# Tag Info

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 ...

8

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 ...

5

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. ...

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

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

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

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

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 ...

2

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. ...

2

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 ...

2

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 ...

2

I'm not quite sure about your background so I'll try to give you a gentle introduction: The article you were reading was probably about Bose-Einstein-Condensation. This is a phenomenon where all the atoms (with bosonic statistics) sit in the ground state of some external confining potential (e.g. a magneto optical trap) at finite temperature. The important ...

2

I recently updated the wikipedia article on statistical ensembles which might be relevant. Basically, in classical physics the probability distribution for the state of a system is written as an integral over position and momentum as in your equation. It turns out to be necessary to choose an arbitrary unit of action (energy times time) in order to define ...

1

It is not a monopole, is just an artifact. Note that any addition of dipolar contributions of magnetic field (spin, coils, magnets…) must be dipolar or of a higher order magnetic field, in no case it could be monopolar. It is to say, there is not possible to construct a monopolar field as a superposition of dipolar (or higher order) contributions. The ...

1

In the specific case of slowing light with a Bose-Einstein condensate there will be a limit because the slowing of the light is due to an interaction of the light with the BEC to form a polariton. If you put too much energy in you'll destroy the BEC and it will stop slowing the light. Offhand I don't know what the limit is, but it will be a very small amount ...

1

It depends what you call "one state". With only one species, the Fock state basis is of the form $|n_1,n_2,n_3\rangle$ which gives the number of particle on the sites $1$, $2$, $3$. This is one state of the system (even though it is not a eigenstate). In the case with two species, one can trivially generalize the notation, with a basis ...

1

Bose-Einstein condensation is based on the indistinguishability and wave nature of particles, which are both basic concept of quantum mechanics. If you want to define Bose-Einstein condensation in one sentence, you can say it is the occupation of the lowest quantum state of the external potential by a large fraction of bosons forming the system. Particles ...

1

For a bose gas where particle number is not conserved, e.g. blackbody photons, indeed $\mu=0$. How does that work? Well, as you approach zero frequency, the number of blackbody photons gets higher and higher -- yes, you bet, it approaches infinity. BUT nevertheless the total collective energy of those low-frequency photons gets lower and lower. (As you ...

1

Feshbach resonance is important to BEC because it allows the adjustment of the interaction between atoms. At low energy regime, the BEC dynamics can be described by the mean field Gross–Pitaevskii equation: $$i \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2 \psi + V\psi + \frac{4\pi \hbar^2 a_s}{m}|\psi|^2 \psi \tag{1}$$ The $\psi$ is ...

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 ...

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