Solution to Problem Set 2

The Hubbard model in the atomic limit is given by,

H = −μ ∑_{σ} f_{σ}^{†}f_{σ} + Uf_{↑}^{†}f_{↑}f_{↓}^{†}f_{↓}.  (1)

The system has four possible states, namely |0 ⟩ (no particle), | ↑ ⟩ (up electron), | ↓ ⟩ (down electron), and | ↑↓ ⟩ (doubly occopied). The grand canonical partition function for the system is given by,

Z = Tre^{−βH}
= 1 + 2e^{βμ} + e^{β(2μ−U)}.  (2)

1. To calculate the retarded Green’s function and the spectral function we will first calculate the temperature Green’s function and then take appropriate analytic continuation.

The temperature Green’s function is defined as,

g_{σ}(τ) = −⟨ T_{τ}f_{σ}(τ)f_{σ}^{†}(0) ⟩
= −\frac{1}{Z} Tr (e^{−βH}T_{τ}f_{σ}(τ)f_{σ}^{†}(0)) .

The Fourier transform in terms of fermionic Matsubara frequency ω_{n} = (2n + 1)π/β, is

g_{σ}(iω_{n}) = ∫_{0}^{β} dτg_{σ}(τ)e^{iω_{n}τ}
= −\frac{1}{Z} ∫_{0}^{β} dτe^{iω_{n}τ} ∑_{n} ⟨ n |e^{−βH}f_{σ}(τ)f_{σ}^{†}(0)| n ⟩.  (3)

Since,

∑_{n} ⟨ n |e^{−βH}f_{σ}(τ)f_{σ}^{†}(0)| n ⟩ = ∑_{n} ⟨ n |e^{−βH}e^{τH}f_{σ}e^{−τH}f_{σ}^{†}| n ⟩
= ⟨ 0 |e^{−βH}e^{τH}f_{σ}e^{−τH}f_{σ}^{†}| 0 ⟩ + ⟨ σ̄ |e^{−βH}e^{τH}f_{σ}e^{−τH}f_{σ}^{†}| σ̄ ⟩
= e^{τμ} + e^{βμ+τ(μ−U)},  (4)

we get from equations (3) and (4),

g_{σ}(iω_{n}) = \frac{1}{Z} \{\frac{e^{βμ} + 1}{iω_{n} + μ} + \frac{e^{β(2μ−U)} + e^{βμ}}{iω_{n} + μ − U} \} .  (5)

Then, after the analytic continuation iω_{n} → ω + iη, we have,

G_{σ}^{R}(ω) = \frac{1}{Z} \{\frac{e^{βμ} + 1}{ω + μ + iη} + \frac{e^{β(2μ−U)} + e^{βμ}}{ω + μ − U + iη} \} .  (6)

The spectral function is given by,

A_{σ}(ω) = −\frac{1}{π} ImG_{σ}^{R}(ω)
= (\frac{e^{βμ} + 1}{Z}) δ(ω + μ) + (\frac{e^{β(2μ−U)}}{Z} ) δ(ω + μ − U).  (7)

In the non-interacting case the spectral function has a single pole with weight one, i.e, A_{σ}(ω) = δ(ω + μ). The effect of interaction is to produce two poles with finite energy gap U. The total spectral weight, which is always one, is shared between the two poles.

2. The Hubbard operators are defined as

X_{↑0} ≡ f_{↑}^{†}(1 − n_{↓}) = | ↑ ⟩⟨ 0|
X_{d↓} ≡ f_{↑}^{†}n_{↓} = | ↑↓ ⟩⟨ ↓ |.  (8)

Their conjugate operators are repectively,

X_{0↑} ≡ f_{↑}(1 − n_{↓}) = |0 ⟩⟨ ↑ |
X_{↓d} ≡ f_{↑}n_{↓} = | ↓ ⟩⟨ ↑↓ |.  (9)

Now, it is easy to see that

f_{↑}^{†} = X_{↑0} + X_{d↓} = | ↑ ⟩⟨ 0| + | ↑↓ ⟩⟨ ↓ |.  (10)

3. Similarly, we have,

f_{↓}^{†} = X_{↓0} + X_{d↑} = | ↓ ⟩⟨ 0| + | ↑↓ ⟩⟨ ↑ |.  (11)

4. The anticommutator

{X_{↑0}, X_{0↑}} = {f_{↑}^{†}(1 − n_{↓}), (1 − n_{↓})f_{↑}}
= (1 − n_{↓})^{2} {f_{↑}^{†}, f_{↑}}
= (1 − n_{↓}),  (12)

since (1 − n_{↓}) is either zero or one. Thus,

⟨ \{X_{↑0}, X_{0↑}\} ⟩ = 1 − ⟨ n_{↓} ⟩ = \frac{e^{βμ} + 1}{Z},  (13)

is the average occupation of a ↓-hole. We also note, from equation (7), that this is also the spectral weight for the pole corresponding to single occupancy.

Similarly, the anticommutator {X_{d↓}, X_{↓d}} = n_{↓}. Thus,

⟨ \{X_{d↓}, X_{↓d}\} ⟩ = ⟨ n_{↓} ⟩ = \frac{e^{β(2μ−U)}}{Z} ,  (14)

is the average occupation of a ↓-electron. Again from equation (7) we note that this is the spectral weight for the pole corresponding to double occupancy.

5. To evaluate the commutator [H, X_{0↑}], it is enough to consider the state | ↑ ⟩, since all other states give zero when acted on by X_{0↑}. We have [H, X_{0↑}]| ↑ ⟩ = μ|0 ⟩ = μX_{0↑}| ↑ ⟩. Thus, [H, X_{0↑}] = μX_{0↑}, and

⟨ {X_{↑0}, [H, X_{0↑}]} ⟩ = μ⟨ {X_{↑0}, X_{0↑}} ⟩ = μ (1 − ⟨ n_{↓} ⟩) .  (15)

Comparing with equation (7) we find that the energy coefficient in the above expression gives the pole for the singly occupied state.

Similarly, we can show that [H, X_{↓d}] = (μ − U)X_{↓d}, and

⟨ {X_{d↓}, [H, X_{↓d}]} ⟩ = (μ − U) ⟨ n_{↓} ⟩.  (16)

Here the energy coefficient gives the pole for the doubly occupied state.

6. Comment

To put parts (1), (4) and (5) into proper perspective we will re-evaluate the retarded Green’s function using the Hubbard operators. From definition,

G_{σ}^{R}(t) = −iθ(t)⟨ {f_{σ}(t), f_{σ}^{†}(0)} ⟩
= −iθ(t)⟨ {X_{0σ}(t) + X_{σ̄d}(t), X_{σ0}(0) + X_{dσ̄}(0)} ⟩
= −iθ(t) [⟨ {X_{0σ}(t), X_{σ0}(0)} ⟩ + ⟨ {X_{σ̄d}(t), X_{dσ̄}(0)} ⟩] .

Here σ̄ is the spin opposite to σ. In the last line above, we have used the fact that crossterms are zero. Now, from our discussion in part (5), we know that [H, X_{0σ}] = μX_{0σ} and that [H, X_{σ̄d}] = (μ − U)X_{σ̄d}. Then, X_{0σ}(t) = e^{iμt}X_{0σ}(0), and X_{σ̄d}(t) = e^{i(μ−U)t}X_{σ̄d}(0). We can write,

G_{σ}^{R}(t) = −iθ(t) [e^{iμt}⟨ {X_{0σ}, X_{σ0}} ⟩ + e^{i(μ−U)t}⟨ {X_{σ̄d}, X_{dσ̄}} ⟩] .

From our discussion in part (4) we know that ⟨ {X_{σ0}, X_{0σ}} ⟩ = 1 − ⟨ nσ̄ ⟩ and that ⟨ {X_{σ̄d}, X_{dσ̄}} ⟩ = ⟨ nσ̄ ⟩. So, finally we have,

G_{σ}^{R}(t) = −iθ(t) {(1 − ⟨ nσ̄ ⟩)e^{iμt} + ⟨ nσ̄ ⟩e^{i(μ−U)t}} .  (17)

Taking the Fourier transform of the above equation we get,

G_{σ}^{R}(ω) = \frac{1 − ⟨ nσ̄ ⟩}{ω + μ + iη} + \frac{⟨ nσ̄ ⟩}{ω + μ − U + iη} .  (18)

This is nothing but equation (6) in a more compact form.