1
$\begingroup$

Today when my professor was deriving this equation: $$\frac{\mathrm d\langle A\rangle}{\mathrm dt}=\frac{i}{\hbar}\langle\left[H,\,A\right]\rangle+\left\langle\frac{\partial A}{\partial t}\right\rangle, $$ he treated the corresponding operator of the observable $A$ as if it was a function and took its time derivative and it shows up in the last term of the equation as can be seen.

What I don't get is what does it mean to take the time derivative of an operator? If the operator is a multiplicative function, then I can understand it, but suppose the operator was $\mathrm d/\mathrm dx$, what does it mean to take the time derivative of a $\mathrm d/\mathrm dx$ operator?

$\endgroup$
  • 1
    $\begingroup$ It's just how you'd expect by ordinary calculus. The time derivative of $d/dx$ is just zero, but the time derivative of $t \, d/dx$ is $d/dx$. $\endgroup$ – knzhou Jan 29 at 12:55
  • 3
    $\begingroup$ If you want, you can think of operators as matrices, in which case the time derivative is taken by taking the time derivative of the elements of the matrix. $\endgroup$ – knzhou Jan 29 at 12:55
  • $\begingroup$ Related: physics.stackexchange.com/q/9122/2451 and links therein. $\endgroup$ – Qmechanic Jan 29 at 13:04
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Why is (d/dx)(0)=d/dx and not (d/dx)(1)=d/dx? $\endgroup$ – Achilles' Advisor Feb 1 at 16:27
  • $\begingroup$ I don't get the question... Are you saying it should be the case or that some equation says that? $\endgroup$ – Lorenz Mayer Feb 1 at 17:48
  • $\begingroup$ I feel like it should be the case $\endgroup$ – Achilles' Advisor Feb 5 at 13:11
  • $\begingroup$ In the way i defined it (as is usual), the relation between an operator $A$ and its time-translate $A(t)$ is $A(t) = U_t^* A U_t$, so that $A(0) = U_0^* A U_0$. You may furthermore check that $U_0$ is the identity. $\endgroup$ – Lorenz Mayer Feb 6 at 14:50
1
$\begingroup$

Actually if you look carefully one is (mostly) taking derivatives of average values of an observable. These average values are certainly real functions so it makes sense to “keep track of the average $\langle A\rangle$ as a function of time” and then take the derivative of this function.

As to $\langle \partial A/\partial t\rangle$, it is possible to have an operator that is explicitly time dependent. This may occur for instance when a potential changes explicitly with time. In this case, it makes sense to think of the derivative of this operator, and then take the average value of this derivative.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.