I Electron Indistinguishability and Repulsion

[Dario56]

The PEP is an axiom of QM. More generally, it's the anti-symmetry of the total wavefunction. It doesn't follow from indistinguishability. Likewise, for bosons the symmetry of the wavefunction is an axiom.

These axioms of QM can be proved for fermions and bosons using QFT - but that's a different ball game.

When I was learning about PEP, it was actually described in a manner I said here. As far as I understand it, since fermions (and bosons) are indistinguishable, probability density function and expectation values of observables should be independent of any exchange between identical particles. This can only be done if wavefunction stays the same under exchange (bosons) or flips the sign (fermions).

When you say PEP is axiomatic, do you mean it is a first principle arrived upon from experiment (like Schrödinger equation)?

 

[Dario56]

One should also stress that indistinguishability has nothing to do with "interacting" or "non-interacting" particles. It's just the Hilbert space of the many-body system, where the Hamiltonian (including the interactions) is defined.

Yes, this is what I claimed in the main post and also in comments. But, I was somewhat confused with how can Hartree product even satisfy the Schrödinger equation. We've gotten that out of the way here.

 

[PeroK]

When you say PEP is axiomatic, do you mean it is a first principle arrived upon from experiment (like Schrödinger equation)?

An axiom is something you assume to be true as a fundamental postulate of the theory. You cannot prove the PEP from the SDE. Symmetric, anti-symmetric and wavefunctions that are neither can all be solutions to the SDE. You need another axiom (PEP) to force antisymmetry on fermionic wavefunctions.

 

[TeethWhitener]

That holds, but it is strange that such an inaccurate model (which has no ground in reality) can even satisfy Schrödinger equation since Schrödinger equation is based on experiment and not on fictitious systems.

The time evolution of any quantum mechanical system must satisfy the Schrodinger equation. There’s nothing about the Schrodinger equation itself that describes electrons specifically. To specify electrons, you have to make sure the wavefunction is antisymmetric (among other things).

 

[vanhees71]

When I was learning about PEP, it was actually described in a manner I said here. As far as I understand it, since fermions (and bosons) are indistinguishable, probability density function and expectation values of observables should be independent of any exchange between identical particles. This can only be done if wavefunction stays the same under exchange (bosons) or flips the sign (fermions).

When you say PEP is axiomatic, do you mean it is a first principle arrived upon from experiment (like Schrödinger equation)?

That's not true. The indistinguishability of particles only implies that the arbitrary exchanges of two particles form a symmetry group of the system, i.e., the quantum theory of a system of $N$ indistinguishable particles must admit a unitary representation of the symmetric group $S(N)$. That at the end you end up only with the trivial (bosons) and the "alternating" (fermions) representation in 3 or more spatial dimensions is due to additional topological properties concerning the exclusion of the case that two (or more) particles cannot be at the same place. In two spatial dimensions you can have arbitrary realizations of that symmetry:

M. G. G. Laidlaw and C. M. DeWitt, Feynman Functional
Integrals for Systems of Indistinguishable Particles, Phys.
Rev. D 3, 1375 (1970),
https://link.aps.org/abstract/PRD/v3/i6/p1375

 

[Dario56]

An axiom is something you assume to be true as a fundamental postulate of the theory. You cannot prove the PEP from the SDE. Symmetric, anti-symmetric and wavefunctions that are neither can all be solutions to the SDE. You need another axiom (PEP) to force antisymmetry on fermionic wavefunctions.

Well, yes, but shouldn't axiom itself be proven by experiment? Isn't the fact that we differentiate between bosons and fermions also derived from experiment? One has higher probability to be found in the same quantum state while the other has zero probability. Isn't this fact derived experimentally?

 

[Dario56]

That's not true. The indistinguishability of particles only implies that the arbitrary exchanges of two particles form a symmetry group of the system, i.e., the quantum theory of a system of $N$ indistinguishable particles must admit a unitary representation of the symmetric group $S(N)$. That at the end you end up only with the trivial (bosons) and the "alternating" (fermions) representation in 3 or more spatial dimensions is due to additional topological properties concerning the exclusion of the case that two (or more) particles cannot be at the same place. In two spatial dimensions you can have arbitrary realizations of that symmetry:

M. G. G. Laidlaw and C. M. DeWitt, Feynman Functional
Integrals for Systems of Indistinguishable Particles, Phys.
Rev. D 3, 1375 (1970),
https://link.aps.org/abstract/PRD/v3/i6/p1375

So, you say physical observables can change after interchange of identical particles? If that is correct, what have I've been studying for quite some time.

 

[PeroK]

Well, yes, but shouldn't axiom itself be proven by experiment? Isn't the fact that we differentiate between bosons and fermions also derived from experiment? One has higher probability to be found in the same quantum state while the other has zero probability. Isn't this fact derived experimentally?

Why are you changing the subject to experimental confirmation of QM postulates? I can't follow your thinking at all.

 

[Dario56]

Why are you changing the subject to experimental confirmation of QM postulates? I can't follow your thinking at all.

Well, we made a digression from the original question since you've mentioned that PEP is an axiom, so I asked some questions which made us go off topic. This question can be considered closed for that manner.

 

[hutchphd]

This question can be considered closed for that manner.

How does one experimentally interchange identical particles? This would seem problematical.
Anyhow I, too, no longer understand your point. There are many solutions to many differential equations that have no known physical significance. Also there are no absolutely complete physical theories (and perhaps cannot be). So your point is...

 

[vanhees71]

So, you say physical observables can change after interchange of identical particles? If that is correct, what have I've been studying for quite some time.

If you have a symmetry, you cannot recognize a change when just considering a symmetry transformation between states. E.g., if you rotate a sphere around its center nothing changes due to the symmetry of the sphere under rotations around its center.

 

[Dario56]

If you have a symmetry, you cannot recognize a change when just considering a symmetry transformation between states. E.g., if you rotate a sphere around its center nothing changes due to the symmetry of the sphere under rotations around its center.

Sphere analogy has sense, but I am still not sure what does that imply in terms of change in physical observables of expectation values when electrons are exchanged?

As far as I know, there should be no change in expectation values (because of indistinguishability) and only way this can be true is if the wavefunction either stays the same after the exchange (bosons) or if it flips the sign (fermions). The fact that electrons go into fermion category is known from experiment or observation, if I am correct (I may be wrong here).

 

[vanhees71]

If you have only two particles there are indeed only two irreducible representations of the $S_2$, the trivial one and the one changing the sign of the wave function. For more than two particles you have more irreducible representations for $S_N$. All that must be fulfilled is that this $S_N$ is a symmetry group. Then expectation values of observables or, more generally, probabilities for the outcome of measurements don't change when the particles are interchanged, and that's all you need to describe the indistinguishability of the particles.

 

[Dario56]

If you have only two particles there are indeed only two irreducible representations of the $S_2$, the trivial one and the one changing the sign of the wave function. For more than two particles you have more irreducible representations for $S_N$. All that must be fulfilled is that this $S_N$ is a symmetry group. Then expectation values of observables or, more generally, probabilities for the outcome of measurements don't change when the particles are interchanged, and that's all you need to describe the indistinguishability of the particles.

Yes, that is what means to be indistinguishable. So, I got that right.

 

[PeterDonis]

it is strange that such an inaccurate model (which has no ground in reality) can even satisfy Schrödinger equation since Schrödinger equation is based on experiment and not on fictitious systems.

The Schrodinger equation, while it was inspired by experimental results, is still just a mathematical model and has plenty of solutions that don't describe anything we actually observe. That's true of every mathematical model in physics.