# Are composite particles, like atoms, identical bosons/fermions?

1. Aug 27, 2014

### atat1tata

I have seen only two arguments for the fact that composite particles, like protons, nuclei, or even Helium-4 atoms, are identical and can be considered bosons or fermions according to their total spin.

The first, in Feyman's lectures [third volume, 4-2]. It is said that if the composite particles are far apart and there is not an appreciable probability of exchanging any of the internal particles, we can consider them as simple objects. In a scattering experiment, either there is no exchange or all the internal particles are exchanged at the same time [why?]. So, for the four fermions of the alpha particle, interchanging them brings a plus sign and the particle is a boson.

The second, in many textbooks. It is reasoned that in the cases we consider the internal degrees of freedom are locked, since excitations require higher energy steps then provided by thermal energy. So the composite particle can be thought as a fundamental particle. Then you can apply the spin-statistics theorem.

Actually I don't understand both of these arguments, I find them too qualitative. I am asking if there is a better, quantitative, justification, that explains also the extent of the approximation.

2. Aug 27, 2014

### Staff: Mentor

They are qualitative because the point they show is qualitative - if you collide two high-energetic alpha particles, they won't behave as bosons, they will behave as collection of nucleons or even collections of quarks and gluons. Treating them as point-like bosons is valid only for low-energetic processes, where both arguments are perfectly valid.

3. Aug 27, 2014

### atat1tata

Sure, but, I am afraid I don't understand what is the probability of exchanging the particles. What does it mean? If it is dependent on the distance, why can a whole gas of atoms be considered a BEC, with no regard (or maybe there is?) to the total volume of the gas?

Moreover, what does it mean to lock the internal degrees of freedom in QM? When will that approximation fail?

I have thought for a while about these question, but I did not come up with a satisfying answer.

4. Aug 27, 2014

### Staff: Mentor

This exchange considered in the statistics is more like a mathematical operation - how does the common wave function change if we swap the particles? You can do that independent of their distance, but to see the effects of the different statistics the particles have to overlap with their "wave function parts". That means large BECs have to have a large number of particles, or be very cold.

It fails when the available energy is sufficient to reach excited states.

5. Aug 28, 2014

### atat1tata

But can there be a way to derive this from an equation, if not in QM, at least in QFT?

Last edited: Aug 28, 2014
6. Aug 28, 2014

### atyy

Do you want to derive that the internal degrees of freedom are locked?

Or do you want to derive that if the internal degrees of freedom are locked, then spin-statistics theorem can be applied?

7. Aug 28, 2014

### haael

Helium-4 can form Bose-Einstein condensate and thus is a boson, right? Despite it's made from fermions.

One thing I still don't understand is how the constitutent fermions "forget" their quantum state. When we have a B-E condensate, then all He atoms are in the same state, th nucleons or quarks can not be in the same state. The fact that quarks form a composite particle somehow allows them to shield their fermion nature and let them form a condensate.

This is a thing I don't understand.

8. Aug 28, 2014

### atat1tata

Actually both, since I don't quite grasp the meaning of "locked degrees of freedom" and its representation in the state space

9. Aug 28, 2014

### atyy

Maybe something like http://arxiv.org/abs/1402.5159? The state in Eq A4 of the appendix seems to start off in a basis in which the antisymmetry under electron exchange is seen, then the basis is changed so that symmetry under positronium exchange is seen. They also state their assumption about which degrees of freedom are in play "For a sufficiently cold and dilute gas, two Ps atoms can only interact through an overall s-wave interaction."

Another place to start looking might be http://arxiv.org/abs/0706.3360v2, where one has a gas of fermions, then depending on the interaction between the fermions, one gets either a Bose-Einstein condensate or Cooper pairs. "The BCS limit of ordinary Fermi superfluidity, the Bose-Einstein condensation (BEC) of dimers and the unitary limit of large scattering length are important regimes exhibited by interacting Fermi gases."

Another review on the BCS-BEC crossover is http://arxiv.org/abs/cond-mat/0404274. Just after Eq 1, they comment on the BCS variational wave function: "As the strength of the attractive pairing interaction U (< 0) between fermions is increased, this wavefunction is also capable of describing a continuous evolution from BCS like behavior to a form of Bose Einstein condensation."

Last edited: Aug 28, 2014
10. Aug 28, 2014

### akhmeteli

You may wish to look at a book by Lipkin called "Quantum mechanics" (https://www.amazon.com/Quantum-Mechanics-Approaches-Selected-Physics/dp/0486458938 ). He shows that the commutation relations for creation/annihilation operators of composite particles containing two fermions are approximately canonical (commutation relations for bosons) unless the composite particle density is very high.

Last edited by a moderator: May 6, 2017
11. Aug 29, 2014

### Avodyne

I don't think we need all this high-tech stuff to understand this issue.

Consider two hydrogen atoms, far apart, each in its ground state. For simplicity, ignore spin. (Including spin would not change the conclusion.) Let

$x_{e1}$ be the position of electron #1
$x_{p1}$ be the position of proton #1
$x_{e2}$ be the position of electron #2
$x_{p2}$ be the position of proton #2

Further define the relative positions

$r_{11}=x_{e1}-x_{p1}$
$r_{22}=x_{e2}-x_{p2}$
$r_{12}=x_{e1}-x_{p2}$
$r_{21}=x_{e2}-x_{p1}$

and the center-of-mass positions

$\overline{x}_{11}=(m_e x_{e1} + m_p x_{p1})/(m_e+m_p)$
$\overline{x}_{22}=(m_e x_{e2} + m_p x_{p2})/(m_e+m_p)$
$\overline{x}_{12}=(m_e x_{e1} + m_p x_{p2})/(m_e+m_p)$
$\overline{x}_{21}=(m_e x_{e2} + m_p x_{p1})/(m_e+m_p)$

Start by considering the wave function

$\psi_0(r_{11})\psi_0(r_{22})\chi(\overline{x}_{11},\overline{x}_{22})$

where $\psi_0(r{})$ is the ground-state wave function of hydrogen, and $\chi(\overline{x}_{11},\overline{x}_{22})$ is a wave function for the two center-of-mass coordinates. We won't need to specify it, but it could be just a product of two gaussians, one for each CM coordinate, each centered on a different point (widely separated).

However, this wave function does not satisfy the required symmetry properties. We must antisymmetrize on both $x_{e1}\leftrightarrow x_{e2}$ and $x_{p1}\leftrightarrow x_{p2}$. This yields the wave function

$\Psi(x_{e1},x_{p1},x_{e2},x_{p2})$
$={1\over2}[\psi_0(r_{11})\psi_0(r_{22})\chi(\overline{x}_{11},\overline{x}_{22}) -\psi_0(r_{21})\psi_0(r_{12})\chi(\overline{x}_{21},\overline{x}_{12}) -\psi_0(r_{12})\psi_0(r_{21})\chi(\overline{x}_{12},\overline{x}_{21}) +\psi_0(r_{22})\psi_0(r_{11})\chi(\overline{x}_{22},\overline{x}_{11})]$
$={1\over2}\psi_0(r_{11})\psi_0(r_{22})\left[\chi(\overline{x}_{11},\overline{x}_{22}) +\chi(\overline{x}_{22},\overline{x}_{11})\right]-{1\over2}\psi_0(r_{21})\psi_0(r_{12})\left[\chi(\overline{x}_{21},\overline{x}_{12}) +\chi(\overline{x}_{12},\overline{x}_{21})\right]$

Look at the last line. The ground-state wave functions imply that the overall wave function is very small unless either electron #1 is close to proton #1 and electron #2 is close to proton #2, or electron #2 is close to proton #1 and electron #1 is close to proton #2. Only one of these can be the case if the protons are far apart from each other. In the first case, the wave function is well-approximated by

$\Psi(x_{e1},x_{p1},x_{e2},x_{p2})\simeq {1\over2}\psi_0(r_{11})\psi_0(r_{22})\left[\chi(\overline{x}_{11},\overline{x}_{22}) +\chi(\overline{x}_{22},\overline{x}_{11})\right]$

We see that the wave-function for the CM coordinates of each atom, in brackets, is symmetric on exchange of the two CM coordinates.

In the second case, the wave function is well-approximated by

$\Psi(x_{e1},x_{p1},x_{e2},x_{p2})\simeq -{1\over2}\psi_0(r_{21})\psi_0(r_{12})\left[\chi(\overline{x}_{21},\overline{x}_{12}) +\chi(\overline{x}_{12},\overline{x}_{21})\right]$

Once again, we see that the wave-function for the CM coordinates of each atom, in brackets, is symmetric on exchange of the two CM coordinates.

Note that this symmetry would not be present if one of the atoms had a different wave function than the other. (I leave it as an exercise to check this.)

So that's it. I claim to have derived the symmetry of the wave function on the CM positions of the two atoms from the antisymmetry of the wave function on the positions of their constituents, when the atoms are in the same state, and widely separated.

Last edited: Aug 29, 2014