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While reading Quantum Mechanics Book by Sakurai, I found the time-dependent Schrodinger equation for Unitary Operator.

$$i\hbar \frac{\partial}{\partial t}\mathcal{U}(t,t_0)=H\mathcal{U}(t,t_0).$$

The solution to the above equation for time-independent Hamiltonian operator is given that

$$\mathcal{U}(t,t_0)=\exp\left[\frac{-iH(t-t_0)}{\hbar}\right].$$

Can anybody explain how it is coming? $U(t,t_0)$ is an operator, not a function.

Reference -- Modern Quantum Mechanics (Sakurai) - 2nd Edition - Chapter 2 (Page No. - 70)

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  • $\begingroup$ What do you mean how it is coming? You mean where does that solution comes from? Just put it in Schrodinger equation (i.e take its derivative with respect to the time, and multiply it by $i\hbar$, which gives $H\mathcal{U}$) or are you asking this from physical point of view? $\endgroup$ – Paradoxy Aug 16 at 12:19
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$\mathcal{U}(t,t_0)$ is still an operator. The exponent has to be understood as a Taylor expansion.

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  • $\begingroup$ For an unbounded operator, the Taylor expansion does not exist. $\endgroup$ – DanielC Aug 17 at 18:33
  • $\begingroup$ Okay, how is it called instead? $\endgroup$ – Jan2103 Aug 17 at 18:35
  • $\begingroup$ The exponential of a self-adjoint operator is defined through „functional calculus”, not through a series expansion, because of domain (convergence) issues. $\endgroup$ – DanielC Aug 17 at 18:58
  • $\begingroup$ But if we have to solution to the time dependent equation, the scattering operator, which is a unitary operator, for example is given via $$ \mathcal{U} = T \exp{- i \int \mathcal{H}_\text{int}}$$ with the time ordering operator $T$ and the interaction Hamiltionian $\mathcal{H}_\text{int}$. And this exponential function is symbolic for $$\mathcal{U} = 1 - i \int dt' \mathcal{H}_\text{int}(t') + (-i)^2 T \int dt' \int dt'' \mathcal{H}_\text{int}(t') \mathcal{H}_\text{int}(t'') + ...$$ which is a series expression. $\endgroup$ – Jan2103 Aug 17 at 20:11
  • $\begingroup$ That „exp” is formal or symbolic. Dyson series is directly definable, iff the Hamiltonian is bounded at all times. See Reed & Simon, Vol.II, p. 282, Th. X.69 $\endgroup$ – DanielC Aug 17 at 21:13
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The operator $U(\Delta t)$ acts on the initial wavefunction (say at $t=0$) to give the wavefunction at any later time $t$. As you have shown in the question $$ U(\Delta t) = e^{-iH\Delta t}$$ Let's say the initial wavefunction was $\psi (x,0)$ then the wavefunction after $\Delta t$ time has elapsed will be given by $$ \psi (x,t) = U(t) \psi (x,0)$$ Or $$ \psi (x,t) = e^{-iH t}\psi (x,0)$$ The problem one has here is that the operator is in the exponent. The solution is to consider the Mc Lauren series. To make it easier let's say that $\Delta t$ is very small. Hence $$ \psi (x,t) = \psi(x,0)-itH\psi(x,0)$$ This is how time evolution operator acts on wavefunctions. For more and more accurate descriptions higher order terms can be included.

Hope this helps.

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