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?
Bose-Einstein Condensate – What Happens to the Electrons of BEC Cold Atoms?
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First, let me say I'm not sure what is meant by a BEC with $T\gt 0$. Condensation is a finite temperature phenomenon, which occurs due to the presence of pair-wise interactions (generally attractive, but pairing can happen even for repulsive potential) in a many-body system. For instance, in a superconductor below some critical temperature $T_c$, electrons with opposite momenta and spin (s-wave pairing) pair up to form a bound state called a Cooper pair.
The ground state of the unpaired electron gas for $T \gt T_c$ is characterized by the Fermi energy $E_F$. After condensation, the many-body system has a new ground state at energy $E_{bcs} = E_F - \Delta $, where $\Delta \sim k_B T_c$ is the binding energy of a Cooper pair. $\Delta$ is also known as the gap.
For $T\lt T_c$ all the electrons are not paired up due to thermal fluctuations. However, the number of unpaired electrons as a fraction of the total number of electrons (the condensate fraction) goes as $N_{free}/N_{all} = 1- (T/T_c)^\alpha$, where $\alpha\gt 0$. The number of free electrons drops rapidly as $T$ is decreased below $T_c$. In lab setups, BEC's generally undergo some form of evaporative cooling to get rid of particles with energies greater than $\Delta$. At this point the condensate can be treated as a gas of interacting (quasi)particles (cooper pairs) with an approximate hard-core repulsion.
So the gas, before and after condensate formation, is always at finite temperature! This is reflected, for instance, in the dependance of the condensate fraction on $T$ as mentioned above.
The mean-field solutions for low-energy excitations of the condensate are given by the Gross-Pitaevskii equation(GPE):
$$ \left( - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2} + V(r) + \frac{4\pi\hbar^2 a_s}{m} |\psi(r)|^2 \right) \psi(r) = \mu \psi(r) $$
where $a_s$ is the scattering length for the hard-core boson interaction, with $a \lt 0$ for an attractive interaction and $a \gt 0$ for a repulsive interaction.
Presumably one should be able to construct a canonical ensemble with solutions $\psi(k)$ ($k$ being a momentum label of the above equation), but this is by no means obvious because of the non-linearity represented by the $|\psi(r)|^2$ term. Here a "zero temperature" state would correspond to a perfect BEC with no inhomogeneities, i.e. the vacuum solution of the GPE. However, the entire system is at some finite temperature $T \lt T_c$ as noted above. The resulting thermal fluctuations will manifest in the form of inhomogeneities in the condensate, the exact form of which will be determined by the solutions of the GPE.
Of course, the GPE's regime of validity is that of dilute bose gases ($l_p \gg a_s$ - the average interparticle separation $l_p$ is much greater than the scattering length). For strong coupling I do not know of any similar analytical formalism. If I had to take a wild guess I'd say that the strong-coupling regime could be made analytically tractable by mapping it to a dual gravitational system, but that's another story altogether.
As $l_p$ approaches $a_s$ from above, the GPE breaks down and it will have singular solutions for any given $T$ and these are likely the singularities that you are referring to.
Reference: The single best reference I can suggest is Fetter and Walecka's book on many-body physics. I'm sure you can find more compact sources with a little effort. But generally the brief explanations leave one wanting for a comprehensive approach such as the one F&W provides.
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As an experimentalist, I might not be the person best suited to answer this, but I'll give it a try.
The wavefunction is going to be difficult to visualize because in general it is a complex function. If you want to 'see' sqrt(-1), I suggest you resort to drugs, lots and lots of drugs. But as for physical interpretation, Born tells us that the amplitude of the wavefunction squared will give you the real probability distribution. So the wavefunction represents a probability amplitude, and the random nature arises in the measurement process.
When making a BEC, you extract energy from the atoms in a few ways. Usually, you start with laser cooling (in a magneto-optical trap, or MOT) that captures and cools about 1e9 atoms to about 100 microkelvin. From there, you often need to employ other tricks to beat the so-called doppler limit set by the linewidth or your laser. Another limit is the "recoil limit" that says you will never cool an atom to have less energy than the photon it (randomly) emits. So you turn the lasers off, and trap these very cold atoms in other ways only to cool them further. Either magnetic fields, or far-off-resonance, focused lasers (optical traps) are used to capture the atoms, but now we are not scattering lots of light off them like we were in the MOT. In both cases you are altering the energy landscape experienced by the cloud, and this causes it to seek the lowest energy. Lower the trap walls, and the hottest (fastest) atoms will escape, leaving the remaining (slower) atoms to thermalize to a lower temperature. This is called evaporative cooling. In magnetic traps, it is often done with radio frequencies which couple your atoms to different Zeeman sublevels, changing their potential energy, and ejecting them from the scene. In far detuned optical traps, they just lower the laser power.
As the atoms get colder, something remarkable happens. Their wave nature starts to emerge. The de Broglie wavelength gets longer as you get colder, until eventually, particle wavefunctions start overlapping with their neighbors. At this point, the atomic statistics are no longer well described by the good-ole Maxwell-Boltzmann distro, and we have to resort to Bose-Einstein statistics. (This of course assumes you are working with bosons, or integer spin particles.) What does that mean? Basically it means that particles start to favor occupying the same state, or 'mode'.
To determine the wavefunction of a BEC, you could start with the Schroedinger equation, where now you simply replace the single particle wave function with a many particle wavefunction that is built as a tensor product of the individual particles. This turns out to be a poor approach because the particles interact. They are constantly bumping into each other, and this manifests as a non-linear (i.e. density dependent) interaction term in the Hamiltonian. So we can far better approximate the situation with the Gross-Pitaevskii equation. If you want to model a BEC, the GPE might be a good place to start.
One of the striking features of QM in my mind is that the description of a particle is inextricably linked to its environment. It's right there in the GPE via the potential energy term, V. The solutions of this equation will be your wavefunction. When you confine atoms in a harmonic trap, one that looks like a parabola, your wavefunction will reflect this. Indeed it is not a gaussian you see, but an inverted parabola (see Thomas-Fermi approx.). It will look different in different traps. The residual gaussian is composed of thermal atoms, still described by the M-B picture.
I should note that when atoms are in the ground state, they still cannot be said to have "no motion". The uncertainty principle tells us that they will always smear out over some tiny volume of phase space. This is called the zero point motion.
The phase is a tricky subject. You are right that the book seems to have described it poorly. BECs are often said to be phase coherent, which is to say across the whole cloud, the phase will be the same. If it were not so, you would necessarily see striations since some of the atoms would destructively interfere with their neighbors. This is basically an excitation, and so energetically unfavorable. I might have just made a theorist lose her lunch with that explanation, but this whole thing is just meant to be a low level intro.
Good luck, and keep asking questions.
Best Answer
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:
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.