Let's consider the hydrogen atom Hamiltonian $$H = - \frac{1}{2}\Delta - \frac{1}{r}$$ The solution for the corresponding time-dependent Schrodinger equation is the following: $$\psi = \psi (t = 0){e^{ - i\int {Hdt} }} = \psi (t = 0){e^{ - iEt}}$$ But imagine that we don't know the eigenvalues and have to explicitly integrate the expression $ - i\int {Hdt} $ and since $H$ is time-independent then we have a differential operator in the exponential function $${e^{ - i\left( { - \frac{1}{2}\Delta - \frac{1}{r}} \right)t}}$$

How is it supposed to be treated and calculated? Or is the question incorrect?

  • $\begingroup$ Integrating $H$ is probably not the most straightforward way to solve the Schrodinger equation directly. Why not just solve $-i\hbar \frac{\partial\psi}{\partial t} = H\psi$? $\endgroup$ Commented Oct 19, 2018 at 13:59
  • $\begingroup$ I just was interested in this way of solving $\endgroup$ Commented Oct 19, 2018 at 14:30
  • $\begingroup$ So one problem with your question is that is not clear what the exponential means. One might define it as a power series, but if this makes any sense is very unclear to me. One way is to diagonalize the Hamilton operator, and then define the exponential as the sum over the projectors on the eigenvectors times the exponential of the eigenvalues. Maybe the easiest way is to use the differential equation, as pointed out by @probably_someone . One may also use path-integration methods. $\endgroup$ Commented Oct 19, 2018 at 14:47
  • $\begingroup$ Thanks for your answer. I name $e^x$ as the exponential function. Maybe it is bad to name it like that $\endgroup$ Commented Oct 19, 2018 at 14:53
  • $\begingroup$ @JamesFlash It's not the name that's the problem, it's the definition of such an operator. Typically these operators are defined by their power series expansion; however, are we sure that such an expansion will actually produce something sensible? At this point, usually you have to assume that perturbation theory applies (i.e. assume that your Hamiltonian deviates only a little from a well-known exact solution). In that case, the terms of the power series expansion get less significant with higher order, and you can safely ignore all but a finite number of terms. $\endgroup$ Commented Oct 19, 2018 at 14:58


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