Elementary Introduction to the Hubbard Model I. INTRODUCTION The Hubbard model o®ers one of the most simple ways to get insight into how the interac- tions between electrons can give rise to insulating, magnetic, and even novel superconducting e®ects in a solid. It was written down in the early 1960's and initially applied to understand- ing the behavior of the transition metal monoxides (FeO, NiO, CoO), compounds which are antiferromagnetic insulators, yet had been predicted to be metallic by methods which treat strong interactions less carefully. Over the intervening years, the Hubbard model has been applied to the understanding of many systems, from `heavy fermion' systems in the 1980's, to high temperature super- conductors in the 1990's. Indeed, it is an amazing feature of the model that, despite its simplicity, its exhibits behavior relevant to many of the most subtle and beautiful properties of solid state systems. The Hubbard model has been studied by the full range of analytic techniques developed by condensed matter theorists, from simple mean ¯eld approaches to ¯eld theoretic methods employing Feynman diagrams, expansions in the degeneracy of the number of ‘flavors' (spin, orbital angular momentum), etc. It has also been extensively attacked with numerical methods like diagonalization and quantum monte carlo. The objective of these notes is to provide an introduction to the Hubbard model and to a few of the most simple ways in which it is solved. We begin with a discussion of the second quantized operators using which the Hubbard model is written. II. CREATION AND DESTRUCTION OPERATORS Creation and destruction operators ay; a are familiar from the treatment of the harmonic oscillator. We will spend some time reviewing their properties, which parallel those of the operators in the Hubbard model in many ways, in this more familiar setting. The harmonic oscillator creation and destruction operators are de¯ned in terms of the position and momentum operators, a^ = m! x^ + i 1 p^ 2~ 2m!~ p q m! 1 a^y = x^ i p^ 2~ ¡ 2m!~ p q From [p;^ x^] = i~, it is easy to show that these operators obey the commutation relations, ¡ [a;^ a^y] = 1 and that the Hamiltonian is, 1 2 1 2 2 1 H^ = p^ + m! x^ = ~!(a^ya^ + ): 2m 2 2 ^ 1 The `number operator' is de¯ned to be n^ = a^ya^, so that H = ~!(n^ + 2 ). Exercise 1: Verify the commutation relation [a;^ a^y] = 1: 1 Exercise 2: Verify the expression for H^ in terms of the creation and destruction operators. Start with the usual expression for the Hamiltonian, H^ = p^2=2m + m!2x^2=2. Exercise 3: Figure out the commutation relations [n^; a^y] and [n^; a^]. The ground state of the quantum oscillator is written as 0 and has the properties that, j i a^ 0 = 0 j i ! H^ 0 = 0 j i 2 j i The excited states are built up by applying the creation operator repeatedly to the ground state. a^y n = pn + 1 n + 1 j i j i and obey the formula, 1 H^ n = ~!(n + ) n j i 2 j i Exercise 4a: Using the form for H^ in terms of the creation and destruction operators, and the commutation relations of Exercise 3, show that the state obeying a^ 0 is indeed an j i eigenstate of H^ . 2 m!x =2~ Exercise 4b: Show that the state 0 with components ª0(x) = x 0 = e¡ in the position representation (that is, injthei basis of eigenstates of the hpositionj i operator) obeys ~ @ a^ 0 = 0. Do this by using the form for a^ is terms of x^ and p^ and the fact that p^ = i @x in thej iposition basis. This helps connect the creation/destruction operator method with the wavefunctions for the eigenstates of the harmonic oscillator obtained in more conventional treatments. Exercise 4c: Compute the wavefunctions ª1(x) = x 1 and ª2(x) = x 2 of the ¯rst two h j i h j i m!x2=2 excited states of the simple harmonic oscillator by using ª (x) = x 0 = e¡ ~ and the 0 h j i expression for a^y in terms of x^ and p^. Exercise 5: Show that the states obtained by acting multiple times with a^y on 0 are j i eigenstates of H^ . One often refers to a^ and a^y as `boson' creation and destruction operators, for reasons that are motivated by the result of Exercises 6 and 7 below. Note that henceforth I will be setting ~ = 1. I will also set Boltzmann's constant kB = 1. ¯H^ Exercise 6: The partition function of a quantum mechanical system is Z = Tr e¡ . Here H^ is the Hamiltonian, ¯ = 1=T where T is the temperature, and the trace operation `Tr' has its usual meaning as the sum of the diagonal elements. In other words, one chooses any ¯H^ ¯H^ complete set of states ® and then Tr e¡ = ® e¡ ® . Show that for the Harmonic j i ®h j j i P 2 ¯!=2 ¯! oscillator Z = e¡ =(1 e¡ ). Use the occupation number states n as your complete set. ¡ j i Exercise 7: The ¯nite temperature expectation value of any quantum mechanical operator 1 ¯H^ ¯! A^ is given by A^ = Z¡ Tr[ A^ e¡ ]. Show that n^ = 1=(e¡ 1), for the quantum harmonic oscillator.h i This is the Bose-Einstein distributionh i function.¡ The Hubbard model is written in terms of `fermion' creation and destruction operators. These operators di®er in several respects from the operators a^y; a^ for a single harmonic oscillator. Perhaps most confusing is a conceptual di®erence: the fermion operators in the Hubbard model are not introduced in terms of familiar position and momentum operators. Rather they stand on their own. Also, instead of just one creation and one destruction operator, in the Hubbard model there is a set of such operators, which are distinguished by attaching indices j and σ. Thus we write c^jyσ and c^jσ. As we shall see shortly, the index j labels the spatial lattice site and the index σ labels the electron spin (up or down). As a consequence, the occupation number states are no longer characterized by a single number n, as for a single harmonic oscillator, but instead by a collection of occupation numbers njσ. One writes such states as n1σ n2σ n3σ : : : : Because these operators are meant tojdescribe fermions,ithey are de¯ned to have certain anticommutation relations. (The anticommutator of two operators A;^ B^ is de¯ned to be f g A^B^ + B^A^.) c^j ; c^y 0 = ±j l± 0 f σ lσ g ; σ,σ c^y ; c^y 0 = 0 f jσ lσ g c^j ; c^l 0 = 0: f σ σ g An immediate consequence of these anticommutation relations is the Pauli principle: the maximum occupation of a particular site with a given spin is 1. Exercise 8: Show that the Pauli exclusion principle is a consequence of the anticommutation relation amongst the fermion creation operators by considering j = i and showing that c^y 1 = 0. Here 1 is the state with one electron on site j and with spin σ. jσj i j i Note that while a fermion creation operator annihilates a state which already has a fermion in it, its action on the empty state is c^y 0 = 1 , which looks just like the bosonic j i j i operator a^y. Besides the Pauli principle, the anticommutation relations also ensure that the particles are fermions, that is, their wave function changes sign when two electrons with di®erent labels j are exchanged. Exercise 9: In Exercise 6 we showed that the partition function of H^ = !a^ya^ is Z = ¯! 1 (1 e¡ )¡ (eliminating the zero point energy term). Show that the partition function of ¡ ¯² the analogous fermion Hamiltonian H^ = ²c^yc^ is Z = (1 + e¡ ). Exercise 10: In statistical mechanics, one can work in either an ensemble of ¯xed particle number, or one in which the particle number is allowed to vary, controlled by the chemical potential ¹. More precisely, the term ¹N^ is added to H^ , wehere N^ is the total particle number. Explore this idea within the¡context of the preceding problem by considering H^ = (² ¹)c^yc^. Plot n^ as a function of ¹ for di®erent values of ¯. ¡ h i 3 Because of the anticommutation relation of the fermion operators, one needs to spec- ify a convention for the relation between a state like 1 0 1 0 0 : : : and the vacuum j i state vac = 0 0 0 0 0 : : : . The two possibilities, 1 0 1 0 0 : : : = cy cy vac and j i j i j i 1 3j i 1 0 1 0 0 : : : = c3y c1y vac di®er by a sign. Either de¯nition is ¯ne, but in all subsequent manipulationsj i whatevjer coni vention was chosen must be followed consistently. Exercise 11: Suppose the index labeling the fermion operators runs over eight possible values. Figure out the occupation number state which results from the following applications of the indicated creation and destruction operators. Choose as your convention that a given occupation number state is formed by acting on the vacuum state with the lowest indices at the right of the string of creation operators. (This corresponds to the second of the two choices discussed above.) c^y c^y c^y vac = ? 5 4 2 j i c^y c^y c^y vac = ? 2 4 5 j i c^y c^y 1 1 0 0 0 0 0 1 = ? 4 5 j i c^y c^y 1 1 0 0 1 0 0 1 = ? 4 5 j i c^y c^ 0 1 0 0 1 0 0 1 = ? 1 2 j i c^y c^ 0 1 0 0 1 0 0 1 = ? 1 4 j i c^y c^ 0 1 0 0 1 0 0 1 = ? 1 5 j i III.