# Fermionic interaction potentials

Are there any examples of fermionic particles or quasiparticles for which the interaction potential is a globally smooth function? i.e. no singularities or branch points.

As an example, in Flügge's Practical Quantum Mechanics, problem 148 has two repulsive particles on a circle. This is supposed to model the two helium electrons in the ground state. The equation he gives is

$$-\frac{\hbar^2}{2mr^2}\left(\frac{\partial^2 \psi}{\partial x_1^2}+\frac{\partial^2 \psi}{\partial x_2^2}\right)+V_0\cos(x_1-x_2)\psi=E\psi$$

I don't quite follow why this potential does not have a singularity when $x_2\rightarrow x_1$. Are there other such examples?

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To clarify, you want a physical system for which there is a hamiltonian $H$ which is a very good approximation over some energy-range and whose term fourth order in fermions, $\rho(r_1)V(r_1-r_2)\rho(r_2)$, has $V(r)$ a $C^{\infty}$ function over all of space? – BebopButUnsteady Jun 27 '11 at 14:40
Yes. Ideally, $V(r_1-r_2)$ is $C^\infty$. At the very least is there any such model for fermionic particles where $V(r_1-r_2)$ is continuous at $r_1=r_2$? – Greg von Winckel Jun 27 '11 at 15:03
I'm not sure I completely understand the question, but if the two electrons have their spin degrees of freedom in a singlet, then the spatial wavefunction is symmetric under exchange of 1 & 2. There are no nodes in the ground-state wavefunction, so an effective potential doesn't have to introduce any singularities. – wsc Jun 27 '11 at 16:14

Well there's no particular reason for a textbook problem to actually model a physical system... But one can certainly write something like this as a completely valid approximation. Take Flugge's example, with He3 so its fermionic[fn.2]. Say the size of the atoms is very small, much smaller than the scales on which the ground state wavefunction varies, which is reasonable enough.

Now there should really be a term $V_{repulse}(x_1-x_2)\psi$ where $V_{repulse}$ gets really big when $|x_1-x_2|\rightarrow 0$ to capture the fact that you can't put the two atoms on top of each other [fn.2]. But this is going to be really short ranged, almost zero if $|x_1-x_2|$ is significantly bigger then the size of the atom. On the other hand we know that $\psi(x_1,x_2)\rightarrow 0$ when $x_1 \rightarrow x_2$. So in precisely the region where $V_{repulse}$ would matter, $\psi$ is basically zero. So we can basically ignore $V_{repulse}\psi$. More exactly, the term is proportional to (size of atom)/(size of circle) squared, which could be very small.

So you don't always have to include a repulsive term. It can actually be quite negligible, even though it seems like a fact you can't ignore.

[fn. 1] There are also times when you can ignore the repulsive interactions of bosons, although its not suppressed like the fermions.

[fn. 2] Its not really true that it should diverge as $x_1\rightarrow x_2$. If you really got the two atoms on top of each they would stop behaving like pointlike atoms, so your model would stop being applicable, rather than anything going to infinity.

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On reading again it seems maybe you want the potential to go to infinity to enforce Pauli exclusion. If that's the case, then the answer is that the Pauli exclusion is a purely geometrical fact, and has nothing to do with potentials. – BebopButUnsteady Jun 27 '11 at 20:58
On further reading I'm not sure what the question is, so you should clarify it. – BebopButUnsteady Jun 27 '11 at 22:17
I have written a spectral code for computing eigenstates of 1D fermion systems with arbitrary confinement and interaction potentials. I am looking for model problems to test the code on and have already tried solving the (no spin) $n$-particle problem $$\left\{-\frac{\hbar^2}{2m}\nabla^2 + \sum\limits_{j=1}^n V_{ext}(x_j) + \sum\limits_{k=j+1}^n V_{int}(x_j-x_k)\right\}\psi(\mathbf{x})=E\psi(\mathbf{x})$$. When I use Coulomb interaction for $V_{int}$, the method converges quadratically. I am looking for problems with smooth $V_{int}$ to see if the convergence improves. – Greg von Winckel Jun 28 '11 at 6:23
Why not just put in some smooth interaction, say $\frac{1}{(x^2 +1)^2}$ and see if it converges? Or if you're looking for an analytically solved model to compare against, there's a fairly extensive set of 1D fermionic systems that are analytically tractable. The confining potential will probably make things difficult, but if you make the scale of the interaction much smaller than the confinement you can probably get something to work. – BebopButUnsteady Jun 28 '11 at 14:34
I have indeed tried a Lorenzian potential and observed spectral convergence. My hope, however, was not to try an arbitrary smooth potential, but one that is used in practice to model something. Some analytically solvable 1D fermionic systems would be of interest anyway. Do you have a reference for any of those? – Greg von Winckel Jun 29 '11 at 8:00

Apparently the Gaussian Effective Potential is used a fair amount. This would be something like $$V(x_1,x_2)=V_0 \exp(-\alpha(x_1-x_2)))$$. Thanks for the responses.

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In atomic physics effective 1D soft-core Coulomb potentials are routinely used for the interaction between particles, particularly when external fields are present. For example, for the interaction between an electron and a (space-fixed) proton at the center of coordinates: $$V(x) = - \frac{1}{\sqrt{x^2 + \epsilon^2}}$$ where atomic units are used and $\epsilon$ is usual fitted or taken as one. It is much simpler to integrate the time-dependent Schrödinger equation for this potential.

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