# Meaning of time derivative of an operator

One comment pointed out that the derivative $$\frac{\partial}{\partial t} \frac{d}{dx} = 0$$. In a certain sense however $$\frac{d}{dt} \frac{d}{dx} \neq 0$$. I will try to explain this further.

Let's think first about kinematics. Then we have a Hilbert space $$\mathscr{H}$$ and some operator $$A$$ on it. For example the Hilbert space could be the space of square-integrable functions and $$A$$ the derivative, $$\mathscr{H} = L^2(\mathbb{R})$$ and $$A = \frac{d}{dx}$$.

If we want to study time evolution, we have to prescribe an one-parameter group of unitaries, $$U_t$$. We may now ask for the orbits of operators under this group of unitaries, that is, we might be interested in

$$A(t) := U_t^{-1} A U_t \ .$$

We might additionally study observables which we let parametrically depend on $$t$$; for example we may multiply an operator with a function $$f(t)$$. I will ignore this in the following, but it is not hard to take into account. Take a time evolution operator

$$U_t = \exp(- i H t) \ .$$

Then we may compute the time derivative of $$A(t)$$:

$$\frac{d A(t)}{dt} = \frac{d }{dt} \left(e^{ i t H} A e^{-i t H}\right) = (i H) e^{ i t H} A e^{-i t H} + e^{ i t H} A e^{-i t H} ( - i H) = i [H,A(t)] \ .$$

Let's consider the case of $$A = \frac{d}{dx}$$ in more detail, and let's choose

$$H = - \frac{1}{2} \frac{d^2}{d x^2} + \frac{1}{2} x^2 \ ,$$

which is of course the harmonic oscillator. Now we want to see what

$$A(t) = \left(\frac{d}{dx}\right)(t)$$

is. Note that this is no longer the derivative! The connection it has to the derivative is that

$$\left(\frac{d}{dx}\right)(0) = \frac{d}{dx} \ .$$

The equation of motion for $$A$$ is

$$\frac{d A(t)}{dt} = i [ H(t), A(t)] = \frac{i}{2} \left[ x(t)^2, \left(\frac{d}{dx}\right)(t) \right] \ .$$

Here i used that $$H(t) = H$$. In this formula we may now use that for two operators $$B(t),C(t)$$ that are obtained from operators $$B,C$$ by conjugating with an unitary $$U_t$$, it holds that

$$[B(t),C(t)] = ([B,C])(t) \ .$$

Hence

$$\left[ x(t)^2, \left(\frac{d}{dx}\right)(t) \right] = \left(\left[ x^2, \frac{d}{dx}\right] \right)(t) = 2 x(t) \ .$$

Playing the same game with $$x(t)$$, we get the set of coupled equations

$$\frac{d \left(\frac{d}{dx}\right)\!(t)}{d t} = i x(t) \ , \\ \frac{d x(t)}{dt} = i \left(\frac{d}{dx}\right)\!(t) \ .$$

which may be easily solved to give

$$\left(\frac{d}{dx}\right)\!(t) = \cos(t) \frac{d}{dx} + i \sin(t) x \, \\ x(t) = \cos(t) x + i \sin(t) \frac{d}{dx} \ .$$