How do constraints affect the Hamiltonian in mean-field approximation?

[CKtalon]

Please correct me if I make any mistakes along the way.

Suppose we have a simple tight-binding Hamiltonian
$H=\sum_i \epsilon _i c_i^\dagger c_i - t\sum_{\langle i j\rangle} c_i^\dagger c_j + h.c.,$
In half-filling systems, we tend to impose a constraint such that each site has only one electron on average, i.e.,
$\langle c^\dagger_i c_i\rangle = 1$

Suppose I were to ensure that this constraint is in the Hamiltonian, I add a Lagrange multiplier (which ends up to be a chemical potential?)
$H=\sum_i \epsilon _i c_i^\dagger c_i - t\sum_{\langle i j\rangle} c_i^\dagger c_j + h.c. - \mu_i (c^\dagger_i c_i-1),$

Now my confusion comes in, how does the constraint work here, since the 1 is a constant, and can be effectively ignored to obtain a Hamiltonian like
$H=\sum_i (\epsilon_i - \mu_i) c_i^\dagger c_i - t\sum_{\langle i j\rangle} c_i^\dagger c_j + h.c.,$
so how can I ensure that my constraint is really doing anything in mean-field (MF) since I could have the constraint be any other number?

I'm working only in the mean-field approximation.
I received a suggestion to square the constraint, and hence end up with a quartic term, that can be solved perturbatively. However, it seems at zeroth (first?) order in mean-field, the entire term disappears

$(c^\dagger_i c_i-1)^2 = c_i^\dagger c_i c_i^\dagger c_i -2 c_i^\dagger c_i +1$

Under MF, with the equation $AB \approx \langle A\rangle B + A\langle B\rangle - \langle A\rangle \langle B \rangle,$

$2\langle c_i^\dagger c_i \rangle c_i^\dagger c_i - \langle c_i^\dagger c_i \rangle \langle c_i^\dagger c_i \rangle -2 c_i^\dagger c_i +1 = 2 c_i^\dagger c_i - 1\cdot 1 - 2c_i^\dagger c_i +1=0$

I hope someone can relieve some of my confusion.
Specifically, I am working with slave fermions and want to impose $\langle f_i^\dagger f_i \rangle=2$ for a spin, $S=1$ system, and with itinerant electrons with $\langle c_i^\dagger c_i \rangle=1$

 

[Mute]

Please correct me if I make any mistakes along the way.

Suppose we have a simple tight-binding Hamiltonian
$H=\sum_i \epsilon _i c_i^\dagger c_i - t\sum_{\langle i j\rangle} c_i^\dagger c_j + h.c.,$
In half-filling systems, we tend to impose a constraint such that each site has only one electron on average, i.e.,
$\langle c^\dagger_i c_i\rangle = 1$

Suppose I were to ensure that this constraint is in the Hamiltonian, I add a Lagrange multiplier (which ends up to be a chemical potential?)
$H=\sum_i \epsilon _i c_i^\dagger c_i - t\sum_{\langle i j\rangle} c_i^\dagger c_j + h.c. + \mu_i (c^\dagger_i c_i-1),$

Now my confusion comes in, how does the constraint work here, since the 1 is a constant, and can be effectively ignored to obtain a Hamiltonian like
$H=\sum_i (\epsilon_i - \mu_i) c_i^\dagger c_i - t\sum_{\langle i j\rangle} c_i^\dagger c_j + h.c.,$
so how can I ensure that my constraint is really doing anything in mean-field (MF) since I could have the constraint be any other number?

But it's not just 1, it's $\mu_i$. You've introduced a new set of variables into your Hamiltonian to enforce your constraint. You can't treat them as constants. Varying the Hamiltonian with respect to a $\mu_i$ should enforce the constraint.

I'm working only in the mean-field approximation.
I received a suggestion to square the constraint, and hence end up with a quartic term, that can be solved perturbatively. However, it seems at zeroth (first?) order in mean-field, the entire term disappears

$(c^\dagger_i c_i-1)^2 = c_i^\dagger c_i c_i^\dagger c_i -2 c_i^\dagger c_i +1$

Under MF, with the equation $AB \approx \langle A\rangle B + A\langle B\rangle - \langle A\rangle \langle B \rangle,$

$2\langle c_i^\dagger c_i \rangle c_i^\dagger c_i - \langle c_i^\dagger c_i \rangle \langle c_i^\dagger c_i \rangle -2 c_i^\dagger c_i +1 = 2 c_i^\dagger c_i - 1\cdot 1 - 2c_i^\dagger c_i +1=0$

I hope someone can relieve some of my confusion.
Specifically, I am working with slave fermions and want to impose $\langle f_i^\dagger f_i \rangle=2$ for a spin, $S=1$ system, and with itinerant electrons with $\langle c_i^\dagger c_i \rangle=1$

I don't see how squaring the constraint would have helped your original problem - you'll still produce a constant term that you would have neglected along your original line of reasoning. But, since I don't think it's appropriate to disregard the $\mu_i$ terms, you should be fine with the linear constraint.

 

[CKtalon]

Thanks for the reply. So the next usual step is to go to Fourier space, so my Hamiltonian and chemical potential becomes something like
$H = \sum_k [-2t\cos ka + (\epsilon_k - \mu_k)]c_k^\dagger c_k + \alpha\mu_{k=0},$
where we had the constraint $c^\dagger_i c_i - \alpha$.

Wouldn't the 2nd $\mu_{k=0}$ term be considered a constant shift of the energy bands then (albeit only at $k=0$, which leads to some discontinuous effects?)

The term in the $\epsilon_k-\mu_k$ doesn't inherit the constraint value, $\alpha$.

Also, how does one vary with respect to $\mu_k$ to find the ground state energy? It seems I could arbitrarily set $\mu_k$ as big as possible (hence enforcing the constraint), but yet doesn't seem really physical to me...

I guess I'm just confused over this constraint business.

I refer to a similar part in Griffiths QM, pg 237,238 (Chapter 5.4.3).
In Eq 5.88 and 5.89, we vary $G$, but we see that $\frac{\partial G}{\partial N_n}$ is independent of $N$ which was the constraint. So how is this constraint enforced in the first place?