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

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    – user9660
    Commented Apr 16, 2015 at 15:58
  • $\begingroup$ 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. $\endgroup$
    – Jens
    Commented Apr 16, 2015 at 16:35
  • $\begingroup$ Shouldn't it be $\exp(\partial_x)\left\{x^2\right\} = 1 + 2x + x^2$? $\endgroup$
    – 2012rcampion
    Commented Apr 16, 2015 at 16:35
  • $\begingroup$ @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 $H$). $\endgroup$
    – SquareRoot2
    Commented Apr 16, 2015 at 16:40
  • $\begingroup$ So then you want something like this, right? Complex valued 2+1D PDE Schroedinger equation, numerical method for NDSolve? $\endgroup$
    – Jens
    Commented Apr 16, 2015 at 16:42

1 Answer 1

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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] *)
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  • $\begingroup$ (+1), but I think this part is a duplicate of Exponential of a Differential Operator $\endgroup$
    – Jens
    Commented Apr 16, 2015 at 16:45
  • $\begingroup$ @Jens Should I move this answer there? $\endgroup$
    – 2012rcampion
    Commented Apr 16, 2015 at 16:59
  • $\begingroup$ 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. $\endgroup$
    – Jens
    Commented Apr 16, 2015 at 17:03

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