Does Adding or Removing an Electron Change a Helium-3 Atom to a Boson?

[hilbert2]

TL;DR Summary

How does the number of electrons affect the boson or fermion nature of ions

In the theory of superfluidity, $^{4}$He atoms are seen as weakly interacting pointlike bosons, with an integer total spin. Does a $^3$He atom also become a boson if I add or remove one electron to give $^3$He$^+$ or $^3$He$^-$ ?

I'm trying to calculate the ground state wave functions of systems of two ions in a 2D or 3D potential well similar to those in this thread. The diffusion monte carlo code can't properly antisymmetrize a fermion system so I have to choose ions with an integer spin. I also assume that in this case most of the interaction is from the total charge on each ion and not from Van der Waals forces and the like.

I already did one calculation for two fictitious particles with charge $+2e$ and mass $m_e$ in a 2D square potential well of depth $V=1500$ and side length $L=3.0$. The program gives an output file with 5 columns that contain the two coordinates of each particle and the value of the wave function. Converting this to a 2-column file with first column containing interparticle distance and the second one containing the square of the wave function, a scatter plot with no connecting lines between data points looks like this:

This looks much like I expected, and a radial distribution function like $P(r) = c_0 re^{-c_1 r} + c_2$, with the $c_k$ constants, can be fitted in this data to capture the main features.

 

[dRic2]

I never took an advanced QM course, but this question was once addressed in a very colloquial way by a professor of mine. He said that you have to see whether electronic spin couples with the nuclear one: in order to estimate the effect of coupling one can look at the fine-splitting of the energy levels of the electrons due to spin coupling with the nucleus and compare it with the thermal excitation $k_B T$. So I assume that if $k_B T >> \Delta E$ you can neglect the coupling hence you might consider $^3He^+$ a boson. But superfluidity occurs at very low temperatures so maybe it is not possible to neglect coupling.

 

[hilbert2]

In the quantum dot-like system I'm simulating, there's no requirement of extremely low temperature, so I think $^3$He$^+$ would then be a boson.

 

[dRic2]

I hope someone more competent than me shows up and gives his opinion too.

 

[DrClaude]

In the theory of superfluidity, $^{4}$He atoms are seen as weakly interacting pointlike bosons, with an integer total spin. Does a $^3$He atom also become a boson if I add or remove one electron to give $^3$He$^+$ or $^3$He$^-$ ?

Yes. The important property is the behaviour of the wave function under the permutation operator $\hat{P}$ that permutes identical particles. Since the nucleus of ³He and electrons are fermions, when you permute two ³He atoms, you are permuting 3 fermions, each permutation changing the sign of the wave function, so in total you get a minus sign and the atom is a fermion. If you remove or add an electron, the total number of fermions will be even and the entire ion behaves as a boson.

 

[DrClaude]

I never took an advanced QM course, but this question was once addressed in a very colloquial way by a professor of mine. He said that you have to see whether electronic spin couples with the nuclear one: in order to estimate the effect of coupling one can look at the fine-splitting of the energy levels of the electrons due to spin coupling with the nucleus and compare it with the thermal excitation $k_B T$. So I assume that if $k_B T >> \Delta E$ you can neglect the coupling hence you might consider $^3He^+$ a boson. But superfluidity occurs at very low temperatures so maybe it is not possible to neglect coupling.

I'm seen such an argument about coupling before, but I don't get it. Atoms for which the nucleus has no spin ($I=0$) and for which there can be no spin coupling between the nucleus and the electrons can still be classified as bosons or fermions.

 

[PeterDonis]

I don't get it. Atoms for which the nucleus has no spin ($I=0$) and for which there can be no spin coupling between the nucleus and the electrons can still be classified as bosons or fermions.

The case of zero nuclear spin is covered by the argument: that case is just $\Delta E = 0$, which obviously satisfies $k_B T >> \Delta E$.

 

[hilbert2]

Thanks, all.

The DMC code also assumes that all particles have mass $m_e$, but it's clearly possible to absorb the mass difference to the energy terms:

$\displaystyle-\frac{\hbar^2}{2m_e}\nabla^2 \psi (x_1 ,y_1 ,x_2 ,y_2 ) + V(x_1 ,y_1 ,x_2 ,y_2 )\psi(x_1 ,y_1 ,x_2 ,y_2 ) = E \psi (x_1 ,y_1 ,x_2 ,y_2 )$

$\displaystyle\Rightarrow -\frac{\hbar^2}{2m_{{\scriptscriptstyle^{3}He}}}\nabla^2 \psi (x_1 ,y_1 ,x_2 ,y_2 ) + \frac{m_e}{m_{{\scriptscriptstyle^{3}He}}}V(x_1 ,y_1 ,x_2 ,y_2 )\psi (x_1 ,y_1 ,x_2 ,y_2 ) = \frac{m_e}{m_{{\scriptscriptstyle^{3}He}}}E \psi (x_1 ,y_1 ,x_2 ,y_2 )$

with the del operator defined here as

$\displaystyle\nabla^2 = \frac{\partial^2}{\partial x_{1}^{2}} + \frac{\partial^2}{\partial y_{1}^{2}} + \frac{\partial^2}{\partial x_{2}^{2}} + \frac{\partial^2}{\partial y_{2}^{2}}$

Edit: as an additional question, is it known to be possible to get two ions like Na$^+$ and Cl$^-$ orbit each other in some high excited state without transferring an electron and becoming neutral atoms? If it's a sodium and chlorine ion, it could even be stable w.r.t. electron transfer, but something like Li$^+$Li$^-$ would probably convert to atoms much faster.

 

[DrClaude]

The case of zero nuclear spin is covered by the argument: that case is just $\Delta E = 0$, which obviously satisfies $k_B T >> \Delta E$.

Sorry, but I don't get what you are trying to say here. What does $T$ or $\Delta E$ have to do with the fact that an atom is a boson or a fermion?

 

[dRic2]

$k_BT >> \Delta E$ is a condition to see whether coupling is relevant or not. It doesn't tell you per se if the resulting atom will be a boson or a fermion. If I=0 there is no coupling so I assume you just sum the spins. If there is coupling I have no simple argument to discern whether an atom could be understood as a boson or a fermion.

This is how I always understood the argument, tell if I'm wrong

 

[DrClaude]

$k_BT >> \Delta E$ is a condition to see whether coupling is relevant or not.

Relevant to what? The bosonic/fermionic nature of atoms is a question for QM, not for thermodynamics. It will of course affect what statistics to apply, but this is of secondary nature.

If there is coupling I have no simple argument to discern whether an atom could be understood as a boson or a fermion.

As I wrote in #5, the important is the behaviour of the wave function under the permutation of identical particles.

 

[mfb]

If they are in different spin states, are they still identical? This might be the reason for this argument?

 

[dRic2]

Relevant to what? The bosonic/fermionic nature of atoms is a question for QM, not for thermodynamics.

If there is coupling you should have a term in the Hamiltonian of the system like $H_{coupling}$. If you neglect it then the Hamiltonian is different, thus the wavefunction of the system is different. That condition should tell you under what circumstances the new wavefunction is good enough to account for what you observe. For example, suppose you neglect coupling and the wavefunction turns to be symmetric then it's a boson. Then suppose that you no longer can neglect coupling, maybe this time it turns out antisymmetric... Then it behaves like a fermion. I don't know if what I said makes sense, but, as I see it, the answer whether an atom behaves like a boson or a fermion depends on which approximation you make to write its wavefunction.

As I wrote in #5, the important is the behaviour of the wave function under the permutation of identical particles.

Ok, this is true, but if the wavefunction does not factorize (because there is coupling) it's not easy to see a priori what will happen under a permutation. At least for me, but I'm weak in QM so it's probably just me.

Btw I've found the lecture notes where my professor addresses the question:

I stressed approximately, becuase the concept of composite particle requires some comments. We said that, to judge whether an atom behaves as a fermion or as a boson, we just have to consider the sum of the spins of its elementary constituents, which eventually boils down to see whether the number of neutron is odd or even. Yet, how do we justify this procedure? Consider for instance two separate atoms of the less common isotope of helium: $^3He$. Each of them has just a single neutron hence it is a fermion: but why can't we consider both atoms as a part of a single composite object, which would in turn be a boson? And what is the difference between two $^3He$ atoms and a deuterium molecule, which is made of two $^2H$ atoms, both fermions, and doesbehave as a boson ? You may say: "That's trivial! The deuterium molecule is kept together by strong covalent bonds, whereas the two $^3He$ atoms can freely fly apart, so they cannot be considered as a single entity". This is surely true, and highlights the role played in deciding whether a composite particle is a fermion or a boson, by the spatial and energy scales on which we observe it. So, $^4He$ atoms behave as composite bosons, but if you try to squeeze them together in the same region of space, their electrons prevent it by fully recovering their separate "fermionic" individuality.

PS: since now it is a published book, should I put the reference or something ? Is everything ok with copyrights ?

PSS: I'm a newbie to physics in general while you are all experienced people so I'm not trying to convince you, I'm just sharing what my professor told me and trying to lear at the same time.

PSSS: What if an atom/nucleus is not in the ground states ? It can happen due to thermal agitation or by exciting it with an external field (EM for example), so you have to take in account at which temperature you are observing the system.

 

[PeterDonis]

What does $T$ or $\Delta E$ have to do with the fact that an atom is a boson or a fermion?

I was responding to this statement of yours:

Atoms for which the nucleus has no spin ($I=0$) and for which there can be no spin coupling between the nucleus and the electrons can still be classified as bosons or fermions.

In other words, you didn't understand how an argument about coupling of nuclear spins and electrons can apply to an atom with no nuclear spin. I was simply pointing out that "no nuclear spin" just means "no coupling between nuclear spin and electrons", and in the language of the argument you said you didn't understand, that just means $\Delta E = 0$, and that case is in fact covered by the argument.

 

[PeterDonis]

the important is the behaviour of the wave function under the permutation of identical particles.

But what counts as "identical particles"? If you have two He-3+ ions, but one has its electron in the ground state and the other has its electron in an excited state due to thermal fluctuations, the electrons aren't identical.