Skip to main content
edited body
Source Link
Roger V.
  • 65k
  • 7
  • 69
  • 215

Let me make a few observations, which hopefully will lead to better understanding of this issue:

  • The metallic Hamiltonian (non-interacting Fermi gas) conserves particle number, i.e., $H$ commutes with $N$, which is why adding this term merely changes the energy origin. This is not tehthe case for the BCS Hamiltonian.
  • The main point of adding the chemical potential terms is actually the convenience of measuring the energy from the ground state. This is most obvious in electrons-and-holes representation, where the annihilation operators annihilate the ground state.
  • The ground state for the BCS Hamiltonian is the BCS ground state, which does not have a fixed number of particles.
  • We do not have a priori knowledge of the chemical potential (except in theoretical calculations) - in practice we know the carrier density, and calculate $\mu$ as a fitting parameter (in expression for the particle density it does not look anymore as a Lagrange multiplier, but it is a matter of terminology).

Let me make a few observations, which hopefully will lead to better understanding of this issue:

  • The metallic Hamiltonian (non-interacting Fermi gas) conserves particle number, i.e., $H$ commutes with $N$, which is why adding this term merely changes the energy origin. This is not teh case for the BCS Hamiltonian.
  • The main point of adding the chemical potential terms is actually the convenience of measuring the energy from the ground state. This is most obvious in electrons-and-holes representation, where the annihilation operators annihilate the ground state.
  • The ground state for the BCS Hamiltonian is the BCS ground state, which does not have a fixed number of particles.
  • We do not have a priori knowledge of the chemical potential (except in theoretical calculations) - in practice we know the carrier density, and calculate $\mu$ as a fitting parameter (in expression for the particle density it does not look anymore as a Lagrange multiplier, but it is a matter of terminology).

Let me make a few observations, which hopefully will lead to better understanding of this issue:

  • The metallic Hamiltonian (non-interacting Fermi gas) conserves particle number, i.e., $H$ commutes with $N$, which is why adding this term merely changes the energy origin. This is not the case for the BCS Hamiltonian.
  • The main point of adding the chemical potential terms is actually the convenience of measuring the energy from the ground state. This is most obvious in electrons-and-holes representation, where the annihilation operators annihilate the ground state.
  • The ground state for the BCS Hamiltonian is the BCS ground state, which does not have a fixed number of particles.
  • We do not have a priori knowledge of the chemical potential (except in theoretical calculations) - in practice we know the carrier density, and calculate $\mu$ as a fitting parameter (in expression for the particle density it does not look anymore as a Lagrange multiplier, but it is a matter of terminology).
Source Link
Roger V.
  • 65k
  • 7
  • 69
  • 215

Let me make a few observations, which hopefully will lead to better understanding of this issue:

  • The metallic Hamiltonian (non-interacting Fermi gas) conserves particle number, i.e., $H$ commutes with $N$, which is why adding this term merely changes the energy origin. This is not teh case for the BCS Hamiltonian.
  • The main point of adding the chemical potential terms is actually the convenience of measuring the energy from the ground state. This is most obvious in electrons-and-holes representation, where the annihilation operators annihilate the ground state.
  • The ground state for the BCS Hamiltonian is the BCS ground state, which does not have a fixed number of particles.
  • We do not have a priori knowledge of the chemical potential (except in theoretical calculations) - in practice we know the carrier density, and calculate $\mu$ as a fitting parameter (in expression for the particle density it does not look anymore as a Lagrange multiplier, but it is a matter of terminology).