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.


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


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

  • $\begingroup$ 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. $\endgroup$ – MBolin Sep 29 '18 at 13:16

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