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10

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 ...


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

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 ...


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

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 ...


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 ...


1

You call the condensate 'scalar' when the atoms are spin-0. When instead atoms have a non trivial spin you talk about 'spinor condensates'.


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 ...



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