# Modifying finite difference solution to Schrodinger eqn to account for fermion/boson effects

I have been playing with an implementation of Visscher's explicit method for solving the time dependent Schrodinger equation (Are there simple ways to numerically solve the time-dependent Schödinger equation?)

Presumably if I want to simulate two identical, non-interacting particles I need to duplicate the simulation array for each particle.

How do I then modify the computation to take into account Fermion/Boson effects, i.e. that Fermions can't occupy the same space (ignoring spin), while Bosons are "encouraged" to do so?

In short, if you have two identical particles, and your wave function is $\psi(x_1,x_2)$, then $$\psi(x_1,x_2) = \pm \psi(x_2,x_1),$$ and this is the relation you should impose on the discretized wave function. You will get a vector of values $\psi_{jk}$ (where $x_j$ are the points in discretized space), and so when you write down the numerical method applied to $\psi_{jk}$, you also require that $\psi_{jk}=\pm \psi_{jk}$, and adjust the numerical method to take account of this (anti-)symmetry.
Also, bear in mind that for non-interacting particles, the PDE separates, and the solution will be of the form $$\frac12\big(\psi_1(x_1)\psi_2(x_2)\pm \psi_1(x_2)\psi_2(x_1)\big),$$ where each $\psi_{i}$ solves the one-particle Schrodinger equation.
• With nobody to discuss textbooks with I have a hard time understanding so thanks for your input :) I'll rephrase the last bit. For two distinguishable non interacting particles, presumably $\psi(x_1,x_2)=\psi_1(x_1)\psi_2(x_2)$. But for indistinguishable particles this is not true. Is that correct? Oct 8 '14 at 13:48
• @SideshowBob Yes, you can check that by writing down the PDE satisfied by $\psi(x_1,x_2)$ and seeing that it separates. Try physics.stackexchange.com. Oct 8 '14 at 14:12