24 1.Second Quantization one obtains for the commutators with the kinetic energy: r). -会-vx-刘p4-4o the potential energy: U(x(x)(x).(x)] =2U(x)(-6③(x-x)(x)=-U(x)(x), and the interaction [/r'txetxvx,xexe.] =专trotxex,exjVx,xe(xex =rx{±6(x”-x)ptx)-ptxK-} ×V(x',x")(x")(x' -()V(xx)(x)v(x). eo)and (are to proced fou the e In this last ond =干(x)(x",the ese expressions give the equation of motion (1.5.15)of the field operator,which is also known as the field equation. The equation of motion for the adjoint field operator reads: -c. -(xt(x.t)V(x.x)v(x'.t). (1.5.18) where it is assumed that V(-V(). If (1.5.15)is multiplied from the left by (x,t)and (1.5.18)from the right by (x,t),one obtains the equation of motion for the density operator ax)-(6+)-()ep2-(we, and thus (x)=-j(x) (1.5.19) t density defined in(1.5.11).Equation(1.5.19)
24 1. Second Quantization one obtains for the commutators with the kinetic energy: d3x 2 2m[∇ ψ†(x )∇ ψ(x ), ψ(x)] = d3x 2 2m(−∇ δ(3)(x − x) · ∇ ψ(x )) = 2 2m ∇2ψ(x) , the potential energy: d3x U(x )[ψ†(x )ψ(x ), ψ(x)] = d3x U(x )(−δ(3)(x − x)ψ(x )) = − U(x)ψ(x) , and the interaction: 1 2 d3x d3x ψ†(x )ψ†(x)V (x , x)ψ(x)ψ(x ), ψ(x) = 1 2 d3x d3x [ψ†(x )ψ†(x), ψ(x)]V (x , x)ψ(x)ψ(x ) = 1 2 d3x d3x ±δ(3)(x − x)ψ†(x ) − ψ† (x)δ(3)(x − x) × V (x , x)ψ(x)ψ(x ) = − d3x ψ†(x )V (x, x )ψ(x )ψ(x). In this last equation, (1.5.17) and (1.5.5c) are used to proceed from the second line. Also, after the third line, in addition to ψ(x)ψ(x ) = ∓ψ(x )ψ(x), the symmetry V (x, x ) = V (x , x) is exploited. Together, these expressions give the equation of motion (1.5.15) of the field operator, which is also known as the field equation. The equation of motion for the adjoint field operator reads: iψ˙ †(x, t) = − − 2 2m ∇2 + U(x) ψ†(x, t) − d3x ψ†(x, t)ψ†(x , t)V (x, x )ψ(x , t), (1.5.18) where it is assumed that V (x, x )∗ = V (x, x ). If (1.5.15) is multiplied from the left by ψ†(x, t) and (1.5.18) from the right by ψ(x, t), one obtains the equation of motion for the density operator n˙(x, t) = ψ† ψ˙ + ψ˙ †ψ = 1 i − 2 2m ψ† ∇2ψ − ∇2ψ† ψ , and thus n˙(x) = −∇j(x), (1.5.19) where j(x) is the particle current density defined in (1.5.11). Equation (1.5.19) is the continuity equation for the particle-number density.
1.6 Momentum Representation 25 1.6 Momentum Representation 1.6.1 Momentum Eigenfunctions and the Hamiltonian The momentum representation is particularly useful in translationally in- wariant syster s.形 conside f din The are normanzed s ome pk(x)=eikx/VV (1.6.1 with the volume V=L By assuming periodic boundary conditions ek(+)=elkr,etc. (1.6.2a the allowed values of the wave vector k are restricted to k=2(2）m=0m=0%=0士 (1.6.2b) The eigenfunctions(1.6.1)obey the following orthonormality relation: rpi(x)pw(x)=6xk (1.6.3) In order to represent the Hamiltonian in second-quantized form,we need the matrix elements of the operators that it contains.The kinetic energy is proportional to pg(-V）pPx=kkk2 (1.6.4a and the matrix element of the single-particle potential is given by the Fourier transform of the latter: i(xU(x)px(x)d-Uw-k (1.6.4b) For two-particle potentials V(x-x')that depend only on the relative coor- dinates of the two particles,it is useful to introduce their Fourier transform Va=re-iaxV(x), (1.6.5a and also its inverse vx)=F∑aea (1.6.5b)
1.6 Momentum Representation 25 1.6 Momentum Representation 1.6.1 Momentum Eigenfunctions and the Hamiltonian The momentum representation is particularly useful in translationally invariant systems. We base our considerations on a rectangular normalization volume of dimensions Lx, Ly and Lz. The momentum eigenfunctions, which are used in place of ϕi(x), are normalized to 1 and are given by ϕk(x)=eik·x/ √ V (1.6.1) with the volume V = LxLyLz. By assuming periodic boundary conditions eik(x+Lx) = eikx, etc. , (1.6.2a) the allowed values of the wave vector k are restricted to k = 2π nx Lx , ny Ly , nz Lz , nx = 0, ±1,... ,ny = 0, ±1,... ,nz = 0, ±1,... . (1.6.2b) The eigenfunctions (1.6.1) obey the following orthonormality relation: d3xϕ∗ k(x)ϕk (x) = δk,k . (1.6.3) In order to represent the Hamiltonian in second-quantized form, we need the matrix elements of the operators that it contains. The kinetic energy is proportional to ϕ∗ k −∇2 ϕkd3x = δk,kk2 (1.6.4a) and the matrix element of the single-particle potential is given by the Fourier transform of the latter: ϕ∗ k (x)U(x)ϕk(x)d3x = 1 V Uk−k. (1.6.4b) For two-particle potentials V (x − x ) that depend only on the relative coordinates of the two particles, it is useful to introduce their Fourier transform Vq = d3xe−iq·xV (x) , (1.6.5a) and also its inverse V (x) = 1 V q Vqeiq·x . (1.6.5b)
26 1.Second Quantization For the matrix element of the two-particle potential,one then finds (p'.k'lV(x-x)lp.k) =Vdrdre-ipxekV(x-)eikex -2∫产∫eexw =房∑gV6-p4g+poV6-k-atko (1.6.5c Inserting(1.6.5a,b,c)into the general representation(1.4.16c)of the Hamil tonian yields H-∑k 工2am+Ek-ata+元∑hdg4n q.p.k (1.6.6) The creation operators of a particle with wave vector k(i.e.in the statek are denoted by ak and the annihilation operators by ak.Their commutation relations are [ak,akl±=0，ak,akl±=0and[ak,a]±=dikk. (1.6.7) The interaction term allows a pictorial interpretation.It causes the annihila- tion of two particles with wave vectors k and p and creates in their place two particles with wave vectors k-q and p+q.This is represented in Fig.1.1a. The full lines denote the particles and the dotted lines the interaction po- tential Va.The amplitude for this transition is proportional to Va.This dia- k-qi-q2 p+q+qz a) particles
26 1. Second Quantization For the matrix element of the two-particle potential, one then finds p , k | V (x − x )|p, k = 1 V 2 d3xd3x e−ip ·xe−ik ·x V (x − x )eik·x eip·x = 1 V 3 q Vq d3x d3x e−ip ·x−ik ·x +iq·(x−x )+ik·x +ip·x = 1 V 3 q VqV δ−p+q+p,0V δ−k−q+k,0. (1.6.5c) Inserting (1.6.5a,b,c) into the general representation (1.4.16c) of the Hamiltonian yields: H = k (k)2 2m a† kak + 1 V k,k Uk−ka† kak + 1 2V q,p,k Vqa† p+qa† k−qakap. (1.6.6) The creation operators of a particle with wave vector k (i.e., in the state ϕk) are denoted by a† k and the annihilation operators by ak. Their commutation relations are [ak, ak ]± = 0, [a† k, a† k ]± = 0 and [ak, a† k ]± = δkk . (1.6.7) The interaction term allows a pictorial interpretation. It causes the annihilation of two particles with wave vectors k and p and creates in their place two particles with wave vectors k − q and p + q. This is represented in Fig. 1.1a. The full lines denote the particles and the dotted lines the interaction potential Vq. The amplitude for this transition is proportional to Vq. This diak p k − q p + q Vq a) k p k − q1 p + q1 Vq1 Vq2 b) k − q1 − q2 p + q1 + q2 Fig. 1.1. a) Diagrammatic representation of the interaction term in the Hamiltonian (1.6.6) b) The diagrammatic representation of the double scattering of two particles
1.6 Momentum Representation 27 grammatic form is a usefl way of representing the perturbation-theoretica scription of such sse The double scattering of two particles represented as shown in Fig.1.b,where all intermediate states. 1.6.2 Fourier Transformation of the Density The other operators considered in the previous section canalso bee the rep in th enta on.As an importar we shall lo defined by ia=rn(x)e-ix=ru(x)v(x)e-iax (1.6.8) From (1.5.4a,b)we insert w=∑=∑ea 1 (1.6.9 which yields i=r7∑∑e-ipxopeikxoxe-inx】 p k and thus,with (1.6.3),one finally obtains na=∑吨p+a (1.6.10) We have thus found the Fourier transform of the density operator in the momentum representation. The occupation-number operator for the state Ip)is also denoted by ip=afap.It will always be clear from the context which one of the two meanings is meant.The operator for the total number of particles(1.3.13)in the momentum representation reads N=∑吨p (1.6.11) 1.6.3 The Inclusion of Spin Up until now,we have not explicitly considered the spin.One can think of it as being included in the previous formulas as part of the spatial degree s The hat on the operator,as used here for and previously for the occupation- number operator,will only be retained where it is needed to avoid confusion
1.6 Momentum Representation 27 grammatic form is a useful way of representing the perturbation-theoretical description of such processes. The double scattering of two particles can be represented as shown in Fig. 1.1b, where one must sum over all intermediate states. 1.6.2 Fourier Transformation of the Density The other operators considered in the previous section can also be expressed in the momentum representation. As an important example, we shall look at the density operator. The Fourier transform of the density operator8 is defined by nˆq = d3xn(x)e−iq·x = d3xψ†(x)ψ(x)e−iq·x . (1.6.8) From (1.5.4a,b) we insert ψ(x) = 1 √ V p eip·xap, ψ†(x) = 1 √ V p e−ip·xa† p , (1.6.9) which yields nˆq = d3x 1 V p k e−ip·xa† peik·xake−iq·x , and thus, with (1.6.3), one finally obtains nˆq = p a† pap+q . (1.6.10) We have thus found the Fourier transform of the density operator in the momentum representation. The occupation-number operator for the state |p is also denoted by nˆp ≡ a† pap. It will always be clear from the context which one of the two meanings is meant. The operator for the total number of particles (1.3.13) in the momentum representation reads Nˆ = p a† pap . (1.6.11) 1.6.3 The Inclusion of Spin Up until now, we have not explicitly considered the spin. One can think of it as being included in the previous formulas as part of the spatial degree 8 The hat on the operator, as used here for ˆnq and previously for the occupationnumber operator, will only be retained where it is needed to avoid confusion
28 1.Second Quantization of freedom x.If the spin is to be given explicitly,then one has to make the replacements(x)→中.(x)and ap一ape and,in addition,introduce the sum over o,the 2 component of the spin.The particle-number density,for example,then takes the form nx)=∑(x)a冈 (1.6.12 p.e The Hamiltonian for the case of a spin-independent interaction reads: H=∑(ga+U6) +与∑rx)og(6xV(x,xa(x,(x),(6.13) the corresponding form applying in the momentum representation. For spin-fermions,the two possible spin quantum numbers for the z component of S are+.The spin density operator S(x)=∑6x-xa)S。 (1.6.14a n= is,in this case, s(x)=∑x)x) (1.6.14b a.a where are the matrix elements of the Pauli matrices. The commutation relations of the field operators and the operators in the momentum representation read: [(x,。(x]±=0，(x,(x)】±=0 (1.6.15) [b(x),bd(x】±=6ao6(x-x) and aka,ak'g±=0，ag,at,l±=0，[aka,ag]±=dkk6aa.(1.6.l6) The equations of motion are given by h品0=（-p+)x （2m +∑/gx,9VKxK,e,xa617
28 1. Second Quantization of freedom x. If the spin is to be given explicitly, then one has to make the replacements ψ(x) → ψσ(x) and ap → apσ and, in addition, introduce the sum over σ, the z component of the spin. The particle-number density, for example, then takes the form n(x) = σ ψ† σ(x)ψσ(x) nˆq = p,σ a† pσap+qσ . (1.6.12) The Hamiltonian for the case of a spin-independent interaction reads: H = σ d3x( 2 2m ∇ψ† σ∇ψσ + U(x)ψ† σ(x)ψσ(x)) + 1 2 σ,σ d3xd3x ψ† σ(x)ψ† σ (x )V (x, x )ψσ (x )ψσ(x) , (1.6.13) the corresponding form applying in the momentum representation. For spin- 1 2 fermions, the two possible spin quantum numbers for the z component of S are ± 2 . The spin density operator S(x) = N α=1 δ(x − xα)Sα (1.6.14a) is, in this case, S(x) = 2 σ,σ ψ† σ(x)σσσψσ (x), (1.6.14b) where σσσ are the matrix elements of the Pauli matrices. The commutation relations of the field operators and the operators in the momentum representation read: [ψσ(x), ψσ (x )]± = 0 , [ψ† σ(x), ψ† σ (x )]± = 0 [ψσ(x), ψ† σ (x )]± = δσσ δ(x − x ) (1.6.15) and [akσ, akσ ]± = 0, [a† kσ, a† kσ ]± = 0, [akσ, a† kσ ]± = δkkδσσ . (1.6.16) The equations of motion are given by i ∂ ∂tψσ(x, t) = − 2 2m ∇2 + U(x) ψσ(x, t) + σ d3x ψ† σ (x , t)V (x, x )ψσ (x , t)ψσ(x, t) (1.6.17)