Solutions to selected problems from Giuliani, Vignale : Quantum Theory of the Electron Liquid

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1 Solutions to selected problems from Giuliani, Vignale : Quantum Theory of the Electron Liquid These problems have been solved as a part of the Independent study class with prof. Neepa Maitra. Compiled by Juraj Tekel City University of New York Graduate Center, PhD. program in physics Spring term The universal functional F [n] as a Legendre transform. Let E[V ] be the ground state energy of an electronic system, regarded as a functional of the external potential V ( r). (a) Show that this functional is concave, that is, for any pair of potentials V 1 ( r) and V ( r) and for any α [0, 1] one has E[V ] αe[v 1 ] + (1 α)e[v ] where V ( r) = αv 1 ( r)+(1 α)v ( r). Also show, that E[V ] is monotonically decreasing, that is, if V 1 ( r) V ( r) for all r, then E[V 1 ] E[V ]. (b) Show that the ground state density n( r) is given by the functional derivative δe[v ] δv ( r). (c) Finally show, that the Legendre transform of E[V ] with respect to V ( r), which exist by virtue of concavity, is just the universal functional F [n]. by Solution. Let us recall the definitions of the used terms. The energy of the system is given E[V, n] = F [n] + V ( r)n( r)d r (7.1.1) with the universal functional defined as F [n] = min ψ n( r) ψ ˆT + Ĥe e ψ (7.1.) (a) We insert V ( r) = αv 1 ( r) + (1 α)v ( r) into the definition of E[V ] to get E[V ] = F [n] + [αv 1 ( r) + (1 α)v ( r)]n( r)d r = αf [n] + (1 α)f [n] + αv 1 ( r)n( r)d r + (1 α)v ( r)n( r)d r = E[V 1 ] + (1 α)e[v ] (7.1.3) 1

2 For V 1 ( r) V ( r) we get V 1 ( r)n( r)d r V 1 ( r)n( r)d r (7.1.4) since n( r) 0 for all r and thus for any F [n] we get (b) Since F [n] is universal for any V ( r), thus δf δv E[V 1 ] E[V ] (7.1.5) δe[v ] δv ( r) = 0, we get 1 = n( r) (7.1.7) (c) We again need just to recall the definition of the Legendre transform with respect to variables x i, which is A new = A old x i y i (7.1.8) where y i are the new variables and A old, A new is untransformed and transformed function respectively. In the case of functionals and the continuous index i r the sum turns into a integral and we get E new [n] = E old [V ] V ( r) }{{} n( r) }{{} old variable new variable d r = F [n] (7.1.9) 7. Hohenberg-Kohn potential for a one-electron system. Given a strictly positive one-electron density n( r), such that n( r)d r = 1, find an expression for the one-electron potential V ( r) that yields n( r) in the ground-state. Solution. We can take two different approaches to this problem, an easy one and a more general one. First, since we are looking for the ground state, we need a non-negative wave-function, so we can take ψ( r) = n( r). Inserting this into the Schrodinger equation Hψ = Eψ and treating it as an equation for an unknown potential with a known wave function, we get We can also use the general expression V ( r) = n( r) + const (7..1) m n( r) δf [n] V ( r) = δn( r) (7..) 1 Here we recall the definition of the functional derivative δf [f(r)] F [f(r) + δf(r)] = F [f(r)] + δf(r)dr (7.1.6) δf(r)

3 where for a single particle H e e = 0 and thus F [n] = ψ ˆT ψ = d r n( r) n( r) = m m ( d r n( r)) (7..3) where we have used (f g) = f g+f g and the boundary condition on the wave function, which is ψ 0 and ψ 0 as r. Now we need to compute the functional derivative of this expression. We take n n + δn so n n + 1 δn n and we get F [n + δn] = m = m [ d r d r = F [n] m ( )] n 1 δn + = n [ ( ) ( ) ( ) δn n + n + n d r ( ) δn n = F [n] + n m ( ( )) ] δn = n d r n n δn (7..4) where we have used the same trick once again and we have neglected the term containing (δn). We can see, that recalling the definition of the functional derivative (7.1.6) we get the same result (7..1). Note, that we could a very similar calculation from he expression F [n] = m d r n( r) n( r) (7..5) By the law of the conservation of difficulty, we would still have to do the trick twice and we would still have to neglect one term containing (δn). 7.3 Example for non-v-representable density in a one-electron system. Show that the one-electron density n( r) = ( r ) e r π 3/ a 3 a a (7.3.1) where a is an arbitrary length scale, cannot be the ground-state density of an electron subjected to a potential that is everywhere finite, but can be realized in an excited state of such potential. Identify this potential. Solution. If this density is to be the ground state, the wave function needs to be ψ( r) = If we now use the result of the previous problem r π 3/ a 5 re a (7.3.) V ( r) = n( r) (7.3.3) m n( r) Discussion what this exactly means and why this is so will follow later in the solution of this problem. 3

4 use the expression for the Laplace operator in spherical coordinates = 1 r r r, carry out r all the derivatives and correct all the errors we get the following expression for the potential ( V ( r) = r m a 5 4 a + ) + conts (7.3.4) r what troubles us is the last term, since the first one is well known potential of an isotropic 3D harmonic oscillator and thr second term is just a constant that could be thrown away. However the third term is non-constant and is singular at the origin. Thus, we have shown that if (7.3.1) is considered as a ground state density, the potential corresponding to it is singular. What to do now? As the problems suggests, we will try to treat the density as an excited state density. What does this mean? Every wave function in the form ψ = ne iα( r), with α a real function of r, reproduces the density n. However, for a non-zero (non-constant) phase, this state is excited, since the wave function has nodes. We,however, still can insert such a wave function into the Schrodinger equation and compute the corresponding potential m ( ne iα( r) ) + V ( r) ne iα( r) = E ne iα( r) [ ( n ( α) + i α + α )] n + E = V ( r) (7.3.5) m n n The imaginary part of the potential should vanish. It does indeed, as we can see using the continuity equation n + j = 0. Since we have the stationary state, we get using the definition t of the quantum mechanical current (ψ ψ ψ ψ) = ψ ψ ψ ψ = 0 e iα[ ne iα i( α)( n)e iα ne iα ( α) i ne iα α ] = e iα[ ne iα + i( α)( n)e iα + ne iα ( α) + i ne iα α ] As we can see, the imaginary term in (7.3.5) vanishes and we get [ ] V ( r) = n ( α) m n ( α)( n) + n α = 0 (7.3.6) (7.3.7) We have already computed the first part of this expression, so we get for the density (7.3.1) ( V ( r) = const + r m a + ) 4 r m ( α) (7.3.8) To make the potential finite everywhere, we need ( α) = r (7.3.9) Obviously, α is a function of r = r only, thus we get α = r α(r) = log(r) + C (7.3.10) 4

5 A constant term in the phase is irrelevant, so we can set C = 0. This way, we got the potential of a harmonic oscillator, which has state r ψ( r) = π 3/ a 5 re a e i log(r) (7.3.11) as one of the excited states. We can see, that it will have an infinite number of nodes as r Exchange-correlation potential for a one-electron system. Show, that the exchange-correlation potential associated with a one-electron density such that n( r)d r = 1 is equal to the negative of the Hartree potential. Solution. Here, we just need to understand the definition of the exchange-correlation potential, which is F [n] = ψ ˆT + H e e ψ = T S [n] + E H [n] + E xc [n] (7.4.1) For a single particle, obviously H e e = 0, thus F [n] = T S [n] and we are left with which yields V xc = V H. E xc [n] = E H [n] (7.4.) 7.5 Exchange-correlation potential for a two-electron system. Show, that the exchange-correlation potential associated with a two-electron density such that n( r)d r = is given by V xc ( r) = n( r) m n( r) e r r n( r )d r V ext ( r) (7.5.1) where V ext ( r) is the external potential that yields n( r). Solution. Similarly as in previous problem we have F [n] = ψ ˆT + H e e ψ = T S [n] + E H [n] + E xc [n] (7.5.) Now, we take the functional derivative if the whole equation to obtain δf [n] δn }{{} V [n] = δt S[n] } δn {{} V KS + δe H [n] } δn {{} V H + δe xc [n] (7.5.3) } δn {{} V xc where we have used the definitions of corresponding potentials. We know, that V H = e n( r ) r r d r and V is just the external potential. We are left to calculate only the Kohn-Sham part of the potential. We know, that the density can be expressed as a sum over the Kohn-Sham orbitals n( r) = φ α ( r, s) (7.5.4) α s 5

6 where the first sum is over all occupied orbitals and the second over the spin states. The particles are non-interacting, thus they occupy the lowest energy level in two different spin states. We get the potential solving the Kohn-Sham equation m φ α + V KS φ α = ε α ψ α (7.5.5) for the first orbital φ 1 ( r, s = 1, ) = n( r)/, which yields (see problem 7.) V KS = n (7.5.6) m n Putting all these results together, we get the desired expression V xc ( r) = n( r) e m n( r) r r n( r )d r V ( r) (7.5.7) 7.16 Different time-dependent potentials can produce the same density. Consider two non-interacting electrons in 1 dimension. A possible initial state is the Slater determinant, Φ(x, x ) formed from φ 1 e x / and φ xe x /, i.e. the two lowest energy states of a particle in a harmonic oscillator potential with force constant k = 1. Another possible initial state is the Slater determinant Φ(x, x ) formed from the orthonormal orbitals φ 1 (x) = 1 + f(x)φ (x)φ 1 (x) φ (x) = 1 f(x)φ 1(x)φ (x) (7.16.1) where f(x) = c(56x 4 19x + 1)e x and c is an arbitrary constant with values between 0.17 and (a) Show, that these two states have the same density and the same time derivative of the density at the initial time (t = 0). (b) Let n( r, t) be the density of the state that evolves from Φ under time-dependent potential V ( r, t) and let Ṽ ( r, t) be the potential that generates the same density starting from the Φ (according to van Leeuwen s theorem such potential exists). Show, that Ṽ differs from V by more than a time-dependent constant. Solution. First, we need to make sure that our orbitals are normalized, i.e. dx φ = 1. This yields φ 1 (x) = 1 / / π 1/4 e x, φ (x) = π 1/4 xe x (7.16.) After this, we find out that orbitals φ 1, φ are properly normalized. The resulting states are thus Φ(x, x ) = 1 [φ 1 (x)φ (x ) φ 1 (x )φ (x)] = 1 e x +x (x x) π Φ(x, x ) = 1 [ φ 1 (x) φ (x ) φ 1 (x ) φ (x)] (7.16.3) 6

7 (a) For the density, we have 3 n(x) = dx Φ(x, x ) (7.16.4) This way we get n(x) = π e x dx e x (x + x xx ) = 1 π (x + 1)e x (7.16.5) For ñ we get ñ(x) = φ 1 (x) dx φ (x ) + φ (x) dx φ1 (x ) φ 1 (x) φ (x) dx φ1 (x ) φ (x ) (7.16.6) The first two integrals are equal to unity due to the normalization, the third integral vanishes, since we are integrating product of even ( φ 1 (x )) and odd ( φ (x )) functions over a symmetric interval. Finally, we get ñ(x) = 1 ] [e x [1 + f(x)φ π (x)] + x e x [1 f(x)φ 1(x)] = and both initial densities are equal. = 1 π e x (x + 1) (7.16.7) To compute the time derivatives of the densities, we use the continuity theorem n t = j (7.16.8) with the current 4 ĵ(x) = Im dx Φ (x, x ) Φ(x, x ) (7.16.9) In both these equations =. Now, since the orbitals we are dealing with are real, we x conclude that the integral will be also real and thus its imaginary part vanishes for both cases. Thus, we got n t = ñ 0 t ( ) 0 3 This can be derived from the density operator ˆn( r) = i δ( r ˆ r i ) using n(x) = Φ ˆn( r) Φ or by understanding the notion of Φ as a probability density. In both cases, we use the (anti)symmetry of the wave function. 4 This can be derived similarly as density form ĵ = ( i [ˆp iδ(r ˆr i ) + δ(r ˆr i )ˆp i ])/ 7

8 (b) In this part, we use an expression (17) from [1]. This expression gives explicit equation for potential Ṽ, that makes states Φ evolve to the same density n, as does potential V the state Φ. In our case and notation, it is x [ n ( V (x, t) Ṽ (x, t) ) + t(x, t) x ] = 0 ( ) where t(x, t) = ( )(ρ 1 (x, x ) ρ (x, x )) x=x (7.16.1) with ρ 1 (x, x ) = φ 1 (x)φ (x ) + φ (x)φ 1 (x ) is an off diagonal one-electron reduced matrix. There is a missing interaction term, which is however zero in our case. For the general expression, derivation and comments see the paper. For our particular orbitals (7.16.1) we get [ ρ 1 (x, x ) = φ 1 (x )φ (x) 1 +φ 1 (x)φ (x ) [ x = +x π xe 1 [ + π x e x +x 1 ] 1 + f(x )φ (x ) 1 f(x)φ 1(x) [ ] f(x)φ (x) 1 f(x )φ 1(x ) 1 + f(x ) π x e x 1 f(x) 1 π e x ] 1 + f(x) π x e x 1 f(x ) 1 π e x ] ( ) We now need just to compute (7.16.1). After few attempts to take the derivatives we, for the sake of our own sanity, use a computer algebra program to evaluate the expression. This is the Mathematica script I have used to compute this expression Clear[x,y,f,P1,P,TP1,TP,DR] f[x_]=c(56x^4-19x^+1)(\[exponentiale])^(-x^); P1[x_]=(\[Pi])^(-1/4)(\[ExponentialE])^(-x^/); P[x_]=()^(1/)(\[Pi])^(-1/4)x(\[ExponentialE])^(-x^/); TP1[x_]=(1+f[x](P[x])^)^(1/)P1[x]; TP[x_]=(1-f[x](P1[x])^)^(1/)P[x]; DR[x_,y_]=P1[x]P[y]+P1[y]P[x]-TP1[x]TP[y]-TP1[y]TP[x]; u=d[d[dr[x,y],{x,}]-d[dr[x,y],{y,}],x]/; y=x; FullSimplify[u] It should work when copy-pasted to the Mathematica just as written, both from the pdf document and the source tex file. The resulting expression is a ridiculous mess. Let us 8

9 therefor continue further with t as some function of x. From ( ) we get n ( V (x, t) Ṽ (x, t) ) + t(x, t) = C(t) x ( V (x, t) Ṽ (x, t) ) = C(t) t(x, t) π ( ) (x + 1)e x If we take this equation at time t = 0, we ll find the initial difference between the two potentials. Thus, if they are to differ by only a time dependent constant, the right hand side of this equation needs to vanish 5. Thus, t needs to be a constant, so then we can choose C(0) to make the whole expression vanish. However, the obtained expression could not be further from a constant, which means that the gradient of difference of potentials V and Ṽ is non-zero, thus these differ by more than a time-dependent constant. We can see, that for states having a different ρ 1, it is fairly impossible to generate a constant t. This is reasonable, since if we want to get the same density from two different states, we expect to need different potentials to do so Density and current density determine one electron state. (a) Show, that the ratio of the current density to the density j( r) n( r) purely longitudinal (i.e., it is the gradient of a scalar field). in a one-electron system is (b) Show that two one-electron states, that have the same density and the same current density must perforce coincide up to a global time-dependent phase. Solution. (a) We have n( r) = ψ ( r)ψ( r) and j( r) = i m [ψ ( r) ψ( r) ψ( r) ψ ( r)], so we get j( r) n( r) = i m [ψ ( r) ψ( r) ψ( r) ψ ( r)] ψ ( r)ψ( r) = i m log ( ψ( r) ψ ( r) ) = i m ( log ψ( r) log ψ ( r)) = (7.17.1) we now write the wave function in terms of the phase and the magnitude as ψ( r) = R( r)e is( r), which yields j( r) n( r) = i m log eis( r) = S( r) (7.17.) m which is the gradient of a scalar field, namely of the phase of the wave function. 5 In more than one dimension, it could be equal to a curl of some function, but we will not concern ourselves with this general case. 9

10 (b) We denote by index 1 quantities associated with one wave function and with with the other. We immediately have j 1 ( r) n 1 ( r) j ( r) n ( r) = 0 [ S 1 ( r) S ( r)] = 0 S 1 ( r) S ( r) = c(t) (7.17.3) and the phases of the wave functions can differ only by a non-space-dependent function. Since they produce the same density their amplitudes must coincide and we get desired statement ψ 1 ( r) = ψ ( r)e ic(t) (7.17.4) References [1] Maitra and Burke, Phys. Rev. A, (001) 10