# Tag Info

## Hot answers tagged exchange-interaction

36

One good piece of evidence that all particles of a given type are identical is the exchange interaction. The exchange symmetry (that one can exchange any two electrons and leave the Hamiltonian unchanged) results in the Pauli exclusion principle for fermions. It also is responsible for all sorts of particle statistics effects (particles following the ...

8

Exchange interaction is an addition to other interactions between identical particles caused by permutation symmetry. This addition is a result of specific form of multi-particle wave function. It gives no contribution to Hamiltonian unlike "usual" interactions but appears as an additional term in equations for single-particle wave functions (e.g. ...

6

How about $\Psi(x_1,x_2) \equiv (x_1 -x_2)e^{-(x_1-x_2)^2}$?

6

BebopButUnsteady answer is obviously correct. Actually most wave functions cannot be written as single Slater determinants and certainly no energy eigenstates of N-fermion systems (atom, molecules or solids) interacting by two-body potentials (like the Coulomb potential). The Slater determinant is just a nice simple approximation to ground state ...

4

My main question is why don't they affect other photons (unless colliding), shouldn't there be an attraction or repulsion by the exchange force? Or is it because I would need a QFT treatment in that case? Photons are quantum mechanical entities, so yes, it is a QFT case. Any interaction between two photons goes through exchange diagrams. Two photon ...

3

Firstly the case why fermions 'repel' each other is simple, it is due to Pauli exclusion principle which states that no two fermions can have the same set of quantum numbers. Therefore if you try to confine them to a single state, then they exert a Fermi pressure and try to 'repel' each other. Bosons don't obey the Pauli exclusion principle by the vitue of ...

2

Interactions have to conserve angular momentum and spin is angular momentum, so trivially "Yes". In your hypothetical with two uncharged (I assuming you mean the electric charge here) particles there will be no electromagnetic repulsion because there will be no electromagnetic interaction. But Hartree-Fock is used for interacting systems...

2

Assuming that: by "a perfect DFT solution" you mean the set of Kohn-Sham orbitals for the exact (presently unknown) denistiy functional that existance of which Hohenberg and Kohn have proven; you are asking whether the individual energies of these "perfect" Kohn-Sham orbitals can be used as estimates of exact ionization and/or excitation energies, the ...

1

Hole is not a particle. Microscopically, hole is a quasiparticle perturbation of many-electron system. And what is called electron in semiconductors is also a perturbation in the same many-electron system. Exchange interaction within this many-electron system results in a complicated interaction between quasiparticles which live there. Which is called ...

1

There's a repulsive van der Waals force between the atoms. This shows up in the second quantized formalism as H = \int d^3x \left[ \frac{\hbar^2}{2m} \nabla \Psi^\dagger \cdot \nabla \Psi - \mu\Psi^\dagger \Psi \right] + \int d^3x\, d^3y\, \frac{1}{2}V\left( \vec{y} - \vec{x} \right) \Psi^\dagger\left( \vec{x} \right) \Psi^\dagger\left( \vec{y} \right) ...

1

It's a good question. The answer is that the bound on the density is given by the requirement that the interactions between the bosons have to remain weak for the Bose-Einstein condensate to exist. In practice, the helium-4 atoms have to be further away from each other than their radius. Why it is so? Well, if you're talking about the bosons occupying the ...

1

"The nonrelativistic theory has no necessary relation between spin and statistics, as you can understand from the example of a nonrelativistic 2d spinless fermion (imagine 2d confined electrons with a huge magnetic field in the z-direction which forces the spin to always be along the z-axis). The electrons behave as spinless particles with respect to the 2d ...

1

The short answer is that Feynman diagrams definitely do not represent the trajectories of particles in spacetime; they are simply a way of writing down formulae that would otherwise be pretty hard to remember. The long answer is below. Quantum Field Theory, which is the mathematical formalism behind particle physics, dictates that when we scatter two ...

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Neutrino can interact only by exchange of electroweak boson. So in each reaction with neutrino $W^\pm$ or $Z$ bosons must be involved. Also, Standard Model neutrino is assumed to be massless, so there is defined handedness: neutrino is left-handed and antineutrino is right-handed. Consequence of it is that left-handed neutrino will interact only with ...

1

Photons do not interact with each other, in QED, because there is no such interaction term in the Lagrangian or Hamiltonian. The only interaction term is the electromagnetic field A (4 vector) with charged particles. There are also deep reasons for that. An interaction of two photons would be depicted in QFT as nonlinear in A. The square term would require ...

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