16

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 ...


14

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 state 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 ...


14

In the context of ultracold Fermi gases, a BEC-BCS crossover means that by tuning the interaction strength (the s-wave scattering length), one goes from a BEC state to a BCS state without encountering a phase transition (thus the word "crossover"). It is also useful to know that the BEC state is a Bose-Einstein condensate of two-atom molecules, while the ...


13

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 ...


11

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. ...


11

Interesting question. The "third law" of thermodynamics actually states that The entropy of a perfect crystal at absolute zero is exactly zero. Since many systems in nature crystallize at some point when their temperature is lowered, this statement is often misinterpreted as: The entropy of a physical system at absolute zero is exactly zero. (wrong!) ...


10

Short answer: the terminology is not always used consistently. A superfluid and a BEC are certainly logically different. Superfluidity refers to the ability of some liquid to flow without any viscosity, and some associated properties, such as the ability to "expel" the angular momentum of its container, which is known as the Hess–Fairbank effect*. In other ...


10

I believe this question is a reference to the superfluid vacuum theory. First and foremost, I believe the Gross-Pitaevskii equation in the problem statement is incorrect. It seems to have mixed up the time dependent and independent case. There should not be a $-\mu$ term on the right hand side. Before I move on to my answer, I want to also point out that ...


9

You're right, there's some confusion due to the way this paper has been publicized. Negative effective mass The 'negative mass' referred to in the paper is effective mass. The idea is that while every fundamental constituent of a physical system has a known, nonnegative mass, the effective degrees of freedom of the system may behave as if they have a ...


9

Essentially, you are trying to prove that your initial Hamiltonian can be rewritten as $$H = r a^\dagger a + sI\tag{1}$$ where $r$ and $s$ are reals and $a^\dagger$ and $a$ satisfy bosonic ccrs and there is a vacuum vector for $a$. This is a much stronger condition than diagonalizability! However, it is possible to prove that the requirement above cannot ...


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 http://www.tcm.phy.cam.ac.uk/...


8

Short answer: Bosons all collapse to the ground state, since there is no restriction on the number of particles that can occupy a given state. You can assign $0$ the ground state energy, or any other number really since you can't really measure its energy, but only differences between energy levels. But it's customary to choose $0$ (the only time you need ...


8

Yes. Bose-Einstein condensation was experimentally achieved in systems of: Rubidium atoms (first experimental realization, 2001 Nobel prize) Potassium atoms Cesium atoms Lithium atoms Sodium atoms Exciton polaritons Photons Phonons (?) If you are using atoms, they must behave like bosons (see also here).


7

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 ...


7

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 London's ...


7

The amount of heat added to the system is the integral of the specific heat wrt temperature: $$ Q = \int C(T)dT $$ So in the link you give it's just the area under this graph: Although it's true that the specific heat tends to infinity at the lambda point it does so sufficiently suddenly that the area under the graph remains finite. That means the amount ...


7

To determine the upper limit on chemical potential for a gas of $\mathcal N$ bosons, look at the form of the Bose distribution in the grand canonical ensemble with $\langle N \rangle = \mathcal N$. When using the GCE, it's easiest to work at chemical potential $\mu$ and to then choose $\mu(\mathcal N)$ so that $\langle N\rangle(\mu)=\mathcal N$. Each state $...


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 ...


6

Fact: quantization of gravity is still at a research stage, only effective quantized field theories are used. Fact:gravastars ( had to look it up) belong to one proposal for quantizing gravity, which is not particularly dominant in the research . A gravastar is an object hypothesized in astrophysics as an alternative to the black hole theory by Pawel O. ...


6

The temperature of a BEC formed from a dilute atomic gas (e.g. Rb87) isn't determined by the ambient radiation field, as the vast majority of the photons don't interact with the atoms. BECs are also produced inside ultra high vacuum vessels, which have a vacuum much better than near-Earth orbit, so the ambient pressure isn't the reason either. A satellite-...


6

TL;DR: The Gross-Pitaevskii equation is only applicable for very weakly-interacting bosons. At $a=\infty$ the gas displays universal physics. Strictly speaking, the Gross-Pitaevskii equation (GPE) is only valid for $$na^3 \ll 1,$$ where $n$ is the density of particles and $a$ is the $s$-wave scattering length. As it is a mean-field theory, one has to look ...


6

How can a scattering process have bound states? We are familiar with bound states from our everyday experience. For example, two hydrogen atoms interact through the Coulomb force. This leads to the formation of a bound state, namely, the hydrogen molecule. The most simple model of this situation is the square-well potential. This potential has a ...


6

The 1st order density matrix, $$ \rho({\bf x}, {\bf x'}) = \langle \hat\psi^\dagger({\bf x'})\hat\psi({\bf x}) \rangle $$ is a Hermitian operator since $\rho({\bf x}, {\bf x'}) = \rho^*({\bf x'}, {\bf x})$ and its diagonal entries, $$ \rho({\bf x}, {\bf x}) = \langle \hat\psi^\dagger({\bf x})\hat\psi({\bf x}) \rangle = n({\bf x}) $$ give the number density ...


6

If $\langle 0|\phi (x) |0\rangle = c $, then $\phi $ is related to the quantum field $\Phi $, whose $|0\rangle $ is the vacuum state, by means of $$ \phi (x) = \Phi (x) + cI\:.$$ In other words, $$\phi (x) = U \Phi (x) U^*$$ where $$U = e^{b (a_0 -a^*_0)}$$ for some constant $b $ depending on $c $ and where $a_0$ and $a^*_0$ are the annihilation and ...


5

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 ...


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

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 ...


5

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 ...


5

When the chemical potential is 0 the extra free energy needed to add or remove a particle to the system is 0(i.e $\mu=\frac{dA}{dN}=0$. So particles can leave and enter the system without changing the (free) energy. In A BEC all particles have condensed to the ground state of the system. Particles entering or leaving the system will be added to the ground ...


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