# Tag Info

11

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 ...

6

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 ...

4

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 ...

4

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 ...

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 ...

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

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

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. ...

3

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 ...

2

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 ...

2

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 ...

2

Any particle in a system (like a photon or an electron) can be in many different energy state. You may be familiar with the energy states of atoms, these are the states occupied by electrons orbiting the nucleus. But electron are fermions and they don't want to be together (in fact they can't). In order to minimize the energy of the atom any additional ...

1

I believe, that the derivation is wrong... If you assume a translationally invariant state, such that $G^1(r, r') = G^1(r - r')$ then you can get the result. Rewrite the exponential as $p r - p' r' = p( r- r') + r'(p - p')$. Since, in this case, the left-hand-side of Eq. (2.27) can only depend on $r - r'$ it must be such that $p = p'$ from the second term. ...

1

The particles are simply distributed evenly between the degenerate lowest energy states. This is the case for a ideal spinor BEC without Zeeman effects, for example: "Because there are three internal states, the condensation temperature of an ideal spin-1 gas at $p = q = 0$ is reduced to $T_c^{\mathrm{spinor}} = (1/3)^{2/3}T_0$" where "$T_0$ is the ...

1

It is necessary to clarify that a uniform, non interacting Bose gas (considered to be confined in a periodic box) in thermal equilibrium does not have a macroscopic occupation of the zero momentum mode if $d<3$. This is not quite accurate for $d=2$ as macroscopic occupation is achieved at T=0, or rather the critical temperature tends to zero in the limit ...

1

I recently started reading this book: http://www.amazon.com/BCS-BEC-Crossover-Unitary-Lecture-Physics/dp/3642219772 So far I like the organization and pace. But judging by the table of contents it appears to be very detailed and thorough. It is, however, a monograph. But the style is pretty close to a textbook. Plus it has around 150 references at the end ...

1

Because water is liquid at much too high a temperature. Helium is only superfluid near absolute zero. To have a superfluid, you need the quantum wavelength of the atoms given the environmental decoherence to be longer than the separation between the atoms, so they can coherently come together.

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