# Representation of the Time Evolution Operator in QM

I was searched about the Time-evolution operator from time $$t_0$$ to $$t$$, denoted as $$U(t,t_0)$$ used in quantum mechanics, and which has the formula for a time-independent Hamiltonian operator $$\hat H$$:

$$U(t,t_0)=\exp(-i\hat H t/\hbar)$$

The argument (what I've not seen completed) is like this: $$U(t,t_0)$$ commutes with $$\hat H$$ and the time-partial derivative, so writing the Schrödinger equation and then applying $$U(t,t_0)$$ to the eq.

$$i\hbar{\partial\over\partial t}\psi(t_0)=\hat H\psi(t_0)$$ $$U(t,t_0)i\hbar{\partial\over\partial t}\psi(t_0)=U(t,t_0)\hat H\psi(t_0)$$ $$i\hbar{\partial\over\partial t}U(t,t_0)\psi(t_0)=\hat HU(t,t_0)\psi(t_0)$$ $${\partial\over\partial t}U(t,t_0)\psi(t_0)=-{i\over\hbar}\hat HU(t,t_0)\psi(t_0)$$

and this is the further I've seen, and then comes the claim that

$${\partial\over\partial t}U(t,t_0)=-{i\over\hbar}\hat HU(t,t_0)$$

because the $$\psi(t_0)$$ is fixed in time, so $$\partial_t\psi(t_0)=0$$, and so

$$\left({\partial\over\partial t}U(t,t_0)\right)\psi(t_0)=\left(-{i\over\hbar}\hat HU(t,t_0)\right)\psi(t_0)$$

and the claim above follow from that, then one "solves" the ODE of $$\partial_t U(t,t_0)$$ and get the formula for the operator. My questions are two:

1. If $$\partial_t\psi(t_0)=0$$, then that doesn't mean that the Schrödinger equation is time-independent?
2. If the time-shift operator for a shift of $$\tau$$-units can be writen as $$\exp(\tau\partial_t)$$, then Would it make more sense the argument below?

$$\psi(t+\tau)=e^{\tau{\partial\over\partial t}} \psi(t)$$

with the identity $$\partial_t=-i\hat H/\hbar$$, we replace above:

$$\psi(t+\tau)=e^{-i\tau\hat H/\hbar} \psi(t)$$ $$\Rightarrow U(t+\tau,t)=\exp(-i\tau\hat H/\hbar)$$

First of all, you have a typo, I think you meant to write $$U(t,t_0)=\exp(\frac{i}{\hbar}(t-t_0)\hat{H}).$$ A short disclaimer: I am going to ignore stuff like regularity/dense domain issues here while answering these questions.
1)The evolution operator perspective is, that you treat a solution to a PDE like this as a map $$\psi:(-t,t) \to \mathcal{H},$$ where $$\mathcal{H}$$ is your respective Hilbert space (for example $$L^2(\mathbb{R}^n)$$) and $$\hat{H}$$ is an operator acting on that Hilbert space. Your solution is then given by $$\psi(t)=U(t,t_0)\psi(t_0).$$ In this case you are basically looking at the initial value problem $$\begin{cases} i \hbar\partial_t \phi =\hat{H}\phi \\ \phi(t_0)=\psi(t_0) \end{cases}$$ and $$\phi(t_0)=\psi(t_0) \in \mathcal{H}$$ is just a fixed element in the Hilbert space and hence independent of time. Think of it like that: In the "normal" ODE cases, an initial condition is also just a vector in the Hilbert Space $$\mathbb{R}^n$$ (or $$\mathbb{C}^n$$), which does not depend on time.
2)Your evolution operator $$\exp(\frac{i}{\hbar}(t-t_0)\hat{H})$$ in this case is indeed a "timeshift operator" in some sense. It has the semigroup property $$U(t+\tau,t_0)\psi(t_0)=U(t+\tau,t)U(t,t_0)\psi(t_0).$$ and you don't need to use the timeshift operator $$\exp(\tau \partial_t)$$, but it is a nice sanity check.
• If $U(t,t_0)=\exp(i(t-t_0)\hat H/\hbar)\psi(t_0)$, then $U(t, t_0)\psi(t_0)=\exp(i(t-t_0)\hat H/\hbar)\psi(t_0)\psi(t_0)$? Commented Sep 12, 2023 at 18:17
• I fixed it - the operator itself, notation was a bit ambiguous. Now we have $\psi(t)=U(t,t_0)\psi(t_0)$, that should clarify things. Commented Sep 12, 2023 at 20:49
• You need to write $t-t_0$ in your operator $U(t,t_0)$. $U(t,t_0)$ takes a vector in your Hilbert space as an input and if you take $\psi(t_0) \in \mathcal{H}$ as your input vector, then $U(t,t_0)\psi(t_0)$ solves your Schrödinger equation. Commented Sep 14, 2023 at 18:44