Appendix A Second Quantization The most remarkable fact about a system of fermions is that no more than one fermion can occupy a quantum particle state (Pauli’s exclusion principle). For bosons no such restriction applies. That is, any number of bosons can occupy the same state. We shall formulate the second quantization in which creation and an- nihilation operators associated with each quantum state are used. This formalism is extremely useful in treating of many-boson and/or many-fermion systems. It is indispensable for the development of the superconductivity and quantum Hall effect theories. A.1 Boson Creation and Annihilation Operators The quantum state for a system of bosons (or fermions) can most conveniently be { } represented by a set of occupation numbers na with na being the numbers of bosons (or fermions) occupying the quantum particle-states a. This representation is called the occupation number representation or simply the number representation. For bosons, the possible values for na are zero, one, two, or any positive integers: = , , , ··· . na 0 1 2 (A.1) The many-boson state can best be represented by the distribution of particles (circles) in the states (boxes) as shown in Fig. A.1. Let us introduce operators na (without prime) whose eigenvalues are given by 0, 1,2,....Weassumethat [na, nb] ≡ nanb − nbna = 0. (A.2) It is convenient to introduce complex dynamic variables η and η† instead of di- rectly dealing with the number operators n. We attach labels a, b,...tothedynamic variables η and η† associated with the states a, b, . and assume that η and η† satisfy the following Bose commutation rules: 227 228 Appendix A Second Quantization Fig. A.1 A many-boson state is represented by a set of boson numbers {n j } occupying the state {p j } p 2 p 1 p0 p1 p2 , n 2 0, n 1 0, n 0 2, n1 0, n 2 1, η ,η† = δ , η ,η = η† ,η† = . [ a b] ab [ a b] [ a b] 0 (A.3) Let us set η† η ≡ = † , a a na ( na) (A.4) † which is Hermitian. Clearly, na ≡ ηaηa satisfy Equation (A.2). We shall show that na has as eigenvalues all non-negative integers. Let n be an eigenvalue of n (dropping the suffix a) and |n an eigenket belonging to it. From the definition n| η†η |n =n n|n . (A.5) Now n| η†η |n is the squared length of the ket η|n and hence n| η†η |n ≥0. (A.6) Also by definition n|n > 0; hence from Equations (A.5) and (A.6), we obtain n ≥ 0, (A.7) with the case of equality occurring only if η|n =0. (A.8) Consider now [η, n] ≡ [η, η†η] . We may use the following identities: A.1 Boson Creation and Annihilation Operators 229 [A, BC] = B[A, C] + [A, B]C , (A.9) [AB, C] = A[B, C] + [A, C]B, and obtain [η, η†η] = η†[η, η] + [η, η†]η = η, ηn − nη. (A.10) Hence nη |n =(ηn − η) |n =(n − 1)η |n . (A.11) Now if η |n = 0, then η |n is, according to Equation (A.11), an eigenket of n belonging to the eigenvalue n − 1. Hence for non-zero n, n − 1 is another eigenvalue. We can repeat the argument and deduce that if n − 1 = 0, then n − 2is another eigenvalue of n. Continuing this argument, we obtain a series of eigenvalues n, n −1, n −2,...,whichcanterminateonly with the value 0 because of inequality Equation (A.7). By a similar process, we can show from the Hermitean conjugate of Equation (A.10): nη† − η†n = η† that the eigenvalue of n has no upper limit (Problem A.1.1). Therefore, the eigenvalues of n are non-negative integers: 0, 1, 2,....(q.e.d) Let |φa be a normalized eigenket of na belonging to the eigenvalue 0 so that |φ =η† η |φ = . na a a a a 0 (A.12) By multiplying all these kets |φa together, we construct a normalized eigenket: |⌽0 ≡|φa |φb ··· (A.13) which is a simultaneous eigenket of all n belonging to the eigenvalues zero. This ket is called the vacuum ket. It has the following property: ηa|⌽0 =0 for any a. (A.14) Using the commutation rules (A.3) we obtain a relation (dropping suffix a) η(η†)n − (η†)n η = n(η†)n −1, (A.15) which may be proved by induction (Problem A.1.2). Multiply Equation (A.15) by † η from the left and operate the result on |⌽0 . Using Equation (A.14) we obtain η†η(η†)n |φ =n(η†)n |φ , (A.16) indicating that (η†)n |ψ is an eigenket belonging to the eigenvalue n. The square length of (η†)n |φ is 230 Appendix A Second Quantization φ|ηn (η†)n |φ =n φ|ηn −1(η†)n −1|φ =···=n!. (A.17) We see from Equation (A.11) that η|n is an eigenket of n belonging to the eigenvalue n −1. Similary, we can show from [n,η†] = η† that η†|n is an eigenket of n belonging to the eigenvalue n + 1. Thus operator η, acting on the number eigenket, annihilates a particle while operator η† creates a particle. Therefore, η and η† are called annihilation and creation operators, respectively. From Equations (A.16) and (A.17) we infer that if n1, n2,...areanynon-negative integers, − / † † ··· 1 2 η n1 η n2 ···|⌽ ≡| , , ··· (n1! n2! ) ( 1) ( 2) 0 n1 n2 (A.18) is a normalized simultaneous eigneket of all the n belonging to the eigenvalues n1, n2, .... Variousketsobtainedbytakingdifferent n form a complete set of kets all orthogonal to each other. Following Dirac [1], we postulate that the quantum states for N bosons can be represented by a symmetric ket |α(1) |α(2) ···|α(N) ≡|α α ···α , S a b g a b g S (A.19) where S is the symmetrizing operator: 1 S ≡ √ P (A.20) N! P and P are permutation operators for the particle-indices (1, 2, ..., N). The ket in Equation (A.19) is not normalized but −1/2 (n1! n2! ···) |αaαb ···αg S ≡|{n} (A.21) is a normalized ket representing the same state. Comparing Equations (A.21) and (A.18), we obtain |α α ···α = η† η† ···η† |⌽ . a b g S a b g 0 (A.22) The unnormalized symmetric kets |αaαb ···αg S for the system can be con- η† η† ···η† |⌽ structed by applying N creation operators a b g to the vacuum ket 0 .So far we have tacitly assumed that the total number of bosons is fixed at N .Ifthis number is variable, we can easily extend the theory to this case. Let us introduce a Hermitean operator N defined by = η† η = , N a a na (A.23) a a A.2 Observables 231 the summation extending over the whole set of boson states. Clearly, the operator N haseigenvalues0,1,2,...,andtheket|αaαb ···αg S is an eigenket of N belonging to the eigenvalue N . We may arrange kets in the order of N , i.e., zero-particle state, one-particle states, two-particle states, ...: |⌽ ,η† |⌽ ,η† η†|⌽ , ··· . 0 a 0 a b 0 (A.24) These kets are all orthogonal to each other. Two kets referring to the same number of bosons are orthogonal as before, and two referring to different numbers of bosons are orthogonal because they have different eigenvalues N . By normalizing the kets, we get a set of kets like (A.21) with no restriction on {n}. These kets form the basic kets in a representation where {na} are diagonal. Problem A.1.1. (a) Show that nη†−η†n = η†, by taking the Hermitian-conjugation of Equation (A.10) and also by using Equations (A.9). (b) Use this relation and obtain a series of eigenvalues n, n + 1, n + 2, ..., where n is an eigenvalue of n. Problem A.1.2. Prove Equation (A.15) by mathematical induction. Hint: use Equa- tions (A.9). A.2 Observables We wish to express observable physical quantities (observables) for the system of identical bosons in terms of η and η† These observables are by postulate symmetric functions of the boson variables. An observable may be written in the form: y( j) + z(ij) +···≡Y + Z +··· , (A.25) j i j where y( j) is a function of the dynamic variables of the jth boson, z(ij) that of the dynamic variables of theith and jth bosons, and so on. ≡ ( j) ( j) |α( j) First, consider Y j y . Because y acts only on the ket of the jth boson, we obtain y( j) |α(1) |α(2) ···|α( j) ··· x1 x2 x j = |α(1) |α(2) ···|α( j) ··· α( j)|y( j)|α( j) . x1 x2 a a (x j ) a (A.26) α( j)| ( j) |α( j) ≡ α | |α The matrix element a y x j a y x j does not depend on the parti- cle index j since we consider identical bosons. Summing Equation (A.26) over all 232 Appendix A Second Quantization j’s and applying operator S to the result, we obtain SY |α(1) |α(2) ··· x1 x2 = S |α(1) |α(2) ···|α(1) |α( j) ··· α | y |α . x1 x2 x1 a a x j j a (A.27) Because Y is symmetric, we can replace SY by YSfor the lhs. After straightforward calculations, we obtain from Equation (A.27) Y η† η† ···|⌽ = η† η† ···η† η† η† ···|⌽ α | y |α x1 x2 0 x1 x2 x j−1 a x j+1 0 a x j j a = η† η† η† ···η† η† ···|⌽ δ α | y |α .