# I Interpreting These '2 Particles in a Box' Plots

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1. Feb 5, 2017

### tomdodd4598

Hey there,

I am familiar with the mathematics of multi-particle systems, but have now moved on to trying to plot them. I was a little ambitious and thought I'd attempt to somehow plot energy eigenfunctions of two particles in a 2-D box. Obviously I immediately ran into the issue that there are actually four coordinates in the wave function rather than two.

Looking for ideas, I quickly came along the Wikipedia article on identical particles, and found these two plots:
https://en.wikipedia.org/wiki/File:Symmetricwave2.png
https://en.wikipedia.org/wiki/File:Asymmetricwave2.png

All that's said about them are that they are the plot of the "(anti)symmetric wave function for a fermionic/bosonic 2-particle state in an infinite square well potential".

I thought that it may be a 1-D box, with the two axes being the positions of the two particles, but if this is indeed a plot of two particles in a 2-D box, what exactly is being plotted?

2. Feb 5, 2017

3. Feb 5, 2017

### blue_leaf77

You have the alternative of plotting the density function which is a function of 2 coordinates in 2D case.
Yes, the two plots are for 1D infinite well potential, not 2D.

4. Feb 6, 2017

### tomdodd4598

https://en.wikipedia.org/wiki/Identical_particles
Really? Would I do that by adding the integrals of the probability density over x1, y1 and x2, y2?
Yep, just confirmed this by recreating them. Thanks :)

5. Feb 6, 2017

### blue_leaf77

The density function is the probability density of finding a particle in space. It's given by
$$\rho(\mathbf r) = \int_{\mathbf r_2} \int_{\mathbf r_3} \ldots \int_{\mathbf r_N} |\psi(\mathbf r, \mathbf r_2, \mathbf r_3, \dots \mathbf r_N)|^2 \ d^3\mathbf r_2 \ d^3\mathbf r_3 \dots d^3\mathbf r_N$$
for $N$ number of particles in 3D space. The adaptation to 2D case should be straightforward.

6. Feb 7, 2017

### tomdodd4598

Thanks - I think I've got it.
It took me a little while to get my head around the apparent fact that, if the particles are indistinguishable, the probability distribution for each particle doesn't depend on whether the particles are fermions or bosons. Wasn't expecting that for some reason, but I guess the reason for this is that, if you express the (anti)symmetric wave function as a sum of products of the one particle wave functions, the cross terms that appear when squaring the wave function (possible because the eigenfunctions are real) vanish, as the integral of a product of two orthogonal eigenfunctions is 0.

I guess this is because systems of multiple indistinguishable particles are entangled?

7. Feb 7, 2017

### blue_leaf77

You mean when the wavefunction is of the form $\psi(x_1,x_2) \propto u_a(x_1) u_b(x_2) - u_b(x_1) u_a(x_2)$ (for fermions)?
That's not really the reason, the functions $u(x)$'s appearing above do not have to be orthogonal for the wavefunction to be symmetric or antisymetric. In fact, the form of wavefunction above where $u_n(x)$ is an eigenfunction of the Hamiltonian only holds for non-interacting particle. For interacting particles, the wavefunction is a sum of terms of that form. The real reason why the probability of detecting a particle is the same for every particle is that because of the nature of the wavefunction itself, namely symmetric - $\psi(x_1,x_2) = \psi(x_2,x_1)$ - or antisymmetric $\psi(x_1,x_2) = -\psi(x_2,x_1)$. If you take the modulus square of these functions, the two types of wavefunction are easily seen to have the property $|\psi(x_1,x_2)|^2 = |\psi(x_2,x_1)|^2$.

8. Feb 9, 2017

### tomdodd4598

I understand, but what I mean is that the probability distribution of the position of one of the particles (found using that integral you wrote above) does not depend on whether they are bosons or fermions, and I suppose this is related to the entanglement of the two subsystems.

Last edited: Feb 9, 2017