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7

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 ...


4

The scattering length is basically a crude measure of how much interaction there is, so if you have a cold atomic gas in a trap, and it starts to interact more, then naturally atoms get kicked out of the trap by these interactions. This is then detected by enhanced loss rates. Depending on the setup you can get very weakly bound states (for example Efimov ...


4

Hints to the sought-for formula (16) for $\hat{H}$: Use integration by parts in ${\bf r}$-space to remove derivatives from the Dirac delta distributions, cf. comment by user ACuriousMind. Work on the problem from both ends (15) and (16). Use Leibniz rule $$\tag{*}\nabla^2 (fg)~=~ g\nabla^2 f + f \nabla^2 g+ 2 \nabla f\cdot\nabla g,$$ so that $\nabla$ only ...


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 ...


2

Let us suppose the gas is confined by a harmonic potential. The bosons have, in three dimensions, energy levels $\hbar\omega(n+3/2)$ with degeneracy $n(n+1)/2$. The grand-canonical partition function of level $n$ is (without degeneracy) $$ \xi_n=\sum_{p=0}^\infty \left(\mathrm e^{-\beta \hbar\omega(n+3/2)+\beta\mu(T)}\right)^p$$ where $p$ is the number ...


2

Your equation (1) describes approximately the centre-of-mass (COM) coordinates of every atom = (some nucleons + some electrons) system. Of course there are many other degrees of freedom that are not taken into account in this description. But those degrees of freedom can always be ignored unless they become correlated with the centre-of-mass coordinates. ...


2

The defining feature of a Bose condensate is that the one-body density matrix $$ \rho^{(1)}(\mathbf{r},\mathbf{r}^{\prime}) = \langle \Psi^{\dagger}(\mathbf{r})\Psi(\mathbf{r}^{\prime})\rangle,$$ has at least one eigenvalue that is macroscopically large, i.e. it is of order $N$, with $N$ the number of particles in the system. Here, $\Psi(\mathbf{r})$ is ...


2

Yes, Bose-Einstein condensates are affected by gravity. Most condensates are formed in laser traps and often (especially in the early experiments) the lasers must be turned off to get a good image of the condensate, with the consequence that many images of condensates (again, especially from the early experiments) show them falling. An example (source):


2

Some may exclude superfluid 3He from being a Bose-Einstein condensate because it obeys Fermi-Dirac statistics. However, this viewpoint is also not quite clear cut as the 3He form Cooper pairs which then condense. However, even those pairs do not obey Bose-Einstein statistics but nonetheless condense. Therefore this question is a little murky and Wikipedia ...


2

There are several ways to create Bose-Einstein condensates or systems that behave that way, there are ultracold atomic gases, solid state quasiparticles, and even photon condensates. Since you are obviously interested in ultracold atomic gases, I am going to cite Experimental methods of ultracold atomic physics by Kurn and Thywissen: The material must ...


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. ...


2

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 ...


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 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

I think that classical particles won't occupy the same same states, as if they are rigid like billiard balls, they can't occupy the same position. But if you drop this constraint and put them in a potential well, they probably would do something similar to a BEC (all occupy the lowest energy level as you drop temperature). However, the difference in a BEC ...


1

This is a very good question. It turns out that the phase transition occurs precisely when the chemical potential becomes equal to zero (assuming that the ground state energy is at zero). The order parameter in the BEC is the "macroscopic wave function" or rather the square root of the single-particle reduced density matrix. The broken symmetry is usually ...


1

"Classical particle", or atoms do condensate. At high temperature $T$, the atoms are far away in the so called gas phase. When the temperature decrease, they will undergo a phase transition and condense to liquid. At even lower temperature, it becomes solid and the atoms are closer together. All of these three phase has clear phase transition temperature. ...


1

There are several ways to destroy a Bose-Einstein condensate. The most common is temperature, which is why BECs are all low-temperature phenomena. For instance, helium becomes superfluid when a large fraction of the atoms enter the same quantum state, which happens around $\mathrm{2\,K = \frac16\,meV}/k$, so apparently the first excited state in fluid helium ...


1

I think the answer should be "no", as they are phenomena happening in two different sectors. That is, Bose-Einstein condensation involves the center-of-mass degrees of freedom of each atom. On the other hand, radioactive decay pertains to the internal interactions among constituent subatomic particles.


1

In principle, it is very simple and straightforward. The problem is to map out the region where the integer filling state is the ground state. Suppose you have $L$ sites. Take $N=L$ particles, find its ground state energy, which is denoted as $E_g(L)$. Note that here the Hamiltonian does not contain the $\mu $ term. Do it again for $N=L+1$, the ground ...



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