How do we justify the chemical potential term in a Hamiltonian of interacting fermions?

Question

Consider a noninteracting fermi gas of electrons. If we know the chemical potential it makes sense that the Hamiltonian is

$$\sum_{|k| > k_f} E_kc_k^{\dagger}c_k +\sum_{|k| < k_f}E_k c_kc_k^{\dagger},$$

where $E_k= |\epsilon_k - \mu|$, $\epsilon_k$ is the energy of the state with momentum $k$ and $\mu$ is the chemical potential. We're just counting holes below the fermi surface and electrons above.

This is obviously different from $\sum_k \epsilon_k c_k^{\dagger}c_k$. All we have done above is impose our knowledge of the chemical potential and use our knowledge of the corresponding ground state. We have just shifted our total energy by the energy of the state when all single-electron states below the chemical potential are filled and those above are empty. Nothing exciting going on here.

For single-particle fermions this seems like a sensible thing to do.

What I've been confused about is using this technique with interacting fermions where we don't have simple, single particle states. Consider the effective superconducting Hamiltonian,

$$H=\sum_k \epsilon_k c_k^{\dagger}c_k - \sum_k \Delta^{*} c_{-k \downarrow}c_{k \uparrow}+\Delta c_{k \uparrow}^{\dagger}c_{-k \downarrow}^{\dagger}$$

The standard technique (e.g see Annett, Superconductivity, Superfluids and Condensates and many others) is to subtract a term $\mu \hat{N} = \mu \sum_k c_k^{\dagger}c_k $, then solve for eigenstates. I don't follow the reasoning for this. My instinct would be to diagonalise before subtracting this term to get fermionic Bolgoliubov particles. Then treat these noninteracting fermionic particles as the single fermions discussed initially. I see that this doesn't lead to anything interesting around the fermi surface and as this is superconductivity, must be wrong.

I should add that I'm happy with the use of Lagrange multipliers when trying to find the ground state of a many body system with the constraint of constant particle number. Simple Lagrange multiplier knowledge can be used to show that the Lagrange multiplier used in this case is the chemical potential. But this seems (to me at least) unrelated to subtracting $\mu \hat{N}$ from a many body hamiltonian and solving for excited states.

Answer 1

Having thought a bit more about this I'm going to try and offer an answer.

I said in the question,

"My instinct would be to diagonalise before subtracting this term to get fermionic Bolgoliubov particles."

Nothing wrong with diagonalising and subsequently changing basis, but what then? We want to solve for many body states that have a specific expectation value of $\hat{N}$, the number of electrons in our system. We don't know how to fill up these knew fermionic states. After all, each state is a linear sum of a hole and an electron. If we have $N$ electrons we can't just fill up $N$ of these new states. So my initial instinct is just wrong.

Now, as I said in the question, if we wanted to find the ground state of the orginal Hamiltonian with a constraint on the electron number, then we must simply find the minimum of $<\phi|H_0 - \mu \hat{N}|\phi>$ along with the equation $<\phi|\hat{N}|\phi>=N$, where $H_0$ is the original Hamiltonian before we start to consider how many particles we have. This is just Euler Lagrange stuff and the knew term constrains the expected particle number to be what we want, $N$.

Now, $H_0 - \mu \hat{N}$ is Hermitian. If we just constrain particle number, find its eigenstates, pick the ones with wanted expected particle number and then pick the lowest energy state from these we have done exactly what we wanted.

However I think this only works because its the ground state. I don't see any reason why this should work for higher energy states.

Now, we can diagonalise $H_0 - \mu \hat{N}$ to look like

$\sum_k E_k b^{\dagger}_k b_k$

where $b^{\dagger}_k$ creates a fermionic Bolgoliubov particle.

The ground state, $|BCS>$, thus has none of these particles. So we then just say that the other states must be made up of these particles on top of this state(our vaccuum).

However, it is not clear just from this whether $b^{\dagger}_k |BCS>$ will be an eignestate of the original Hamiltonian, $H_0$.

Answer 2

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).