# Expression of the time evolution operator $\hat U(t)$ in terms of the hamiltonian [closed]

I'm trying to derive the expression of the time evolution operator, $$\hat U$$, in terms of the Hamiltonian of a system, $$\hat H$$. This operator $$\hat U$$ is defined so that $$\Psi(x,t)=\hat U(t)\Psi(x,0)$$.

My attempt at a solution

I have substituted $$\Psi(x,t)$$ by $$\hat U(t)\Psi(x,0)$$ in the time-dependent Schrödinger equation:

$$\hat{H} \Psi(x,t)=i \hbar \frac{\partial}{\partial t}\Psi(x,t)$$

$$\hat{H}\left(\hat{U}(t) \psi(x, 0)\right)=i \hbar \frac{\partial}{\partial t}\left(\hat{U}(t) \psi(x, 0)\right)$$

At this point I make two assumptions of which I am not very sure: $$(1)$$ I assume that $$\psi(x, 0)\neq0$$, and that I can divide by it the equation, and $$(2)$$ I assume such a thing as dividing by an operator can be done. This leads us to

$$\frac{1}{\hat U} \frac{\partial \hat U}{\partial t}=- \frac {i} {\hbar}\hat H$$

Integrating, and supposing that the constant of integration can be taken to be $$1$$, I get the expression we where looking for:

$$\hat{U}(t)=\exp \left(-\frac{i t}{\hbar} \hat{H}\right)$$

The result I get is correct, but is the process I have followed valid?

• This is done in 100s of QM textbooks... Commented Aug 31, 2020 at 15:47

## 1 Answer

This process is valid, but indirectly so. Like, if you were my student and you handed it to me I would be very worried that you had a “magical” attitude towards the mathematics involved, whereas this is something more than just an “umbral” similarity between two different mathematical domains.

What is actually happening is that you have a time-independent Hamiltonian $$\hat H$$ and therefore it has stationary eigenstates $$\hat H |n\rangle = E_n |n\rangle$$. Under the Schrödinger equation these then gain a time dependence which multiplies them by a rotating phase factor, $$|n(t)\rangle = e^{-i E_n t/\hbar} |n\rangle$$ so that for an arbitrary other state if $$|\Psi(0)\rangle = \sum_n c_n |n\rangle$$ then $$|\Psi(t)\rangle = \sum_{n=0}^\infty c_n~e^{-iE_n t/\hbar} |n\rangle.$$This operator is justifiably then $$e^{-i\hat H t/\hbar};$$ that is what you get when you do a Taylor series.

So, what you are doing here is you are performing this derivation in a basis in which the Hamiltonian is diagonal. That is the missing aspect of your argument which indirectly makes the whole thing work; it is that if you look at this equation $$i \hbar \frac{\partial U}{\partial t} = \hat H U$$in the basis in which $$\hat H$$ is diagonal then a diagonal $$U$$ suffices to solve it, as the product of two diagonal matrices is diagonal: and furthermore each term on the diagonal is a separate differential equation $$i \hbar ~\dot U_n(t) = E_n ~U_n(t)$$ with boundary condition $$U_n(0) = 1.$$ It is valid because it is obviously true in one particular basis.

It’s also worth pointing out that the order $$\hat H U$$ does not matter here but it does matter when $$\hat H = \hat H(t)$$ is no longer constant over time, in which case you get a term which is often written, since we don't have a great notation for continuous products, as $$U(t) = \mathcal T \exp\left[-\frac{i}{\hbar} \int_0^t\mathrm d\tau~\hat H(\tau) \right],$$ the symbol $$\mathcal T$$ meant to remind us that this is to be interpreted as a time-ordered series of products $$U(t) = \lim_{\delta t\to 0} e^{-i\hat H(t-\delta t)\delta t/\hbar}~e^{-i\hat H(t-2\delta t)\delta t/\hbar}\dots e^{-i\hat H(\delta t)\delta t/\hbar}e^{-i\hat H(0)\delta t/\hbar}.$$ Per the other question you asked, the adjoint of this operator does not just swap out $$+i$$ for $$-i$$ but also it must reverse the time ordering of these terms for an anti-time-ordering $$\bar{\mathcal T}.$$

In particular in an interaction picture we split the Hamiltonian into some “easy” part and some “interacting part” $$\hat H = \hbar (\eta_0(t) + \xi(t))$$ and then we try to invent the easy-evolution operator $$u$$ as, $$u(t)=\mathcal T \exp\left[ -i \int_0^t \mathrm d\tau~\eta_0(\tau)\right].$$Now since $$u u^\dagger = 1$$ by construction, we can insert it into all of our expectation values to get a sort of quantum coordinate transform, mapping $$\hat A \to \tilde A = u^\dagger \hat A u$$ while mapping $$|\Psi(t)\rangle \to |\tilde \Psi(t)\rangle = u^\dagger |\Psi(t)\rangle$$ obeying the new equations, $$i \frac{\mathrm d\tilde A}{\mathrm dt} = -[\eta, \tilde A] + \frac{\partial \tilde A}{\partial t}, \\ i \frac{\mathrm d\hphantom{t}}{\mathrm dt} |\tilde\Psi(t)\rangle = \tilde\xi~ |\tilde\Psi(t)\rangle.$$ Doing this derivation correctly absolutely requires that the order flip when calculating $$\mathrm d/\mathrm dt(u^\dagger),$$ so that the $$\eta_0$$ comes out on the right hand side of the operator. This requires keeping straight in your head that $$u^\dagger$$ is anti-time-ordered and so $$u^\dagger(t + \delta t)\approx u^\dagger(t) e^{i \eta_0(t) \delta t}$$ appears on the right of the operator and can be expanded out to first-order on that side.