Silvestru Posted December 7, 2017 Posted December 7, 2017 I understand how multiple photons may potentially share the same quantum state but I find it more difficult to accept this for bosons that have mass like the Z and W bosons. (Bosons all have integer spin - This might be the explanation for my question but I would need some clarification). I am sure it is just my limitation and I do understand that the PEP does not concern mass but I would appreciate if someone could give me more details on how this can be possible.
Mordred Posted December 7, 2017 Posted December 7, 2017 (edited) You have essentially identified the main aspect of boson's having integer spin. All boson's can share the same quantum state and space. While fermions essentially take up space if in the same quantum state. The fermion family is what comprises what is referred to as matter while the boson's are attributed as force guage bosons. The details of why this works with the Pauli exclusion principle gets rather technical and requires an understanding of the principle quantum numbers and how they affect the probability amplitude functions. Edited December 7, 2017 by Mordred 1
swansont Posted December 7, 2017 Posted December 7, 2017 Fermions have antisymmetric wave functions, while Bosons are symmetric. The implication of that is that the wave function of two indistinguishable Fermions would vanish. Thus, they must be in some different quantum state in order to exist. Bosons don't have this limitation. http://hyperphysics.phy-astr.gsu.edu/hbase/pauli.html 1
Mordred Posted December 7, 2017 Posted December 7, 2017 LOL I was just about to type that in when you posted. Saves me the time lol
MigL Posted December 7, 2017 Posted December 7, 2017 Fermions can also 'pair up' so that they exhibit integer spin and, as such, obey Bose-Einstein statistical rules.
Silvestru Posted December 8, 2017 Author Posted December 8, 2017 (edited) 20 hours ago, MigL said: Fermions can also 'pair up' so that they exhibit integer spin and, as such, obey Bose-Einstein statistical rules. So how are these fermions that pair up distinguishable from bosons? Quote The Bose–Einstein statistics apply only to those particles not limited to single occupancy of the same state—that is, particles that do not obey the Pauli exclusion principle restrictions. Such particles have integer values of spin and are named bosons. Are you referring to Cooper pairs? Edited December 8, 2017 by Silvestru
swansont Posted December 8, 2017 Posted December 8, 2017 52 minutes ago, Silvestru said: So how are these fermions that pair up distinguishable from bosons? They aren't. They are bosons Quote Are you referring to Cooper pairs? Or atoms. Rb-87 is one isotope that can form a Bose-Einstein condensate. 37 protons, 37 electrons, and 50 neutrons.
Silvestru Posted December 8, 2017 Author Posted December 8, 2017 7 minutes ago, swansont said: Or atoms. Rb-87 is one isotope that can form a Bose-Einstein condensate. 37 protons, 37 electrons, and 50 neutrons. So that means certain atoms such as the one mentioned by you, cooled to temperature close to absolute 0 can share the same quantum state with a photon. That's very interesting. I have a lot of reading to do.
swansont Posted December 8, 2017 Posted December 8, 2017 24 minutes ago, Silvestru said: So that means certain atoms such as the one mentioned by you, cooled to temperature close to absolute 0 can share the same quantum state with a photon. That's very interesting. I have a lot of reading to do. Not sure what you mean by sharing the state with a photon, but the atoms in a BEC are in a single quantum state.
Silvestru Posted December 11, 2017 Author Posted December 11, 2017 On 12/8/2017 at 5:38 PM, swansont said: Not sure what you mean by sharing the state with a photon, but the atoms in a BEC are in a single quantum state. Yeah, that's what I meant to say
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