M1 franco-allemand "Ingénierie des polymères" & M1 "Matériaux et nanosciences"

Exam in quantum mechanics

December 2017

duration of the exam session: 2h

Neither documents nor calculators are allowed.The grading scale might be changed.

1. Questions about the lectures (6 points)

a) [3 pts] Let H denote the Hamiltonian operator of the hydrogen atom and �–(x, y, z) = e≠–r2

a trial

wavefunction where r =

x2 + y2 + z2 and – > 0 . Give the explicit expression for H. Let E(–) =

È�–|H|�–ÍÈ�–|�–Í . Is there any value of – such that E(–) equals the exact ground-state energy E0 of the

hydrogen atom ? Justify your answer. If not, how can we find the value of – such that E(–) is as close

as possible to E0 ?

b) [3 pts] What is the purpose of both Hartree–Fock and Hückel methods ? What is the main advantage

of the former over the latter ? Is the Hartree–Fock approach in principle exact ? Justify your answers.

2. Problem I: the Heisenberg inequality and the harmonic oscillator (12 points)

According to the Heisenberg inequality, the standard deviations �x =

ÒÈ�|x2|�Í ≠ È�|x|�Í2 and

�px =

ÒÈ�|p2

x|�Í ≠ È�|px|�Í2 for the position x and momentum px of a particle described by a quan-

tum state |�Í are such that

�x �px Ø ~/2. (1)

In this exercise, we consider a particle with mass m attached to a spring of constant k moving along the x

axis. The corresponding (so-called one-dimensional harmonic oscillator) Hamiltonian reads

H =p

2x

2m+

1

2mÊ

2x

2, (2)

where Ê =

Úk

m. It can be shown that, by introducing the so-called annihilation operator a defined as

1

follows,

a =1Ô2

3ÚmÊ

~ x +iÔ

m~Êpx

4, where i

2= ≠1, (3)

and its adjoint a†

(referred to as creation operator), the Hamiltonian in Eq. (2) can be rewritten as

H = ~Ê

3N +

1

2

4, (4)

where N = a†a is the so-called counting operator. By using the commutation rule

Ëa, a

†È

= aa† ≠ a

†a = 1, (5)

it can finally be shown that the eigenvalues n of the counting operator are integers (n = 0, 1, 2, . . .) and that

the associated orthonormalized eigenvectors

Ó|�nÍ

Ô

n=0,1,2,...are connected by the relation

a†|�nÍ =

Ôn + 1|�n+1Í. (6)

a) [2 pts] Show that

x =

Û~

2mÊ

1a

†+ a

2and px = i

Ûm~Ê

2

1a

† ≠ a

2. (7)

Conclude from Eq. (6) that È�n|x|�nÍ = 0 = È�n|px|�nÍ.

b) [1 pt] Explain why, according to Eq. (4), the energies of the one-dimensional harmonic oscillator are

En = ~Ê

3n +

1

2

4and the corresponding eigenstates are |�nÍ with n = 0, 1, 2, . . .

c) [1 pt] Deduce from question 2. b) and Eq. (2) that, for a given eigenstate |�nÍ, the expectation value of

p2x is obtained from the one of x

2as follows,

È�n|p2x|�nÍ = m~Ê(2n + 1) ≠ m

2Ê

2È�n|x2|�nÍ. (8)

d) [0.5 pt] In order to determine the expectation value of x2

for |�nÍ, we propose to introduce a real

variable ⁄ and the ⁄-dependent Hamiltonian

H(⁄) =p

2x

2m+

⁄

2mÊ

2x

2. (9)

Its normalized eigenvectors and associated eigenvalues are denoted |�n(⁄)Í and En(⁄), respectively.

What is the connection between H(⁄) and the problem we are interested in ?

2

e) [2.5 pts] Explain why En(⁄) =

e�n(⁄)

---H(⁄)

---�n(⁄)

f. Prove the Hellmann–Feynman theorem,

dEn(⁄)

d⁄=

K

�n(⁄)

-----ˆH(⁄)

ˆ⁄

-----�n(⁄)

L

, (10)

and conclude that+�n(⁄)

--x2--�n(⁄),

=2

mÊ2dEn(⁄)

d⁄.

f) [1 pt] Explain why, according to Eqs. (2) and (9), En(⁄) =Ô

⁄~Ê

3n +

1

2

4. Hint: introduce the ⁄-

dependent frequency Ê(⁄) = ÊÔ

⁄, rewrite H(⁄) in terms of Ê(⁄) and compare the expression with the

one in Eq. (2). Conclude from question 2. b).

g) [1 pt] Conclude from questions 2. d), e), and f) that È�n|x2|�nÍ =~

mÊ

3n +

1

2

4.

h) [1 pt] Deduce from questions 2. c) and g) that È�n|p2x|�nÍ = m~Ê

3n +

1

2

4.

i) [2 pts] Verify from questions 2. a), g) and h) that the solutions to the Schrödinger equation for the one-

dimensional harmonic oscillator fulfill the Heisenberg inequality in Eq. (1). What is remarkable about

the ground state |�0Í ?

3. Problem II: one-dimensional harmonic oscillator in the presence of a uniform and static

electric field (4 points)

Let us consider a particle with charge q and mass m that is attached to a spring of constant k and

that moves along the x axis. In the presence of a uniform and static electric field of intensity E , the total

Hamiltonian operator varies with E as follows, H(E) = H ≠ qE x, where the operators H and x are defined

in Eqs. (4) and (7), respectively.

a) [2 pts] Let us introduce the E-dependent creation and annihilation operators, a†(E) = a

† ≠ qEÊ

Ô2m~Ê

,

and a(E) = a ≠ qEÊ

Ô2m~Ê

. Show that

Ëa(E), a

†(E)

È= 1 and that the E-dependent Hamiltonian can be

rewritten as H(E) = ~Ê

3a

†(E)a(E) +

1

2

4≠ q

2E2

2mÊ2 .

b) [1 pt] Conclude from the introduction of Problem I that the exact energies of the one-dimensional

harmonic oscillator in the presence of the electric field are En(E) = ~Ê

3n +

1

2

4≠ q

2E2

2mÊ2 . Hint: let

|�n(E)Í be an eigenvector of N(E) = a†(E)a(E). Explain why, according to question 3. a), the associated

eigenvalue is an integer n and conclude.

c) [1 pt] Explain briefly why, for this particular system, perturbation theory through second order is exact

for the energy.

3

Exam - Quantum Mechanics -Ml

- 2017 - 2018 1/2

Problem I

a) any,÷[V¥ : + vigor]→£t=iE[¥fit+ft÷of*pit)←

RER

⇒' ' at 'Eiµ¥i.

E. M ⇒

late. EYE 's

1^

) 'd t.ae =- ftp.go.TK

Hms leading to I =V¥w(£tt£) and pn=ilFEw ( Eta )

Consequently < 4n1x44n ) = YEW ( h4n1EH4n > + < 4n1a^l4n > ) and < 4n1pil4n > =iV¥w ( Hnl 'a'H4n )

where < 4h16'H4n ) = VTTT < 4n1Yn+ , > and < 4n1El% > =/, 'a'4n/4n)*= <%1EH4n)*=o

-'' 4M£14 ))

=o -°

( orthonwmd basis

of eigenvectors)and therefore { ynlnyyn ) = < q| panky =o c)

since # 14h ) = En 14N ) ⇒ { Ynl 'H' 14h ) = En thing

b) IIIYN > =kw N' Mn ) + k¥14 ) = £4 n +

£ ) 14h )or

' etiralenthyi ztm t4nlPI44n ) + lzmw44n1k44n )I En

=4n ) t 4

eigenvectorthus leading to < Ynlpiyyn > =2m[

Kw ( ntz ) - Emw4kpi4h ))eigeuraheen ( weigenstah )

( energy )⇒

{ ynlpiy %) = mtw ( Zn + D. m2wk4n1oi44n >

d) IT( A =D = #← Hamiltonian we are interested in .

e)

'

A' (a) 14nA ) ) = En(H14n6y ) ⇒ En6Y= k Ynlx ) I # (d) 14nA ) ) since < 4nA ) 14nA) ) =1 th .

Therefore dd±jnH) = { dd±yH)l

# (d) 14nA) )

thus leading to

da¥nN=⇐m×[tarantulas */

µ , ,qµ ,;#*""%"d¥¥ + < 4n( d) 1 IEH) 14nA ) )+ HnMls¥¥) 14nA ) > ↳

ddg¢4n(a) 14nA )>]=aEa[ 1) =02×

.

Thus we obtain the Hellmann . Feynmantheorem { IT(d)

4n4y1dg4yHDdaIanM-ae4nlA12gEtfH14nlxDsinaogtIlH-tzmw2I5itamesdaEgnM-HnlDlt2mosi44ntDLngyq-Dmegniralenthht4nWloi44nkn-2zwzdzEyHt1fjtTCN-PzImttzm@cxy2oi2whoeucg-yIwi.e.wEy-xw2.It

means that # ( a ,EMI# 4) HEY

" " )

Enld )

is nothing but the Hamiltonian ofthe harmonic oscillator with frequency WCd) .

As a

result ,its energies are quantized as films

,En (d) = KWH) ( ntlz )

where helm or, equivalently,

En (d) =VItW ( h + E) .

g) da±InlD= ttztwcnt

ftp.ydaELH/a.jtEktE)taHnlIllx44nH=D)=nIitEfn+E) this leading to <kpi4%)=hnz(n+E )

4h ! KTh ) { ftp.iyyn ) = m£w(2n+D .

m2w2. Lew ( ht £ ) = mkw (24+1) . mhIL(2n+D=mtw(n+£)=(4n1pI214n ) .

i ) For 14 ) = 1147 ,IE = < 4n1I44n ) - < 4h 154145 and ApE= Hnlpiihltn ) - < Ynl pick )2 ⇒ an = HEEP and Apx .tn#zT

. .

o0

thus leading to A x. 1pm =|/¥w .YIN ( n+±)=tkt£ ) = Axttpx ⇒ Axapx

= he + nt >, HE .

Conclusion : indeed,

14N ) fulfills the Heisenberg inequality . Interestingly ,

the lower bound HE is reached exactly when h=o i. e . when the particle is in the ground Stat

ofthe harmonic oscillator

.

Problem II 2/2

a) [ nice,

# g) = iicesaky . atleast =(£- Teng'¥)(£±gf÷g¥a/

meets .fi#atElETFaEi ).

±

'

a' Et- qe E

. QI Et +

942¥www.t#.*+EIEIEEEIeiIIIseJ'

¥'

Moreover,

tw ( Etcgalg + is ) =tw(at 'a' + b)

.

,t4j9,÷=⇒l£t+£) + to -921 =

E- QEI + 9212

,

, \ Zmtsw3 - ZMWZ

1 -:

qe . # they2M€

qqjib) According to Eg . AO

,

the eigenrahes of #E)'

aye) =

'

N'(e)

are integers n ( n=o, 1,2 , . . . ) . If we denote 14n( g) the associated

thus leading to # (E) = kw ( EHqE( e) + z ) - 9212

eigenvectorsthe "

'

A(E) 14N (g) = kw RICE )1%( g)zmw2

'

thug 14nA ) ) -912 IKCL ) )

2mW2

c)A perturbation expansion of

the energy ( where the electric field is treated as < perturbation =§w(n +

E)-

922=2) 14h (E) )

i. e .E is

"small

"

) would read as follows,

-

Entry = E- ( to ) +

En' " E +

¥nc4E2+

EINE

'+ . . . . .

where TENN 1 dkEt4 Esl E)- ⇒ an

1

E~=( nttglhw2=0

k ),

1

The latter expression should be compared with the exact expression ( which is valid even when th electric field is strong ! )

Enter) = ( nttlhw -921 → II "=o

, Ena = . 292g

,

and Ej4=o for k ),3 !2mW

-

Conclusion:

in this particular can,

the energy

obtained though second order is exact.