Solution to problem set 3

1. SU(N) Anderson Impurity model

We consider impurity electrons on a single site hybridizing with a bath of conduction electrons. The electrons carry SU(N) spin. We consider the limiting case where the on-site Coulomb repulsion (U) at the impurity site is infinitely large. Thus, at most a single electron can occupy the impurity site (multiple occupancy is not allowed). In this limit, the allowed states of the impurity site are |0⟩ (i.e., no electron), and |m⟩, where m = (1, ⋯, N) (i.e., a single electron with SU(N) spin m). The Hubbard operators are defined as X_{pq} = |p⟩⟨q|, where p, q = (0, 1, ⋯, N). In this restricted subspace creation operator for an impurity electron f_{n}^{†} ≡ |n⟩⟨0| = X_{n0}. Similarly, f_{n} ≡ X_{0n}. The completeness relation in this subspace (of the impurity level) is given by X_{00} + ∑_{n} X_{nn} = 1. Note, that the Hubbard operators satisfy the relation [X_{pq}, X_{rs}]_{±} = δ_{qr}X_{ps} ± δ_{ps}X_{rq}. This is unlike standard commutation or anticommutation of bosons and fermions. The Hamiltonian for the system is given by

H = H_{b} + H_{f} + H_{I},

where H_{b} = ∑_{k,n} ϵ_{k}c_{kn}^{†}c_{kn} describes the bath of conduction electrons, H_{f} = ϵ_{f} ∑_{m} f_{m}^{†}f_{m} = ϵ_{f} ∑_{m} X_{mm} describes the impurity electrons, and H_{I} = ∑_{km}(V_{k}c_{km}^{†}X_{0m} + V_{k}^{∗}X_{m0}c_{km}) is the hybridization between the impurity electrons and the conduction electrons.

We consider the retarded Green’s function ⟨⟨f_{n}f_{n}^{†}⟩⟩(ω) = ⟨⟨X_{0n}X_{n0}⟩⟩(ω). The corresponding spectral function is given by

ρ_{ff^{†}}(ω) = −\frac{Im}{π}G_{ff^{†}}(ω + iδ).

The first moment is given by ⟨{X_{0n}, X_{n0}}⟩ = ⟨X_{00} +X_{nn}⟩. We can write ⟨X_{00}⟩ = 1−⟨n_{f}⟩, where ⟨n_{f}⟩ = ∑_{n}⟨X_{nn}⟩ is the expectation of electron occupancy at the impurity site. Also, ⟨X_{nn}⟩ = ⟨n_{f}⟩/N. Thus,

⟨\{X_{0n}, X_{n0}\}⟩ = 1 − (\frac{N − 1}{N}) ⟨n_{f}⟩.  (1)

Since,

∫_{−∞}^{∞} dωρ_{ff^{†}}(ω) = ⟨{X_{0n}, X_{n0}}⟩,

the first moment gives the total spectral weight of the f-band.

For the calculation of the second moment we get

[H, X_{0n}] = −ϵ_{f}X_{0n} − ∑_{k} V_{k}^{∗}c_{kn}X_{00} − ∑_{km} V_{k}^{∗}c_{km}X_{mn}.

Then,

⟨{[H, X_{0n}], X_{n0}}⟩ = −ϵ_{f}⟨{X_{0n}, X_{n0}}⟩ − ∑_{k,m≠n} V_{k}^{∗}⟨c_{km}X_{m0}⟩.  (2)

Since,

∫_{−∞}^{∞} ωρ_{ff^{†}}(ω)dω = −⟨{[H, X_{0n}], X_{n0}}⟩,  (3)

the centre of the f-band is given by

\frac{∫_{−∞}^{∞} ωρ_{ff^{†}}(ω)dω}{∫_{−∞}^{∞} dωρ_{ff^{†}}(ω)} = ϵ_{f} + \frac{1}{⟨\{X_{0n}, X_{n0}\}⟩} ∑_{k,m≠n} V_{k}^{∗}⟨c_{km}X_{m0}⟩.

Presumably (since we cannot conclusively say anything about the line shape) the spectral function is peaked at this energy. The second term is the shift in the energy of the f-level due to hybridization. For the third moment we get

⟨{[H, X_{0n}], [H, X_{n0}]}⟩ = ⟨{−ϵ_{f}X_{0n} − ∑_{vk} V_{k}^{∗}c_{kn}X_{00} − ∑_{km} V_{k}^{∗}c_{km}X_{mn} , ϵ_{f}X_{n0}

+ ∑_{k} V_{k}c_{kn}^{†}X_{00} + ∑_{km} V_{k}c_{km}^{†}X_{nm}}⟩

= −ϵ_{f}^{2}⟨{X_{0n}, X_{n0}}⟩ − ϵ_{f} ∑_{k,m≠n} V_{k}^{∗}⟨c_{km}X_{m0}⟩ + ϵ_{f} ∑_{k,m≠n} V_{k}⟨c_{km}^{†}X_{0m}⟩

− ∑_{k} |V_{k}|^{2} + ∑_{k_{1}k_{2},m_{1}m_{2}} V_{k_{1}}^{∗}V_{k_{2}}⟨c_{k_{2}m_{2}}^{†}c_{k_{1}m_{1}}X_{m_{1}m_{2}}⟩

− ∑_{k_{1}k_{2},m} V_{k_{1}}^{∗}V_{k_{2}}⟨c_{k_{2}m}^{†}c_{k_{1}m}X_{nn}⟩.  (4)

We can show that

⟨\{[H, X_{0n}], X_{n0}\}⟩ = \frac{1}{Z} ∑_{nm} (e^{−βE_{n}} + e^{−βE_{m}}) (E_{n} − E_{m}) ⟨n|X_{0n}|m⟩⟨m|X_{n0}|n⟩
= −⟨{X_{0n}, [H, X_{n0}]}⟩.  (5)

This is due to time translation invariance of the correlation function in equilibrium. Using this result we get,

− ∑_{k,m≠n} V_{k}^{∗}⟨c_{km}X_{m0}⟩ = ∑_{k,m≠n} V_{k}⟨c_{km}^{†}X_{0m}⟩.  (6)

The SU(N) spin at the impurity site can be written as S = ∑_{m_{1}m_{2}} Γ_{m_{1}m_{2}}^{M}X_{m_{1}m_{2}}, where Γ^{M} are the generators of SU(N) group, with M = (1, ⋯, (N^{2} − 1)). The spin of the conduction electrons in the momentum basis is similarly given by s_{k_{2}k_{1}} = ∑_{m_{1}m_{2}} c_{k_{2}m_{1}}^{†}Γ_{m_{1}m_{2}}^{M}c_{k_{1}m_{2}}. The generators of the group satisfy the completeness relation ∑_{M} Γ_{m_{1}m_{2}}^{M}Γ_{m_{3}m_{4}}^{M} = δ_{m_{1}m_{4}}δ_{m_{2}m_{3}}− \frac{1}{N} δ_{m_{1}m_{2}}δ_{m_{3}m_{4}}. Using this we can show that

∑_{k_{1}k_{2}} V_{k_{1}}^{∗}V_{k_{2}}⟨s_{k_{2}k_{1}}·S⟩ = ∑_{k_{1}k_{2},m_{1}m_{2}} V_{k_{1}}^{∗}V_{k_{2}}⟨c_{k_{2}m_{1}}^{†}c_{k_{1}m_{2}}X_{m_{2}m_{1}}⟩−\frac{1}{N} ∑_{k_{1}k_{2},mm′} V_{k_{1}}^{∗}V_{k_{2}}⟨c_{k_{2}m}^{†}c_{k_{1}m}X_{m′m′}⟩.

(7)

Thus, we can write

⟨{[H, X_{0n}], [H, X_{n0}]}⟩ = −ϵ_{f}^{2}⟨{X_{0n}, X_{n0}}⟩ − 2ϵ_{f} ∑_{k,m≠n} V_{k}^{∗}⟨c_{km}X_{m0}⟩ − ∑_{k} |V_{k}|^{2}
+ ∑_{k_{1}k_{2}} V_{k_{1}}^{∗}V_{k_{2}}⟨s_{k_{2}k_{1}} · S⟩.  (8)

We identify the last term in the above equation as a spin-spin correlation function. The sum rule for the third moment is given by

∫_{−∞}^{∞} ω^{2}ρ_{ff^{†}}(ω)dω = ⟨{[H, [H, X_{0n}]], X_{n0}}⟩ = −⟨{[H, X_{0n}], [H, X_{n0}]}⟩.  (9)

The width of the impurity band (standard deviation of a distribution, in general) is given by

\frac{∫_{−∞}^{∞} ω^{2}ρ_{ff^{†}}(ω)dω}{∫_{−∞}^{∞} dωρ_{ff^{†}}(ω)} − (\frac{∫_{−∞}^{∞} ωρ_{ff^{†}}(ω)dω}{∫_{−∞}^{∞} dωρ_{ff^{†}}(ω)} )^{2} =

−\frac{1}{⟨\{X_{0n}, X_{n0}\}⟩} ∑_{k_{1}k_{2}} V_{k_{1}}^{∗}V_{k_{2}}⟨s_{k_{2}k_{1}} · S⟩ − (\frac{1}{⟨\{X_{0n}, X_{n0}\}⟩} ∑_{k,m≠n} V_{k}^{∗}⟨c_{km}X_{m0}⟩)^{2} (10)

Originally the f-level was a delta function peak at ϵ_{f}. Due to hybridization there is a shift in the energy level, and also a broadening of the band.

2. f-sum rule

The retarded current-current correlation function is defined by

Π_{jj}^{R}(ω) = ⟨⟨j, j⟩⟩(ω) = \frac{1}{Z} ∑_{nm} (e^{−βE_{n}} − e^{−βE_{m}}) |⟨n|j|m⟩|^{2} \frac{1}{ω + E_{n} − E_{m} + iη} . (11)

The corresponding spectral function is

ρ_{jj}(ω) = −\frac{1}{π} ImΠ_{jj}^{R}(ω)
= \frac{1}{Z} ∑_{nm} (e^{−βE_{n}} − e^{−βE_{m}}) |⟨n|j|m⟩|^{2} δ(ω + E_{n} − E_{m})
= ∫_{−∞}^{∞} \frac{dt}{2π} e^{iωt}⟨[j(t), j]⟩.  (12)

In terms of the spectral function, the real part of the conductivity (which is what enters the f-sum rule) is given by,

σ(ω) = −\frac{1}{V} \frac{1}{ω} ImΠ_{jj}^{R}(ω) = \frac{π}{V} ρ_{jj}(ω).  (13)

We define the polarization operator P such that \frac{∂P}{∂t} = j. Then,

ρ_{jj}(ω) = ∫_{−∞}^{∞} \frac{dt}{2π} e^{iωt}⟨[\frac{∂P}{∂t} , j]⟩
= −iω ∫_{−∞}^{∞} \frac{dt}{2π} e^{iωt}⟨[P (t), j]⟩
= −iωρ_{P j}(ω),  (14)

where

ρ_{P j}(ω) = ∫_{−∞}^{∞} \frac{dt}{2π} e^{iωt}⟨[P (t), j]⟩.  (15)

Then,

∫_{−∞}^{∞} dωσ(ω) = −\frac{πi}{V} ∫_{−∞}^{∞} dωρ_{P j}(ω)
= −\frac{πi}{V}⟨[P (0), j]⟩.  (16)

The current operator is given by j = ∑_{i=1}^{N}(ep_{i})/m, where p_{i} is the momentum of the i-th particle. The polarization operator is given by P = ∑_{i=1}^{N} er_{i}, where r_{i} is the position of the i-th particle. Since,

[P, j] = \frac{e^{2}}{m} ∑_{ij}[r_{i}, p_{j}] = \frac{Ne^{2}i}{m},

(in units where h̄ = 1), we finally get the f-sum rule as

∫_{−∞}^{∞} \frac{dω}{π} σ(ω) = \frac{ne^{2}}{m} ,  (17)

where n = N/V is the density of particles.