Second Quantization Method for Bosons

Transcription

1 Second Quantization Method for Bosons Hartree-Fock-based methods cannot describe the effects of the classical image potential (cf. fig. 1) because HF is a mean-field theory. DFF-LDA is not able either to describe the long range polarization of the electron gas because of the localization of the exchange-correlation hole (in the LD approximation) at the position of the electron. Figure 1: Normal-incidence IPS of a clean Au(110)-(1 2) and Ag(110)-(1 1) displaying the n 1, 2 image states on the gold surface and n 2 on the silver surface (G. Binnig et al. Phys. Rev. Lett. 55, 991 (1985)). The long range correlated movement of electrons in solids in response to the presence of an external electron can be described via: Electron-hole pair excitations Plasmon excitation The plasmons and electron-hole pairs are quantized and they behave like bosons (zero spin). To describe them theoretically we can, therefore, use the quantum theory of the harmonic oscillator.

2 Plasmons: the harmonic quantum oscillator in the second quantization approach Requirements for the description of complex systems within the harmonic oscillator model: The interaction forces between the particles vary linearly with the separation between the particles. The potential energy is a quadratic function of the interparticle distance V (x) = 1 2 kx2. Applications of the harmonic oscillator model: molecule vibrations electromagnetic field (photons) crystal vibrations (phonons) bosons (particles with integer spin) elementary excitations in the solid state (plasmons, spin waves) spectroscopic transitions Cooper pairs (rough approximation), He 4 nuclei

3 Figure 2: The harmonic oscillator: potential energy as a function of the displacement. V (x) = 1 2 kx2 The meaning of the coordinate depends on the practical problem. The interaction force: F (x) = d V (x) = kx dx (k-force constant) is then a linear function of the coordinate. Consider the motion of a particle of mass m bound elastically to its position of equilibrium. As particle coordinate q we take the displacement from the position of equilibrium (fig. 2), let the corresponding momentum be p. The classical equation of the oscillator periodic motion in one dimension: F (q) = kq

4 m d2 q = kq dt 2 q(t) = Asinωt d 2 q = Aω 2 sinωt = ω 2 q dt 2 mω 2 q = kq k = mω 2 ω: the frequency of the oscillator k: the force constant F (q) = mω 2 q V (q) = q 0 F (q)dq V (q) = m 2 ω2 q 2 = k 2 q2 if at the equilibrium position V (q = 0) = 0 The classical Hamilton function: H = T + V = p2 2m + m 2 ω2 q 2 (1) Replace the momentum p by an operator: p = h i to enter the quantum mechanical description with the help of Schrödinger s equation: d 2 d dq Hψ(q) = Eψ(q) h2 2m dq 2ψ(q) + m 2 ω2 q 2 ψ(q) = Eψ(q) (2)

5 The commutation relation of the operators of momentum and position coordinate: pq qp = [p, q] = h i (3) is valid for any function which can be differentiated. h i d dq (qψ) q h i h i ψ + q h i d dq ψ q h i d dq ψ = h i ψ d dq ψ = h i ψ Introduce a dimensionless coordinate ξ in eqn. (2) defined by: q = h mω ξ dξ dq = mω h The operator of momentum can be rewritten as: p = h d i dq = h d i dξ p = h i dξ dq mω h d dξ (4)

6 Rewrite Schrödinger s equation introducing eqn. (4) in eqn. (1) as: hω 2 d2 dξ + 2 ξ2 ψ(ξ) = Eψ(ξ) (5) If we were now able to treat the operators ξ and d dξ as ordinary numbers, it might well appear that the operator on the left-hand side of eqn. (5): d2 dξ 2 + ξ2 could be regarded as an expression of the type ( α 2 + β 2 ) ( α + β)(α + β) Just as an experiment, let us replace the left-hand side of eqn. (5) by: hω 1 d 2 dξ + ξ d dξ + ξ 1 2 ψ Multiplying out the brackets, but taking care to keep the operators ξ and d dξ in the right order, we obtain: = hω 2 hω 1 d 2 dξ + ξ d dξ + ξ d2 dξ + 2 ξ2 ψ(ξ) hω ψ d dξ ξ ξ d ψ(ξ) (6) dξ Here the first expression is the same as the left-hand side of eqn. (5).

7 The commutation relation: d dξ ξ ξ d dξ = 1, (7) retrieved by introducing q = p = h i h mω ξ d dq (8) in the commutation relation eqn. (3), is used to reduce the second term of eqn. (6) to: hω 2 ψ Proof of the commutation relation eqn. (7): h i pq qp = h i q = h mω ξ d dq = mω d h dξ p = h d i dq = h mω i h mω d h h dξ mω ξ h ξ h mω mω i h d dξ ξ ξ d dξ d dξ d dξ = h i = 1 q.e.d. (9)

8 New operators: 1 d 2 dξ + ξ = b (10) 1 d 2 dξ + ξ = b (11) As eqn. (6) and the left-hand side of Schrödinger s equation (5) differ by the term hω 2 ψ, we introduce the displaced energy: E = E hω 2 (12) Schrödinger s equation (5) can finally be replaced by: hωb bψ = E ψ. (13) Commutation relation for the operators b and b, bb b b by substituting the operators according to eqs. (10) and (11): bb b b = 1 d 2 dξ + ξ 1 d 2 dξ + ξ 1 d 2 dξ + ξ 1 d 2 dξ + ξ = d dξ ξ ξ d dξ = 1 bb b b = 1 (14) taking into account eqn. (7). Operators satisfying commutation relation eqn. (14) are called Bose operators.

9 Construct eigenstates of the harmonic oscillator with the aid of operators b and b and the commutation relation eqn. (14): assume that the energy of the quantum mechanical oscillator is always positive and has a lower bound: ψ o is the state with the lowest energy value E o. Multiplying equation: hωb bψ o = E oψ o (15) from the left by the operator b, we obtain: hω(bb b)ψ o = E obψ o (16) hω(1 + b b)bψ o = E obψ o (17) hωb b(bψ o ) = (E o hω)(bψ o ). (18) Hence, bψ o is a new eigenfunction with the eigenvalue E o hω, contrary to the assumption that ψ o is the lowest state. This contradiction can be resolved only if: bψ o = 0 (19) d dξ + ξ ψ o = 0, (20) a differential equation for ψ o with the solution: ψ o (ξ) = Ce ξ2 /2 (21) C: normalization costant of ψ o.

10 Harmonic oscillator wavefunctions ψ n (x) = N n H n (x)e x2 2, n = 0, 1, 2... Hermite polynomials: H n (x) = ( 1) n e x2 ( d dx )n e x2 N n = (απ n n!) 1 2 α = h mω ω = k m Table 1: n H n (ξ) ξ 2 4ξ ξ 3 12ξ 4 16ξ 4 48ξ ξ 5 160ξ ξ 6 64ξ 6 480ξ ξ (a) (b) Figure 3: The wavefunctions of a harmonic oscillator for n up to 6. (a) even values, (b) odd values.

11 Solve eqn. (13): hωb bψ = E ψ without using differential equations, and thus try to find the wavefunctions and energy values using nothing but algebra. Multiply eqn. (13) by b on the left: ( hωb b b)ψ = E b ψ (22) b b = bb 1 hωb (bb 1)ψ = E b ψ (23) hωb b(b ψ) = (E + hω)(b ψ). (24) Clearly if ψ is an eigenfunction, b ψ is also an eigenfunction, and its eigenvalue is larger than the eigenvalue of the former function by an amount hω. This means that the operator b transforms the eigenfunction ψ in another eigenfunction b ψ which has a higher energy by the amount hω, i.e. b creates an energy quantum hω and is therefore called a creation operator. Apply the operator b to the ground state wavefunction ψ o n times to obtain the n-th excited eigenfunction: ψ n = (b ) n ψ o (25) whose eigenvalue is larger than the eigenvalue of the ground state eigenfunction by an amount n hω.

12 Figure 4: Creation and annihilation operators of bosons have the effect of an upward or downward movement along the ladder of states Similarly, we can multiply eqn. (13) by b on the left: hωbb bψ = E bψ (26) hωbb (bψ) = hω(1 + b b)(bψ) hωb b(bψ) = (E hω)(bψ) (27) meaning that, if ψ is an eigenfunction, then bψ is also an eigenfunction, with hω lower energy than the eigenenergy of ψ. This is, of course, true if ψ is not the lowest energy state. When b operates on ψ the eigenfunction is transformed into a new eigenfunction with hω lower energy, i.e. b annihilates a quantum of energy, so that b is called an annihilation operator (fig. 4).

13 The free constant coefficient is chosen to warrant the normalization of ψ n (cf. Appendix 1): + ψ n(ξ)ψ n (ξ)dξ = 1. (28) Appendix 1: The normalization factor of the oscillator function We want to show that the normalization factor N n of the n-th eigenfunction of a harmonic oscillator, given that ψ o is already normalized: ψ n = N n (b ) n ψ o (29) equals: N n = 1 n! (30) This will be proved by induction: 1. Assumption eqn. (30) is certainly correct for n = 0 in line with the assumption for normalized ψ o 2. Assume eqn. (30) has been proved for n n o, i.e. ψ no has already been normalized. 3. We show that it is also valid for n = n o + 1 ψ no +1 = N no +1b ψ no (31) where N no +1 is a constant to be determined and ψ no is assumed normalized. The equation for normalization is: ψ no +1 ψ no +1 = 1 (32) By substituting ψ no 1 we have: N no +1 2 b ψ no b ψ no = N no +1 2 ψ no bb ψ no = N no +1 2 ψ no (b b + 1)ψ no = N no +1 2 (n o + 1) ψ no ψ no (33) The last equation follows from the fact that ψ no is an eigenfunction of the operator b b with eigenvalue equal to n o. Hence: N no +1 2 (n o + 1) = 1 (34)

14 and, if we choose N no +1 to be real, then: N no +1 = 1 no + 1 (35) The validity of the first line of eqn. (33) follows from the following considerations. We consider two wavefunctions of the coordinate ξ: χ(ξ) and φ(ξ), which tend to zero at infinity, and form the integral: + (bφ )χdξ = 1 [( + ξ + d ) ] φ χdξ (36) 2 dξ By integrating the second term by parts: ( ) + d + dξ φ χdξ = φ χ + φ d dξ χdξ which gives (with the first term in the last equation equal to zero, because the wavefunctions of the harmonic oscillator vanish at infinity): [( 1 + ξ + d ) ] φ χdξ = dξ 2 The operator in the brackets is identical to b : ( 1 ξ d ) 2 dξ Hence: + In a similar way we can prove that: + and, by using the method n times, that: + (bφ )χdξ = (b φ )χdξ = [(b ) n φ ]χdξ = + + [( φ ξ d ) ] χ dξ (37) dξ b. (38) + φ (b χ)dξ (39) φ (bχ)dξ (40) φ (b n χ)dξ (41) Using eqn. (35) for the normalization constant of the n o + 1 wavefunction we have: ψ no +1 = 1 (n o + 1)! (b ) n o+1 ψ o (42) Similarly, the following formulae can be proved: ψ n+1 b ψ n = n + 1 (43) ψ n 1 bψ n = n (44) ψ m ψ n = δ nm (45)

15 The normalized eigenfunctions: ψ n = 1 n! (b ) n ψ o. (46) What are the corresponding eigenvalues? Compare this with the general equation Obviously E o = 0. bψ o = 0 (47) b bψ o = 0 (48) hωb bψ = E ψ. (49) As the n-fold operation of b on ψ o increases the energy E o n times by hω, the energy value corresponding to ψ n is: E n = n hω (50) or, in terms of the original energy scale on which Schrödinger s equation (2) is based: E n = E n + hω 2 E n = n + 1 hω, (51) 2 which corresponds to Schrödinger s equation for the n-th excited state: hω(b nb n )ψ n = E n ψ n. (52)

16 Eqn. (50) may be taken as meaning that in the n-th state there are n energy quanta of magnitude hω. By applying b to ψ n, the eigenfunction is changed to another eigenfunction with hω higher energy, i.e. the number of quanta is increased by one, one additional quantum is created by applying the creation operator b. Schrödinger s equation for the n-th excited state: hωb b(b ) n ψ o hωb bψ n = E n(b ) n ψ o = n hωψ n b bψ n = nψ n (53) and we realize that the number of energy quanta n in the n-th excited state is the eigenvalue of the operator b b. How to create the higher wavefunctions of the harmonic oscillator with the help of the creation operators b? b = 1 d 2 dξ + ξ b = mω 2 h q h 2mω d dq (54) ψ 1 = b ψ o = N o b e mω = N o mω 2 h q + = N 2mω o h 2 h q2 h mω 2mω h q e mω 2 h q2 qe mω 2 h q2 (55)

17 Then ψ 1 is normalized. Generally: where H n are the Hermite polynomials. mω ψ n = N n H n 2 h qe mω 2 h q2 (56) The operation of b b on the boson wavefunctions: ψ n = 1 n! (b ) n ψ o = 1 (n 1)! 1 n (b ) n 1 b ψ o = 1 n b ψ n 1 b ψ n 1 = nψ n b ψ n = n + 1ψ n+1 (57) Similarly for b: bψ n = nψ n 1 (58) Then: b bψ n = nb ψ n 1 = n nψ n b bψ n = nψ n (59) Therefore the operator b b is called the number operator, it gives as its eigenvalue the number of energy quanta in the wavefunction ψ n.

18 The second quantization formalism for several identical independent, uncoupled oscillators. The separate oscillators with coordinates q k, distinguished by the suffix k, vibrate independently. The masses m k and frequences ω k may differ. The classical Hamilton function of the system is the sum of the Hamilton functions of the individual oscillators: H = k H k H k = 1 2m k p 2 k + m k 2 ω2 kq 2 k Quantizing, by substituting the classical momentum p k by the quantum mechanical operator and carrying out the same transformations as for the single oscillator, we obtain the Hamilton operator for the system of identical uncoupled oscillators: H = k hω k b kb k + 1. (60) 2 k 1 2 hω k: zero point energy, assumed to be the energy zero. Schrödinger s equation for N uncoupled oscillators: ( hω1 b 1b 1 + hω 2 b 2b hω N b Nb N ) Φ = EΦ (61)

19 Commutation relations: 1. For a single harmonic oscillator: b k b k b kb k = 1 (62) 2. Commutation relations satisfied by b k and b k with different indexes k and k : b k b k b k b k = 0 for k k Proof: and b k b k = 1 ( ) ( 1 + ξ k ) + ξ k 2 ξ k 2 ξ k = 1 ( 2 + ξ k ξ k + ) ξ k ξ k 2 ξ k ξ k ξ k ξ k b k b k = 1 ( 2 + ξ k ξ k ) ξ k + ξ k 2 ξ k ξ k ξ k ξ k (63) (64) As ξ k commutes with ξ k (k k ), we obtain: b k b k b k b k = 0 for k k q.e.d. (65) b k b k b k b k = δ kk δ kk = = 0 for k k (66) = 1 for k = k b k b k b k b k = 0 (67) b kb k b k b k = 0. (68)

20 Many-boson wavefunction Eqs. (66-68) are the commutation relations for Bose operators, referring to independent bosons: The order of creating or annihilating bosons in different states does not matter (eqs. (67, 68): b kb k = b k b k b k b k = b k b k. Bosons do not comply with Pauli s exclusion principle: there may exist many bosons in the same quantum state. The total many-boson wavefunction need not be expressed as a Slater determinant. If Φ o is the lowest energy many-boson state, then: b l Φ o = 0 for l = 1, 2 N (69) E o = 0 (70) Repeated application of the creation operators b j will generate a general eigenstate of the hamiltonian of a system of non-interacting bosons eqn. (60) in the form: Φ n1,n 2,,n N = 1 n1!n 2! n N! (b 1) n1 (b N) n N Φ o = N P P ψ n 1 1 ψ n 2 2 ψ n N N (71)

21 the expression under the square root ensuring normalization. The energy, corresponding to this many-boson wavefunction is: E = hω 1 n 1 + hω 2 n 2 + hω N n N + zero point energy (72) where zero point energy = 1 2 ( hω 1 + hω hω N ) (73) When referring to bosons we can reformulate the statements and reinterpret them in the following way: Instead of saying b k creates an energy quantum hω k by the k-th harmonic oscillator, we say b k creates a boson-particle with energy hω k. Instead of saying that the k-th harmonic oscillator is in its n-th excited state, we say there exist n bosons with energy hω k = hν k.

22 The harmonic oscillator in an external field. The displaced harmonic oscillator as a model for elementary excitations in solids The hamiltonian of the harmonic oscillator in an external field: d 2 H = h2 2m dq + m 2 2 ω2 q 2 + W = H o + W (74) H o : the hamiltonian of the non-perturbed harmonic oscillator, W : potential energy due to the external force F = dw dq, which is assumed to be a constant, independent of time. Therefore W is approximated as a linear function of the oscillator coordinate q: W = C 1 q (75) The respective quantum mechanical operator can be elaborated remembering that the operator b + b has the meaning of an operator of the particle coordinate. Using the definitions eqs. (10) and (11), we realize that: b + b = 2ξ = 2mω h q. (76) Since b + b is an operator of the dimensionless coordinate ξ, the operator W can be expressed with the help of b + b taking care of the dimensions (q = C 2 (b + b)): W = C 1 C 2 (b + b). (77)

23 The dimension of C 1 C 2 should be energy, therefore as a proportionality constant we choose something, which has the dimension of energy, namely γ hω: W = γ hω(b + b) (78) with γ a dimensionless proportionality constant (cf. fig. 5). The operator of the potential energy, resulting from the action of the external force: W = γ hω(b + b) = 2γ hωξ. (79) Figure 5: Behaviour of the potential of the displaced harmonic oscillator. undisplaced oscillator ( ). Potential of the

24 Schrödinger s equation for the harmonic oscillator in an external field, using an expression equivalent to eqn. (52) for the hamiltonian of the non-perturbed harmonic oscillator H o : H o = hω(b b ), (80) looks like: Hψ = Eψ (81) hω(b b )ψ + γ hω(b + b)ψ = Eψ (82) [ b b + γ(b + b) ] ψ = Ē hω ψ 1 2 ψ (83) [ b b + γ(b + b) ] ψ = (λ 1 )ψ (84) 2 Define new creation and destruction operators: b = b + γ (85) with the commutation relation: since γ is just a number. Proof: b = b + γ (86) [ b, b] = 1 (87) b b = (b + γ)(b + γ) = b b + γ(b + b) + γ 2 b b = bb + γ(b + b) + γ 2 [ b, b ] = [b, b] = 1 q.e.d. (88)

25 b = b γ (89) b = b γ (90) Substituting into Schrödinger s equation (84) yields: [ b b + γ 2 γ( b + b) + γ( b + b) 2γ 2 ]ψ = (λ 1 )ψ (91) 2 [ b b γ 2 ]ψ = (λ 1 )ψ (92) 2 b bψ = (λ + γ 2 1 )ψ (93) 2 hω( b b)ψ = (E + γ2 hω 1 2 hω)ψ (94) The eigenenergy of the harmonic oscillator in an external field: [ hω( b b) hω γ2 hω]ψ = Eψ (95) The last equation is exactly Schrödinger s equation for the harmonic oscillator, however, with an energy shifted by the value γ 2 hω. This means that the solutions for the harmonic oscillator, which experiences an external constant force, can be constructed from the eigenvalues and the eigenfunctions of the non-perturbed harmonic oscillator.

26 1. The eigenvalues of the shifted oscillator are shifted to lower energies by the value γ 2 hω (cf. fig. 5): E n = hω(n γ2 ) n = 0, 1, 2 (96) 2. Displacement of the equilibrium position due to the external potential. Since, according to eqn. (76): ξ = 1 2 (b + b) q = ξ = h mω ξ = h 2mω (b + b) mω h q (97) we can evaluate the coordinate of the shifted oscillator q using instead the operators b and b q = = = h 2mω ( b + b) h 2mω (b + b + 2γ) h 2mω (b + b) + 2 h mω γ q = q + 2 h mω γ ξ = mω h q ξ = mω h q + 2γ = ξ + ξ (98) The result is a harmonic oscillator which is displaced with respect to the non-perturbed oscillator at a distance 2 h mω γ (ξ = 2γ).

27 [ bf(b ) f(b )b ] ψ o = γf(b )ψ o 3. The eigenfunctions of the displaced oscillator expressed as linear combinations of the eigenfunctions of the non-perturbed oscillator. (cf. eqn. (42) in Appendix 1): ψ displ o = n C nψ n = n C n n! (b ) n ψ o (99) ψ displ o = f(b )ψ o (100) with f(b ) = n C n n! (b ) n (101) Applying the annihilation operator b to the ground state wavefunction of the displaced oscillator yields zero: bψ displ o = bf(b )ψ o = 0 [ bf(b ) + γf(b ) ] ψ o = 0 (102) We replace the zero at the right-hand side of the last equation with its equivalent: bψ o = 0 bψ o = f(b )bψ o = 0 (103) and, after some reorganization of the terms, we obtain: [ b, f(b ) ] ψ o = γf(b )ψ o (104)

28 The last equation can be rewritten using eqn. (111) from Appendix 2, namely: as: [ b, f(b ) ] = f(b ) b f(b ) b ψ o = γf(b )ψ o (105) Appendix 2: Some useful formulae bb = 1 + b b b(b ) 2 = b + b bb = b + b + (b ) 2 b = 2(b ) 1 + (b ) 2 b b(b ) 3 = 2b b + (b ) 2 bb = 2(b ) 2 + (b ) 2 + (b ) 3 b = 3(b ) 2 + (b ) 3 b b(b ) n = n(b ) n 1 + (b ) n b (106) [ b, (b ) n] = n(b ) n 1 (107) Similarly: [ b, (b ) n] = (b ) n [ b, (b n ) ] = nb n 1 = (bn ) b b (108) (109)

29 We can generalize eqs. (108) and (109) to any function f of b or b, i.e. f(b ) or f(b). which is equivalent to: bf(b ) f(b )b = f(b ) b (110) b f(b) f(b)b = f(b) b (111) [ b, f(b ) ] = f(b ) b (112) [ b, f(b) ] = f(b) b (113)

30 The solution of equation: f(b ) b ψ o = γf(b )ψ o (114) resulting from the considerations which lead to eqn. (117) in Appendix 3, is: Appendix 3: Solution to eqn. (105): f(b ) = N o e γb. (115) f(b ) b = γf(b ) f(b ) f(b ) = γ b lnf(b ) = γb + C f(b ) = e γb e C (116) f(b ) = N o e γb (117)

31 ψ displ o = f(b )ψ o The wavefunction of the lowest lying state of the displaced oscillator: According to eqs. (99) and (101): ψ displ o = N o e γb ψ o (118) f(b ) = n C n n! (b ) n and ψ displ o = n C n n! (b ) n ψ o. Equation (115) allows an expansion of f(b ) in a Taylor series: f(b ) = N o 1 γb + 1 2! γ2 (b ) 2 1 3! γ3 (b ) 3 ± (119) ψ displ o = N o 1 γb + 1 2! γ2 (b ) 2 1 3! γ3 (b ) 3 ± ψ o (120) The comparison of eqs. (100) and (120) yields: C n = ( 1) n 1 N o n! n! γn (121) The expansion coefficients of the displaced oscillator wavefunction: C n = ( 1) n γn n! N o (122) N o = e γ 2 2 (123)

32 Plasmons: harmonic oscillators in an external field The creation and annihilation of collective excitations of the electron gas (plasma) in a crystal due to the electrostatic field of an external electrical charge can be treated within the second quantization approach described so far. The potential energy due to an external field was up to now described with the operator W : W = γ hω(b + b) by introducing the creation and destruction operators for the displaced oscillator: b = b + γ b = b + γ For the image potential we make the following choice: W im = B(α + α) Here B corresponds to γ hω in the general formalism. We introduce the plasmon creation α and destruction operators α and the creation and destruction operators for the displaced plasmons, β and β, respectively: β = α + B hω β = α + B hω (124) (125) reducing the task of solving Schrödinger s equation for the plasmons in the field of an external electrostatic charge to the task of solving Schrödinger s equation for the shifted harmonic oscillator. Thus we can make use of and transfer the results (eigenenergies, eigenfunctions) we just received to the new situation: plasmons in an external field.

33 References [1] H. Haken Quantum Field Theory of Solids: an Introduction, North-Holland Publ. Co., Amsterdam-New York-Oxford, second printing [2] P.W. Atkins and R.S. Friedman Molecular Quantum Mechanics, Oxford University Press, Oxford-New York-Tokyo, 3rd ed