Why can fermions occupy only one state and

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Why can bosons occupy more? Surely the reason must be more than the maths? Whats the physical reason?
 
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Davio said:
Why can bosons occupy more? Surely the reason must be more than the maths? Whats the physical reason?

The physical reason is in identical particle properties: their exchanging may not change the system state. So only two possibilities exist: symmetric or antisymmetric wave function transformations. Both cases exist in nature. The antisymmetric states are called fermionic and their statistics is different.

Photons are not charged, there is a superposition principle for EMF, so they can be "together" as one photon of higher strength.
 
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IMHO your question is ill put. We define fermions by their relativistic transformation behavior. This is like saying why don't positive charges repel negative charges. It is part of the definition of the negative charge.

Fact is: With the prevalent spacetime symmetry particles have few transformation behaviors to choose from. (Well ok few sane ones...2 dimensional anyons and parastatistic excluded)
 
Physical reason? Well in some experiment... for exaple you have some target and you bomb that target with electrons. Electrons very quick louse their energy and you can't say is some electron in some other moment the electron you look in some other moment.
If you look two electrons and let's say that state of the system is \psi(\xi_1,\xi_2), where \xi_1 and \xi_2 denoting the three coordinates and spin projection. And define some operator of permutation:
\hat{P}_{1,2}\psi(\xi_1,\xi_2)=\psi(\xi_2,\xi_1)

Eigen problem of this operator is:
\hat{P}_{1,2}\psi(\xi_1,\xi_2)=\lambda\psi(\xi_1,\xi_2)

\Rightarrow \lambda=\pm1

\lambda=-1 - fermions
\lambda=1 - bosons
 
To follow up on Petar's response, because you can't tell one particle from another, because of the probabilistic interpretation of QM, you must write the wavefunction as a superposition of particle one in state one particle with two in state two and particle one in state two and particle two in state one.

In the case of N electrons it is more complicated. Because the permutation operators commute with the Hamiltonian, but not each other only the "totally symmetric" and "total anti-symmetric" combinations of permutations are allowed. The wavefunction that is total anti-symmetric is what we find electrons as and this is the case of Fermions.

There is an excellent section in Richard Liboff's textbook on this. In the third edition it is section 12.3 pages 613-619.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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