University of Strasbourg, second year Master’s level

Exam in advanced quantum chemistry

January 2016

duration of the exam session: 2h

Neither documents nor calculators are allowed.

The grading scale might be changed.

1. Questions about the lectures (10 points)

Give detailed answers to the following questions:

a) [2 pts] What is the main di�erence between variational and non-variational quantum chemical methods

?

b) [2 pts] Explain why the Hartree–Fock (HF) method can be formulated as an orbitals rotation problem.

How can the HF orbitals be optimized ? How can we verify in a configuration interaction (CI) calculation

that the determinants are contructed from HF orbitals ? Does the full CI (FCI) energy depend on the

choice of orbitals ?

c) [1 pt] What is the motivation for using a Coupled-Cluster (CC) ansatz for the wavefunction ?

d) [2 pts] Define the concept of static correlation. What is the appropriate method for its description ?

How is the wavefunction parameterized within this method ? Is it possible to describe excited states

with this method ? Does it provide accurate results ?

e) [1 pt] Does the energy depend explicitly on the electron density ?

f) [2 pts] Is it possible to describe ground-state electronic structures exactly with one single electron

configuration ?

2. Problem: uniform Hubbard dimer (12 points)

Let us consider the Hubbard model Hamiltonian for the H2 molecule which is written in second quanti-

zation as follows,

H(t, U) = ≠t

Cÿ

‡=–,—

1a†

1sA,‡a1sB ,‡ + a†1sB ,‡a1sA,‡

2D

+ Ua†1sA,–a1sA,–a†

1sA,— a1sA,— + Ua†1sB ,–a1sB ,–a†

1sB ,— a1sB ,—, (1)

1

where t > 0 describes the delocalization of the electrons over the two hydrogen atoms while U > 0 models

the repulsion between two electrons localized on the same hydrogen atom (HA or HB). The operators a†1sA,‡

and a†1sB ,‡ create one electron with spin ‡ on the 1s orbital localized on HA and HB, respectively.

a) [2 pts] The orthonormal basis to be considered for a ground-state FCI calculation in this context is

| øA¿AÍ = a†1sA,–a†

1sA,— |vacÍ, (2)

| øB¿BÍ = a†1sB ,–a†

1sB ,— |vacÍ, (3)

| øA¿BÍ = a†1sA,–a†

1sB ,— |vacÍ, (4)

| ¿AøBÍ = a†1sA,— a†

1sB ,–|vacÍ. (5)

Show that, in this basis, the FCI Hamiltonian matrix equals HFCI(t, U) =

S

WWWWWWWU

U 0 ≠t t

0 U ≠t t

≠t ≠t 0 0

t t 0 0

T

XXXXXXXV

.

Hint: apply H(t, U) in Equation (1) to each ket of the basis first and simplify the expressions.

b) [1 pt] How would you show that the ground-state energy equals E(t, U) = 12

1U ≠

ÔU2 + 16t2

2?

c) [1 pt] The normalized ground state associated with E(t, U) can be expressed as follows,

|�(t, U)Í = C1| øA¿AÍ + C2| øB¿BÍ + C3| øA¿BÍ + C4| ¿AøBÍ. (6)

Show that C1 = C2, C3 = ≠C4, C21 + C2

3 = 12 and C1/C3 = ≠E(t, U)

2t.

d) [2 pts] Show from questions b) and c) that, when U/t æ 0, the exact normalized ground state becomes

|�(U/t æ 0)Í = 12

1| øA¿AÍ + | øB¿BÍ + | øA¿BÍ ≠ | ¿AøBÍ

2. Is there any electron correlation in the

state |�(U/t æ 0)Í ? Hint: introduce a†1‡g ,‡ = 1Ô

2

1a†

1sA,‡ + a†1sB ,‡

2, show that |�(U/t æ 0)Í =

a†1‡g ,–a†

1‡g ,— |vacÍ, and conclude.

e) [1 pt] Explain why, when U/t æ +Œ, the ground state becomes

|�(U/t æ +Œ)Í = 1Ô2

1| øA¿BÍ ≠ | ¿AøBÍ

2. Why is this solution referred to as "strongly correlated" ?

f) [2 pts] Let nA =q

‡=–,—È�(t, U)|a†1sA,‡a1sA,‡|�(t, U)Í and nB =

q‡=–,—È�(t, U)|a†

1sB ,‡a1sB ,‡|�(t, U)Í.

What is the physical meaning of nA and nB ? Show that nA = 2C21 + C2

3 + C24 and nB = 2C2

2 + C23 + C2

4 .

Conclude from question c) that nA = nB = 1.

g) [2 pts] Let n1‡g =q

‡=–,—È�(t, U)|a†1‡g ,‡a1‡g ,‡|�(t, U)Í (see question d)). Show that n1‡g = 1

2!nA+nB +

nAB"

where nAB =q

‡=–,—

e�(t, U)

---1a†

1sA,‡a1sB ,‡ + a†1sB ,‡a1sA,‡

2---�(t, U)f

. Show that, according to

2

Figure 1: Occupation of the 1‡g orbital obtained at the FCI level in the Hubbard dimer (top panel) and inthe hydrogen molecule (bottom panel).

0

0.5

1

1.5

2

0 5 10 15 20 25 30

occu

panc

y

U/4t

uniform Hubbard dimer

n1σgn1σu=2−n1σg

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

1 2 3 4 5 6 7 8 9 10

orbi

tal o

ccup

atio

n

bond distance [units of a0]

H2 (FCI, aug−cc−pVQZ)1σg1σu

1σg+1σu

the Hellmann–Feynman theorem ˆE(t, U)ˆt

=K

�(t, U)-----ˆH(t, U)

ˆt

-----�(t, U)L

, we have nAB = ≠ˆE(t, U)ˆt

.

Conclude from question f) that

n1‡g = 1 + 1Û

1 + U2

16t2

. (7)

h) [1 pt] The expression in Equation (7) as well as accurate quantum chemical FCI values for n1‡g in H2

are plotted in Figure 1. Comment on the results in the light of questions d) and e).

3

3€

nG'{

,^\ --*

t\/J-$l i n

id- f i -"Q

,*t e

-+ {l :

+<tSjJ -k q o

'F'-'A7..f\

1'-f JS 1

.'*'-t+ -i14 -.- -. ;.., I _"- (J (<. <J€l {-<

-t-r!((/

i!iAI

t-a:{5+

*-t

t-{F<

(45

xUJ

{-+ --{'<d

I+.*{f-u+

YI<d-{.JT$

^t t*ll+-f+*

.t.-'= l'+r-

(tJIt-(

f\t*\(\td' )ql+-."t I

.1.,+ll

--t I **-*n Io{-t

(('+-l -i

,3i/t r'{-ijT't-:-t -t

:,

I -l\ --a

r -st I {

sll+r t(

4_6t'

YI.t- \

r4-6-

(,

<-<

d,{.

J_ .'

(dO

L\../

J

<-: +-'{g'<{_ iy"

tJI(

tt1-

s

'. 1

t-tln/\4€

-t4{,

(- +-{

J _:4

, (d

-{1 +36<s

.to J

Frbo' f''

lta(< -.

--) -

L ,I

rF-.<L.f

.A a_

{(- 3'

! t to:

J" +- s'

| ((-

,l -1"

,I

€

o,{--r

f-p{

<-

tl>l-.o-$-

(gt

+j<(JI

c4- I -IiLJ

-J\ts(J!33{IJ$lgx

+---l

ab_d

<(-b-

r<t(SF..J t$

I

dl I

+3 t

(dt,+

a+{I

I{-<sr{-+J(sT

Ilttlbt-(-r

va^lILo.,1{

CI(

IJ$iul,la

t'l'rt(3.

€-l€

Jr' IJr-.r rtt

('

r-----10

tD .^ ,t

gt ft

(u /lI

I+

ot\

*scr

.$rltll !

.i., o

J,

-b .b [r(

t, !toUt

0ltL)I

lu,r.jt!I

Itr_-_------).b$

0o

.LJrlog

oll!+5o",+L__-_---l

llr-1sri{.rLJ

E'tJa

<9

ol

iIItUlf

-o

1,:JII3ua

a))\

.JdtuI\,ltlJ

,v

viIatJt'1.(treF)

L.-)

flltrl_

tujot,l

0rtl$lI

st\\,\tu )*+I

Orl

Iri, I

,-\tlt

''-l5

tll*/

1-\.

lJr

i,l$J+vl0ILrt)--/

J. o

'l. rl

U.r+ r-------l

t---1 \rj

a\ltbLt\;5c\J

IJLu++)

Ir,

rJufcJ

L--)qY!,90s

+.JIA/#lJ)

t+J,/4

Lnt5;I

J.r+lr-1,lI

f-----l.t.1,I

tpI5tl)IDt

J,-burt\)I5

'o

$ o\

i\o!l,. \

1",',o

\Dio'rA)

Uloli

r.,

J.!:9+ns

_t

+{(0o+-+o

(--

i/tJ)

M

I5t.3I3

o\.Tl.t:

.6-5

'd-\Yb$-7

{.4'oIi

l<ItC-t!*+c0t(-_hlll!J

lt +

ju- \

r tosto - lnl

(-)^ tl

-r$l.t

du "b

'J .E

* rb

{Ur)+iJ5

t=ItP

${

:- /\

t:r ?

l.+ t

lfG

{r=s-rl-Strr

/\@6f.[nt!ad

.|.to-'T

\'---lt I

\ilr{tl'

t,

f'

-c,tf " t-'9rrrtl rl of.rrr* [ [ii

,t rt

.i"J dJJI.!t'rll

l+J-.i I

ds5ir,

l{;;:.UJ U'd

s.UErlr---_--]d J8.r

k-IJ

sII.:t\,1

J-td'iJ{su

{,-+

L'l<d-{+

.,4<d:

fotl|ilU+olt

f'oJ(rr,

${tl

't3+U+ol)\)tc(iftlstl-It.!J-

Itl

gtuIlr(JU.'\-lt,

f't;6o

Dn!I

U,l11,l..1tv51f^lI

{1"t l.\ulraIt

.jls

oItz.--- Iu \od lt'e

I

9E

f\

r"- t

I -rd l(-3')A,\,+a,!I5-till

6l .tI

t

[rE

r, l

..llcalTI-r Jli.,lI

\l

t{'a(:

IU.$- lr.l

ll3J',UIrdo\)+.l-ofI

\JG,1

L)f)aI

+jf{c+\st'

<d -'b+J.::JNr+

t(sla5*rX(s

d.Jtc.,+

Prl

J+\)+J,ao+.la\)r.J,l

)rao+)e\)+d+U+d,r\J|le3

clYl\JclLJ-Uc.,lld*\l+ctp\J+d-o^ltl

Jj"o+d_o+fl rt,o+c_ollt3f

r.n.o.6

Jt +-

9< '*{ri\

t\'

\) \)

-{++)

f ^

-.<tra

Jl\(f<k{rJ't"tu

'S.3 "

,,,-Ir

n r..

6p -Fr

1-

*l is i;1

t r

--*, ],

f (< {'s.a \,

.--It'{{

t ]lfi

*': 't+--l\

J'oc'/\Z+. v< e

-r A

l-ci.tt

-.{t

e31la-rrF

f ,r*$:

tgJtgvovI-(

llI(<r<3t,

.s

dl<

I

Ic'l-rdtlfd

1f'{.t.cIllI];+

,a(<d+c

33J-'tDILlr{.- [,ru l6sCS'a

llfff

L\dJfl ]- rIrd5" ),".

)

s\rlst

<d- \t\

+i[.(d-

\<<lJ1,-l'Ailc-<c-a

(<

.r .*O

(drO

l-4,;,\t)u{f:t*{tg

rri/'i

*L6J

F

,a

rg6r =

l.u J

l-ILl-,.r[,..i .lJc (------'-'-1

Il.ll-l.llJ,|-,r>

tl

L.S(d:.t+dF((-t-J r,

a

5rr. . t

-?f T'- ls

I

tu:\t4

s{I

:r {-{

JJ \a

-9Yr dt

ig -Fl

i f rl 3

-rt 1:o.1

J rJr

-{s$3i {,iii

3 it

tf.r t e.o-

J !,i

,l'. t.i

*l ,l

I j{ +i-( lE {t_I J ', f,ljrti{

+$llh

Elrl,rvt

+ills-6-e.rb-t-1(Jj