Lasted edited by Andrew Munsey, updated on June 15, 2016 at 1:38 am.
Quantum Mechanics has several strange concepts, that defy our comprehension, because they are concepts that attack our sense of logic.
But because people got use of considering QM as a strange theory, sometimes they see the “strange” even in some concepts of QM that do not need to be interpreted as strange.
It’s the case of the indistinguishibility .
Let’s see why it is not strange.
Consider a body where there are N protons and M electrons, so that M-N=X.
Consider that its electric charge is Q.
The electric charge of each electron is q , and we know that:
Q = X.e ,
and we realize that each electron keeps its individuality, that is, there is no indistinguishibility in this case.
Then why, in the case of the electric charge, the electron keeps its individuality, but in Quantum Mechanics the indistinguishibility occurs?
In order to understand why there is need the concept of indistinguishibility in Quantum Mechanics, we need to understand physically its meaning.
Then let’s analyse, from the viewpoint of the physical meaning, the three known distributions:
Boltzmann’s classical statistics
Consider a gas with N particles (harmonic oscillators) with temperature T, within a vessel.
According to Boltzmann’s classical statistics, the probable number of particles in the status of energy E is given by nBoltz(E):
nBoltz(E) = 1/ea eEkT .......... (1)
In the development of eq. 1 Boltzmann has considered the individuality of each particle.
Let’s see what is the physical meaning of Boltzmann’s assumption.
Consider three particles A, B, C, into a gas of N particles, where A has energy E, and B and C have energies EB and EC respectively.
Suppose that in a certain time the particles A, B, C, have an interaction, and after that their energies become EB , E, and EC respectively .
Although their energy have changed, however the eq. 1 that describe statistically the number of particles with energy E continues to be satisfied. The particles A, B, C keep their individuality.
Consider a system with N bosons with temperature T, within a vessel.
According to Bose’s statistics, the probable number of particles in the quantum status of energy E is given by nBose(E):
nBose(E) = 1/[ea eEkT - 1 ] .......... (2)
In the development of eq. 2 Bose has considered that the particles do not keep their individuality.
Let’s see what is the meaning of Bose’s assumption.
Consider three bosons A, B, C, into a gas of bosons, where A has energy E, and B and C have energies EB and EC respectively.
Suppose that in a time t1 A and B are close. As they are bosons, A and B interact, so that the energy of the three bosons now is E, E, EC .
In a next time t2 , B leaves its partnership with A, and approaches near to C, in order that B and C interact, and now the energies of the three bosons are E, EC , EC .
In a next time t3 , C leaves its parnership with B, and approaches to A, so that the energies of the three bosons is now E, EC , E .
Statistically, the Boltzmann’s eq. 1 cannot be used to describe the number of bosons with energy E. There is need to use the Bose’s eq. 2.
Besides, look their energies along the time:
t1: B has E , and C has EC
t2: B has EC , and C has EC
t3: B has EC , and C has E
Compare the energies of B and C at the times t1 and t3:
t1: B has engery E , and C has energy EC
t3: B has energy EC , and C has energy E
There is not way to know, from the statistical viewpoint, who is B or C. The system behaves as B and C exchange their individuallity, they become indistinguishable.
Consider a system with N fermions with temperature T, within a vessel.
According to Fermi’s statistics, the number of particles in the quantum status of energy E is given by nFermi(E):
nFermi(E) = 1/[ea eEkT + 1 ] ..........(3)
In the development of eq. 3 it is considered that the particles do not keep their individuality.
Let’s see what is the meaning of Fermi’s assumption.
From Pauli’s Principle, two fermions that are close one each other cannot be in the same quantum status of energy E.
Consider two fermions A and B into a gas of fermions, both with energy E.
Suppose that B approaches to A. As they cannot be in the same quantum energy status, then B changes its energy, getting EB.
The same happens with other fermions with energy E when they aproach one each other, and so Boltzmann’s classical eq. 1 cannot be applied, because when two fermions with energy E are close the quantum energy status of the system formed by A and B is changed, so that it influences the quantity of particles with energy E into the total system formed by all the fermions within the vessel. There is need to use the eq. 3.
But why are A and B are indistinguishible?
Consider A and B both with energy E, far away one each other.
Their helical trajectory is the same, since they are in the same quantum energy status.
But when they aproach one each other, and they interact, one of them has to change its helical trajectory.
There is no way to know which one from A or B keeps its helical trajectory. When they interact, A can take the helical trajectory that B had earlier the interaction, while B changes the features of its helical trajectory. Also, B can take the helical trajectory that A, while A changes the features of its helical trajectory. There is no way to distinguish the two fermions.
W. Guglinski, Quantum Ring Theory-foundations for cold fusion, Bäuu Press, 2006