Second Quantization: Quantum Fields

Transcription

1 Second Quantization: Quantum Fields Bosons and Fermions Let X j stand for the coordinate and spin subscript (if any) of the j-th particle, so that the vector of state Ψ of N particles has the form Ψ Ψ(X 1, X 2,..., X N ). The postulate of indistinguishability states that the Hilbert space of particles of the same kind 1 is restricted to the states featuring permutational symmetry: Ψ(..., X i,... X j,...) = ±Ψ(..., X j,... X i,...), X i X j. (1) The sign-plus particles are called bosons and the sign-minus particles are called fermions. Single-particle-based basis states Let {φ s (X)} be an ONB of vectors of states for one particle. Then, the basis in the N-particle Hilbert space can be chosen as follows (fermions and bosons are labelled with the subscript F and B, respectively): Ψ F (X 1, X 2,..., X N ) = 1 N! Ψ B (X 1, X 2,..., X N ) = C φ s1 (X 1 ) φ s1 (X 2 )... φ s1 (X N ) φ s2 (X 1 ) φ s2 (X 2 )... φ s2 (X N ) φ sn (X 1 ) φ sn (X 2 )... φ sn (X N ) nontrv prm, (2) φ s1 (X 1 )φ s2 (X 2 ) φ sn (X N ). (3) In the fermionic case, we have a compact determinant form, the so-called Slater determinant. For bosons, the summation is over all non-trivial 2 permutations of pairs of coordinates. The bosonic normalization constant C can 1 Referred to as identical particles. 2 The permutation is trivial if it does not change the form of the term. This happens if and only if the the corresponding functions φ are the same. 1

2 be conveniently expressed in terms of the occupation numbers of corresponding single-particle states: C 2 = n 1! n 2! n 3! N!. (4) Problem 1. Derive Eq. (4) [after reading the next section where the notion of occupation numbers is introduced]. Fock nomenclature and Fock space With the representations (2) and (3) for the basis states, it is quite clear that each basis state is exhaustively characterized by the set of N single-particle states, {φ s1, φ s2,..., φ sn } involved in the N-particle (anty-)symmetric basis state. For fermions, all the N states must be different. 3 For bosons, there are no restrictions. Extremely powerful is the Fock nomenclature for the above-mentioned sets of single-particle states. In this approach, one globally enumerates all the single-particle states of a given set {φ s } and then uses a string of natural numbers, (n (1), n (2), n (3),...), such that n (s) = N, (5) s=1 to label all the N-particle basis states. The convention is that n (s) = 0, if the state φ s does not participate in the given basis state, while 1 n (s) N stands for the multiplicity of the state φ s. Obviously, for fermions, n (s) can be either 0 or 1. The next important step is to abandon the constraint (5) thus allowing the basis states with different numbers of particles. 4 This way we arrive at the Fock space. Each vector of the Fock space can be represented as a linear combination of the Fock basis states. Furthermore, each Fock basis vector can be viewed as a direct product of single-mode (Fock) basis vectors: n (1), n (2), n (3),... = n (1) n (2) n (3). (6) 3 Otherwise, the Slater determinant is identically equal to zero. 4 By definition, the basis states with different total numbers of particles are orthogonal to each other. 2

3 Factorization (6) is central for casting the theory of many-body identicalparticle systems in the form of quantum field theory (QFT). Elementary subsystems of which the quantum field consists are not the individual particles. Rather, these are the single-particle modes. For each mode s, there is a complete set of orthonormal states { n (s) }, n (s) 1 n (s) 2 = δ (s) n. (7) 1 n(s) 2 Recalling that n (s) = 0, 1 for fermions and n (s) = 0, 1, 2,... for bosons, we see that, nomenclature-wise, the Fock space of fermions can be viewed as a set of two-level systems, while the Fock space of bosons corresponds to that of a set of quantum harmonic oscillators. The analogy between the quantum field of bosons and a set of independent quantum harmonic oscillators becomes essentially direct for a gas of noninteracting particles. 5 Within the QFT picture, the number of particles is an observable rather than a fixed external parameter. The operator of the total number of particles, ˆN, is constructed as a sum of operators of occupation of numbers for each mode: ˆN = ˆn s, (8) s ˆn s n (s) = n (s) n (s). (9) In this formalism, the occupation numbers of the mode s are the eigenvalues of the operator ˆn s. Ideal gas Consider now a noninteracting system. Choose {φ s (X)} to be an ONB of single-particle energy eigenstates, with ɛ s corresponding single-particle energy. Then, in the Fock representation, the Hamiltonian reads: Ĥ = s ɛ sˆn s. (10) Note a qualitative difference between (10) and (8). While (10) requires that (i) the system be ideal and (ii) the set {φ s (X)} be the set of single-particle energy eigenstates, none of the two requirements is relevant to (8). 5 No matter whether the system of bosons is uniform or placed in an external potential. 3

4 Given that, by construction, different operators ˆn s commute with each other, we conclude that the gas of ideal fermions is equivalent to a set of non-interacting two-level systems, while the ideal bosons are equivalent to a set of non-interacting harmonic oscillators. In both cases, the single-particle energies ɛ s play the role of inter-level separations. Gibbs distribution. In particular, the above observation means that the (grand canonical) Gibbs statistical operator 6 factorizes into a product of (grand canonical) Gibbs statistical operators for each single-particle mode: e βĥ = s e βĥ s. (11) Indeed, Ĥ = Ĥ µ ˆN = s Ĥ s, Ĥ s = ɛ sˆn s, ɛ s = ɛ s µ. (12) This way we reduce the problem of equilibrium statistics of ideal fermions or bosons to the equilibrium statistics of (respectively) a two-level system or a quantum harmonic oscillator. In the Fock representation, an explicit expression for the statistical operator for the mode s is e βĥ s = n (s) e β ɛsn(s) n (s). (13) n (s) Up to a difference in the value of ɛ s, we have the same expression for each mode, so that the subscript s becomes redundant and we can write (ˆρ e βĥ s ) ˆρ = n e β ɛn n. (14) n The normalization constant 7 z = Tr ˆρ = e β ɛn (15) has a very deep statistical-mechanical meaning. This is the so-called partition function of the single mode. The partition function, Z, for the whole (ideal gas) system is then nothing but a product (here we restore the mode label) Z = s n z s. (16) 6 For clarity, we omit the normalization factor. 7 The normalization rule is ˆρ ˆρ/z. 4

5 Formula (16) is central for the statistical mechanics of ideal Fermi and Bose gases. 8 Problem 2. Calculate single-mode z as a function of (ɛ, µ, T ). The average occupation number of the mode, is given by n = 1 z Tr ˆρ ˆn = 1 z 1 n = e β ɛ n = e β ɛ 1 n e β ɛn n (Fermions), (17) (Bose). (18) Problem 3. Derive (17) and (18). Hint. The following mathematical trick might be useful. n e λn = e λn. λ n n In the limit β ɛ 1 (that is ɛ T ), the difference between bosons and fermions disappears, since the contribution of n > 1 is exponentially suppressed: ˆρ e ɛ/t 1, n e ɛ/t ( ɛ T ). (19) This ultra quantum (in terms of the statistics of occupation numbers) regime actually corresponds to the regime of Boltzmann gas of classical particles! Problem 4. Show that n s = e ɛs/t corresponds to the Maxwell distribution of classical particles, once s is understood as the momentum eigenvalue, 8 In statistical mechanics, one extracts all the thermodynamic quantities from Z, by simply calculating partial derivatives of the function T ln Z with respect to temperature, or/and chemical potential, or/and volume. 5

6 s p. Note that it is not sufficient to simply identify the exponential with Maxwell s exponential. It is equally important to demonstrate that the summation over s can be replaced with the integration dp; that is, it is crucial to make sure that there is no nontrivial Jacobian associated with going from s(...) to (...)dp. Hint. Since, in a macroscopic system, the answer is independent of the system s shape and the boundary conditions, it makes sense to choose the most convenient ones. Namely, a finite-size box of rectangular shape with periodic boundary conditions. For bosons, there is yet another classical limit. It takes place at ɛ T and corresponds to the classical limit of corresponding harmonic oscillator. In terms of the QFT perspective, this limit corresponds to the classical-field behavior of corresponding mode. Second quantized form of operators I: Statement of the problem Consistently with the distinguishability postulate, all the relevant operators/observables should be permutational symmetric. The class of permutational symmetric operators/observables further splits into single-particle, ˆF (1), two-particle, ˆF (2), etc., subclasses: ˆF (1) = N j=1 ˆf (1) (X j ), ˆF (2) = 1 2 N j=1 N k=1( j) ˆf (2) (X j, X k ),... (20) Second quantization is the formalism allowing one to represent the operators ˆF (1), ˆF (2),..., etc. directly within the Fock space, without explicitly resorting to their N-body X-dependence. Second quantized form of operators II: bosons By explicit calculation, we find that the only non-zero off-diagonal 9 matrix elements for the operator ˆF (1) are the following ones: n (s), (n (r) 1) ˆF (1) (n (s) 1), n (r) = f sr (1) n (s) n (r), (21) 9 That is, connecting two different states. 6

7 f (1) sr = φ s(x) ˆf (1) (X) φ r (X) dx. (22) In the two different Fock states of Eq. (21), we explicitly show only those occupation numbers that are different between the two states. The integration over dx is understood as integration over dr and summation over the spin subscript (if any). For the diagonal matrix elements of the operator ˆF (1), we find ˆF (1) = s f (1) ss n (s). (23) A crucial observation about Eqs. (21) (23) is that (i) the explicit dependence on N disappears and (ii) the X-dependence enters only the single-particle matrix element (22). Problem 5. Derive Eqs. (21) (23). Creation and annihilation operators. The annihilation operator, ˆb s, (of the mode s) is defined as ˆbs n (s) = n (s) n (s) 1. (24) n (s) 1 ˆb s n (s) = n (s). (25) The creation operator, ˆb s, (of the mode s) is defined as a Hermitian conjugate to ˆb s. We thus have n (s) ˆb s n (s) 1 = n (s) 1 ˆb s n (s) = n (s), (26) ˆb s n (s) = n (s) + 1 n (s) + 1. (27) From (24) and (27) we then readily see that ˆb s ˆbs = ˆn (s), (28) ˆbsˆb s = ˆn (s) + 1. (29) With (24), (27), (28), and (29), we find the commutation relations 10 [ˆb s, ˆb r ] = 0, [ˆb s, ˆb r] = 0, (30) 10 It is very instructive to compare those relations to the Poisson brackets of complex canonical variables in classical mechanics, replacing Hermitian conjugation with complex conjugation and replacing the commutator with the Poisson bracket. 7

8 [ˆb s, ˆb r] = δ sr. (31) Now we can straightforwardly check that, in terms of the operators {ˆb s } and {ˆb s} acting directly in the Fock space, we have ˆF (1) = s,r f (1) sr ˆb s ˆb r. (32) The few-body interactions have similar structure. In particular, ˆF (2) = 1 2 s,r,q,t f (2) sr;qt ˆb s ˆb r ˆb q ˆbt, (33) where f (2) sr;qt = φ s(x 1 ) φ r(x 2 ) ˆf (2) (X 1, X 2 ) φ q (X 2 ) φ t (X 1 ) dx 1 dx 2. (34) As a by-product of establishing relations (32), we can readily establish a law of transformation of creation and annihilation operators when changing the single-particle ONB generating the Fock basis. Indeed, suppose we go from one single-particle basis to the other by the unitary transformation φ s = r U sr φr, φs = r U rsφ r, (35) where U is the unitary matrix of the transformation. Then, the matrix elements (22) get transformed with the matrix U, and for the form (32) to stay invariant, 11 the creation/annihilation operators have to be transformed as ˆb s = U srˆ b r, ˆ b s = Ursˆb r, (36) r r ˆbs = r U srˆ br, ˆ bs = r U rsˆbr. (37) Field operator. The field operator, ˆψ(X), is introduced as follows: ˆψ(X) = s ˆbs φ s (X). (38) 11 Note that Eq. (32) was derived for any basis. 8

9 It is also useful to introduce ˆψ (X) [ ˆψ(X)] + = s ˆb s φ s(x). (39) A prominent feature of the operator ˆψ(X) is that it is invariant with respect to the transformation (35) (37). Also worth mentioning are the commutation relations following directly from the definition (38): [ ˆψ(X 1 ), ˆψ(X 2 )] = [ ˆψ (X 1 ), ˆψ (X 2 )] = 0, (40) [ ˆψ(X 1 ), ˆψ (X 2 )] = s φ s (X 1 )φ s(x 2 ) = δ(r 2 r 1 ) δ σ1 σ 2, (41) where we recall that X (r, σ), with σ the spin subscript (for spineless particles, δ σ1 σ 2 should be omitted). With the field operator, the expressions for ˆF (1), ˆF (2),..., etc. can be cast into a compact, basis-invariant (quantum-field-theoretical) form: ˆF (1) = ˆψ (X) ˆf (1) (X) ˆψ(X) dx, (42) ˆF (2) = 1 2 ˆψ (X 1 ) ˆψ (X 2 ) ˆf (2) (X 1, X 2 ) ˆψ(X 2 ) ˆψ(X 1 ) dx 1 dx 2. (43) The particle (number) density operator, ˆn(r), has the form ˆn(r) = σ ˆψ (X) ˆψ(X). (44) Indeed, in the original N-particle coordinate representation, the operator ˆn(r) has the ˆF (1) -type form with a delta-functional ˆf (1) : ˆn(r) = N δ(r r j ). (45) j=1 The rest follows from (42). Integrating ˆn(r) over r, we are supposed to get the total number of particles. Hence, the operator of the total number of particles is ˆN = ˆψ (X) ˆψ(X) dx. (46) 9

10 Second quantized form of operators III: fermions In the case of fermions, we proceed similarly to what we have done for bosons. Here, however, we have to deal with one (annoying) circumstance. Namely, the fact that the sign of the basis wave function (2) depends on the order in which single-particle functions are listed in the determinant. Swapping any two of them, φ s φ r, changes the global sign of the wave function (2). Hence, to unambiguously fix the sign of the matrix elements between different Fock states, we have to fix the signs of the functions (2). To this end, we rely on the above-introduced global enumeration of all the modes, s = 1, 2, 3,..., and require that corresponding single=particle functions in the determinant (2) be ordered is such a way that s 1 < s 2 < s 3 <... < s N. (47) With this enumeration, the following integer-valued function defined for a given Fock basis state proves relevant { max(s,r) 1 Θ(s, r) = Θ(r, s) = q=min(s,r)+1 n(q), if s r > 1, (48) 0, otherwise. The meaning of Θ(s, r) is very simple. It is the sum of the occupation numbers of all the modes (if any) between the modes s and r, presuming the ordering of modes by the above-mentioned enumeration. An analog of (21) then is 1 s, 0 r ˆF (1) 0 s, 1 r = f sr (1) ( 1) Θ(s, r), (49) where 0 s stands for n (s) = 0 and 1 s stands for n (s) = 1. It is not surprising that, apart from the sign factor, the result (49) corresponds to (21), with the occupation numbers not exceeding unity. To verify the sign factor in (49), observe that, if the two Fock basis functions between which the matrix element is calculated were related to each other by simply replacing φ s φ r in the Slater determinant (2), then no sign would occur. 12 In accordance with our ordering convention, the state 0 s, 1 r can be obtained from the state 1 s, 0 r in two steps: (i) replacing φ s φ r and (ii) assigning the sign ( 1) Θ(s, r) to compensate for the wrong 13 position of the row [φ r (X 1 ), φ r (X 2 ),..., φ r (X N )] in the determinant (2). 12 Because each term in the bra state would have the same sign as its counterpart in the ket state. 13 Speaking generally. 10

11 For the diagonal element, we get the same expression (23). Creation and annihilation operators for fermions. The annihilation and creation operators (of the mode s), â s and â s, are defined as operators having (only) the following non-zero matrix elements 0 s â s 1 s = 1 s â s 0 s = ( 1) γs, (50) γ s = { s 1 r=1 n (r), if s > 1, 0, if s = 1. (51) There is a qualitative difference with the bosonic case, where the annihilation and creation operators are defined for each single-particle mode individually. This is not true for fermions. Now the sign of the matrix element depends on the occupation numbers of the other modes, and the action of creation/annihilation operators of different modes is not independent! The crucial properties of creation/annihilation operators are as follows. and 1 s, 0 r â s â r 0 s, 1 r = ( 1) Θ(s, r), (52) 1 s, 0 r â r â s 0 s, 1 r = ( 1) Θ(s, r), (53) â s â s = ˆn (s), (54) â s â s = 1 ˆn (s). (55) Combining (52) and (54) with (49) and (23) [it is readily seen that (23) works for Fermions as well], we arrive at the fermionic analog of (32): ˆF (1) = s,r f (1) sr â s â r. (56) Combining (52) and (53), we find the following anti-commutation relation â s â r + â r â s = 0 (s r). (57) From (54) and (55) we see that â s â s + â s â s = 1, (58) 11

12 so that (57) and (58) can be combined into a single relation â s â r + â r â s = δ sr. (59) Similarly, starting from the definition (50), we can establish yet another set of anti-commutation relations implying, in particular, â s â r + â r â s = 0, â s â r + â râ s = 0, (60) â s â s = 0, â s â s = 0. (61) Problem 6. Starting from the definition (50), derive (52) through (55) and also (60). It is not only Eq. (56) that exactly reproduces its bosons counterpart. The same situation takes place for all the formulas derived for bosons in terms of creation and annihilation operators, as well as in terms of the field operators ˆψ(X) = s â s φ s (X), ˆψ (X) = s â s φ s(x), (62) with one obvious exception for the commutation properties of the field operators. Given the anti-commutative properties of fermionic creation and annihilation operators, we readily see that fermionic analogs of bosonic commutators of field operators are the following anti-commutators (below {Â, ˆB} stands for  ˆB + ˆBÂ): { ˆψ(X 1 ), ˆψ(X 2 )} = { ˆψ (X 1 ), ˆψ (X 2 )} = 0, (63) { ˆψ(X 1 ), ˆψ (X 2 )} = s φ s (X 1 )φ s(x 2 ) = δ(r 2 r 1 ) δ σ1 σ 2. (64) The only subtlety is that now because of the anti-commutativity one should be extremely careful with the order of operators in fermionic analogs of Eqs. (33) and (43). Being aware of this circumstance, we made sure that the order of bosonic operators in Eqs. (33) and (43) is such that their fermionic 12

13 analogs are obtained simply by replacing ˆb â, no minus sign emerging. Problem 7. Based on (anti-)commutation relations for creation/annihilation operators of bosons and fermions, prove Eqs. (40), (41), (63), and (64). The transformation (35) of the single-particle ONB implies corresponding (the very same as for bosons) transformation of the creation/annihilation operators: â s = U srˆã r, ˆã s = Ursâ r, (65) r r â s = r U srˆã r, ˆã s = r U rs â r. (66) Problem 8. Make sure that the transformations (36) (37) and (65) (66) preserve corresponding (anti-)commutation properties of the creation/annihilation operators. Axiomatic quantum field theory We introduce quantum field operators ˆψ(X) = s â s φ s (X), ˆψ (X) [ ˆψ(X)] + = s â s φ s(x), (67) such that {φ s (X)} is a certain single-particle ONB, and the (annihilation and creation) operators {â s } and {â s (â s ) + } are postulated to obey either bosonic or fermionic (anti-)commutation rules: [â s, â r ] = 0, [â s, â r] = δ sr (bosons), (68) {â s, â r } = 0, {â s, â r} = δ sr (fermions). (69) The axioms imply corresponding (anti-)commutational properties of the field operators. The representational invariance of the axiomatic quantum field theory is established by the observation that the transformation (36) (37), on one hand, preserves the bosonic and fermionic (anti-)commutational properties, and, on the other hand, preserves the structure of the field operators, 13

14 if accompanied by corresponding transformation (35) of the single-particle ONB. The Fock space is defined as a space in which the creation/annihilation operators act. A special axiom postulates the existence of the vacuum state, 14 vac, as the (unique) state featuring the following property: â s vac = 0, s. (70) The Fock ONB then is defined as the set of the following vectors obtained by acting with the creation operators on the vacuum state: n (1), n (2), n (3),... = (â 1) n(1) (â 2) n(2) (â 3) n(3) n (1)! n (2)! n (3)! vac. (71) The integers n (s) = 0, 1, 2,... are called occupation numbers of the mode s. A special theorem (Problem 9) establishes the fact that the states (71) are orthonormalized. For fermions, the anticommutation axiom (69) implies â sâ s = 0, so that the occupation numbers can be either 0 or 1 (Pauli principle), simplifying (71) to n (1), n (2), n (3),... = [ (â 1) n(1) (â 2) n(2) (â 3) n(3) ] vac (fermions). (72) With fermions, one can further simplify notation by mentioning only nonzero occupation numbers: 1 s1, 1 s2,..., 1 sn = â s 1 â s 2 â s N vac, s 1 < s 2 <... < s N. (73) For bosons, it is very instructive to observe that the axioms imply that the Fock space and the algebra of creation/annihilation operators reduce to those of independent harmonic oscillators. Problem 9. Prove that (i) each state (71) is normalized to unity and (ii) any two different states (71) are orthogonal. Hint. It is helpful to derive and then use the following formula â s (â s) m = m(â s) m 1 + (â s) m â s (bosons). (74) 14 Not to be confused with the zero vector. 14

15 The relations (24) through (29) and (50) through (55) are derived from axioms. Problem 10. Perform the above-mentioned derivations. 15