I found Veit's description admirably clear, except for the sentence An arbitrary wave function can be symmetrized or antisymmetrized, as needed for bosons or fermions. I feel like I'm lacking context for the operation of symmetrization/antisymmetrization, but I'm not sure if I can explain what I mean by context. I'm guessing Veit means something like "If you symmetrize it, you'll get a wave-function that's just as descriptive of reality; the universe won't notice the difference." Or "If you symmetrize it, you'll still get a solution to the relevant equation, but if it isn't unitary then it won't correspond to anything we see in the universe." Veit, can you clarify? Jim Propp On Tuesday, June 24, 2014, Veit Elser <ve10@cornell.edu> wrote:
I always cringe a little bit when people refer to particles being "in the same quantum state”. Were I a better citizen, I would edit wikipedia something along these lines:
In a multi-particle quantum system the individual particles do not have a “state" assigned to them, in the same sense as some property like mass or charge. There is always just a single quantum “state”, and this state describes the entire system. A quantum state of two particles, for example, is a single wave function psi(x1,x2), where x1 and x2 are the positions of the particles. There is a small class of wave functions for which one can say there is an assignment of particles to states. This is when the multi-particle wave function factorizes; wave functions in this class, for two particles, would be written as psi1(x1) psi2(x2). However, wave functions with this structure are never the states of definite energy (solutions of the time-independent Schroedinger equation) when the particles interact.
The property of being bosons or fermions does not restrict how particles are assigned to states, but restricts the permutation symmetry of the wave function for the system. The wave function for two bosons has the property psi(x1,x2) = psi(x2,x1), and for fermions, psi(x1,x2) = -psi(x2,x1). An arbitrary wave function can be symmetrized or antisymmetrized, as needed for bosons or fermions. Antisymmetrizing the product wave function psi(x1)psi(x2) yields no wave function at all (the null vector of linear algebra) — it is for this reason that it is sometimes said fermions cannot be in the same state.
-Veit
On Jun 23, 2014, at 10:53 PM, Henry Baker <hbaker1@pipeline.com <javascript:;>> wrote:
"An important characteristic of bosons is that their statistics do not restrict the number that can occupy the same quantum state"
https://en.wikipedia.org/wiki/Boson
Ok, suppose that I want to _weigh_ one boson. I guess that most bosons are pretty light, so I'll have to weigh a bunch of identical bosons (in "the same" quantum state) & divide by their number.
Q: Does this work? Is there any additional energy required to hold them together that would show up as additional mass? Or does an individual boson weigh more singly than when together in "the same" quantum state?
How big is the biggest boson, anyway?
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