Time evolution of a wave packet from the time-independent Schroedinger equation [duplicate]

Starting off with the time-independent Schroedinger equation (TISE)

$\quad \quad -\frac{\hbar^2}{2m} \nabla^2 \psi + V(r, \theta) \psi = E\, \psi,$

I would like to study the time evolution of an initial wave packet $\psi(0)$ dropped in the potential $V$. In particular, I want to evaluate $\langle \theta \rangle$ as a function of the time $t$.

The TISE for the potential $V$ does not have any exact solution in terms of any known functions. The wave packet $\psi (0)$ is expressed as a linear combination of a certain class of functions (infinite in number), which form a basis, but are NOT eigenstates of the Hamiltonian.

I noticed a similar question which uses the time-ordering operator. In that context, my questions were as follows:

• Given a Hamiltonian in differential form (with a particularly nasty $\nabla^2$ in polar coordinates), how does one convert it to its matrix form using Mathematica, considering that the basis used to express $\psi(0)$ has infinitely many functions?

• Is there some way to implement $\exp \big(\partial/ \partial \theta \big)$ directly as a function operating on an argument in Mathematica? As suggested by my attempts, Mathematica does not seem to recognise the dangling derivative in the exponential.

Having an answer to Part 1 of my question would enable me to generalise this answer to the problem at hand. However, I would certainly welcome any efficient approach to the direct computation of $\langle \theta \rangle (t)$ that you may suggest.

Edit

Just to prevent confusion about the second part of the question, by $\exp \big(\partial/ \partial \theta \big)$, I mean

$\quad \quad \exp \big(\partial/ \partial \theta \big) = 1 + \frac{\partial}{\partial \theta} + \frac{1}{2!}\frac{\partial^2}{\partial \theta^2} + \ldots$

So $\exp \big(\partial/ \partial x \big) (x^2) = 1 + 2x + x^2 \ne \exp(2x)$.

marked as duplicate by Jens, bbgodfrey, Öskå, m_goldberg, ciaoApr 17 '15 at 20:03

• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – user9660 Apr 16 '15 at 15:58
• This is a time-independent problem, and your first link is to a time-dependent solution. They are very different problems. Also the link doesn't use the time-ordering operator (although it's mentioned), it just discretizes time. If you want to a time-dependent 2D problem, see this link. – Jens Apr 16 '15 at 16:35
• Shouldn't it be $\exp(\partial_x)\left\{x^2\right\} = 1 + 2x + x^2$? – 2012rcampion Apr 16 '15 at 16:35
• @Jens, Starting with a solution of the time-independent problem, one can propagate it in time using $\exp (- \mathrm{i} H t / \hbar)$ to obtain $\psi(t)$, so the problems are related. If you want, equivalently, you can find $\psi (t)$ directly from the time-dependent equation and compute $\langle \theta \rangle$. Note: I'm not trying to solve the TISE, I'm just trying to evolve a given wave packet which is NOT necessarily a solution of the TISE (read as eigenstate of $H$). – SquareRoot2 Apr 16 '15 at 16:40
• So then you want something like this, right? Complex valued 2+1D PDE Schroedinger equation, numerical method for NDSolve? – Jens Apr 16 '15 at 16:42

Try this exponential derivative operator:

expD[f_, x_] :=
Module[{x0},
Sum[SeriesCoefficient[f, {x, x0, i}], {i, 0, \[Infinity]}] /. {x0 -> x}
]


Examples:

expD[x^2, x]
(* (1 + x)^2 *)

expD[Sin[x], x]
(* Sin[1 + x] *)

expD[Exp[x], x]
(* Exp[1 + x] *)

• (+1), but I think this part is a duplicate of Exponential of a Differential Operator – Jens Apr 16 '15 at 16:45
• @Jens Should I move this answer there? – 2012rcampion Apr 16 '15 at 16:59
• It does seem like your answer belongs to that linked question, since this one mentions the exponential only as a part of the problem... but actually, the answer for the special case here is trivial: your operator, applied to any function $f(x)$, is simply equivalent to $f(x+1)$ by the Taylor formula. – Jens Apr 16 '15 at 17:03