Hot answers tagged

6

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 ...


5

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 ...


5

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 ...


4

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 ...


4

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 ...


3

If operators $a_j$, $a^\dagger_j$ correspond to the system's orthonormal natural orbitals $\phi_j({\bf x})$, such that $$ \int{d{\bf x}\; \phi^*_j({\bf x})\phi_k({\bf x})} = \delta_{jk}, \;\;\;\sum_j{\phi^*_j({\bf x})\phi_j({\bf x'})} = \delta({\bf x} - {\bf x'})\\ \hat\psi({\bf x}) = \sum_j{\phi_j({\bf x})\;a_j},\;\;\; a_j = \int{d{\bf x} \;\phi^*_j({\bf ...


3

It is not possible to get a Bose-Einstein condensate of photons in three-dimensional equilibrium. Since the photons have no mass gap and no chemical potential, they can just be absorbed by the walls. In this example, however, the experimenters used a gas that was out of equilibrium, with different effective temperatures for the motion in different ...


3

An ensemble of interacting particles will, over time, develop entanglement between widely separated parts*, so this is similar to asking whether an interacting system can still be a BEC. The short answer is yes, but a subtlety is that various authors define BEC in slightly different ways. One way of defining BEC, as I mention in a recent answer, is the ...


3

In the Bogoliubov transformation, it is the case that $$u_{-p} = u^*_p,$$ and $$v_{-p} = v^*_p,$$ so your result is correct. In fact, if you move on to basically the next equation, they assume that both $u$ and $v$ are real, in which case $$u_{-p} = u^*_p = u_p,$$ as mentioned by Mark Mitchison in a comment below this answer. And incidentally, yes it is ...


3

You can think of the chemical potential as the amount of free energy needed to add one additional particle to the system. Because the ground state of a BEC is degenerate and can hold an infinite number of particles, there's no energy cost to add another particle to that state. So, $\mu = 0$.


3

Optimizing the MOT beams Since the maximal force created by each beam on the atoms is limited by the saturation intensity $I_\text{sat}$ : $$\tag{1} F_\text{max}=\frac{\Gamma}{2}\hbar k\quad\text{when}\quad\frac{I}{I_\text{sat}}\geq 1 $$ you will always want to have enough power in the beams of your MOT so that you hit the saturation regime. Here, $k$ ...


2

Fundamentally it is that the $1/N!$ for the classical system only correctly compensates for overcounting of indistinguishable states if the particles are always in different states. For a system of Bosons at low temperature, where it is quite likely that many particles are in the same state, this breaks down. For a very understandable introduction to this ...


2

A Bose-Eistein condensate is typically a very, very dilute gas. While it's big enough to see in principle, in practice it doesn't scatter light strongly enough to be visible. There are exceptions to this. Some BECs have been made that interact strongly with light. However I think that even in these cases they'd be invisible because the light levels used in ...


2

One can avoid the concept of symmetry breaking in this context, to avoid "non-conservation of the particle number". People have devised way to do that, see for example http://arxiv.org/pdf/cond-mat/0105058v1.pdf. However, all these approaches gives the same results than standard Bogoliubov-like methods in the thermodynamic limit. This is not too ...


2

When you diagonalize $H_\text{BdG}$, you find the eigenvalues and the eigenvectors. Let us assemble these (column) vectors into a unitary matrix $U$ (this is always possible since $H_\text{BdG}$ is Hermitian), so that $U^\dagger H_\text{BdG} U=E$ is diagonal (i.e. the eigenvalues). Define $\Gamma_k = U^\dagger \psi_k$, then $ H=\frac{1}{2} \Gamma_k^\dagger ...


2

The Landau criterion is not in itself a criterion for superfluidity, but a criterion for the breakdown of superfluidity. Indeed, if applied to insulators or ordinary fluids, it would tell you that all of these are also superfluid... What the Landau criterion tells you is the velocity of the superfluid flow at which excitations are created from the ...


1

Well, the trite answer is "no" because every bit of matter we see around us is in an entangled state. It's the normal classical world.


1

We -first- exhaust the chemical potential, and -then- start occupy the lowest state. What is the reasoning for this? That "first"/"then" isn't the logic used in deriving the Bose-Einstein condensate, but it is an intuitive picture of what's going on. So, we have for our grand canonical potential something like: $\Omega=T \sum_{i} ...


1

Check out this paper: Improved magneto-optic trapping in a vapor cell. K.E. Gibble, S. Kasapi and S. Chu. Opt. Lett. 17 no. 7, 526 (1992), PSU eprint. You'll notice that the rate at which atoms below the capture velocity enter the volume of your cooling region increases with the beam diameter squared. There is also a plot of the number of atoms ...


1

I will try to address the first point raised by the OP, i.e. the occurrence of spontaneous symmetry breaking in Bose-Einstein condensation. The free boson gas is described by the hamiltonian: $$ H_V=\int_V\frac{d^sx}{2m}\big|\nabla\phi(x)\big|^2. $$ The ground state satisfies $H_V\Psi_0 = 0,\ \forall V$ and hence $\nabla \phi(x)\Psi_0=0,\ \forall x.$ ...


1

The transition between the two phases is called (not unreasonably!) a phase transition, and phase transitions come in two flavours: first order and second order. First order phase transitions (generally) have a sharp transition temperature. Steam condensing to water is a first order phase transition and as you've pointed out occurs at 100ºC (at one ...


1

Superfluid helium finds an application as a coolant in superconducting systems (http://link.springer.com/chapter/10.1007%2F3-540-45542-6_4#page-1 )


1

What is to be understood I guess is why the equality holds. (and maybe what is q, r being a dummy variable anyway.) I'm a little bit rusty on the exact relations and numerical factors, but I'll give the idea: one does a change of variable in order to use the following identity: $$ \delta (\mathbf{p}) = \int_V e^{i\mathbf{p}\cdot \mathbf{x}}\ d \mathbf{x}$$ ...


1

I dug around in the literature a bit and found that this formulation (semiclassical Bose-Hubbard plus Langevin-type dissipation) has been studied before. Here is the relevant reference: http://arxiv.org/abs/1304.5071. What you are trying to do is derive their equation (9). You probably missed it because they refer to their model as the discrete nonlinear ...


1

No, most bosonic atoms (all except H and He) form solids. In fact, most of the BECs studied experimentally with ultracold atomic gases are metastable. The true ground state is a solid.


1

I was confused about this point too when I first learned about Feshbach resonance. Eventually I realized that my confusion had more to do with scattering theory than Feshbach resonance per se, so I'll focus on that. The important point to remember is that scattering length is really about the phase shift. If at low energy you get the same s-wave phase ...


1

It means that all atoms are in the ground state, then since potential energy is defined up to a constant you can say ground state has zero energy



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