Recent content by loandbehold

I'm assuming by "fermionic condensates" you're talking about the experiments with ultracold atoms? If so, I've taken a stab at answering the question below. If not, then I hope you find it interesting anyway... Basically, to see the connection you need to understand something called the...

What do mean by a "solution at r=0"? You have a Schrodinger equation, with a Coulomb potential. You solve it. You get a set of discrete eigenstates and eigenvalues, the lowest of which is the ground state, which is stable. Just in the same way as one would for a harmonic oscillator, or pretty...

Try going to google and typing in "plum pudding model". This brings up a load of hits. I think the first one might have what you are looking for.

Not really. Surely if l=0 then you just have the -1/r term, which diverges to minus infinity at r=0, not at r=1.

I wasn't "forgetting". I neglected the angular parts of the wavefunction specifically because, as I stated in my post, I was solving for the ground state, where the wavefunction is spherically symmetric. So the centrifugal barrier term isn't there. This can be seen by setting l=0 in the...

Huh?!...I thought that the Bohr radius just drops out when you solve the Schrodinger equation with the Coulomb potential. Why do you need "other interactions" in there as well? Just write down the usual time independent Schrodinger equation for the hydrogen atom: -\frac{\hbar^2}{2\mu}...

I really don't understand what you mean. For the electron to decay and lose energy, there must be a lower energy state for it to go to. So, if the system is in it's ground state, then by definition it is in its lowest energy state, and therefore it can't decay. What am I missing here? And...

Probably not, but I don't understand what your objection is either. For a hydrogen atom the way the problem is usually presented is to solve the Schrodinger equation for a Coulomb potential V(r) \propto -1/r. Doing this gives rise to a set of eigenstates and eigenenergies, the lowest of which...

Yes, that's exactly what should happen in that model. The reason why it doesn't is because an electron in an atom isn't really a point particle but is better described as a "spread out" wavefunction, with an energy that has to be one of certain discrete values ("energy levels"). Are you going...

Some of the roots are going to be complex, so the way I would tackle the problem is to rewrite your number in the form: z=(1+\sqrt{2}){\rm e}^{2\pi ni}, where n=0,1,2,... Then taking the fifth root gives: z^{1/5}=(1+\sqrt{2})^{1/5} {\rm e}^{2\pi ni/5}, which you can write in the...

Actually, it's much colder- something like 10^{-7} K is more typical. As for how they did it, take a look at http://www.colorado.edu/physics/2000/bec/index.html . which provides an easy introduction.

How about "Experimental realization of Shor's quantum factoring algorithm using nuclear magnetic resonance," L. M. K. Vandersypen et al, Nature 414, 883 (2001). http://www.nature.com/nature/links/011220/011220-2.html

Well, one of the most common examples of superfluid is liquid Helium-4 cooled to below 2.17K. I don't know exactly what the "biggest quantity" is, but if you've ever seen dewar flasks of liquid nitrogen you can probably guess that it's a fair amount, and certainly macroscopic. No, photography...