Silly idea of BEC in any temperature

[luxiaolei]

Hi, all, as the title mentioned, its a really silly idea:

A box of fermions,at any give temperature, theoretically, and occasionally, number of particles might have zero velocities due to collisions in a very short time period.

Then in this time period, (although its short, let's put it on a theoretical point of view), those particles has zero(or very low) temperature, and if treat those particles as a whole, they could satisfy BEC condition, i.e. form composed bosons, and share one state.

Can this really happen?

 

[eljose79]

Hello, thank you for your interesting question. While it may seem like a silly idea at first, it is actually a concept that has been explored by scientists in the field of quantum mechanics. This phenomenon is known as "quantum degeneracy" and it can occur in systems of fermions, such as atoms, at extremely low temperatures.

At very low temperatures, the individual particles in a system can lose their individual identities and start behaving as a collective whole. This is due to the principles of quantum mechanics, which state that particles can exist in multiple states at the same time. In a system of fermions, this can lead to the formation of a "superparticle" or "composite boson" that behaves like a single particle with a single state. This is known as Bose-Einstein condensation (BEC).

While this phenomenon has been observed in laboratories, it is important to note that it requires extremely low temperatures, close to absolute zero, and careful control of the system. It is also a very short-lived state, as any slight increase in temperature or disturbance in the system can cause the particles to "break out" of their collective state and return to their individual identities.

In summary, while it is theoretically possible for a group of fermions to form a BEC state due to temporary collisions at low temperatures, it is a highly complex and delicate phenomenon that requires precise conditions to occur. I hope this helps answer your question.