What Happens to Matter Near Absolute Zero and the Role of Bosons?

[physicsnoob12]

i am not exactly sure what happens when the temperature almost reaches absolute zero how things react to this drop and what are boson ( sorry for spelling errors)

 

[G01]

Bosons are the names for types of particles that follow Bose-Einstein statistics. Basically, bosons are particles with integer spin.

In contrast, particles with half integer spin, like electrons and protons, are fermions and follow Fermi-Dirac statistics.

Many particle systems of bosons must maintain a wave function that is symmetric with respect to particle exchange. This forces bosons to follow Bose-Einstein statistics.

If you take a system of bosons and put all the bosons into the ground state at once (something you can't do with fermions due to the Pauli exclusion principle), you get what is known as a Bose-Einstein condensate.

 

[dydxforsn]

Essentially at low temperatures bosons will "overlap" each other's wave functions and form a sort of macroscopic wave function (it's one of the few instances in which quantum mechanical effects can be seen on a large scale.) Bosons will do this because the energy of configurations in which they share the ground state (ground state meaning the lowest energy state of anyone of the particles) is the least. Thus if the energy of the particles is the least it can obtain low temperatures, which is the main condition for Bose-Einstein Condensation.

Light is a good example of a boson, and what you see when you view light is a macroscopic wavefunction for the individual photons contained inside (I believe, I may have misunderstood Feynmen about all this though).

Oh, and I suppose I should add that boson are just a group of particles that share the property of having a full integer spin. Although this may sound like an odd property to be talking about at first when it comes to some sort of particle classification, spin in quantum is actually somewhat of a strange concept. For now, just know that particles have this thing called "spin" (don't take the word too literally though..) that can distinguish them from one another. Also note that it is a fundamental theorem from quantum mechanics that the Pauli exclusion principle doesn't apply to bosons. Thus the boson particles can share the ground energy state and form these condensates as previously mentioned.

I'm no expert, but again, I believe all this stuff to be true.I actually have a question for everyone while we're on the subject. I remember in my undergraduate statistical mechanics class when our professor briefly mentioned that superfluid helium wasn't actually a Bose-Einstein Condensate. Can someone tell me more on the differences and exactly what he was talking about? I was always a little confused by what he meant.. It's possible that he mis-spoke as well.

 

[physicsnoob12]

thanks makes sense now.

 

[Cthugha]

Light is a good example of a boson, and what you see when you view light is a macroscopic wavefunction for the individual photons contained inside (I believe, I may have misunderstood Feynmen about all this though).

Photon wavefunctions pose some theoretical problems. However, all photons inside the coherence volume of the light are completely indistinguishable and this volume can become rather macroscopic for lasers.

I actually have a question for everyone while we're on the subject. I remember in my undergraduate statistical mechanics class when our professor briefly mentioned that superfluid helium wasn't actually a Bose-Einstein Condensate. Can someone tell me more on the differences and exactly what he was talking about? I was always a little confused by what he meant.. It's possible that he mis-spoke as well.

The BEC formalism is suitable and intended for a noninteracting gas of bosons. This is a pretty good description for rubidium BECs and similar stuff. Superfluid helium is a liquid and therefore shows larger interactions between the particles it consists of. Therefore you get some state best described in a framework like the two-fluid theory by Landau. Due to these strong interactions not all of the particles condense, but only some fraction. The magnitude of this fraction depends on the actual temperature, but can be pretty small - even less than 10 %. Therefore you do not have an ideal BEC, but a coexistence of condensed and uncondensed helium.

 

[dydxforsn]

Oh, I remember similar talks now, thanks for the reply Cthugha :D

 

[nnnm4]

At about 1K the superfluid fraction of helium is about 100%. You're right in saying that the BEC formalism applies to noninteracting gas, but 4He is a bose liquid and the interactions can be treated as causing a shift in energy levels, while maintaining a 1-to-1 correspondence with non interacting states. The helium atoms are then replaced by quasiparticles that are dressed in the interaction, and these quasiparticles can condense like a gas.

 

[Cthugha]

This is true, but it should be stressed that it is only the superfluid fraction that approaches 100%, not the condensate fraction. Due to strong depletion of the ground state caused by the interactions, the condensate fraction is much less, on the order of 10% and the excitation spectrum becomes well populated.

 

[nnnm4]

Absolutely, the superfluid fraction contains both the condensate and it's excitations.

 

[genneth]

This is true, but it should be stressed that it is only the superfluid fraction that approaches 100%, not the condensate fraction. Due to strong depletion of the ground state caused by the interactions, the condensate fraction is much less, on the order of 10% and the excitation spectrum becomes well populated.

I think this is an important distinction which clearly (as the other posters in this thread has noted) confuses people.

The basic idea of a Bose-Einstein Condensate is that you have non-interacting bosons (in this case meaning they take, statistically, the Bose-Einstein distribution in equilibrium) which nevertheless condense --- this is novel because normally gasses condense due to long-range attractive interactions (usually some dipolar interaction)! In the case of a BEC, no long range attractive interaction is needed, but purely because of quantum statistics.

On the theoretical side, a BEC is necessarily a single-particle theory, i.e. it describes a state of many particles as a simple (symmetrised) product of single particle states.

Real systems have interactions, which cause their real many-body quantum state to *not* be expressible in such a simple fashion, and thus in theory strictly speaking it does not really make sense to call them BECs because no Bose-Einstein distributions are in sight.

Nevertheless, for weakly interaction atomic gasses, one can show (both theoretically and experimentally) that the actual state is very close (large overlap) with the simple BEC state. Thus we can approximate it, and talk about the atoms in them as if they are essentially non-interacting. The self-consistency check is the "condensate fraction" (which is an awful term, because it doesn't measure anything of the sort) --- a non-interacting BEC would have a condensate fraction of 100% at 0 kelvin, and it changes reasonably smoothly with interaction strength.

Nevertheless, one intuitively feels that actually a dilute weakly interacting Bose gas and superfluids proper (strong interactions) are quite similar --- for one thing in theory there is no phase transition as you change the interaction strength! This is possible, but unfortunately is quite mathematical and not really appropriate to go into here. For the specialists, I will mention that the best unifying view I have come across is to consider the spectral distribution of the single-particle density operator --- the canonical reference should be Leggett's book on Quantum Liquids.