Second Quantization∗ Jörg Schmalian May 19, 2016 1 The harmonic oscillator: raising and lowering operators Lets first reanalyze the harmonic oscillator with potential m!2 V (x) = x2 (1) 2 where ! is the frequency of the oscillator. One of the numerous approaches we use to solve this problem is based on the following representation of the momentum and position operators: r x = ~ ay + a b 2m! b b r m ! p = i ~ ay − a : (2) b 2 b b From the canonical commutation relation [x;b pb] = i~ (3) follows y ba; ba = 1 y y [ba; ba] = ba ; ba = 0: (4) Inverting the above expression yields rm! i ba = xb + pb 2~ m! r y m! i ba = xb − pb (5) 2~ m! ∗Copyright Jörg Schmalian, 2016 1 y demonstrating that ba is indeed the operator adjoined to ba. We also defined the operator y Nb = ba ba (6) which is Hermitian and thus represents a physical observable. It holds m! i i Nb = xb − pb xb + pb 2~ m! m! m! 2 1 2 i = xb + pb − [p;b xb] 2~ 2m~! 2~ 2 2 1 pb m! 2 1 = + xb − : (7) ~! 2m 2 2 We therefore obtain 1 Hb = ! Nb + : (8) ~ 2 1 Since the eigenvalues of Hb are given as En = ~! n + 2 we conclude that the eigenvalues of the operator Nb are the integers n that determine the eigenstates of the harmonic oscillator. Nb jni = n jni : (9) y Using the above commutation relation ba; ba = 1 we were able to show that p a jni = n jn − 1i b p y ba jni = n + 1 jn + 1i (10) y The operator ba and ba raise and lower the quantum number (i.e. the number of quanta). For these reasons, these operators are called creation and annihilation operators. 2 second quantization of noninteracting bosons While the above results were derived for the special case of the harmonic oscil- lator there is a similarity between the result 1 E = ! n + (11) n ~ 2 for the oscillator and our expression X Efnpg = "pnp (12) p for the energy of a many body system, consisting of non-interacting indistin- guishable particles. While n in case of the oscillator is the quantum number label, we may alternatively argue that it is the number of oscillator quanta in the oscillator. Similarly we can consider the many body system as a collection 2 of a set of harmonic oscillators labelled by the single particle quantum number p (more generally by p and the spin). The state of the many body system was characterized by the set fnpg of occupation numbers of the states (the number of particles in this single particle state). We the generalize the wave function jni to the many body case jfnpgi = jn1; n2; :::; np; :::i (13) and introduce operators p bap jn1; n2; :::; np; :::i = np jn1; n2; :::; np − 1; :::i y p bap jn1; n2; :::; np; :::i = np + 1 jn1; n2; :::; np + 1; :::i (14) That obey h y i bap; bap0 = δp;p0 : (15) It is obvious that these operators commute if p 6= p0. For p = p0 follows y p bapbap jn1; n2; :::; np; :::i = np + 1bap jn1; n2; :::; np + 1; :::i = (np + 1) jn1; n2; :::; np; :::i (16) and y p y bapbap jn1; n2; :::; np; :::i = npbap jn1; n2; :::; np − 1; :::i = np jn1; n2; :::; np; :::i (17) y y which gives bapbap − bapbap = 1. Thus the commutation relation follow even if the operators are not linear combinations of position and momentum. It also follows y nbp = bapbap (18) for the operator of the number of particles with single particle quantum number P y p. The total number operator is Nb = p bapbap. Similarly, the Hamiltonian in this representation is given as X y Hb = "pbapbap (19) p which gives the correct matrix elements. We generalize the problem and analyze a many body system of particles with single particle Hamiltonian p2 bh = b + U (r) (20) 2m b which is characterized by the single particle eigenstates bh jφαi = "α jφαi : (21) 3 α is the label of the single particle quantum number. We can then introduce the occupation number representation with jn1; n2; :::; nα; :::i (22) h y i and corresponding creation and destruction operators baα; baα0 = δα,α0 . We can then perform a unitary transformation among the states X X jβi = Uβα jαi = jαi hαjβi (23) α α The states jβi are in general not the eigenstates of the single particle Hamil- tonian (they only are if Uβα = hαjβi = δαβ). We can nevertheless introduce creation and destruction operators of these states, that are most naturally de- fined as: X baβ = hβjαi baα (24) α and the corresponding adjoined equation y X ∗ y baβ = hβjαi baα: (25) α This transformation preserves the commutation relation (see below for an ex- ample). We can for example chose the basis β as the eigenbasis of the potential. Then holds in second quantization X y Ub = hβ jU (r)j βi aβaβ (26) β and we can transform the result as X 0 y Ub = hαjβi hβ jU (r)j βi hβjα i baαbaα0 β,α,α0 X 0 y = hα jU (r)j α i baαbaα0 (27) α,α0 0 3 It holds of course hα jU (r)j α i = d rφα (r) U (r) φα0 (r). In particular, we can chose jβ´i = jri such that hβjαi = hrjαi = φα (r). In this case we use the notation bar = b(r) and our unitary transformations are X b(r) = φα (r) baα α y X ∗ y b (r) = φα (r) baα (28) α 4 The commutation relation is then δα,α0 h 0 i X ∗ 0 h y i b(r) ; b(r ) = φα (r) φα0 (r ) baα; baα0 α,α0 X ∗ 0 X 0 = φα (r) φα (r ) = hrjαi hαjr i α α = hrjr0i = δ (r − r0) (29) and it follows Ub = d3rU (r) by (r) b(r) (30) ˆ Similarly holds for the kinetic energy 2 ~ 3 3 0 2 0 y 0 Tb = − d rd r r r r b (r) b(r ) 2m ˆ 2 ~ 3 3 0 y 2 0 0 = − d rd r b (r) r δ (r − r ) b(r ) 2m ˆ 2 ~ 3 y 2 = − d r b (r) r b(r) (31) 2m ˆ Thus we find X y H = "αbaαbaα α 2 2 3 y ~ r = d r b (r) − + U (r) b(r) (32) ˆ 2m With the help of the field operators b(r) and by (r) is it possible to bring the many body Hamiltonian in occupation number representation into the same form as the Hamiltonian of a single particle. 2.1 Example 1: a single particle We consider the most general wave function of a single spinless boson: 3 y j αi = d rφα (r) b (r) j0i (33) ˆ where j0i is the completely empty system. Let the Hamiltonian be 2 2 3 y ~ rr H = d r b (r) − + U (r) b(r) (34) ˆ 2m It follows 2 2 3 3 0 y ~ rr 0 y 0 H j αi = d r d r b (r) − + U (r) φα (r ) b(r) b (r ) j0i ˆ ˆ 2m 2 2 3 3 0 y ~ rr 0 y 0 = d r d r b (r) − + U (r) φα (r ) b (r ) b(r) j0i ˆ ˆ 2m 2 2 3 3 0 y ~ rr 0 0 + d r d r b (r) − + U (r) φα (r ) δ (r − r ) j0i(35) ˆ ˆ 2m 5 The first term disappears since b(r) j0i = 0 for the empty state. Performing the integration over r0 gives 2 2 3 y ~ rr H j αi = d r b (r) − + U (r) φα (r) j0i ˆ 2m 2 2 3 ~ rr y = d r − + U (r) φα (r) b (r) j0i (36) ˆ 2m Thus, we need to find the eigenvalue of and eigenfunction of 2r2 −~ + U (r) φ (r) = " φ (r) (37) 2m α α α to obtain 3 y H j αi = "α d rphiα (r) b (r) j0i = "α j αi : (38) ˆ Thus, for a single particle problem we recover the original formulation of the "first quantization". The function φ (r) in Eq.33 is therefore the wave function of the single particle problem. y P ∗ y y 3 y Using b (r) = φα (r) baα follows baα = d rφα (r) b (r) and our above α ´ wave function is nothing but y j αi = baα j0i (39) Applying the Hamiltonian to the wave function in this basis is obviously giving the same answer. X y y H j αi = "α0 baα0 baα0 baα j0i = "a j αi (40) α0 3 Second quantization of interacting bosons Next we analyze the formulation of particle-particle interactions within the sec- ond quantization. We consider a two body interaction Vb that has, by definition, matrix elements that depend on the states of two particles. Thus the expression for a single particle where X 0 y Ub = hα jUj α i baαbaα0 (41) α,α0 will be determined by a matrix elements of the kind: 0 0 3 3 0 ∗ ∗ 0 0 0 hαγ jV j α γ i = d rd r φ (r) φ (r ) V (r; r ) φ 0 (r ) φ 0 (r) : (42) ˆ α γ α γ In general there will be a two particle basis jαγi where the interaction is diagonal Vb jαγi = Vαγ jαγi (43) 6 where Vαγ = hαγ jV j αγi. In this basis we can proceed just like for the interac- P y tion Ub, where the operator was given by α hα jUj αi baαbaα.