Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constraints in Hamiltonian

  1. Jul 19, 2011 #1
    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]
    Last edited: Jul 20, 2011
  2. jcsd
  3. Jul 19, 2011 #2


    User Avatar
    Homework Helper

    But it's not just 1, it's [itex]\mu_i[/itex]. 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 [itex]\mu_i[/itex] should enforce the constraint.

    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 [itex]\mu_i[/itex] terms, you should be fine with the linear constraint.
  4. Jul 19, 2011 #3
    Thanks for the reply. So the next usual step is to go to fourier space, so my Hamiltonian and chemical potential becomes something like
    [itex]H = \sum_k [-2t\cos ka + (\epsilon_k - \mu_k)]c_k^\dagger c_k + \alpha\mu_{k=0},[/itex]
    where we had the constraint [itex]c^\dagger_i c_i - \alpha[/itex].

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

    The term in the [itex]\epsilon_k-\mu_k[/itex] doesn't inherit the constraint value, [itex]\alpha[/itex].

    Also, how does one vary with respect to [itex]\mu_k[/itex] to find the ground state energy? It seems I could arbitrarily set [itex]\mu_k[/itex] 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 [itex]G[/itex], but we see that [itex]\frac{\partial G}{\partial N_n}[/itex] is independent of [itex]N[/itex] which was the constraint. So how is this constraint enforced in the first place?
    Last edited: Jul 20, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook