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Physics
Quantum Physics
Understanding second quantization
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[QUOTE="vanhees71, post: 6038903, member: 260864"] If the Hilbert space you describe is a single-particle Hilbert space (which is a proper subspace of the Fock space the annihilation and creation operators are defined on), then what you have there is a single-particle operator. The potential you describe is then an external potential (e.g., due to an electrostatic potential acting on charged particles). For an interaction potential like the Coulomb potential between two charged particles, the corresponding operator consists of two annihilation and two creation operators. It's thus a two-particle operator, which comes about because the Coulomb potential describes interactions between all possible pairs of particles in a many-body system. This "ladder" of operators goes on in principle, i.e., you can also have three-body, four-body, etc. operators. An example is nuclear physics, where you describe the strong interaction by an effective low-energy description of QCD, e.g., using chiral perturbation theory. This leads to many-particle forces in the many-nucleon system, i.e., to strong correlations between two and more particles, necessary to describe, e.g., nuclei as many-body bound states of nulceons (protons and neutrons). [/QUOTE]
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Understanding second quantization
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