# Bose Einstein condensate

1. Jul 17, 2007

### jet10

A few questions about BEC:

A) When these bosons condensate, many particles fill the ground state. I read from textbooks that this would imply that they occupy the same position.
-1- experiments are done with Rb atoms, which consist of protons, neutrons and electrons. Wouldn't the electrons and protons in the atoms repel each other respectively? Can the particles really occupy the same position?
-2- what happens to the electromagnetic and strong interaction? Wouldn't the particles form molecules or different nucleus if the particles are so close to each other?

B) How did they detect the position of the condensate? Can the wavefunction^2 be measured?

2. Jul 17, 2007

### ZapperZ

Staff Emeritus
First of all, condensing into the "same state" need not necessarily imply the "same position".

Secondly, what you are asking for is the situation for a composite boson. This require a bit more in-depth study on not just the bosonic state, but also the fermionic state that make up the composite boson. The easiest way to illustrate this is using the Cooper pairs that make up the supercurrent condensate.

Each Cooper pair naively can be describe via a singlet spin-up and spin-down pair, but with opposite momentum, i.e. something like this:

$$|k_1, up> - |-k_1, down>$$

However, in the Fermi sea, you can have almost an infinite set of these k's. So another pair may have

$$|k_2, up> - |-k_2, down>$$

etc...

Now, the composite pair may all condensed into a single BE state, but each of the electrons making up all of those pairs will have some unique momentum state. So the electrons still obey the FD statistics, even when they are part of a bosonic conglomerate.

Similar description applies to LH3 and the systems you described.

Zz.

Last edited: Jul 17, 2007
3. Jul 18, 2007

### jet10

I am not very familiar with theory of Cooper pairs. Maybe I just have to read it up. It is a little hard for me to imagine it.

About a composite boson: Let's say you have $$|k_1, up> - |-k_1, down>$$ and $$|k_2, up> - |-k_2, down>$$. They are two bosons or Cooper pairs made up of 2 electrons respectively. The second boson is in the same state as the first only if k2 = k1, am I right? In order to do that, I will need another pair of electrons that have k1 and -k1. But the Pauli Principle would forbid the existence of an additional pair of these states in the system. How can I get two Cooper pairs to be in the same state then?

4. Jul 18, 2007

### ZapperZ

Staff Emeritus
Ah. You are confusing "single-particle" states with "two-particle states".

What I wrote for the states of the fermions are single-particle states. So each fermion must occupy a unique single-particle states. However, once you form a pair of anything (not just bosons) and they are now coupled together, you can no longer treat each particle individually. You now have to do a 2-particle state. This has to be solved differently and depends on what situation you are dealing with.

In the superconductivity case, you can naively construct a wavefunction that is a linear combination of all the pairs, i.e. you sum up all the k's. This is the "single state", because you have one coherent wavefunction.

Zz.

5. Jul 18, 2007

### Gokul43201

Staff Emeritus
A condensate of alkali metal atoms (like Rb) is a condensate of "composite bosons". Each atom is made up of an even number of fermions, making it "look" like a boson to another atom that is far away from it. If the atoms get close enough to each other that they can "distinguish" the individual fermions in the other atom, then we are no longer in the limit of a weakly-interacting Bose gas.

In most experiments done with alkali atoms, the density of the gas is actually very small (several orders of magnitude less dense than air at STP), and the individual atoms are pretty far away from each other (as expected).

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