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

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 ...


4

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 ...


3

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 ...


3

At constant pressure the volume of an ideal gas is given by Charles' law: $$ V \propto T $$ and this law tells us that when the temperature $T$ falls to zero the volume $V$ also becomes zero. But no gas is ideal and real gases show all sorts of non-ideal behaviour. For example real gases liquify then solidify as the temperatue falls. Real gases deviate ...


3

It seems to me that you just cannot tell the difference between a Bose condensate and nothing in this case. What will change if you add some photons or phonons with zero energy to the system? No characteristics of the system will change. So it seems to me we have no criterion to decide if there is a Bose condensate in this case, and what's more important, it ...


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 ...


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

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, $\mu = 0$.


2

Since the non-interacting condensate is a pathological situation (it is not a superfluid), I will assume that by "traditional" you mean a (perhaps extremely) weakly interacting condensate. I will denote the repulsive interaction strength (the T-matrix) by $g>0$. For simplicity, I will describe the situation at very low temperatures. The elementary ...


2

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

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 ...


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

When cooled to around 2.18K - the lambda point - liquid helium enters a superfluid phase. This is similar to a BEC, but remember that strictly speaking, BEC deals with bosons in the gas phase. In this case, since the helium is in the liquid phase, there are significant interactions between He atoms not present in the theory of a non-interacting gaseous BEC. ...


1

The photons in the experiment are confined in a cavity, which gives effective mass to photons. With that, you could calculate the critical temperature as usual. Although it is non-trivial to distinguish between lasing and Bose-Einstein condensation, they claim that they see "thermalization" via a lot of absorption and emission events with dye molecules, ...


1

In the field of multi-component condensates, the single mode approximation (SMA) means that different dipole states are assumed to share the same spatial wave function. Thus, there are no dipolar textures. SMA is well justified when the inter-component (e.g., spin-dependent or dipole) interactions are much weaker than the interactions independent of the ...


1

The short answer is that you have to work with particle densities, namely, $$n=N/V,$$ where $N$ is the number of particles, and $V$ is the volume of your system. The long answer is as follows. Working in the thermodynamic ensemble with a fixed chemical potential $\mu$, and knowing that the Bose-Einstein distribution gives the average number density of ...


1

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 ...


1

Let me try to formulate the question more precisely and then give my answer. The ``phase'' $\theta$ of a BEC is introduced as $\langle \psi_N|\hat{a}|\psi_{N+1}\rangle = |\psi|e^{i\theta}$, where $|\psi_N\rangle$ is the ground state being occupied by $N$ bosons. Because the occupation number of the condensed state $|\psi_N\rangle$ is of the order $N$, the ...


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

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

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


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

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

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 )



Only top voted, non community-wiki answers of a minimum length are eligible