# QM: Time evolution with $H = H(t)$

In order to calculate time evolution in QM we use Schrödinger equation \begin{align*} i \partial_t |\psi\rangle_t = H(t) | \psi\rangle_t. \end{align*} If $$H\neq H(t)$$ then \begin{align*} i \partial_t |\psi\rangle_t = H(0) | \psi \rangle_t \end{align*} and we can expand the state in its Taylor series \begin{align*} | \psi \rangle_t & = |\psi\rangle_0 + t \, \partial_t |\psi\rangle_t \Big|_{t=0} + \frac{1}{2} \, t^2 \, \partial_t^2 |\psi\rangle_t \Big|_{t=0} + ... \\ & = |\psi\rangle_0 + (-i t H(0)) | \psi \rangle_t \Big|_{t=0} + \frac{1}{2} (-itH(0))^2 | \psi \rangle_t \Big|_{t=0} + ...\\ & = |\psi\rangle_0 + (-i t H(0)) | \psi \rangle_0 + \frac{1}{2} (-itH(0))^2 | \psi \rangle_0 + ...\\ & = e^{-itH(0)}| \psi \rangle_0. \end{align*} So far so good. But now we consider $$H=H(t)$$. My question is: why can't you do the same? Even if now you have $$i \partial_t |\psi\rangle_t = H(t) | \psi \rangle_t$$ instead of $$i \partial_t |\psi\rangle_t = H(0) | \psi \rangle_t$$, you still have \begin{align*} \partial_t |\psi\rangle_t \Big|_{t=0} = (-i H(t)) |\psi\rangle_t \Big|_{t=0} = (-iH(0)) |\psi\rangle_0. \end{align*} According to this you would always get the same time evolution operator: \begin{align*} | \psi \rangle_t & = e^{-itH(0)}| \psi \rangle_0, \end{align*} both for time independent and time dependent operator.

Of course I realize this doesn't make sense for $$H=H(t)$$, because it implies that the state at any point is only given by the state and the Hamiltonian at $$t=0$$, and according to Schrödinger's equation the Hamiltonian "drives" the state at each time. So I just want to know why you can't expand the state in a Taylor series for $$H=H(t)$$. My guess is that, for some reason, in an isolated system the state is "analytic" and equal to its Taylor series, while for a non isolated system the Taylor series only converges in a neighbourhood of the point, and the correct formula is \begin{align*} | \psi \rangle_{t+\Delta t} & = e^{-itH(t)}| \psi \rangle_t + \mathcal{O}( \Delta t^2), \end{align*} which leads to the general time evolution operator.

Or maybe it has nothing to do with "analyticity" and it's just somehting silly I'm not seeing right now.

## 2 Answers

The reason why your argument doesn't work for time-dependent Hamiltonian is that $$\left. \partial_t^2 |\psi(t)\rangle \right|_{t=0} = \left. \partial_t (\partial_t |\psi(t)\rangle) \right|_{t=0} = \left. \partial_t (-\mathrm i H(t) |\psi(t)\rangle) \right|_{t=0} \neq \left. (-\mathrm i H(t))^2 |\psi(t)\rangle \right|_{t=0}$$ The time evolution is still analytic, as long as the function $$H(t)$$ is.

The correct way to do it instead is using a time-ordered exponential $$|\psi(t)\rangle = \mathbf{T} \mathrm e^{-\mathrm i \int_{t_0}^t H(\tau)\, \mathrm d\tau} |\psi(t_0)\rangle ,$$ which is defined via the Dyson series. (From your question, I assume that you know this already, so I won't write more about it -- feel free to ask if you have more questions!)

The main reason why you can't do those sorts of naïve manipulations is that in general, if $$H = H(t)$$, hamiltonians evaluated at different times may not commute. If the hamiltonian is time-dependent but $$[H(t_1), H(t_2)] = 0$$ for all $$t_1$$ and $$t_2$$, then you could still analitically solve Schrödinger equation to give

$$|ψ⟩_t=e^{−i\int_0^tdt'H(t')/\hbar}|ψ⟩_0.$$

When hamiltonians at different times don't commute, however, you need to introduce the Dyson time-ordering operator.

• Okay but my question is not about the formula for time evolution when $H=H(t)$, no matter if the commutator is zero or not. My question is about why can't you write the state as a Taylor expansion. – MBolin Sep 29 '18 at 13:16