- #1
CKtalon
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Please correct me if I make any mistakes along the way.
Suppose we have a simple tight-binding Hamiltonian
[itex]H=\sum_i \epsilon _i c_i^\dagger c_i - t\sum_{\langle i j\rangle} c_i^\dagger c_j + h.c.,[/itex]
In half-filling systems, we tend to impose a constraint such that each site has only one electron on average, i.e.,
[itex]\langle c^\dagger_i c_i\rangle = 1[/itex]
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?)
[itex]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),[/itex]
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
[itex]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.,[/itex]
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
[itex](c^\dagger_i c_i-1)^2 = c_i^\dagger c_i c_i^\dagger c_i -2 c_i^\dagger c_i +1[/itex]
Under MF, with the equation [itex]AB \approx \langle A\rangle B + A\langle B\rangle - \langle A\rangle \langle B \rangle,[/itex]
[itex] 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[/itex]
I hope someone can relieve some of my confusion.
Specifically, I am working with slave fermions and want to impose [itex]\langle f_i^\dagger f_i \rangle=2[/itex] for a spin, [itex]S=1[/itex] system, and with itinerant electrons with [itex]\langle c_i^\dagger c_i \rangle=1[/itex]
Suppose we have a simple tight-binding Hamiltonian
[itex]H=\sum_i \epsilon _i c_i^\dagger c_i - t\sum_{\langle i j\rangle} c_i^\dagger c_j + h.c.,[/itex]
In half-filling systems, we tend to impose a constraint such that each site has only one electron on average, i.e.,
[itex]\langle c^\dagger_i c_i\rangle = 1[/itex]
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?)
[itex]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),[/itex]
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
[itex]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.,[/itex]
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
[itex](c^\dagger_i c_i-1)^2 = c_i^\dagger c_i c_i^\dagger c_i -2 c_i^\dagger c_i +1[/itex]
Under MF, with the equation [itex]AB \approx \langle A\rangle B + A\langle B\rangle - \langle A\rangle \langle B \rangle,[/itex]
[itex] 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[/itex]
I hope someone can relieve some of my confusion.
Specifically, I am working with slave fermions and want to impose [itex]\langle f_i^\dagger f_i \rangle=2[/itex] for a spin, [itex]S=1[/itex] system, and with itinerant electrons with [itex]\langle c_i^\dagger c_i \rangle=1[/itex]
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