# Construction of propagator for time-dependent hamiltonian

In deriving a general propagator to the time-dependent ($$H = H(t)$$) Hamiltonian problem, Shankar works to first order in $$\Delta = T/N$$ (a small time interval for large $$N$$) and argues that by integrating the Schrodinger Equation over the interval $$\Delta$$ we get:

$$|\psi(\Delta)\rangle \approx |\psi(0)\rangle + \Delta \frac{d}{dt}|\psi(0)\bigg|_{t=0} = |\psi(0)\rangle - \frac{i\Delta}{\hbar}H(0)|\psi(0)\rangle = \left(1-\frac{i\Delta}{\hbar}H(0)\right)|\psi(0)\rangle \approx e^{-i\Delta H(0)/\hbar}|\psi(0)\rangle.$$

Then by integrating over each individual interval of width $$\Delta = T/N$$ gives:

$$|psi(t)\rangle = \prod_{n=0}^{N-1}\exp[-\frac{i\Delta}{\hbar}H(n\Delta)]|\psi(0)\rangle.$$

So far, I'm on board. However, he says that we can't simply turn the product into a sum in the exponent and take a limit as $$N \rightarrow \infty$$, get a nice integral representation and call it day. He says we can't do this $$[H(t_1), H(t_2)] \neq 0$$ so by Baker-Campbell-Hausdorff there will be additional terms. However, in the limit as $$N \rightarrow \infty$$ and so $$\Delta \rightarrow 0$$, can we not say that each $$H(n\Delta)$$ will commute with its neighbor $$H((n+1)\Delta)$$, and so argue that the integral representation is valid?

Let's call the evolution operator $$U(t_2, t_1)$$ for convenience.

It is true that when $$\Delta \rightarrow 0$$, $$U(t + \Delta, t) \sim e^{- i \hbar^{-1} \Delta H}.$$

It is also true, like you noted, that $$\lim_{\Delta \rightarrow 0} \left[ U(t + 2 \Delta, t + \Delta); U(t + \Delta, t) \right] = 0.$$

However, there will be terms proportional to $$\Delta$$ in the commutator: $$\left[ U(t + 2 \Delta, t + \Delta); U(t + \Delta, t) \right] = A(t) \Delta + \mathcal{O}(\Delta^2).$$

Now imagine "collapsing" the string of exponentials using BCH. You will pick up $$e^{A(t) \Delta}$$ every time you collapse two neighboring exponentials into one. You will have $$N$$ such terms, all of them with a different $$A(t)$$.

The product of these terms is (with further corrections coming from BCH, which are $$\mathcal{O}(\Delta^2)$$) $$e^{\sum A(t) \Delta} \sim e^{A (t_2 - t_1)},$$ where I assumed that all $$A(t)$$ are of the same order. This does not go to $$1$$ in the $$\Delta \rightarrow 0$$ limit.

Now this may be an illustration of a fallacy in your argument, but it is by no means a proof of anything. The actual proof comes from comparing the answer you would get if you assumed noncommutativity doesn't influence the outcome to the right answer. The right answer is

$$U(t_2, t_1) = \mathcal{T} \exp \left( -i \hbar^{-1} \intop_{t_1}^{t_2} H(t) dt \right),$$

while you would get a similar formula with an ordinary exponential instead of the time-ordered one. These are demonstrably different operators.