Tutorials of Theoretical Condensed Matter 2017-2018
Benoıt Doucot, Benoıt Estienne and Laura Messio
The Hubbard model on two sites
The Hubbard model describes spin 1/2 fermions hopping on a lattice according to the following tight-bindingHamiltonian
HHub = −t∑
σ∈{↑,↓}
∑〈ij〉
(c†iσcjσ + c†jσciσ) + U∑i
(ni − 1)2 − µ∑i
(ni − 1), ni = ni↑ + ni↓
where U > 0 is the repulsive interaction, t the hopping and µ the chemical potential. The sum < ij >stands for a sum over i, j nearest neighbour sites on some lattice (in two dimensions this could be a squarelattice for instance). As usual for fermions we have{
ciσ, c†jσ′
}= δi,j δσ,σ′{
ciσ, cjσ′}
= 0{c†iσ, c
†jσ′
}= 0
Particle conservation and U(1) symmetry
1. Let N =∑
i ni be the total number of particles. Show without any calculation that [N , HHub] = 0(as a homework exercice, this can be checked by working through the algebra). This means thatN is a conserved quantity. What is the expression for the corresponding current ? Give the localconservation equation. What would be an exemple of a (physical) tight-binding Hamiltonian withoutparticle conservation ?
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Correction.
• All the terms in the Hamiltonian conserve the number of particles, so HHub commutes with N .
• Physically this simply means that charge is conserved. Since it is a conserved quantity, we canwrite a conservation equation (in the continuum it would be of the form dρ
dt + div~j = 0). In theHubbard model the local density at site i is ni. In the Heisenberg picture we have
ni(t) = e−iHHubtnieiHHubt
and it obeys
d
dtni(t) = −i[HHub, ni(t)] = it
∑j,<j,i>
∑σ
(c†jσciσ − c
†iσcjσ
)The right hand side is the lattice version of div(~j). It is a sum over all nearest neighbours of i :each edge < i, j > carries a current Jij (oriented from i to j)
Jij = it(c†iσcjσ − c
†jσciσ
)and the equation of conservation is
d
dtni(t) +
∑j,<j,i>
Jij = 0
• The typical exemple of a non-particle conserving Hamiltonian is that of BCS superconductivity,which contains terms in cc and c†c† :
HBCS = −t∑
σ∈{↑,↓}
∑〈ij〉
(c†iσcjσ + c†jσciσ) +∑k
∆kck↓c−k↑ + ∆kc†−k↑c
†−k↓
2. Check that the Hamiltonian is invariant under
cjσ → e−iθcjσ, c†jσ → eiθc†jσ
Is such a transformation allowed ? Such a symmetry is called a global U(1) symmetry. Why is itcalled a U(1) symmetry ? And why global ?
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Correction.By inspection the Hamiltonian is clearly invariant : each term creates as many particlesas it annihilates. Such a symmetry is called U(1) (or sometimes S1) symmetry because the groupacting here is the group of complex numbers of modulus one (i.e. phases). More generally U(n) is thegroup of n× n unitary matrices.This transformation is allowed since the new creation and annihilation operators have the correctanticommutation relations (we will see in the next question that we are actually dealing with a unitarytransformation). In fact one can even consider a more generic situation : that of a local transformation
cjσ → djσ = e−iθjcjσ, c†jσ → d†jσ = eiθjc†jσ
in which the Hamiltonian changes to
HHub = −t∑
σ∈{↑,↓}
∑〈ij〉
(ei(θj−θi)d†iσdjσ + e−i(θj−θi)d†jσdiσ) + U∑i
(ni − 1)2 − µ∑i
(ni − 1)
This is nothing but a gauge transformation.
3. Let U be the unitary operator U = eiθN . What are UcjσU† and Uc†jσU
† ? Show the equivalencebetween the global U(1) symmetry of question 2. and the particle conservation of question 1.
Correction.We have
eiθNcjσe−iθN = e−iθcjσ, eiθNc†jσe
−iθN = eiθc†jσ
Global U(1) symmetry means [eiθN , H] = 0 which is clearly equivalent to [N , H] = 0.
SU(2) symmetry
1. What are the (global) SU(2) generators in second quantized form ?
Correction.
Sz =1
2
∑i
(ni↑ − ni↓) , S+ =∑i
c†i↑ci↓, S− =∑i
c†i↓ci↑
It is a straightforward exercice to check that they obey the correct commutation relations [S+, S−] =2Sz and [Sz, S±] = ±S±.
2. Is the Hubbard model SU(2) symmetric ?
Correction.Yes, it is. Clearly [Sz, HHub] = 0. The only non-trivial part in showing that [S+, HHub] =0 is the kinetic term (indeed [S+, ni] = 0). Using [AB,C] = A{B,C} − {A,C}B, one can first derive
[S+, c†iσ] = δσ↓c†i↑, [S+, ciσ] = −δσ↑ci↓
From this we get [S+, c†iσcjσ + c†jσciσ] = 0, and we are done.
Note that the U(1) and SU(2) symmetries can be combined into a U(2) symmetry
c†jσ →∑σ′
Vσσ′c†jσ′ , V ∈ U(2)
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Indeed a generic matrix V in U(2) can be decomposed as
V = eiθW, W ∈ SU(2)
and is therefore the product of U(1) and a SU(2) unitary transformation.
Particle-hole conjugation
1. For spinless particles, particle-hole conjugation can be defined as a linear, unitary operator Γ suchthat
Γc†iΓ† = ci
(homework : does such an operator exist ? is it unique ? What is Γ|0〉 ?)
Correction.Knowing that Γc†iΓ† = ci is not enough to define Γ. Indeed suppose we have such an
operator Γ, then Γ′ = eiφΓ also works. So we do not have unicity. Another way to think of thisambiguity is to try to compute Γ|0〉. This state is fully occupied since it is annihilated by all the c†i ,so it has to be of the form (we suppose there are N one-body states) :
Γ|0〉 = eiφc†N · · · c†2c†1|0〉
but the phase eiφ has to be specified. Once this phase is chosen though, we know Γ|0〉 and the actionon any state can be computed. For instance
Γc†ic†j |0〉 = cicjΓ|0〉 = eiφcicjc
†N · · · c
†2c†1|0〉
So we see that- up to the choice of phase eiφ - we have unicity of the operator Γ.But we have not yet proven that the operator we just defined consistently obeys Γc†iΓ
† = ci. Whileit’s not very difficult to do so from the previous expression (we know how it acts on each state), it is
a bit tedious. A cleaner way is to actually construct this operator in terms of the cj and c†j . Define
Γj = eiϕjc†j + e−iϕjcj
It is easy to check that Γj is unitary, and that ΓjcjΓ† = e2iϕjc†j . Then notice that the operator
Γ = Γ1Γ2 · · ·ΓN
indeed obeys ΓcjΓ† = (−1)N−1e2iϕjc†j . So we have existence (by setting eiϕj = iN−1 for instance).
If we ignore the SU(2) symmetry of the Hubbard model, we can define particle-hole conjugation as
Γc†iσΓ† = ciσ
Is the µ = 0 Hubbard model invariant under Γ ?
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Correction.Let’s compute ΓHHubΓ†. Since
ΓniσΓ† = 1− niσ
the interaction term U∑
i(ni − 1)2 is indeed invariant.
Γ∑i
(ni − 1)2Γ† =∑i
(ni − 1)2
The chemical potential term −µ∑
i(ni − 1) changes sign
Γ∑i
(ni − 1)Γ† = −∑i
(ni − 1)
but that has to be expected : if µ is the energy cost of removing an electron, then −µ is energy costof removing a hole (i.e adding an electron). So far so good.
But the kinetic term −t∑
σ∈{↑,↓}∑〈ij〉(c
†iσcjσ + c†jσciσ) is not invariant :
Γ(c†iσcjσ + c†jσciσ
)Γ† = −
(c†iσcjσ + c†jσciσ
)it changes sign under Γ.To sum things up, the parameters (t, U, µ) of the Hubbard model become (−t, U,−µ) under theconjugation with Γ. Therefore (on a generic lattice) the Hubbard model is not invariant under particle-hole conjugation (even at µ = 0). But we will see in the following questions that on a bipartite lattice,we do have particle-hole symmetry.
2. We focus on the Hubbard model on a bipartite lattice, i.e. a lattice which can be partitioned into twosublattices A and B, where are all the nearest neighbours of A are members of B. Show that the signof t is unphysical (i.e. one can find a unitary transformation that changes the sign of t).
Correction.The following local U(1) transformation
ciσ → ciσ on the A sublattice, ciσ → −ciσ on the B sublattice
changes the sign of t. This can be achieved with the following unitary transformation
U =∏i∈B
(−1)ni = (−1)NB
We have
U(c†iσcjσ + c†jσciσ
)U † = −
(c†iσcjσ + c†jσciσ
)provided the sites i and j are not in the same sublattice. This is why on a bipartite lattice we canchange the sign of t by acting with U .
3. It follows from the previous two questions that the Hubbard model (at µ = 0) is in fact particle-holesymmetric on a bipartite lattice. To make this more explicit we consider a slightly modified version
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of the particle-hole conjugation defined by:
Γc†iσΓ† = ciσ on the A sublattice,
Γc†iσΓ† = −ciσ on the B sublattice.
Check that at µ = 0 the Hubbard model is indeed invariant under this new Γ. What are the conse-quences on the spectrum and the eigenstates of HHub ?
Correction.
ΓHHubΓ† = −t∑〈ij〉,σ
Γ(c†iσcjσ + c†jσciσ)Γ† + U∑i
Γ(ni − 1)2Γ† − µ∑i
Γ(ni − 1)Γ†,
= t∑〈ij〉,σ
(ciσc†jσ + cjσc
†iσ) + U
∑i
(2− ni↑ − ni↓ − 1)2 − µ∑i
(2− ni↑ − ni↓ − 1),
= −t∑〈ij〉,σ
(c†jσciσ + c†jσciσ) + U∑i
(ni − 1)2 + µ∑i
(ni − 1),
We find that under Γ, the chemical potential changes sign. In particular at µ = 0 (half-filling), wehave particle-hole symmetry
ΓHHubΓ† = HHub.
As the total number of particle N =∑
i ni commute with the Hamiltonian, we can solve the problemin fixed number of particles subspaces. Thus, for a lattice of N sites, the subspaces with n particleshave the same eigenenergies as the one with 2N − n particles and their eigenstates are linked by theparticle-hole transformation.
4. For spin 1/2 particles the particle-hole conjugation we have defined is not very satisfactory since itdoes not conserve the spin (it is straightforward to check that Γ changes Sµ → −Sµ). It is morenatural to define particle-hole conjugation as
Γc†i↑Γ† = ci↓, Γc†i↓Γ
† = −ci↑
(homework : does such an operator exist ? is it unique ?). To understand why this definition is morenatural, compute ΓSiΓ
†.
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Correction.For a single site we can use
Γj =(c†j↑c
†j↓ + cj↑cj↓ − nj + 2nj↑nj↓
)we can check that
ΓjΓ†j = 1, Γjc
†j↑Γ
†j = cj↓, Γjc
†j↓Γ
†j = −cj↑
For N sites we can use
Γ = Γ1Γ2 · · ·ΓN
and we have proven the existence of such an operator. Now for the unicity, we can follow the same reasoning as in the spinless case. We must have
Γ|0〉 = eiφ
c†N↓c
†N↑ · · · c
†1↓c
†1↑|0〉
so we have unicity up to a phase.
Now we have
ΓniσΓ† = Γc†iσciσΓ† = ciσc†iσ = 1− niσ
where σ stands for the spin opposite to σ (i.e. ↑ =↓). It follows that
ΓniΓ† = 2− ni, ΓSzi Γ† = Szi
Likewise
ΓS+i Γ† = Γc†i↑ci↓Γ
† = −ci↓c†i↑ = S+i , ⇒ ΓS−i Γ† = S−i
The choice ensures the conservation of Sxi , Syi and Szi . This modified version of the particle-holeconjugation commutes with the SU(2) symmetry. For instance the total spin along the z direction isnow preserved under particle-hole conjugation.
Putting all this together, we are lead to define the following particle-hole conjugation :
Γc†i↑Γ† = ci↓ and Γc†i↓Γ
† = −ci↑ on the A sublattice,
Γc†i↑Γ† = −ci↓ and Γc†i↓Γ
† = ci↑ on the B sublattice.
We now have
ΓHHub(µ)Γ† = HHub(−µ), ΓSµΓ† = Sµ, ΓN Γ = 2N − N
where N is the number of lattice sites.
Weak coupling and strong coupling regimes
5. Some insights can be gained into the Hubbard model by considering the t = 0 limit, in which thedifferent sites decouple. Since the system is now a collection of independant sites, one just needs tosolve a single site. What are the eigenstates and energies of a single site ? What is the partitionfunction at inverse temperature β ? What is the density ρ = 〈n〉 ? Plot ρ versus µ for various valuesof β. What happens at zero temperature ? How is the particle-hole symmetry manifest ?
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Correction.For a single site we have the following eigenstates
• |0〉 with energy U + µ,
• |↑〉 and |↓〉 with energy 0,
• and |↑↓〉 with energy U − µ.
The partition function is
Z = Tr(e−βH
)= 2
(1 + e−βU coshβµ
)and the density is
ρ− 1 =1
ZTr(
(n− 1)e−βH)
=1
Z
1
β
∂
∂µZ =
1
β
∂
∂µlogZ
which yields
ρ =e−βU sinhβµ
1 + e−βU coshβµ+ 1
Below is the plot of ρ versus µ for U = 4 :
Β=0.5
Β=1.5
Β=100
-15 -10 -5 5 10 15Μ
0.5
1.0
1.5
2.0
Ρ
At zero temperature the mean occupation is discontinuous :
ρ =
0 for µ < −U1 for −U < µ < U2 for µ > U
6. Solve the non-interacting case (U = 0) on a one dimensional lattice of N sites with periodic boundaryconditions. What is the dispersion relation ? Is the particle-hole symmetry manifest ? As a homeworkexercice, what would be the dispersion relation on a two-dimensional N ×N square lattice ?
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Correction.Translation invariance is exploited using Fourier transform:
d†kσ =1√N
∑j
eikj c†jσ, k =2π
Nm, m = 0, 1, · · · , N − 1
It is straightforward to check that
{dkσ, dqσ′} = δkqδσσ′ ,
and we get (up to an additive constant term)
HHub =∑k,σ
(εk − µ)nkσ, εk = −2t cos k
The dispersion relation is spin-independent (this was expected by SU(2) symmetry)
εk = −2t cos k.
The particle-hole symmetry is there (assuming µ = and that N is even!), but is not particularly easyto see. For instance one can check that the empty and full occupied system have the same energy
0 =∑k
εk
If we call A the sites with even positions, Γ acts as follow in Fourier space
ΓciσΓ = (−1)ic†iσ ⇒ Γd†k,σΓ = dπ−k,σ
Since επ−k = −εk, an eigenstate and its particle hole conjugate have the same energy indeed.In two-dimensions we would introduce
d~kσ =1√N
∑~j
ei~k·~j c†~jσ
, ~k =2π
N(m1,m2), mj = 0, 1, · · · , N − 1
and the dispersion relation would be
ε~k = −2t(cos kx + cos ky).
Exact solution for 2 sites
7. We choose to exactly solve the interacting problem for two sites 1 and 2. What symmetries can oneexploit (i.e. what are the good quantum numbers) ?
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Correction.The symmetries (↔ operators commuting with HHub and commuting with each other)that can be used are
• the total particle number N = n1 + n2,
• the total spin along z Sz,
• the total spin S2
• the exchange symmetry between the two sites P (acting as P c1σP† = c2σ, P c2σP
† = c1σ, andP |0〉 = |0〉)
We denote the corresponding quantum numbers by (N,Sz, S, P ). The Hubbard Hamiltonian is blockdiagonal with respect to these quantum numbers. Moreover the particle hole conjugation Γ relate thesector with N particles to the sector with 4−N particles.
(N,Sz, S, P )→ (4−N,Sz, S, P )
We have 24 = 16 states (each fermionic state ↑ or ↓, on site 1 or 2, can be either filled or empty).Thus, we have a 16 × 16 matrix to diagonalize. From the N -symmetry, we block-diagonalize it bydividing the states according to their occupation number:
(a) 1 state with 4 particles.
(b) 4 states with 3 particles
(c) 6 states with 2 particles
(d) 4 states with 1 particle
(e) 1 empty state → E0.
From the particle-hole symmetry, we identify the spectrum of the 4 and 0 particle subspaces, and ofthe 3 and 1 particle subspaces. We have 5 energies less to find.Now we look at the 4 1-particle states. From the Sz-symmetry, we limit the research to the Sz = 1/2subspace. We have 2 energies less to find. From P12, we construct a symmetric and an antisymmetricstate, with energies ES1 and EAS1 .Now we look at the 6 2-particle states. From the Sz-symmetry, we limit the research to the Sz = 0 (4states) and Sz = 1 (1 state) subspaces. One energy is removed this way and ESz=1
2 is characterized.In the Sz = 0 sector, two states are symmetric and two are antisymmetric with respect to P12. Inthe antisymmetric sector, S2 leads to define EAS,S=0
2 and EAS,S=12 . Finally, only two energies E±2 are
undetermined once all symmetries are taken into account.
8. Determine the spectrum and the eigenstates.
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Correction.
• N = 0 sector : There is only one state |0〉, with energy 2U + 2µ.
• N = 1 sector : There are four states : c†1↑|0〉, c†1↓|0〉, c
†2↑|0〉 and c†2↓|0〉. On this subspace the term
U∑
i(ni − 1)2 − µ∑
i(ni − 1) is constant (equal to U + µ) and can be ignored, and we are leftwith the hopping terms. They can be dealt with as in question 4, or using P . The eigenstatesare
P = ±1, Sz =1
2:
1√2
(c†1↑ ± c
†2↑
)|0〉, with energy ∓ t+ U + µ,
P = ±1, Sz = −1
2:
1√2
(c†1↓ ± c
†2↓
)|0〉, with energy ∓ t+ U + µ.
• N = 2 sector : There are six states : 4 states with 1 particle per site (c†2σ′c†1σ|0〉), and 2 states
with a doubly occupied site (c†1↓c†1↑|0〉 and c†2↓c
†2↑|0〉). Let’s arrange them by SU(2) multiplet.
We have
◦ one state with Sz = 1 : c†2↑c†1↑|0〉
◦ four states with Sz = 0 : c†2↑c†1↓|0〉, c
†2↓c†1↑|0〉, c
†1↓c†1↑|0〉 and c†2↓c
†2↑|0〉
◦ one state with Sz = −1 : c†2↓c†1↓|0〉
from which we conclude that there are three singlets (S = 0) and one triplet (S = 1). The
highest weight state of the triplet is of course the state with Sz = 1 : c†2↑c†1↑|0〉. The other ones
are obtained by acting with S−. Therefore the triplet is
c†2↑c†1↑|0〉,
1√2
(c†2↓c
†1↑ + c†2↑c
†1↓
)|0〉, c†2↓c
†1↓|0〉
Note that the triplet is odd under P . Indeed we have Pc1σP = c2σ and Pc2σP = c1σ . Hence
Pc†2↑c
†1↑|0〉 = Pc
†2↑PPc
†1↑P |0〉 = c
†1↑c
†2↑|0〉 = −c
†2↑c
†1↑|0〉
P1√
2
(c†2↓c
†1↑ + c
†2↑c
†1↓
)|0〉 =
1√
2
(c†1↓c
†2↑ + c
†1↑c
†2↓
)|0〉 = −
1√
2
(c†2↓c
†1↑ + c
†2↑c
†1↓
)|0〉
Pc†2↓c
†1↓|0〉 = −c
†2↓c
†1↓|0〉
All these states have the same energy by SU(2) invariance, which is straightforward to evaluate
by acting with H on c†2↑c†1↑|0〉 :
Hc†2↑c†1↑|0〉 = 0
The value 0 is easy to interpret : the hopping terms are frozen by the Pauli principle, and thereare no U or µ terms since we have exactly one particle per site.
Now we are left with the three singlets :
c†1↓c†1↑|0〉,
1√2
(c†2↓c
†1↑ + c†2↑c
†1↓
)|0〉, c†2↓c
†2↑|0〉
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Correction.Acting with H on this subspace would yield a 3 by 3 matrix, which we can diagonalize. But wehaven’t used yet the permutation symmetry P12. We can split the three triplets into two symmetricstate (P = +1) and one antisymmetric state (P = −1) :
◦ P12 = +1 :c†1↓c
†1↑+c
†2↓c†2↑√
2|0〉 and 1√
2
(c†2↓c
†1↑ − c
†2↑c†1↓
)|0〉
◦ P12 = −1 :c†1↓c
†1↑−c
†2↓c†2↑√
2|0〉
On the one hand we have
Hc†1↓c
†1↑ + εc†2↓c
†2↑√
2|0〉 = − t√
2
[(−c†2↑c
†1↓ + c†2↓c
†1↑
)|0〉+ ε
(c†2↓c
†1↑ − c
†2↑c†1↓
)|0〉]
+ 2Uc†1↓c
†1↑ + εc†2↓c
†2↑√
2|0〉
= t(1 + ε)
(c†2↑c
†1↓ − c
†2↓c†1↑
)√
2|0〉+ 2U
c†1↓c†1↑ + εc†2↓c
†2↑√
2|0〉
and the odd statec†1↓c
†1↑−c
†2↓c†2↑√
2|0〉 is indeed an eigenvector, with eigenvalue 2U . To compute the action of the
kinetic part of the Hamiltonian we used
c†2↑c1↑c
†1↓c
†1↑|0〉 = −c
†2↑c1↑c
†1↑c
†1↓|0〉 = −c
†2↑c
†1↓|0〉
c†2↓c1↓c
†1↓c
†1↑|0〉 = c
†2↓c
†1↑|0〉
and
c†1↑c2↑c
†2↓c
†2↑|0〉 = −c
†1↑c2↑c
†2↑c
†2↓|0〉 = −c
†1↑c
†2↓|0〉 = c
†2↓c
†1↑|0〉
c†1↓c2↓c
†2↓c
†2↑|0〉 = c
†1↓c
†2↑|0〉 = −c
†2↑c
†1↓|0〉
And on the other hand
Hc†2↓c
†1↑ − c
†2↑c†1↓√
2|0〉 = −2t
c†1↓c†1↑ + c†2↓c
†2↑√
2|0〉
The only matrix to diagonalize is a 2 × 2 matrix, between the two even statesc†1↓c
†1↑+c
†2↓c†2↑√
2|0〉 and
c†2↓c†1↑−c
†2↑c†1↓√
2|0〉 : (
2U −2t−2t 0
)= U + Uσz − 2tσx
The (un-normalized) eigenvectors and eigenenergies are
(U ±√U2 + 4t2)
c†1↓c†1↑ + c†2↓c
†2↑√
2|0〉 − 2t
c†2↓c†1↑ − c
†2↑c†1↓√
2|0〉, E± = U ±
√U2 + 4t2
• N = 3 sector : The eigenvectors can be obtained form the N = 1 sector using particle-holeconjugation.
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Correction.
Γ1√2
(c†1σ ± c
†2σ
)|0〉, with energy ∓ t+ U − µ
We can the compute
Γc†1↑|0〉 = Γc†1↑Γ†Γ|0〉 = c1↓c
†2↓c†2↑c†1↓c†1↑|0〉 = c†2↓c
†2↑c†1↑|0〉
Γc†1↓|0〉 = c1↑c†2↓c†2↑c†1↓c†1↑|0〉 = −c†2↓c
†2↑c†1↓|0〉
Γc†2↑|0〉 = −c2↓c†2↓c†2↑c†1↓c†1↑|0〉 = c†2↑c
†1↓c†1↑|0〉
Γc†2↓|0〉 = −c2↑c†2↓c†2↑c†1↓c†1↑|0〉 = −c†2↓c
†1↓c†1↑|0〉
and we find the following eigenstates
1√2
(c†2↓c
†2↑c†1↑ ± c
†2↑c†1↓c†1↑
)|0〉, with energy ∓ t+ U − µ
1√2
(c†2↓c
†2↑c†1↓ ± c
†2↓c†1↓c†1↑
)|0〉, with energy ∓ t+ U − µ
• N = 4 sector : The fully occupied state Γ|0〉 = c†2↓c†2↑c†1↓c†1↑|0〉 has energy −2µ
Strong-coupling regime at half-filling : effective Hamiltonian
We now focus on the Hubbard model at half-filling (we work with a fixed number of particles N , N beingthe number of sites).
HHub = −t∑〈ij〉,σ
(c†iσcjσ + c†jσciσ) + U∑i
(ni − 1)2,
In the limit U →∞, this model can be (moderately) simplified, notably into the Heisenberg Hamiltonian :
HHeis = J∑〈ij〉
Si · Sj ,
This effective Hamiltonian can be obtained using perturbation theory (see Appendix).
1. Find the ground state(s) of the Hubbard model for t = 0. What is the ground-state degeneracy ?
Correction.In the ground state, there is exactly one particle per site. It leads to a degeneracy of 2N
with N the number of sites. We have for each site the possibility to place a ↑ or ↓ spin particle.
2. We now consider the regime U � t at half-filling. We want to compute the effective Hamiltonian ofthe Hubbard model (in the subspace of the previous question). Why do we need to go to second orderperturbation theory ? Show that the effective is the Heisenberg model
HHeis =2t2
U
∑〈ij〉
Si · Sj ,
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Correction.We use perturbation theory with
H0 = U∑i
(ni − 1)2, V = −t∑〈ij〉,σ
(c†iσcjσ + c†jσciσ)
At half-filling the ground states of H0 have been obtained at question 1. They can written in spinnotation as |σ1, σ2, · · · , σN 〉, with σi ∈ {↑, ↓}, and they have energy E0 = 0. The excited states are allthe other ones, namely those with a least an empty site or a doubly occupied one.When acting with V on a ground state, we get terms like c†iσcjσ that takes an electron from site j andmove it to site i. Starting from a ground state, we end up with a (linear combination of) state withni = 2 and nj = 0, i.e. a state with energy (under H0) Em = U . Therefore the first order perturbationtheory vanishes, and we need to go to second order. At second order we get
〈σ′1 · · ·σ′N |Heff|σ1, · · ·σN 〉 = −t2∑<i,j>
∑|m〉
∑σσ′
〈σ′1 · · ·σ′N |(c†iσ′cjσ′ + c†jσ′ciσ′)|m〉〈m|(c
†iσcjσ + c†jσciσ)|σ1, · · ·σN 〉
Em
In the sum over |m〉, only two states contribute :
• |m〉 = c†iσcjσ|σ1, · · · , σN 〉
• and |m〉 = c†jσciσ|σ1, · · · , σN 〉
and they both have energy Em = 2U (ni = 2, nj = 0 for the one, and ni = 0, nj = 2 for the other).We end up with
〈σ′1 · · ·σ′N |Heff|σ1, · · ·σN 〉 =− t2
2U
∑<i,j>
∑σσ′
(〈σ′1 · · ·σ′N |c
†jσ′ciσ′c
†iσcjσ|σ1, · · · , σN 〉
+〈σ′1 · · ·σ′N |c†iσ′cjσ′c
†jσciσ|σ1, · · · , σN 〉
)The effective Hamiltonian is therefore
Heff = − t2
2U
∑<i,j>
∑σσ′
(c†jσ′ciσ′c
†iσcjσ + c†iσ′cjσ′c
†jσciσ
)= − t2
2U
∑<i,j>
∑σσ′
(−2c†iσciσ′c
†jσ′cjσ + δσσ′(niσ + njσ)
)where we used
c†jσ′ciσ′c†iσcjσ = ciσ′c
†iσc†jσ′cjσ = −c†iσciσ′c
†jσ′cjσ +
{ciσ′ , c
†iσ
}c†jσ′cjσ = −c†iσciσ′c
†jσ′cjσ + δσσ′njσ
and likewise for c†iσ′cjσ′c†jσciσ. We have now, using ni = nj = 1,
Heff =t2
U
∑<i,j>
∑σσ′
(c†iσciσ′c
†jσ′cjσ −
1
4
)
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Correction.The last step is to rewrite this in terms of spin operators
Szi =1
2(ni↑ − ni↓) , S+
i = c†i↑ci↓, S−i = c†i↓ci↑
We can separate the sum over σ, σ′ into terms where σ = σ′ and terms where σ = −σ′ :∑σσ′
c†iσciσ′c†jσ′cjσ =
∑σ
niσnjσ + S+i S−i + S−i S
+j
On the other hand we have
Szi Szj =
1
4(ni↑ − ni↓) (nj↑ − nj↓) =
1
4
∑σ
(niσnjσ − niσnj−σ) =1
2
∑σ
niσnjσ −1
4
where we used niσ + ni−σ = 1. Therefore∑σσ′
c†iσciσ′c†jσ′cjσ = 2Szi S
zi + S+
i S−i + S−i S
+j +
1
2= 2Si · Sj +
1
2
We get that
Heff =2t2
U
∑<i,j>
∑σσ′
(Si · Sj +
1
8
)
Since U > 0, we expect ferromagnetic order.Note that there is a much shorter way to get this result (without the prefactor, and up to the constantterm) without much calculation, using simple symmetry arguments :
• the effective Hamiltonian is a spin Hamiltonian (empty and doubly occupied orbitals are forbid-den)
• the effective interaction is nearest neighbour only
• it is SU(2) symmetric
• there are only two (linearly independent) nearest-neighbour interactions for spin 1/2 which areSU(2) symmetric : 1 and Si · Sj .
A Perturbation theory
We recall here some results of degenerate perturbation theory for an Hamiltonian H = H0 + λV for λ→ 0.We denote by H0 the ground state subspace, with energy E0. To second order in perturbation theory, theeffective Hamiltonian in H0 is
Heff = E0P + λP V P + λ2PV Q(E0 −QH0Q)−1QV P
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where P is the projector on H0 and Q = 1−P . This can be reformulated as follow. For two states |φ〉, |φ′〉in H0, we have
〈φ′|Heff|φ〉 = E0〈φ′|φ〉+ λ〈φ′|V |φ〉+ λ2∑|m〉/∈H0
〈φ′|V |m〉〈m|V |φ〉E0 − Em
where the sum runs over all eigenvectors |m〉 (of H0) not in H0.
Note : it is a good exercice to derive this formula.
References
[1] Henk Eskes and Robert Eder. Hubbard model versus t - J model: The one-particle spectrum. Phys.Rev. B, 54:R14226–R14229, Nov 1996.
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