# Tag Info

## Hot answers tagged bose-einstein-condensate

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

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

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 ...

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

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

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

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):

1

I will answer your second question because it's the one with which I'm more familiar. The question we're answering is: "Why does current in a superconductor move with no resistance?" To understand this we should first understand why normal metals have nonzero resistivity. Imagine an electron in the metal and suppose it is traveling in some direction. If ...

1

It feels like you are going to fast in your way to think. The difference between the chemical potential and the ground state energy is clear : 1.The ground-state energy of your system is here $\epsilon_\textbf{k}=0$ correponding to the ground-state $|\textbf{k=0}\rangle$, which is macroscopically occupied in a BEC (i.e. $N\sim N_0$). 2.The chemical ...

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.

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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 ...

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

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 ...

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