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I am trying to understand why chemical potential $= U/2$ is considered to be at half-filling in the case of the Hubbard Model Hamiltonian. So when I substitute this in its Hamiltonian, this is the equation I am getting $$ H = K.E-\frac{U}{2}\Sigma [n_{j,up}n_{j,down} + (n_{j,up} - n_{j,down})^{2}]. $$

I am trying to understand how this works graphically and mathematically to see how this incorporates the half-filling condition. I plotted only the second term and setting $n_{j,up}$ to a constant either zero or one. Also, just want to confirm that $n_{j,up}$ can only take 0, 1 - and no intermediate values.

So when I plotted this, I am getting a parabola, at $n_{j,up} = 0$, its the normal parabola open upward with vertex at the origin and for $n_{j,up} = 1$, its shifted by some amount. When I consider half-filling for one-site hubbard model, it can have only four eigenbasis - $ |0\rangle, |up\rangle, |down\rangle, |up, down\rangle$, so the possible states at half-filling are $|up\rangle, |down\rangle$, so since only one of them only possible in this type of filling, then I am getting $\frac{-U}{2}$ added to the energy. Is it because we are adding this energy with the half-factor in it, we are calling this as the half-filling type. I do not understand completely how to interpret the half-filling case from this plot or the expression itself intuitively. I want to understand this in general, as through the form of the Hamiltonian itself. Any assistance would be appreciated. Thanks.

Qmechanic
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CuriousMind
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2 Answers2

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It is cleaner to formalise the problem using second quantisation. For the fermion Hubbard model with Hamiltonian: $$ H = -t\sum c_{i\sigma}^\dagger c_{j\sigma}+U\sum n_{i\uparrow}n_{i\downarrow}+\mu\sum n_{i\sigma} \\ n_{i\sigma} = c_{i\sigma}^\dagger c_{i\sigma} $$ you have a particle-hole symmetry obtained by the following substitution: $$ c_{i\sigma}\to (-1)^ic_{i\sigma}^\dagger $$ This gives the following change in number operators: $$ n_{i\sigma}\to1-n_{i\sigma} $$ So the Hamiltonian is transformed as (ignoring the additive constant): $$ H = -t\sum c_{i\sigma}^\dagger c_{j\sigma}+U\sum n_{i\uparrow}n_{i\downarrow}+(U-\mu)\sum n_{i\sigma} $$ At half filling, the particle hole symmetry leaves the system unchanged. Since the new Hamiltonian is equal to the original one, this imposes: $$ \mu = U-\mu $$ i.e. $\mu = U/2$.

Hope this helps.

LPZ
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Just to spell out the last steps of @LPZ perfect answer: when $\mu=U/2$ the Hamiltonian (and so its ground state if it's non-degenerate) is invariant under the unitary particle-hole symmetry transformation.

As such the expectation value $ \langle n_{i\sigma} \rangle $ must be equal to the expectation value after the transformation. That is

$$ \langle n_{i\sigma} \rangle = 1-\langle n_{i\sigma}\rangle $$

which means $\langle n_{i\sigma} \rangle =1/2$, that is half filling.

lcv
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