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BEC cold atoms occupy the same ground state. But what about the electrons or other fermions of the BEC atoms? Are they in the same state? Do electrons of one atom interact with those of another?

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I think this a pretty deep question that applies broadly to quantum phenomena. I might phrase it this way: How does hierarchy work in all of QM? That is, when a complex object "goes quantum," what is the relationship of its internal components to the rest of the universe? I think in general this gets handled by noting simply that it's OK to ignore the details. However, when you look at effects such Mossbauer, that kind of simple assumption doesn't help much. –  Terry Bollinger Mar 8 '12 at 13:42

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Yes, the electrons are in the same state and yes they interact (in the sense that identical bosons interact to create a BEC). The explanation is kind of involved.


In a collection of identical atoms, it's not possible for us to distinguish between them. This also applies to their electrons. This is true whether or not the atoms are cold enough to be in a BEC state.

Now the rules of QM state that when you have identical particles (electrons or atoms) you have to symmetrize over them. Thus a literal answer to your question "But what about the electrons or other fermions of the BEC atoms? Are they in the same state?" is "Yes, they're in the same state but this doesn't have anything to do with the BEC state.

Your second question: "Do electrons of one atom interact with those of another?" can be answered similarly. If you consider the electrons as identical particles which must be symmetrized over, then it is impossible to distinguish one electron from the other. Therefore they must all be in the same state. So anything you do to "one" electron will effect all of them. Therefore they do interact; their waveforms are shared.


Now let's be more specific about your questions in terms of the nature of the BEC state. Consider just a single atom. It is described by a wave function. As the atom is placed in a cooler environment it loses energy. This loss of energy changes the shape of the wave function. It makes the wave function bigger. The wavelength for the wave function of a gas atom is called the "Thermal de Broglie wavelength". The formula is:
$$\lambda_T = \frac{h}{\sqrt{2\pi m k T}}$$
where $h$ is the Planck constant, $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the mass of the gas atom.

The thing to note about the above formula is that as the temperature gets lower, the wavelength $\lambda_T$ gets bigger. When you reach the point that the wavelengths are larger than the distances between atoms, you have a BEC. The reference book I've got is "Bose-Einstein Condensation in Dilute Gases" by C. J. Pethick and H. Smith (2008) which has, on page 5:

"An equivalent way of relating the transition [to BEC] temperature to the particle density is to compare the thermal de Broglie wavelength $\lambda_T$ with the mean interparticle spacing, which is of the order $n^{-1/3}$. ... At high temperatures, it is small and the gas behaves classically. Bose-Einstein condensation in an ideal gas sets in when the temperature is so low that $\lambda_T$ is comparable to $n^{-1/3}$."

When the wavelengths are longer than the inter-particle distances, the combined wave function for the atoms (which one must symmetrize by the rules of QM) becomes "coherent". That is, one can no longer treat the wave functions of different atoms as if they were independent.


To illustrate the importance of this, let's discuss the combined wave function of two fermions that are widely separated with individual wave functions $\psi_1(r_1)$ and $\psi_2(r_2)$. For fermions, the combined symmetrized wave function is:
$$\psi(r_1,r_2) = (\psi_1(r_1)\psi_2(r_2) - \psi_1(r_2)\psi_2(r_1))/\sqrt{2}.$$ Now the Pauli exclusion principle says that exchanging two fermions causes the combined wave function to change sign. That's the purpose of the "-" in the above equation; swapping $r_1$ for $r_2$ gives you negative of the wave function before the change.

Another, more immediate, way of stating the Pauli exclusion principle is that you can't find two fermions in the same position. Thus we must have that $\psi(r_a,r_a) = 0$ for any fermion wave function where $r_a$ can be any point in space. But for the case of the combined wave function of two waves that are very distant from one another there is no point $r_a$ where both of the wave functions are not zero (that is, the wave functions do not overlap). Thus the Pauli exclusion principle doesn't make a restriction on the combined wave function in the sense of excluding the possibility that both electrons could appear at the same point. That was already intrinsically required by the fact that the two wave functions were far apart.

Now I used the above argument about "far apart fermions" because the Pauli exclusion principle has a more immediate meaning to a lot of people than the equivalent principle in bosons. The same argument applies to bosons, but in reverse. Bosons prefer to be found near one another, however if we write down the combined wave function for two widely separated bosons, there is still a zero probability of finding both bosons at the same location. Again the reason is the same as in the fermion case: $\psi_1(r_b)$ is only going to be nonzero in places where $\psi_2(r_b)$ is already zero. The bosons are too far apart to interact (in the sense of changing the probability of finding both at the same point from what you'd otherwise expect classically).

Another way of saying all this is that in QM, there is no interaction between things except if their wave functions overlap in space. You use the thermal de Broglie wave function to determine how big a wave function has to be. If the atoms are closer than that, they're interacting in the sense of bose (or fermi) condensates.


So let's apply our understanding to the question "do the electrons interact" in a BEC. Consider the $\lambda_T$ formula to the electron. Since $m$ appears in the denominator, replacing the atom with the electron decreases $m$ by a factor of perhaps 3 or 4 orders of magnitude. This increases $\lambda_T$ by perhaps 2 orders of magnitude. Therefore, any gas which is cold enough to be a BEC will be composed of electrons that are much more than cold enough to also be coherent.

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-1, because I see no point in discussing the thermal wavelength for electrons, as far as they are bound. This would make sense in an ionized gas, probably. Also, I do not agree that all electrons are in the same state because the wavefunction is anti-symmetric. If I had a system of two non-interacting electrons in a potential well, I would be able to have them in different states (actually I would have to). I think that this is what the "state" in the questions refers to. –  Peter Kravchuk Jul 5 '13 at 16:06

When you describe BEC as condensation of an ideal Bose gas, you completely neglect interaction between the atoms. This is ok because the density of a BEC is so much lower of the liquid of solid phase of the same species of atoms. Thus the answer to your question is that nothing happens to the electrons, they keep orbiting the nuclei at their undisturbed orbits. The only times their fermionic nature matters for BEC is an occasional collision between the atoms which is not essential for the basic nature of the Bose condensation effect.

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Good question! I'm not an expert in Bose-Einstein condensation but I will try to give an answer.

When temperature is lowered in an ensemble of atoms which have integer spin (so that they are bosons and then they are susceptible to condensate) the wavefunction of each individual atom overlaps with the wavefunction of the other atoms. They form, as I see it, a kind of superatom, with different orbitals (each atom). The electrons are not in the same state because themselves are fermions and that would violate Pauli principle. Indeed, the electrons are in different space-time positions $\mathbf{r}_i,t_i$ with with different quantum numbers $\nu_i$ for each orbital so everything its fine.

The question about interactions is very interesting. I imagine that, as happens in isolated atoms, electron- electron interactions can be tackled roughly with a Thomas-Fermi model for screening but I have never read anything like this. The phenomena of BEC is robust against the interactions and appears even when considering free ball-like bosonic atoms, because is purely statistical effect...

D.

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yeah, i don't know what the accepted answer is going on about, but this is the correct answer. in light of Ken Wilson's recent passing, perhaps we might say that the electrons on the atoms are so deeply bound that they are not present as degrees of freedom in the low energy effective field theory of BEC. ;) –  wsc Jul 5 '13 at 19:15
    
RIP, Dr. Wilson. –  DaniH Aug 25 '13 at 18:48

This is a great question. I won't claim to have the final answer, but I do think there is an important point to be made here.

Let's take the simple case of Rb87. There is a single electron in the valence shell, allowing the atomic structure to be described as "hydrogen-like." If you study the electronic structure, you find that the electron spin interacts with its own angular momentum about the nucleus. This gives rise to different possible orientations of angular momentum and spin with different energies. In other words, an energy splitting, known as fine structure. A further splitting of each of the fine structure lines results from the interaction of the electron's total angular momentum, and nuclear spin. This is called hyperfine structure and it's roughly 3 orders of magnitude weaker than FS (but still quite important). So in the end, the possible states an atom can occupy are given by it's hyperfine energy levels, which in Rb87 are usually separated on the order of 100 MHz. THE reference for line data can be found here.

The point is that the atomic state is all we have. The coupling of electronic and nuclear states means that expressing the bare electron wavefunction becomes fruitless, much in the same way that you would never expect to describe an entangled photon using a single particle wavefunction.

An illuminating example comes from Cooper pairing, which gives rise to superconductivity. Cooper pairs are bosons, and so they can Bose condense. But the wavefunctions of the constituent electrons are now tied to their partners, making it impossible to give the state of just one of these paired electrons.

Just a thought, I hope it helps.

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Commenting here because it's the easiest way to reach you: if you would like to merge two accounts, see this page and follow the instructions there. –  David Z Jul 5 '13 at 21:11

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