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

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


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

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

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