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:

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


1 Answer 1


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.

  • $\begingroup$ If $U(t,t_0)=\exp(i(t-t_0)\hat H/\hbar)\psi(t_0)$, then $U(t, t_0)\psi(t_0)=\exp(i(t-t_0)\hat H/\hbar)\psi(t_0)\psi(t_0)$? $\endgroup$ Commented Sep 12, 2023 at 18:17
  • $\begingroup$ 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. $\endgroup$
    – F. Conrad
    Commented Sep 12, 2023 at 20:49
  • $\begingroup$ Sorry, but I don't see the change, and so my previous question remains the same $\endgroup$ Commented Sep 14, 2023 at 18:28
  • $\begingroup$ 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. $\endgroup$
    – F. Conrad
    Commented Sep 14, 2023 at 18:44
  • $\begingroup$ I think my notation is correct by all this physics.stackexchange.com/questions/196848/… $\endgroup$ Commented Sep 14, 2023 at 19:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .