Proving Bloch's Theorem

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SUMMARY

The discussion centers on proving Bloch's Theorem by demonstrating that a Hamiltonian with periodic potential commutes with the translation operator. The equation T(R)H(r)ψ(r) = H(r+R)ψ(r+R) is established through the periodicity of the Hamiltonian, H(x) = H(x+a). This periodicity allows the translation operator to act on both the Hamiltonian and the wave function, confirming the relationship between the two. The use of ket notation, θ(r) = H(r)ψ(r), further clarifies the action of the translation operator on the system.

PREREQUISITES
  • Understanding of Hamiltonians in quantum mechanics
  • Familiarity with periodic potentials
  • Knowledge of translation operators in quantum mechanics
  • Basic proficiency in ket notation and quantum state representation
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  • Study the derivation of Bloch's Theorem in quantum mechanics
  • Explore the implications of periodic potentials on wave functions
  • Learn about the mathematical properties of translation operators
  • Investigate the role of kets and bra-ket notation in quantum mechanics
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Quantum physicists, students of quantum mechanics, and researchers interested in solid-state physics will benefit from this discussion, particularly those focusing on the mathematical foundations of Bloch's Theorem and periodic potentials.

naele
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One of the more common ways of showing that a Hamiltonian with periodic potential commutes with the translation operator is to write the following (like Ashcroft and Mermin p. 133)

[tex] T(R)H(r)\psi(r)=H(r+R)\psi(r+R)=H(r)T(R)\psi(r)[/tex]

I suspect this might be a dumb question, but what allows us to write [itex]T(R)H(r)\psi(r)=H(r+R)\psi(r+R)[/itex], that is why is the translation operator acting on both the Hamiltonian and the wave, and not just on the Hamiltonian?
 
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I think its because the potential is periodic then the Hamiltonian is too: H(x)=H(x+a), you can then sub this in directly and the translation operator now just acts on psi
 
naele said:
I suspect this might be a dumb question, but what allows us to write [itex]T(R)H(r)\psi(r)=H(r+R)\psi(r+R)[/itex]
Because that is the definition of how the space translation operator acts on a ket.

It may help to write [itex]\theta(r) = H(r) \psi(r)[/itex]. [itex]\theta(r)[/itex] is a ket. What [itex]T(R) \theta(r)[/itex]...It may help more to consider more traditional function notation for what I believe is being written:
[tex](T(R) H \psi)(r) = (H \psi)(r + R).[/tex]
 

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