Time-dependent linear response theory: exact and approximate formulations. Emmanuel Fromager

Transcription

1 June 215 ISTPC 215 summer school, Aussois, France Page 1 Time-dependent linear response theory: exact and approximate formulations Emmanuel Fromager Institut de Chimie de Strasbourg - Laboratoire de Chimie Quantique - Université de Strasbourg /CNRS ISTPC 215 summer school, Aussois, France, June 215

2 June 215 ISTPC 215 summer school, Aussois, France Page 2 Response properties in the time-independent regime In the following we shall refer to Ĥ = ˆT + Ŵee + ˆV ne as the unperturbed Hamiltonian with ground state Ψ. Let us introduce a perturbation operator ˆV with strength ε. The Hamiltonian becomes ε-dependent: Ĥ(ε) = Ĥ + ε ˆV. Example: if the perturbation is a uniform electric field E along the z axis, then E = ε e z and ˆV = ε ẑ where ẑ = dr z ˆn(r) second-quantized notation! Response theory is nothing but perturbation theory formulated for both exact and approximate wavefunctions. Let Ψ(ε) denote the exact normalized ground state of Ĥ(ε) with energy E(ε). Linear and higher-order response functions: ˆV (ε) = Ψ(ε) ˆV Ψ(ε) = Ψ ˆV Ψ + ε ˆV ; ˆV ε2 ˆV ; ˆV, ˆV +...

3 June 215 ISTPC 215 summer school, Aussois, France Page 3 In our example, Response properties in the time-independent regime Ψ ẑ Ψ is the permanent dipole moment along the z axis ẑ; ẑ = α zz is the static polarizability ẑ; ẑ, ẑ = β zzz is the static hyperpolarizability Hellmann Feynman theorem: de(ε) dε = Ψ(ε) Ĥ(ε) ε Ψ(ε) = ˆV (ε) Exact response functions can be expressed as energy derivatives: ˆV ; ˆV = d2 E(ε) dε 2, ˆV ; ˆV, ˆV = d3 E(ε) dε 3

4 June 215 ISTPC 215 summer school, Aussois, France Page 4 Response theory for variational methods In variational methods, the energy is expressed as an expectation value. It depends on both the trial variational parameters λ and the perturbation strength ε: E(λ, ε) = Ψ(λ) Ĥ(ε) Ψ(λ) At the HF level of approximation, λ parameterizes orbital rotations. At the CI level (in the basis of perturbation-independent orbitals), λ contains all the CI coefficients. At the MCSCF level, it contains both orbital rotation and CI coefficients. Stationarity condition: ε, E(λ, ε) λ = λ(ε) λ=λ(ε) In the following, we will use parameterizations such that λ() =. Consequently λ(ε) quantifies the response of the electronic wavefunction to the perturbation.

5 June 215 ISTPC 215 summer school, Aussois, France Page 5 Response theory for variational methods The converged ground-state energy only depends on the perturbation strength: E(ε) = E(λ(ε), ε) The Hellmann Feynman theorem remains valid for approximate variational methods: de(ε) dε = [ ] λ(ε) T E(λ, ε) ε λ + λ=λ(ε) }{{} E(λ, ε) ε λ=λ(ε) thus leading to de(ε) dε = ( ( ) Ψ λ(ε)) ˆV Ψ λ(ε) Conclusion: response functions obtained from variational wavefunctions can be expressed as energy derivatives.

6 June 215 ISTPC 215 summer school, Aussois, France Page 6 Example: exact response theory We denote {Ψ i } i=,1,... the exact orthonormal eigenvectors of the unperturbed Hamiltonian Ĥ with energies {E i } i=,1,... Exact wavefunction parameterization: Ψ(S) = eŝ Ψ where Ŝ = i> S i ( ˆR i ˆR i ) is anti-hermitian, ˆR i = Ψ i Ψ and S {S i } i=1,2,... Linear response function: ˆV ; ˆV = d2 E(ε) dε 2 where E(ε) = E(S(ε), ε). BCH expansion: de(ε) dε = Ψ e Ŝ(ε) ˆV e Ŝ(ε) Ψ = Ψ ˆV Ψ + Ψ [ ˆV, Ŝ(ε)] Ψ Ψ [[ ˆV, Ŝ(ε)], Ŝ(ε)] Ψ +...

7 June 215 ISTPC 215 summer school, Aussois, France Page 7 Example: exact response theory Using the condition S() = leads to ˆV ; ˆV = Definition: V [1] i = Ψ [ ˆV, ˆR i ˆR i ] Ψ Ψ [ ˆV, Ŝ(ε) ε ] Ψ component i of the gradient property vector Usual expression for the linear response function: ˆV ; ˆV = [ ] S(ε) T ε V [1] The linear response of the wavefunction is obtained by differentiation of the stationarity condition with respect to the perturbation strength: d dε [ E(S, ε) S = S=S(ε)]

8 June 215 ISTPC 215 summer school, Aussois, France Page 8 Example: exact response theory BCH expansion for the energy: E(S, ε) = Ψ e ŜĤ(ε)eŜ Ψ = Ψ Ĥ(ε) Ψ + Ψ [Ĥ(ε), Ŝ] Ψ Ψ [[Ĥ(ε), Ŝ], Ŝ] Ψ +... which leads to d dε [ E(S, ε) [ ] S(ε) S = V S=S(ε)] [1] + E [2] ε = where the hessian matrix elements equal E [2] ij = 1 2 Ψ [[Ĥ, ˆR i ˆR ] i, ˆR j ˆR ] Ψ j Ψ [[Ĥ, ˆR j ˆR ] j, ˆR i ˆR ] Ψ i

9 June 215 ISTPC 215 summer school, Aussois, France Page 9 Example: exact response theory In summary: E [2] [ S(ε) ε ] = V [1] [ ] S(ε) ( ) 1V ε = E [2] [1] ˆV ; ˆV = [ ] S(ε) T ε V [1] ˆV ; ˆV = ( V [1]) ( ) T 1V E [2] [1] Conclusion: in order to compute linear response functions, the gradient property vector and the hessian matrix are needed. EXERCISE: Show that V [1] i = 2Ψ i ˆV Ψ, E [2] ij = 2(E i E )δ ij, S i (ε) ε = Ψ i ˆV Ψ, and ˆV ; ˆV = 2 E E i i> Ψ i ˆV Ψ 2 E E i second-order perturbation theory!

10 June 215 ISTPC 215 summer school, Aussois, France Page 1 Some comments before turning to the time-dependent regime Let us return to (approximate) variational methods. X = λ(ε) ε is usually referred to as the linear response vector. Like in the exact theory, the linear response equation writes E [2] X = V [1] In the time-dependent regime, a linear response vector will be obtained for each frequency ω. We will show in the following that the linear response equation writes ( E [2] ωs [2]) X(ω) = V [1] What about non-variational methods such as MP2, CC, CI with ε-dependent HF orbitals? We, in principle, do not have a stationarity condition anymore. How to proceed with the derivation of the response equations then? What about the Hellmann Feynman theorem?

11 June 215 ISTPC 215 summer school, Aussois, France Page 11 Let us denote t the non-variational parameters (CC amplitudes for example). For each perturbation strength, a set of equations has to be solved: f(t(ε), ε) = The non-variational energy is then determined for each perturbation strength: E(ε) = E(t(ε), ε) We introduce the Lagrangian function: L(t, ε, t) = E(t, ε) + t T f(t, ε) and impose the following stationarity conditions: ε, L(t, ε, t) t = = f(t, ε) and L(t, ε, t) t = = E(t, ε) t + t T f(t, ε) t t(ε) t(ε) Note that E(ε) = L(t(ε), ε, t) Hellmann Feynman theorem: de(ε) dε = L(t, ε, t) ε = t(ε),t(ε) E(t, ε) ε + t T (ε) t(ε) f(t, ε) ε t(ε)

12 June 215 ISTPC 215 summer school, Aussois, France Page 12 Time-dependent variational principle Time-dependent Schrödinger equation: Ĥ(t) Ψ(t) = i d dt Ψ(t) Alternative formulation based on a change of phase: Ψ(t) = e i t t Q(t)dt Ψ(t) Ĥ(t) Ψ(t) i d dt Ψ(t) = Q(t) Ψ(t) Connection with the Runge Gross theorem: two local potentials that differ by a real time-dependent function lead to the same time-dependent density for a given initial wavefunction Ψ(t ): Ψ(t ) = Ψ(t ) and n Ψ(t) (r) = Ψ(t) ˆn(r) Ψ(t) = Ψ(t) ˆn(r) Ψ(t) = n Ψ(t) (r) In the particular case of a time-independent Hamiltonian Ĥ, searching for time-independent solutions Ψ(t) = Ψ and Q(t) = E leads to the time-independent Schrödinger equation: Ĥ Ψ = E Ψ

13 June 215 ISTPC 215 summer school, Aussois, France Page 13 Returning to the time-dependent regime, Q(t) is referred to as time-dependent quasienergy. Since d Ψ(t) Ψ(t) = iĥ(t)ψ(t) Ψ(t) iψ(t) Ĥ(t) Ψ(t) =, dt if Q(t) is real then Ψ(t) Ψ(t) = Ψ(t) Ψ(t) = Ψ(t ) Ψ(t ) = 1 and Q(t) = Ψ(t) Ĥ(t) i d dt Ψ(t) The real character of the time-dependent quasienergy can be explicitly connected with the conservation of the norm: Q(t) = Ψ(t) Ĥ(t) Ψ(t) d + i Ψ(t) dt Ψ(t) = Ψ(t) Ĥ(t) Ψ(t) d i Ψ(t) Ψ(t) dt = Q(t) since d Ψ(t) dt Ψ(t) = d d Ψ(t) Ψ(t) Ψ(t) Ψ(t) dt dt }{{}

14 June 215 ISTPC 215 summer school, Aussois, France Page 14 Time-dependent variational principle For a given trial wavefunction Ψ(t), we define the action integral as follows A[Ψ] = t1 t Q[Ψ](t) dt where Q[Ψ](t) = 1 Ψ(t) Ψ(t) Ψ(t) Ĥ(t) i d dt Ψ(t) Note that Q[ Ψ](t) = Q(t). Stationarity condition: δa[ Ψ] = Ĥ(t) Ψ(t) i d dt Ψ(t) = Q(t) Ψ(t) variational formulation non-variational formulation Proof: let us consider variations Ψ(t) Ψ(t) + δψ(t) around the exact solution Ψ(t) with the boundary conditions δψ(t ) = δψ(t 1 ) =.

15 June 215 ISTPC 215 summer school, Aussois, France Page 15 Consequently, the action integral varies as follows: δa[ Ψ] = A[ Ψ + δψ] A[ Ψ] = = t1 t where t1 t δψ(t) Ĥ(t) i d dt Ψ(t) dt + t1 t t1 t Q(t) ( Q[ Ψ + δψ](t) Q[ Ψ](t) ) t1 t ( δψ(t) Ψ(t) + Ψ(t) δψ(t) ) Ψ(t) dδψ(t) dt = dt t1 dt Ψ(t) d Ĥ(t) i dt δψ(t) dt dt d δψ(t) Ψ(t) dt t dt }{{} t1 t d Ψ(t) dt δψ(t) dt thus leading to [ Ψ(t) δψ(t) ] t1 t = δa[ Ψ] = t1 t ( δψ(t) Ĥ(t) i d dt Q(t) Ψ(t) + δψ(t) Ĥ(t) i d ) dt Q(t) Ψ(t) dt

16 June 215 ISTPC 215 summer school, Aussois, France Page 16 Variational principle in adiabatic TD-DFT In time-dependent density-functional theory (TD-DFT), the physical time-dependent Hamiltonian is written as Ĥ(t) = ˆT + Ŵee + dr v(r, t)ˆn(r). }{{} ˆV (t) time-dependent local potential operator In standard TD-DFT, the exact time-dependent exchange-correlation (xc) potential is approximated with the ground-state xc density-functional potential calculated at the time-dependent density (adiabatic approximation): ˆT + ˆV (t) + dr δe Hxc [n ΦKS (t) δn(r) ] ˆn(r) i d Φ KS (t) = Q KS (t) Φ KS (t) dt ΦKS where n ΦKS (t) (r) = (t) ˆn(r) Φ KS (t) time-dependent density n Ψ(t) (r). is an approximation to the exact physical

17 June 215 ISTPC 215 summer school, Aussois, France Page 17 EXERCISE: (1) Show that, within the adiabatic approximation, the Kohn Sham TD-DFT equation is equivalent to the stationarity condition δa adia [ Φ KS ] = where, for a trial wavefunction Ψ(t), A adia [Ψ] = t1 t 1 Ψ(t) Ψ(t) Ψ(t) ˆT + ˆV (t) i d dt Ψ(t) dt + t1 t E Hxc [n Ψ(t) ] dt and n Ψ(t) (r) = Ψ(t) ˆn(r) Ψ(t). Ψ(t) Ψ(t) (2) Within the adiabatic approximation, the equation to be solved in TD range-separated DFT is ˆT + Ŵ lr,µ ee + ˆV (t) + dr δe sr,µ Hxc [ ] n Ψµ (t) δn(r) ˆn(r) i d Ψ µ (t) = Q µ (t) Ψ µ (t). dt Show that it is equivalent to δa µ adia [ Ψ µ ] = where t1 A µ adia [Ψ] = 1 t Ψ(t) Ψ(t) Ψ(t) ˆT + Ŵ ee lr,µ + ˆV (t) i d dt Ψ(t) dt + t1 t E sr,µ Hxc [n Ψ(t)] dt

18 June 215 ISTPC 215 summer school, Aussois, France Page 18 Floquet theory In the following we consider a periodic Hamiltonian with period T : Ĥ(t + T ) = Ĥ(t). Ĥ(t) can be written as a Fourier series: Ĥ(t) = ˆT + Ŵee + ˆV ne + N e iωkt ε x (ω k ) ˆV x, x k= N }{{} ˆV(t) time-dependent perturbation where ω k = 2πk T and ε x (ω k ) is the strength of the perturbation ˆV x at frequency ω k. ˆV x is any kind of (hermitian) operator, not necessarily a one-electron operator even though in practice it usually is. In order to apply TD-DFT, ˆV x should in principle be a (one-electron) local potential operator: ˆV x dr v x (r)ˆn(r)

19 June 215 ISTPC 215 summer school, Aussois, France Page 19 Floquet theory Example 1: in the presence of a dynamic uniform electric field, E(t) = E x (t) e x + E y (t) e y + E z (t) e z = N k= N e iω kt (ε x (ω k )e x + ε y (ω k )e y + ε z (ω k )e z ), the perturbation is ˆV(t) = ˆr. E(t) = ˆx Ex (t) + ŷ E y (t) + ẑ E z (t) thus leading to ˆV(t) = N k= N e iω kt ( ) ε x (ω k ) ˆx + ε y (ω k ) ŷ + ε z (ω k ) ẑ. Comment: note that ˆr is written in second quantization as ˆr = r ˆn(r)dr so that ˆV(t) = r. E(t) ˆn(r) dr local potential operator!

20 Floquet theory Example 2: in the presence of a dynamic uniform magnetic field, B(t) = B x (t) e x + B y (t) e y + B z (t) e z = N k= N e iω kt (b x (ω k )e x + b y (ω k )e y + b z (ω k )e z ), 1 the perturbation equals ˆV(t) = ˆL. B(t) 2 thus leading to ˆV(t) = N k= N e iω kt ( b x (ω k ) ˆL x 2 + b y(ω k ) ˆL y 2 + b z(ω k ) ˆL z 2 ). Comment: note that ˆL can be written as ˆL = i ˆV(t) = i 2 B(t). ( r r ˆn 1 (r, r) ) r =r σ ˆΨ σ(r) r r ˆΨσ (r) dr so that dr non-local potential operator! June 215 ISTPC 215 summer school, Aussois, France Page 2

21 June 215 ISTPC 215 summer school, Aussois, France Page 21 EXERCISE: (1) By using the hermiticity of ˆL show that ˆL = equals r ĵ(r) dr where the current density operator ĵ(r) = 1 2i ( ˆΨ σ (r) r ˆΨ σ (r) ( r ˆΨ σ (r) ) ) ˆΨσ (r) σ (2) Show that the perturbation can be expressed as ˆV(t) = ˆµmag. B(t) where the magnetic dipole moment operator equals ˆµ mag = 1 2 r ĵ(r) dr (3) Explain why TD-DFT is in principle not adequate for modeling such a perturbation. Show that the paramagnetic current density j p (r, t) = Ψ(t) ĵ(r) Ψ(t) would be a better variable to consider (rather than the density).

22 June 215 ISTPC 215 summer school, Aussois, France Page 22 Floquet theory Let us collect all perturbation strengths into the vector ε =. ε x (ω k ). ˆV (t) = ˆV(t) ε x ( ω k ) = ε x (ω k ) The time-dependent wavefunction varies with the perturbation strengths: Ψ(t) Ψ(ε, t) Choice of the phase: we want the time-dependent wavefunction to reduce to the (time-independent) ground-sate wavefunction Ψ in the absence of perturbation, Ψ(ε =, t) = Ψ. In the following, the action integral will be calculated over a period: t = and t 1 = T.

23 June 215 ISTPC 215 summer school, Aussois, France Page 23 Response functions Taylor expansion of the time-dependent expectation value for the perturbation ˆV x : ˆV x (ε, t) = Ψ(ε, t) ˆV x Ψ(ε, t) = Ψ ˆV x Ψ zeroth order + y e iωkt ε y (ω k ) ˆV x ; ˆV y ωk k linear response e i(ω k+ω l )t ε y (ω k )ε z (ω l ) ˆV x ; ˆV y, ˆV z ωk,ω l y,z k,l quadratic response +... We will focuse in the following on the exact and approximate description of the linear response functions ˆV x ; ˆV y ωk.

24 June 215 ISTPC 215 summer school, Aussois, France Page 24 Hellmann Feynman theorem in the time-dependent regime The exact action integral depends both implicitly (through the time-dependent wavefunction) and explicitly (through the perturbation) on the perturbation strengths ε: [ ] A(ε) = A Ψ(ε), ε where A [Ψ, ε] = T 1 Ψ(t) Ψ(t) Ψ(t) Ĥ + ˆV(t) i d dt Ψ(t) dt and Ĥ = ˆT + Ŵee + ˆV ne unperturbed Hamiltonian Ψ(ε, [ ] t) is determined from the variational principle: ε, δa Ψ(ε), ε =

25 June 215 ISTPC 215 summer school, Aussois, France Page 25 Hellmann Feynman theorem in the time-dependent regime Let us consider the variation ε x (ω k ) ε x (ω k ) + dε x (ω k ): ( ) ( ) da(ε) = A ε x (ω k ) + dε x (ω k ) A ε x (ω k ) = [ A[Ψ, ε] ε x (ω k ) dε x (ω k ) + A Ψ(ε) + Ψ(ε) ] [ ] Ψ=Ψ(ε) ε x (ω k ) dε x(ω k ), ε A Ψ(ε), ε }{{} [ ] δa Ψ(ε), ε = thus leading to the Hellmann Feynman theorem da(ε) dε x (ω k ) = A[Ψ, ε] ε x (ω k ) Ψ=Ψ(ε)

26 June 215 ISTPC 215 summer school, Aussois, France Page 26 ˆV(t) ε x (ω k ) = e iω kt ˆVx da(ε) dε x (ω k ) = T e iω kt ˆV x (ε, t) dt Important consequence: response functions can be expressed as action integral derivatives! zeroth order: da(ε) dε x (ω k ) = T e iω kt Ψ ˆV x Ψ dt = T Ψ ˆV x Ψ δ(ω k ) thus leading to Ψ ˆV x Ψ = 1 T da(ε) dε x () Linear response: d 2 A(ε) dε y (ω l )dε x (ω k ) = T e i(ω k+ω l )t ˆV x ; ˆV y ωl dt = T ˆV x ; ˆV y ωl δ(ω k + ω l ) thus leading to ˆV x ; ˆV y ωl = 1 T d 2 A(ε) dε y (ω l )dε x ( ω l )

27 June 215 ISTPC 215 summer school, Aussois, France Page 27 Some general statements before deriving more equations... (Linear) response functions can be expressed as derivatives of the action integral with respect to the perturbation strengths. Such a formulation is convenient for deriving exact and approximate expressions for the response functions. In the latter case, non-variational methods such as Coupled-Cluster (CC) theory can also be considered (Lagrangian formalism). Various (approximate) parameterizations of the time-dependent wavefunction Ψ(ε, t) will lead to various response theories. Variational methods such as HF and MCSCF will be considered in the following. Adiabatic TD-DFT equations (Casida equations) can be obtained similarly. TD linear response CC theory can be derived by means of a Lagrangian formalism (in analogy with time-independent CC response theory).

28 June 215 ISTPC 215 summer school, Aussois, France Page 28 In the exact theory, ˆV x ; ˆV y ωl = 1 T T e iω lt Ψ(ε, t) ε y (ω l ) ˆV x Ψ + }{{} linear response of the wavefunction (first order in perturbation theory) Ψ(ε, t) ε y (ω l ) ˆV x Ψ dt Note that, in the static case, the action integral over T becomes the energy. Consequently, the standard second-order energy correction Ψ ˆV x Ψ (1) is recovered. Linear and higher-order responses of the wavefunction are obtained through differentiations of the stationarity condition with respect to the perturbation strengths: ( d [ ] ) δa Ψ(ε), ε dε y (ω l ) = Ψ(ε, t) ε y (ω l )

29 June 215 ISTPC 215 summer school, Aussois, France Page 29 Wavefunction parameterization Double-exponential parameterization of a trial wavefunction: Ψ(t) = e iˆκ(t) e iŝ(t) Ψ The hermitian operators ˆκ(t) and respectively. Ŝ(t) ensure rotations in the orbital and configuration spaces, Fourier series: ˆκ(t) = l,i e iω lt κ i (ω l )ˆq i + e iω lt κ i ( ω l)ˆq i where ˆq i = Êpq and p > q, Ŝ(t) = l,i e iω lt S i (ω l ) ˆR i + e iω lt S i ( ω l) ˆR i where ˆR i = i Ψ. The time-dependent wavefunction is fully determined from the Fourier component vectors κ i (ω l ) S i (ω l ) Λ(ω l ) = κ i ( ω to be used as variational parameters! l) Si ( ω l)

30 June 215 ISTPC 215 summer school, Aussois, France Page 3 Such a parameterization will enable us to derive an exact response theory when ˆκ(t) = and ˆR i = Ψ i Ψ with i > and k, Ĥ Ψ k = E k Ψ k. HF response theory (RPA) when Ŝ(t) =, Ψ Φ (HF determinant), and ˆq i Êaj (single excitation from the occupied j orbital to the unoccupied a orbital) MCSCF response theory when Ψ Ψ () (MCSCF wavefunction), ˆR i det i Ψ () (rotation within the active space), and ˆq i Êuj, Êaj, Êau.

31 June 215 ISTPC 215 summer school, Aussois, France Page 31 Response properties from adiabatic TD-DFT The HF parameterization enables also to derive standard TD-DFT response equations: for pure exchange functionals, the action integral expression to be used is A adia [Ψ, ε] = T Ψ(t) ˆT + ˆV ne + ˆV(t) i d dt Ψ(t) dt + T E Hxc [n Ψ(t) ] dt for hybrid exchange functionals, the action integral expression to be used is A α adia [Ψ, ε] = T Ψ(t) ˆT + ˆV ne + αŵee + ˆV(t) i d dt Ψ(t) dt + T (1 α)e Hx [n Ψ(t) ] dt + T E c [n Ψ(t) ] dt

32 June 215 ISTPC 215 summer school, Aussois, France Page 32 Response properties from adiabatic TD-DFT The MCSCF parameterization enables also to derive multiconfiguration range-separated TD-DFT equations: the action integral expression to be used is, in this case, A µ adia [Ψ, ε] = T Ψ(t) ˆT + Ŵ ee lr,µ + ˆV ne + ˆV(t) i d dt Ψ(t) dt + T E sr,µ Hxc [n Ψ(t)] dt For sake of generality, we will derive, in the following, response equations for a mixed wavefunction/density-functional variational action integral: A [Ψ, ε] A var [Ψ, ε] = T Ψ(t) Ĥ + ˆV(t) i d dt Ψ(t) dt + T E Hxc [n Ψ(t) ] dt

33 June 215 ISTPC 215 summer school, Aussois, France Page 33 Hellmann Feynman theorem for time-dependent variational methods A var [Ψ, ε] = T Ψ(t) Ĥ + ˆV(t) i d dt Ψ(t) dt + T E Hxc [n Ψ(t) ] dt Let us keep in mind that the wavefunction Ψ is determined from the vector Λ {Λ(ω l )} l The action integral will therefore be denoted A var (Λ, ε) in the following. For any perturbation strength ε, Λ(ε) is obtained from the stationarity condition: A var (Λ, ε) ε, Λ = Λ=Λ(ε) Consequently, the Hellmann-Feynman theorem is fulfilled for the variational action integral A var (ε) = A var (Λ(ε), ε), exactly like in the exact theory: da var (ε) dε = A var(λ, ε) ε Λ=Λ(ε)

34 June 215 ISTPC 215 summer school, Aussois, France Page 34 Linear response functions Therefore, in analogy with the exact theory, ˆV x ; ˆV y ωl = 1 T where d 2 A var (ε) dε y (ω l )dε x ( ω l ) da var (ε) dε x ( ω l ) = = T T e iω lt Ψ(t) ˆV x Ψ(t) e iω lt Ψ e iŝ(t) e iˆκ(t) ˆVx e iˆκ(t) e iŝ(t) Ψ dt dt thus leading to da var (ε) dε x ( ω l ) = T e iω lt [ Ψ ˆVx Ψ + i = T δ(ω l ) Ψ ˆVx Ψ ] [ ] [ ] Ψ ˆVx, ˆκ(t) Ψ + i Ψ ˆVx, Ŝ(t) Ψ it V [1] x Λ(ω l ) +... dt

35 June 215 ISTPC 215 summer school, Aussois, France Page 35 Linear response functions where the gradient property vector is defined as V [1] x = Ψ [ˆq i, ˆV x ] Ψ Ψ [ ˆR i, ˆV x ] Ψ Ψ [ˆq i, ˆV x ] Ψ Ψ [ ˆR i, ˆV x ] Ψ Conclusion: ˆV x ; ˆV y ωl = i V [1] x Λ(ω l ) ε y (ω l ) We now need to derive the linear response equation that is fulfilled by the linear response vector X y (ω l ) = Λ(ω l) ε y (ω l ).

36 June 215 ISTPC 215 summer school, Aussois, France Page 36 Linear response equation d dε y (ω m ) [ A var (Λ, ε) Λ ( ω l ) Λ=Λ(ε) ] = where A var [Λ, ε] = T T T Ψ(t) Ĥ Ψ(t) dt + E Hxc [n Ψ(t) ] dt + Ψ(t) ˆV(t) T Ψ(t) dt + Ψ(t) i d dt Ψ(t) dt }{{}}{{}}{{}}{{} AĤ [Λ] A Hxc [Λ] A ˆV [Λ, ε] A d/dt [Λ]

37 June 215 ISTPC 215 summer school, Aussois, France Page 37 Linear response equation A ˆV [Λ, ε] = x N k= N [ ε x (ω k ) T δ(ω k ) Ψ ˆVx Ψ p + it δ(ω k + ω p ) V [1] x Λ(ω p ) }{{} V x [1] Λ(ω p ) 1 2 Λ ( ω p )V x [1] ] d dε y (ω m ) [ A ˆV (Λ, ε) Λ ( ω l ) Λ=Λ(ε) ] = it 2 δ(ω m + ω l ) V [1] y EXERCISE: Let ˆf(x, t) = e xâ(t) d dt exâ(t). Show that ˆf(1, t) = 1 ˆf(x, t) x dx = dâ(t) dt [ ] dâ(t), dt Â(t) +...

38 June 215 ISTPC 215 summer school, Aussois, France Page 38 Linear response equation Using (e iˆκ(t) e iŝ(t) ) = e iŝ(t) (e iˆκ(t) ddt eiˆκ(t) ) e iŝ(t) e iˆκ(t) d dt e iŝ(t) + e iŝ(t) d dt eiŝ(t) leads to A d/dt [Λ] = T Ψ dˆκ(t) dt + dŝ(t) dt Ψ dt +i T Ψ 1 2 [ dˆκ(t) dt ], ˆκ(t) [ ] dŝ(t) [ ] dˆκ(t), dt Ŝ(t) +, dt Ŝ(t) Ψ dt +... d dε y (ω m ) [ Ad/dt [Λ] Λ ( ω l ) Λ=Λ(ε) ] = T 2 ω l S [2] Λ( ω l ) ε y (ω m )

39 June 215 ISTPC 215 summer school, Aussois, France Page 39 Linear response equation where S [2] = Σ, Σ Σ = = Ψ [ˆq i, ˆq j ] Ψ Ψ [ˆq i, ˆR j ] Ψ Ψ [ ˆR i, ˆq j ] Ψ Ψ [ ˆR i, ˆR j ] Ψ, Ψ [ˆq i, ˆq j ] Ψ Ψ [ˆq i, ˆR j ] Ψ Ψ [ ˆR i, ˆq j ] Ψ Ψ [ ˆR i, ˆR. j ] Ψ

40 June 215 ISTPC 215 summer school, Aussois, France Page 4 Linear response equation Using the BCH expansion leads to AĤ [Λ] = T ] ] Ψ Ψ Ĥ + i [Ĥ, ˆκ(t) + i [Ĥ, Ŝ(t) dt T Ψ 1 2 [[Ĥ, ˆκ(t) ] ], ˆκ(t) ] ] ] [[Ĥ, Ŝ(t)], Ŝ(t) + [[Ĥ, ˆκ(t), Ŝ(t) Ψ dt +... d dε y (ω m ) [ AĤ [Λ] Λ ( ω l ) Λ=Λ(ε) ] = T 2 E[2] Λ( ω l ) ε y (ω m )

41 June 215 ISTPC 215 summer school, Aussois, France Page 41 Linear response equation where E [2] = A B B A, A = B = Ψ [ˆq i, [Ĥ, ˆq j ]] Ψ Ψ [[ˆq i, Ĥ], ˆR j ] Ψ Ψ [ ˆR i, [Ĥ, ˆq j ]] Ψ Ψ [ ˆR i, [Ĥ, ˆR j ]] Ψ, Ψ [ˆq i, [Ĥ, ˆq j]] Ψ Ψ [[ˆq i, Ĥ], ˆR j ] Ψ Ψ [ ˆR i, [Ĥ, ˆq j]] Ψ Ψ [ ˆR i, [Ĥ, ˆR. j ]] Ψ

42 June 215 ISTPC 215 summer school, Aussois, France Page 42 Linear response equation DFT-type contribution: d dε y (ω m ) A Hxc [Λ] Λ ( ω l ) Λ=Λ(ε) = d dε y (ω m ) [ T dt dr δe Hxc[n Ψ(t) ] δn(r) n Ψ(t) (r) Λ ( ω l ) Λ=Λ(ε) ] = T dt dr δe Hxc[n Ψ ] δn(r) d dε y (ω m ) [ nψ(t) (r) Λ ( ω l ) Λ=Λ(ε) ] potential! + T dt dr dr δ2 E Hxc [n Ψ ] δn(r )δn(r) n Ψ(t) (r) n Ψ(t) (r ) Λ ( ω l ) kernel! ε y (ω m ) The "potential" term is simply taken into account with the substitution, Ĥ Ĥ + dr δe Hxc[n Ψ ] δn(r) ˆn(r)

43 June 215 ISTPC 215 summer school, Aussois, France Page 43 Linear response equation Using the expressions n Ψ(t) (r ) = i ε y (ω m ) p e iωpt n [1] (r ) Λ(ω p) ε y (ω m ) and n Ψ(t) (r) Λ ( ω l ) = i 2 e iωlt n [1] (r), the "kernel" contribution can be rewritten as follows, T 2 dr dr δ2 E Hxc [n Ψ ] δn(r )δn(r) n[1] (r)n [1] (r ) }{{} Λ( ω l ) ε y (ω m ) K Hxc kernel matrix

44 June 215 ISTPC 215 summer school, Aussois, France Page 44 Linear response equation Conclusion: in the particular case of wavefunction linear response theory (no DFT contributions), the linear response equations to be solved are (E [2] + ω l S [2] ) Λ( ω l ) ε y (ω m ) = iδ(ω m + ω l ) V y [1] thus leading to and (E [2] ω l S [2] ) Λ(ω l ) ε y (ω l ) = i V y [1] ˆV x ; ˆV y ωl = i V [1] x ( ) 1 Λ(ω l ) ε y (ω l ) = V x [1] E [2] ω l S [2] V y [1]

45 June 215 ISTPC 215 summer school, Aussois, France Page 45 EXERCISE: (1) Show that, in exact response theory, Σ ij = δ ij, ij =, A ij = δ ij (E i E ), and B ij =. (2) Show that V x [1] = Ψ i ˆV x Ψ Ψ ˆV x Ψ i (3) Conclude that ˆV x ; ˆV y ω = i> ( Ψ ˆV x Ψ i Ψ i ˆV y Ψ E i E ω + Ψ i ˆV x Ψ Ψ ˆV ) y Ψ i E i E + ω (4) Using real algebra and the formula + a a 2 + ω 2 dω = π, prove the fluctuation dissipation theorem 2 + Ψ ˆV x ˆVy Ψ Ψ ˆV x Ψ Ψ ˆV y Ψ = 1 π ˆV x ; ˆV y iω (5) The so-called "response function" is defined in Physics as χ(r, r, ω) = ˆn(r); ˆn(r ) ω. Conclude that Ψ ˆn(r)ˆn(r ) Ψ n (r)n (r ) = 1 π + χ(r, r, iω)

46 .9.8 (a) H 2 [n 1 Σ g +, n=2 5] FCI (2 1 + Σ g ) TD MCSCF (2 1 + Σ g ) TD LDA (2 1 + Σ g ) Excitation energy (a.u.) June 215 ISTPC 215 summer school, Aussois, France Page 46

47 Excitation energy (a.u.) (b) H 2 [n 1 Σ g +, n=2 5] FCI (2 1 + Σ g ) TD HF srlda (2 1 + Σ g ) TD MC srlda (2 1 + Σ g ) Interatomic distance (a.u.) E. Fromager, S. Knecht and H. J. Aa. Jensen, J. Chem. Phys. 138, 8411 (213) June 215 ISTPC 215 summer school, Aussois, France Page 47

48 June 215 ISTPC 215 summer school, Aussois, France Page 48

49 June 215 ISTPC 215 summer school, Aussois, France Page 49 Some references J. Olsen and P. Jørgensen, J. Chem. Phys. 82, 3235 (1985). O. Christiansen, P. Jørgensen, and C. Hättig, Int. J. Quantum Chem. 68, 1 (1998). E. Fromager, S. Knecht and H. J. Aa. Jensen, J. Chem. Phys. 138, 8411 (213). E. Hedegård, F. Heiden, S. Knecht, E. Fromager, and H. J. Aa. Jensen, J. Chem. Phys. 139, (213). F. Pawlowski, J. Olsen, and P. Jørgensen, J. Chem. Phys. 142, (215).