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12

I assume you mean the relatively recent phenomenon of Bose-Einstein Condensation in dilute atomic vapors (first produced in 1995 in Colorado). The overall phenomenon of Bose-Einstein Condensation is closely related to superconductivity (in a very loose sense, you can think of the superconducting transition in a metal as the formation of a BEC of pairs of ...


12

Er ... nothing prevents this. That's what a Bose-Einstein condensate is: lots of bosons in the same place and quantum state. You are observing that the sate is not perfectly localized, but that is a consequence of the state not being exactly zero momentum. Ultimately the Heisenberg principle puts a lower limit on how localized they could be. If the bosons ...


10

First and foremost, the BEC systems studied in detail today do not involve the formation of any bonds between atoms. Bose-Einstein Condensation is a quantum statistical phenomenon, and would happen even with noninteracting particles (though as a technical matter, that's impossible to arrange, but you can make a condensate and then manipulate the interactions ...


10

Well, photons are massless. The key is the confinement of photons and molecules in an optical cavity long enough for them to reach thermal equilibrium. A BEC is a state of matter that spontaneously emerges when a system of bosons becomes cold enough that a significant fraction of them condenses into a single quantum state to minimize the system's free ...


10

Let me make quite clear that the recent experiment does NOT imply the detection of a true magnetic monopole. Somehow, in all the excitement, the word "synthetic" was dropped rather quickly from the phrase "synthetic magnetic field". A synthetic magnetic field is a physical quantity that obeys the same equations as a magnetic field, typically realized in ...


8

Bogoliubov proved long, long ago that the condensate is stable against weak interactions. The interactions scatter some fraction of bosons out of the lowest-energy single-particle state ("depleting" the condensate), but off-diagonal long range order remains. For a nice introduction to Bogoliubov's theory see Ben Simon's lectures ...


8

This is a nice puzzle--- but the answer is simple: the composite bosons can occupy the same state when the state is spatially delocalized on a scale larger than the scale of the wavefunction of the fermions inside, but they feel a repulsive force which prevents them from being at the same spatial point, so that they cannot sit at the same point at the same ...


7

The claim is often that there is no condensation in $d<3$. The other answers are correct, but let's be clear, there are actually two assumptions present in the claim: Assume you have $N$ noninteracting bosons in $d$-dimensions in a hypervolume $L^d$ Assume that these bosons have an energy-momentum relationship of $E(p) = Ap^s$. Now, the way we ...


7

Throughout, let's assume that the ground state energy of the system under consideration is zero. Chay Paterson has addressed your question in the case of a gas of bosons in which the number of particles is not conserved, but from the wording of your question, it seems that you're concerned about the case in which the total number of particles is fixed. ...


6

I think what you are talking about is the stimulated emission of radiation. This is part of the process that occurs in a LASER and in fact, gave the LASER it's name - Light Amplification through the Simulated Emission of Radiation. When an atom is in an excited state it can spontaneously decay to a lower energy state with the emission of a photon of a ...


5

This is a good example of when the theoretical rubber meets the proverbial experimental road. The two issues you bring up are actually completely independent. The chemical potential problem is purely kinematic, and may be solved by simply introducing a harmonic trapping potential or any other way to modify the density of states. Mermin-Wagner is more ...


5

In reality you almost always find that the particles prefer to go to the same state because of some tiny energy shifts in the system. For example, a ferromagnetic condensate can have many degenerate states, but there is an energy cost for particles that disagree. These systems will break symmetry by having all the particles choose the same (arbitrary) state. ...


5

Lagerbaer gets it right in the comments: The set of all (properly [anti-]symmetrized) product states is a complete basis. That's it, game over. So the first sum in your series represents every possible state. We could also make (generally overcomplete) bases that consist entirely of products of maximally entangled pairs -- for example the valence bond ...


4

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 ...


4

I will just answer the first part of question: is phonon attraction stronger than Coulomb? Short answer: No. Longer answer: Nothing (in condensed matter) is ever stronger than the Coulomb force. Longest answer: There are two aspects to consider. First is the self-screening of the electrons, which will add a mass term to the photons, giving a Yukawa-esque ...


4

This is really just a comment to dmckee's answer, but it got a bit long for a comment. The problem with your question: what keeps bosons from occupying the same location? is that no particle has a precisely defined position. Remember that when we get down to the sizes of atoms etc particles don't have a position. They are described by a wavefunction ...


4

You can have superfluids that are not BECs and BECs that are not superfluid. Let me quote a text, "Bose-Einstein Condensation in Dilute Gases", Pethick & Smith, 2nd edition (2008), chapter 10: Historically, the connection between superfluidity and the existence of a condensate, a macroscopically occupied quantum state, dates back to Fritz ...


4

You refer to the Landau criterion for superfluidity (there is a separate question whether this is really the best way to think about superfluids, and whether the Landau criterion is necessary and/or sufficient). In a superfluid the low energy excitations are phonons, the dispersion relation is linear $E_p\sim c p$, and the critical velocity is non-zero. In ...


4

You can gain some intuition from looking at the density distribution function in momentum space which for the $|BCS\rangle$ is given by $n_k=v^{2}_k$. In the BCS limit one finds approximately the filled Fermi sphere, while in the BEC limit $n_k\sim 1/(1+[ka]^2)^2$ which is proportional to the square of the Fourier transform of the dimer wave function. For ...


4

After reading this paper, I wracked my brain trying to come up with the perfect analogy. Suffice it to say I failed, so here is my less than ideal answer. The monopole created and referred to in this article is not a true Dirac monopole. It is no more a real monopole than a thermal vacuum testing chamber is outer space. That is, it is an artificially ...


4

Hints to the sought-for formula (16) for $\hat{H}$: Use integration by parts in ${\bf r}$-space to remove derivatives from the Dirac delta distributions, cf. comment by user ACuriousMind. Work on the problem from both ends (15) and (16). Use Leibniz rule $$\tag{*}\nabla^2 (fg)~=~ g\nabla^2 f + f \nabla^2 g+ 2 \nabla f\cdot\nabla g,$$ so that $\nabla$ only ...


3

The linear terms it seems you can handle. As piece of general advice, the meaning of these terms are always clearly if integrate over the momentum coordinates of each of the fields, using delta functions to preserve the value. So the non-linear term would be $$\sum_{q_1,q_2,q_3} g(q_1,q_2,q_3) \psi(q_1)^*\psi(q_2)\psi(q_3)\delta(-q_1+q_2+q_3 -k)$$ Maybe you ...


3

There are different ways to define this phase. In mean-field (low temperature, weak interaction regime), the many-body wave function $\psi(x_1, x_2,...)=\prod_i \Phi(x_i)$ where $\Phi(x)$ is sometimes called the macroscopic wavefunction (because all the bosons are in the same state described by $\Phi$). In the simplest case (homogeneous system), one can ...


3

Yes, they can, an experimental example of that is Bose-Einstein Condensate of fermions. And that is possible because actually they will have the same wave function, in sense that nature no more capable of making any distinguish between them. Regarding everyday life, actually saying that it is bosonic is just a formal statement, in sense that Pauli exclusion ...


3

1) Some of the assumptions of the Gross-Pitaevskii equation (GPE) are: all atoms are in the same condensate wave function, the condensate is at $T=0$, collisions between atoms are sufficiently low energy that the interactions can be well described by the $s$-wave scattering length, so that the interaction can be written ...


3

You cannot have a total vorticity with periodic boundary conditions, since if you take a path around all of your vortices, it will have a non-zero circulation. But you have periodic bc, so you can continuously deform that path to a point, and a point has zero circulation. Mirror images are not quite the same as in electrostatics. We want periodic boundary ...


3

It is possible to think of a Bose Einstein condensate is simply matter in a situation where it is described by a classical field. Any classical field is a BEC of its particle, so electromagnetic radiation is the BEC for photons. Your question is whether there are electromagnetic fields which are thermally stable. This is not true, because there is no photon ...


2

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, ...


2

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 ...



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