{d}{ dt}e^{\hat A(t)} \stackrel{?}{=} \frac{d \hat{A}}{ dt} e^{\hat A(t)}$$ that one might naively hope to satisfy. To see why this does not work, consider the series expansion of the exponential \begin{align*} \frac{d}{dt}e^{\hat A(t)} & = \frac{d}{dt}\sum_{n=0}^\infty \frac{1}{n!} = \sum_{n=0}^\infty \frac{1}{n!} \frac{d}{dt} \hat A^n(t), \end{align*} When we apply the product rule, we get the individual derivatives of each of the operators in the product, at their place within the product $$\begin{equation*} \frac{d}{dt} \hat A^n(t) = \frac{d\hat A}{dt} \hat A^{n-1}(t) +\hat A(t)\frac{d\hat A}{dt} \hat A^{n-2}(t) + \ \dots \ +\hat A^{n-2}(t)\frac{d\hat A}{dt} \hat A(t) +\hat A^{n-1}(t)\frac{d\hat A}{dt} \end{equation*}$$ This can simplify to just $$n\frac{ d\hat A}{dt} \hat A^{n-1}(t)$$, in which case $$\frac{d}{dt}e^{\hat A(t)} = \frac{d \hat A}{dt} \sum_{n=1}^\infty \frac{\hat A^{n-1}(t)}{(n-1)!} = \frac{d\hat A}{dt}e^{\hat{A}(t)}$$, but only under the condition that $$\hat A(t)$$ commute with its derivative $$\left[\frac{ d\hat A}{ dt} , \hat A(t)\right] \stackrel{?}{=} 0$$

In this case, $$A^n$$ is seen as a product of $$A$$ with itself $$n$$ times, instead of $$A \circ A \circ \dots \circ A$$ $$n$$ times.

$$2^{\rm{nd}}$$ example, by Wikipedia:

The expectation value of an observable $$A$$, which is a Hermitian linear operator, for a given Schrödinger state $$\vert\psi(t)\rangle$$, is given by $${\displaystyle \langle A\rangle _{t}=\langle \psi (t)|A|\psi (t)\rangle .}$$ In the Schrödinger picture, the state $$\vert\psi(t)\rangle$$ at time $$t$$ is related to the state $$\vert\psi(0)\rangle$$ at time $$0$$ by a unitary time-evolution operator $$U(t)$$: $${\displaystyle |\psi (t)\rangle =U(t)|\psi (0)\rangle .}$$ In the Heisenberg picture, all state vectors are considered to remain constant at their initial values $$\vert \psi(t)\rangle$$, whereas operators evolve with time according to $${\displaystyle A(t):=U^{\dagger }(t)AU(t)\,.}$$ The Schrödinger equation for the time-evolution operator is $${\displaystyle {\frac {d}{dt}}U(t)=-{\frac {iH}{\hbar }}U(t)}$$ where $$H$$ is the Hamiltonian and $$\hbar$$ is the reduced Planck constant. It now follows that {\displaystyle {\begin{aligned}{\frac {d}{dt}}A(t)&={\frac {i}{\hbar }}U^{\dagger }(t)HAU(t)+U^{\dagger }(t)\left({\frac {\partial A}{\partial t}}\right)U(t)+{\frac {i}{\hbar }}U^{\dagger }(t)A(-H)U(t)\\&={\frac {i}{\hbar }}U^{\dagger }(t)HU(t)U^{\dagger }(t)AU(t)+U^{\dagger }(t)\left({\frac {\partial A}{\partial t}}\right)U(t)-{\frac {i}{\hbar }}U^{\dagger }(t)AU(t)U^{\dagger }(t)HU(t)\\&={\frac {i}{\hbar }}\left(H(t)A(t)-A(t)H(t)\right)+U^{\dagger }(t)\left({\frac {\partial A}{\partial t}}\right)U(t),\end{aligned}}} where differentiation was carried out according to the product rule.

I really don't understand

• Your operator is linear, so with some slight abuse of notation $AB$ and $A\circ B$ are the same thing. Dec 4, 2022 at 9:50

On top of what @Raskolnikov pointed out in their comment, I believe there is another problem here: you seem to be mixing the derivative of a Hilbert space operator $$A:\mathcal H\to\mathcal H$$ and the derivative of objects $$f:\mathbb R\to\mathcal B(\mathcal H)$$ which map into the space of (bounded linear) Hilbert space operators.
• The chain rule $$D[A\circ B](\psi)=D[A](B(\psi))\circ D[B](\psi)$$ which you mentioned first refers to the Fréchet derivative of operators. Here, given any function $$A:\mathcal H\to\mathcal H$$ one looks for the best linear approximation at a given point $$\psi\in\mathcal H$$. More precisely a bounded linear operator $$D_\psi[A]$$ (or $$D[A](\psi)$$ as you wrote) is called the (Fréchet) derivative of $$A$$ at $$\psi$$ if $$\lim_{\|h\|\to 0}\frac{\|A(\psi+h)-A(\psi)-D_\psi[A]h\|}{\|h\|}=0\,.$$ The reason this notion does not really pop up in quantum mechanics -- and thus not in the two examples you cited -- is that the best linear approximation of an operator that is already linear is the operator itself: given $$B\in\mathcal B(\mathcal H)$$ and any $$\psi\in\mathcal H$$ one finds $$D_{\psi}[B]=B$$ because $$\frac{\|B(\psi+h)-B(\psi)-D_\psi[B]h\|}{\|h\|}=\frac{\|B\psi+Bh-B\psi-Bh\|}{\|h\|}=0$$ for all non-zero $$h$$ already. With this it's also easy to verify the chain rule you mentioned: given $$A,B\in\mathcal B(\mathcal H)$$, $$\psi\in\mathcal H$$ one finds $$A\circ B=D_\psi[A\circ B]=D_{B(\psi)}[A]\circ D_\psi[B]=A\circ B\,.$$
• When considering dynamical systems (e.g., in quantum mechanics) a time parameter usually enters the picture. This parameter can then be taken as the input of a function, for example of a solution to a differential equation which describes the dynamics of some quantum system. As you may know if a system in an initial state $$\rho_0$$ is described by a Hamiltonian $$H\in\mathcal B(\mathcal H)$$ the solution $$\rho(t)$$ to the Liouville-von Neumann equation $$\frac{d}{dt}\psi(t)=-\frac{i}\hbar \underbrace{[H,\rho(t)]}_{:=H\rho(t)-\rho(t)H}\qquad\text{with}\qquad \rho(0)=\rho_0$$ is given by $$\rho(t)=e^{-\frac{i}\hbar tH}\rho_0 e^{\frac{i}\hbar tH}$$, $${}^\text{footnote 1}$$. In particular this is a function from the real numbers into (a certain subset of) operators acting on $$\mathcal H$$. Going one level higher the time-evolution operator $$U(t)=e^{-\frac{i}\hbar tH}$$ which contains all the information of how $$\rho_0$$ evolves in time is itself the solution to the differential equation $$\frac{d}{dt}U(t)=-\frac{i}\hbar HU(t)$$ with $$U(0)=\operatorname{id}$$ being the identity operator on $$\mathcal H$$. Either way it "makes sense" to talk about the derivative of the maps $$\rho(t),U(t)$$, etc. because -- while $$\rho(t),U(t)$$ for any $$t$$ are linear operators -- the map $$t\mapsto \rho(t)$$ is in general not linear. To connect this to the examples you gave in your question: both times one considers maps $$:\mathbb R\to\mathcal B(\mathcal H)$$ (e.g., $$t\mapsto e^{\hat A(t)}$$ and $$t\mapsto U(t)$$) and asks about their time-derivative. This is also why the Leibniz rule occurs here: given differentiable curves $$A,B:\mathbb R\to\mathcal B(\mathcal H)$$, when talking about the time-derivative of the product $$AB$$ -- or, equivalently, the composition $$A\circ B$$ -- what one really asks about is the derivative of the map $$AB:\mathbb R\to\mathcal B(\mathcal H)$$ defined via $$(AB)(t):=A(t)\circ B(t)$$. This derivative is given by the Leibniz rule $$D_t[A\circ B]=D_t[A]\circ B+A\circ D_t[B]$$ and not the chain rule, because the input of $$A$$ relevant for the derivative is the time-parameter $$t$$ -- and not some output of the operator $$B(t)$$.
$${}^\text{footnote 1}$$: If $$\rho_0$$ is a pure state $$|\psi_0\rangle\langle\psi_0|$$, then this reduces to the Schrödinger equation with solution $$\psi(t)=e^{-\frac{i}\hbar tH}\psi_0$$.