Tutorials of Theoretical Condensed Matter 2017-2018€¦ · Tutorials of Theoretical Condensed...

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 ?

1

Tutorials 2017–2018

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 ?

M2 ICFP Theoretical Condensed Matter


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)



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 ?



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



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

†.



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 ?



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 ?



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) ?



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.



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〉



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.



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↑


1√2

(c†2↓c

†2↑c†1↓ ± c

†2↓c†1↓c†1↑


• 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 ,



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

)



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



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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