# Calculate the matrix size for ground states BEC using difference with Gross–Pitaevskii equation

The Ground states BEC at $$0$$ temperature can be described by Gross–Pitaevskii equation as

$$(-\frac{\hbar^2}{2m}\nabla^2+V(r)+g|\psi|^2)\psi(r)=\mu\psi(r)$$

We limit the BEC in 2D where $$V(r)=\frac{1}{2}m\omega_x^2r^2$$, $$g$$ was the interaction strength and $$\mu$$ was the chemical potential.

Further, Thomas-Fermi radius $$R$$ was given by $$\mu=\frac{1}{2}m\omega_i^2R^2$$ where $$i$$ was the indice for the dimension $$x,y,z$$.

Also, it's been known that $$\psi(r)=\sqrt{n(r)}*e^{ist}$$ where $$n(r)=\frac{\mu}{g}(1-\frac{r^2}{R^2})$$ at ground states.

During the simulation, I realized that the matrix size somehow was related to the probability sates in the system, and thus was not a "flexible" choice. Rather, it need to be calculated, as somewhat counter intuitive to the classical simulation.

Part I: The approach I used was to use finite difference to estimate $$\Delta x$$ at ground states.

Here's what I did:

I simplified $$\psi(r)=\sqrt{n(r)}$$, i.e. ignore the phase, then replace $$\sqrt{n(r)}\approx (1+r^2/R^2)$$.

I then replaced $$\nabla^2$$ with $$\partial_x^2+\partial_x^2+\partial_z^2$$. Although $$w_z$$ was different than $$w_x=w_y$$, it didn't matter much in approximate in one direction, thus I took $$w_z=w_y=w_x$$ for calculation $$\Delta x$$.

I set $$\vec{r}=(\Delta x,0,0,)$$, and the above equation become

$$-\frac{\hbar^2}{2m}\{ [(1-(\Delta x)^2/R^2)-2+(1-(\Delta x)^2/R^2)]\cdot2+ [(1-(4\Delta x)^2/R^2)-2(1-(\Delta x)^2/R^2)+1] \}/(\Delta x)^2 +\frac{1}{2}m\omega_x^2(\Delta x)^2(1-(\Delta x)^2/R^2) +\sqrt{\frac{\mu^3}{g}}(1-(\Delta x)^2/R^2)^3 =\sqrt{\frac{\mu^3}{g}}(1-(\Delta x)^2/R^2)$$

Where I was able to solve for $$\Delta x$$.

Correction 05APR19: Turns out, this made the simulation much more realistic. Please notice that the above calculation only proved that there's connection between $$\Delta x$$ and $$N$$ number of atoms. However, $$\Delta x$$ it self was not a good estimator. If you really want to use it, $$(\Delta x) ^{2/3}$$ could provide a basic idea of the magnitude of $$E(\Delta x)$$. Where $$(\Delta x) ^{2/3}$$ gave an estimate of 43 $$\mu m$$, and $$\sqrt{(\Delta x)^2\cdot 2/3}$$ gave an estimate of 52.9132 $$\mu m$$, compare to theoretical estimate 51.2949 $$\mu m$$. Notice, although $$\sqrt{(\Delta x)^2\cdot 2/3}$$ had a better physical explanation, and better accuracy for this case. It's not a very stable estimator in many cases.

Part II: Another interesting observation was that, as the number of atoms increase, the Thomas-Fermi radius increase, and $$\Delta x$$ was decreasing. This seemed to imply the connection between classical physics, i.e. as the object size increase, the quantization in space approach continuum:

Number of atom   $$\Delta x$$
%1e5      0.9635076655238579380395061110698
%1e6      0.58284627936979343835811180819135
%3e6      0.4616614182450212192630060919487
%1e7      0.35877452754389916540026981055872
%1e8      0.22300940071850495154563998299695
%1e9      0.13941912041357370288498277025999
%1e23     0.00021760029506920409559645922567376
%1e25     0.00008662698728520975705211778808917


Part III: Further investigation through number of sates: I followed up with another thoughts through discrete mathematics and the ideal of statistical mechanics by doing volume integral, where I obtained the following equation $$4\cdot pi\cdot R^3\cdot (5(\Delta x)^2-3)/15/(\Delta x)^4\cdot \mu/g=N$$, where this equation seemed to be able to give an estimate for $$\Delta x$$ as well.

I checked the results, for some parameters, the simulation results for the obtained Thomas Fermi radius of BEC was 53.76664 $$\mu m$$ and theoretical estimate 51.2949 $$\mu m$$. The error was coming from a "rough" estimation in states, the fact that it's hard to approximate cubic vertices in a sphere when number of grid was finite.

Although both Part I and Part II indicated the need for calculating $$\Delta x$$, the estimation in Part III was completely different, where it predicted an almost constant $$\Delta x$$, even for very few atoms. However, it didn't make much sense, as the $$\Delta x$$ did change. The error may resulted from the ignorance of the changes in probability distribution, as well as the deviation in 2D approximation. Thus, Part III works well only for a very narrow band width of number of atoms. Luckily, the band seem to be at where most experiments were done, and where the approximations worked well.

Here's my question:

1. Was my approximation correct?

2. Why in a quantum simulation, we could not just choose any arbitrary matrix size? Is there any theoretical proof for it?

3. Does the results of fixed $$\Delta x$$ have any implication? as it seemed to confirm my suspicion of the relation between matrix size and probability states function $$g(E)$$, which kind of related matrix size with statistical mechanics.

(senior capstone study initial YC)

• Please note that Stack Exchange posts are version controlled, so there is no need to add "Edit" or "correction" to the post, just seemlessly integrate the new information into the post. The interested parties who want to see the changes made can always look at the edit history. – Kyle Kanos Apr 7 at 19:08