Second quantization. Emmanuel Fromager

Transcription

1 Institut de Chimie, Strasbourg, France Page 1 Emmanuel Fromager Institut de Chimie de Strasbourg - Laboratoire de Chimie Quantique - Université de Strasbourg /CNRS M2 lecture, Strasbourg, France.

2 Institut de Chimie, Strasbourg, France Page 2 Electronic Hamiltonian in first quantization N-electron Hamiltonian within the Born Oppenheimer approximation: Ĥ = ˆT + ˆV ne + Ŵee ˆT = N ˆt(i) where ˆt(i) r i kinetic energy i=1 ˆV ne = N ˆv ne (i) where ˆv ne (i) i=1 nuclei A Z A r i R A electron-nuclei attraction Ŵ ee = 1 2 N ŵ ee (i, j) where ŵ ee (i, j) i j 1 electron-electron repulsion r i r j

3 Institut de Chimie, Strasbourg, France Page 3 One-electron wavefunction Let us start with Schrödinger theory: the quantum state of a single electron is written as Ψ = dr Ψ(r) r where Ψ(r) is the one-electron wavefunction (orbital) and r denotes the quantum state "the electron is at position r". This choice of basis is known as "r representation". Orthonormalization condition: r r = δ(r r) Dirac distribution where " f", dr f(r)δ(r r) = f(r ) i f i δ ii = f i Kronecker delta Pauli theory: the spin of the electron is now considered as an additional degree of freedom. The quantum state of a single electron is then written as Ψ = dr σ=α,β Ψ(r, σ) r, σ where r, α denotes the quantum state "electron at position r with spin up" and r, β corresponds to the state "electron at position r with spin down".

4 Institut de Chimie, Strasbourg, France Page 4 Two-electron wavefunction In the non-relativistic case, a single electron will have a spin σ 0 which is either up or down. The corresponding wavefunction Ψ σ0 can then be written as a spin-orbital Ψ σ0 (r, σ) = Ψ(r)δ σσ0. With the notations X (r, σ) and dx dr, σ=α,β a one-electron quantum state in Pauli theory is simply written as Ψ = dx Ψ(X) X with X X = δ(x X) = δ σ σδ(r r) and, consequently, X Ψ = Ψ(X ). Two-electron case: Ψ = dx 1 dx 2 Ψ(X 1, X 2 ) 1: X 1, 2: X 2 where the two-electron quantum state 1: X 1, 2: X 2 corresponds to "electron 1 in state X 1 and electron 2 in state X 2 ".

5 Institut de Chimie, Strasbourg, France Page 5 Anti-symmetrization principle: a physical two-electron wavefunction should fulfill the condition Ψ(X 1, X 2 ) = Ψ(X 2, X 1 ) thus leading to Ψ = dx 1 dx 2 Ψ(X 2, X 1 ) 1: X 1, 2: X 2 = dx 1 dx 2 Ψ(X 1, X 2 ) 1: X 2, 2: X 1 }{{} ˆP 1 2 Ψ and Ψ = 1 2 ] dx 1 dx 2 [Ψ(X 1, X 2 ) Ψ(X 2, X 1 ) 1: X 1, 2: X 2 }{{} 0 if X 1 = X 2 Conclusion: the anti-symmetrization of the wavefunction ensures that electrons are indistinguishable and that they cannot be in the same quantum state (Pauli principle).

6 Institut de Chimie, Strasbourg, France Page 6 Let { } ϕ K (X) K Slater determinants denote an orthonormal basis of (molecular) spin-orbitals. Two electrons that occupy the spin-orbitals ϕ I (X) and ϕ J (X) will be described by the (normalized) Slater determinant Φ IJ (X 1, X 2 ) = 1 2 ϕ I (X 1 ) ϕ I (X 2 ) ϕ J (X 1 ) ϕ J (X 2 ) ) = (ϕ 1 I (X 1 )ϕ J (X 2 ) ϕ I (X 2 )ϕ J (X 1 ) 2 Note that Slater determinants and, consequently, linear combinations of Slater determinants are anti-symmetric. Therefore, Slater determinants are convenient "building blocks" for computing the electronic wavefunction. Still, we may wonder if we really need this complicated expression obtained from the determinant (obviously things get worse for a larger number of electrons). Another drawback of the current formulation: both Slater determinant and Hamiltonian expressions depend on the number of electrons.

7 Institut de Chimie, Strasbourg, France Page 7 "What is occupied?" rather than "Who occupies what?" Since electrons are indistinguishable, there is no need to know that electron 1 occupies ϕ I electron 2 occupies ϕ J or the other way around... and The important information is that spin-orbitals ϕ I and ϕ J are occupied and the remaining ones are empty. is a formalism that relies on this idea. Let us tell the story... At the beginning, there was "nothing"... vac normalized "vacuum state"... then was introduced the concept of annihilation of an electron occupying ϕ I, that would obviously give zero when applied to the vacuum state: I, â I vac = 0 (rule 1)... and then came the concept of creation of an electron occupying ϕ I : â I vac ϕ I

8 Institut de Chimie, Strasbourg, France Page 8 "What is occupied?" rather than "Who occupies what?"... and then came the idea to create another electron occupying ϕ J : â J â I vac Φ IJ Note that the creation operator â I is the adjoint of the annihilation operator â I. This ensures, in particular, that one-electron and vacuum states are orthogonal: ϕ I vac = â I vac vac = vac â I vac = 0 In order to have a representation that is equivalent to the one used in first quantization, we only need two more rules: I, J, [â I, â J ] + = â I â J + â J â I = 0 (rule 2) I, J, [ ] â I, â J = â Iâ + J + â J âi = δ IJ (rule 3) Notation: the anticommutator of  with ˆB reads ] commutator equals [Â, ˆB =  ˆB ˆBÂ. [ ] â I, â J = [â J, â I ] + = 0 + } [Â, ˆB] {Â, = ˆB =  ˆB+ ˆB while the +

9 Institut de Chimie, Strasbourg, France Page 9 "What is occupied?" rather than "Who occupies what?" Rule 2 contains the indistinguishability of the electrons, â J â I vac = â Iâ J vac, and the Pauli principle, â Iâ I vac = 0. Rule 3 ensures that you can only annihilate what has already been created (!), â I â J vac = δ IJ vac â J âi vac = δ IJ vac It is now very easy to generate representations of Slater determinants for an arbitrary number N of electrons: multiply more creation operators! I 1 I 2... I N 1 I N = â I 1 â I 2... â I N 1 â I N vac 1 [ {ϕii det (X j ) } ] N! 1 i,j N

10 EXERCISE: (1) Show that I 1 I 2... I N 1 I N is normalized. (2) Let us consider another state J 1 J 2... J N 1 J N and assume that at least one of the occupied spin-orbitals (let us denote it ϕ Jk ) is not occupied in I 1 I 2... I N 1 I N. Show that the two states are orthogonal. (3) The "counting" operator ˆN is defined as ˆN = ˆn I I where ˆn I = â IâI. Show that ˆn I I 1 I 2... I N 1 I N = I 1 I 2... I N 1 I N if I = I k 1 k N = 0 otherwise and conclude that ˆN I1 I 2... I N 1 I N = N I 1 I 2... I N 1 I N. (4) Explain why states corresponding to different numbers of electrons are automatically orthogonal. (5) Explain why any normalized state Ψ fulfills the condition 0 Ψ ˆn I Ψ 1. Institut de Chimie, Strasbourg, France Page 10

11 Institut de Chimie, Strasbourg, France Page 11 One-electron operators in second quantization Let ĥ denote a one-electron operator (ˆt + ˆv ne for example): it acts on the one-electron states ϕ I. Resolution of the identity: ϕ I ϕ I = ˆ1, I which leads to the conventional representation ĥ = ˆ1 ĥ ˆ1 = I,J ϕ I ĥ ϕ J ϕ I ϕ J. Second-quantized representation: ĥ I,J ϕ I ĥ ϕ J â IâJ Indeed, ϕ I ĥ ϕ J â IâJ ϕ K = I,J ϕ I ĥ ϕ J â IâJ â K vac = ϕ I ĥ ϕ J â I δ JK vac I,J I,J = I ϕ I ĥ ϕ K ϕ I = ĥ ϕ K

12 Institut de Chimie, Strasbourg, France Page 12 What is convenient is that this second-quantized representation is valid for any number N of electrons: N i=1 ĥ(i) I,J ϕ I ĥ ϕ J â IâJ ĥ The information about N has been completely transferred to the states. It does not appear in the operator anymore. { } EXERCISE: Let us consider another orthonormal basis ϕ K (X) K decompose in the current basis as follows, ϕ P = U QP ϕ Q. Q of spin-orbitals that we (1) Show that the matrix U is unitary (U = U 1 ). (2) Explain why â P = Q U QP â Q and show that I,J ϕ I ĥ ϕ J â Ĩâ J ĥ. (3) Show how the diagonalization of ĥ in the one-electron space leads automatically to the diagonalization in the N-electron space (use exercise page 10).

13 Two-electron operators in second quantization Let ŵ denote a two-electron operator: it acts on two-electron states ϕ I ϕ J = 1: ϕ I, 2: ϕ J. A complete anti-symmetrized basis should be used for describing the two electrons: IJ = 1 ) ( ϕ I ϕ J ϕ J ϕ I 2 â Iâ J vac with I < J. Consequently, any two-electron anti-symmetrized state Ψ shoud fulfill the condition ˆP A Ψ = Ψ where ˆPA = I<J IJ IJ projection operator! Projection of the two-electron operator onto the space of anti-symmetrized states: ŵ A = ˆP A ŵ ˆP A = I<J,K<L IJ ŵ KL IJ KL Institut de Chimie, Strasbourg, France Page 13

14 Two-electron operators in second quantization EXERCISE: Prove that ŵ A 1 2 ϕ I ϕ J ŵ ϕ K ϕ L â Iâ J âlâ K IJKL hint: apply ŵ A and the proposed second-quantized representation to P Q â P â Q vac Conclude. (P < Q). What is convenient is that this second-quantized representation is valid for any number N of electrons and includes the projection onto anti-symmetrized states: 1 2 N ŵ(i, j) 1 2 i j IJKL ϕ I ϕ J ŵ ϕ K ϕ L â Iâ J âlâ K ŵ Institut de Chimie, Strasbourg, France Page 14

15 Summary In summary, the electronic Hamiltonian can be written in second quantization as follows, Ĥ = IJ ϕ I ĥ ϕ J â IâJ IJKL ϕ I ϕ J ŵ ee ϕ K ϕ L â Iâ J âlâ K where ϕ I ĥ ϕ J = dx ϕ I (X) (ĥϕj )(X) one-electron integrals ϕ I ϕ J ŵ ee ϕ K ϕ L = ) dx 1 dx 2 ϕ I (X 1)ϕ J (X 2) (ŵ ee ϕ K ϕ L (X 1, X 2 ) two-electron integrals Note that this expression is also valid for a relativistic Hamiltonian. Two or four-component spinors should be used rather than spin-orbitals in conjunction with the Dirac (Breit) Coulomb Hamiltonian. The standard (non-relativistic) Hamiltonian will be used in the following. Institut de Chimie, Strasbourg, France Page 15

16 Institut de Chimie, Strasbourg, France Page 16 EXERCISE: At the non-relativistic level, real algebra can be used, ϕ I (X) = ϕ iσ (r, τ) = φ i (r)δ στ, ĥ r + v ne(r) and ŵ ee 1 r 1 r 2. Show that { the Hamiltonian, } that is here a spin-free operator, can be rewritten in the basis of the molecular orbitals φ p (r) as follows p Ĥ = p,q h pq Ê pq p,q,r,s ) pr qs (Êpq Ê rs δ qr Ê ps where Ê pq = σ â p,σâq,σ, h pq = φ p ĥ φ q and pr qs = dr 1 dr 2 φ p (r 1 )φ r (r 2 ) 1 r 1 r 2 φ q(r 1 )φ s (r 2 ) = (pq rs).

17 Institut de Chimie, Strasbourg, France Page 17 EXERCISE: For any normalized N-electron wavefunction Ψ, we define the one-electron (1) and two-electron (2) reduced density matrices (RDM) as follows, D pq = Ψ Êpq Ψ and D pqrs = Ψ Ψ ÊpqÊrs δ qr Ê ps. (1) Show that the 1RDM is symmetric and that p, the occupation n p = D pp of the orbital p fulfills the inequality 0 n p 2. Show that the trace of the 1RDM equals N. (2) Explain why the expectation value for the energy Hint: show that D pq = 1 D pqrr. N 1 r Ψ Ĥ Ψ can be determined from the 2RDM. (3) Let us consider the particular case Ψ Φ = are non-zero only in the occupied-orbital space. N/2 i=1 σ â i,σ vac. Explain why both density matrices Show that D ij = 2δ ij and D ijkl = 4δ ij δ kl 2δ jk δ il and...

18 ... deduce the corresponding energy expression: N/2 Φ Ĥ Φ = 2 i=1 h ii + N/2 i,j=1 ( 2 ij ij ij ji ). (4) Let i, j and a, b denote occupied and unoccupied (virtuals) orbitals in Φ, respectively. Explain why Êai and ÊaiÊbj are referred to as single excitation and double excitation operators, respectively. Hint: derive simplified expressions for Φ a i = 1 2 Ê ai Φ and Φ ab ij = 1 2 Ê ai Ê bj Φ with i < j, a < b. Institut de Chimie, Strasbourg, France Page 18

19 Institut de Chimie, Strasbourg, France Page 19 Why "second" quantization? Let us focus on the (one-electron) electron-nuclei local potential operator which, in second quantization, reads N ˆVne = ˆv ne (i) ϕ I ˆv ne ϕ J â IâJ i=1 IJ where ϕ I ˆv ne ϕ J = dx v ne (r)ϕ I (X)ϕ J (X), thus leading to ( ) ( ) ˆV ne dx v ne (r) ϕ I (X)â I ϕ J (X)â J = dr v ne (r) σ ˆΨ (r, σ) ˆΨ(r, σ) ˆV ne I }{{}}{{} J ˆΨ (X) ˆΨ(X) field operators For a single electron occupying the spin-orbital Ψ(X) = Ψ(r, σ), the corresponding expectation value for the electron-nuclei potential energy equals Ψ ˆv ne Ψ = dx v ne (r)ψ (X)Ψ(X) = dr v ne (r) σ Ψ (r, σ)ψ(r, σ).

20 Physical interpretation of the field operators: ˆΨ (X) vac = I ϕ I (X)â I vac = I ϕ I (X) ϕ I = I ϕ I ϕ I X = X = ˆΨ (X) vac, which means that ˆΨ (X) = ˆΨ (r, σ) creates an electron at position r with spin σ. Consequently, the density operator reads in second quantization ˆn(r) = σ ˆΨ (r, σ) ˆΨ(r, σ), and the electron density associated with the normalized N-electron wavefunction Ψ is simply calculated as follows, n Ψ (r) = Ψ ˆn(r) Ψ. Anticommutation rules: [ ˆΨ(X), ˆΨ(X )] + = IJ ϕ J (X)ϕ I (X ) [â J, â I ] + = 0 and [ ˆΨ(X), ˆΨ (X )] + = IJ ϕ J (X)ϕ I (X ) [ ] â J, â I = + IJ ϕ J (X)ϕ I (X )δ IJ = I X ϕ I ϕ I X = X X = δ(x X ). Institut de Chimie, Strasbourg, France Page 20

21 Institut de Chimie, Strasbourg, France Page 21 Model Hamiltonians: example of the Hubbard Hamiltonian h ij t ( δ i,j 1 + δ i,j+1 ) + εi δ ij ij kl Uδ ij δ ik δ lj Ê ik Ê jl δ kj Ê il ˆn iˆn i ˆn i where ˆn i = Êii = ˆn i + ˆn i so that ˆn iˆn i = 2ˆn i ˆn i + ˆn i Ĥ t â i,σâj,σ i,j σ=, }{{} + U ˆn i ˆn i i }{{} + ε iˆn i i }{{} ˆT (hopping) on-site repulsion local potential

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