The Quantum Mechanics Solver 23 pptx

As the magnetic field increases, the energy splitting between the ground state and the first excited state diminishes but it stays much smaller than k B T if ω c is less than ω/ √ 2.. Sinc

1 2 3 4

Lz / h

E / hω

Fig 21.3 Allowed quantum numbers for the couple L z , E

correspond to the degeneracy of an energy level of ˆH0found previously This justifies the fact that {ˆn l , ˆ n r } form a CSCO If two different states would

correspond to the same couple of eigenvalues (n l , n r), the corresponding point

of the diagram would be twofold degenerate, and the degeneracy of the energy

level E N would be larger than N + 1.

21.2.6 We must find in this subspace two eigenvectors of ˆL zcorresponding to the two eigenvalues±¯h A first method for finding these eigenvectors consists

in calculating the action of ˆL z on the vectors of the basis{|n x , n y
} Using

the expression of ˆL z in terms of ˆa x , ˆ a y , , one finds:

ˆ

L z |n x = 1, n y= 0
= i¯h |n x = 0, n y= 1

ˆ

L z |n x = 0, n y= 1
= −i¯h |n x = 1, n y = 0
,

or the 2× 2 matrix to diagonalize: i¯0h −i¯h0

The eigenstates associated to the eigenvalues±¯h are therefore:

(|n x = 1, n y= 0
± i|n x = 0, n y = 1
) / √2 .

Another method consists in starting from the ground state |n l = 0, n r =

0
and letting act on this state: (i) the operator ˆa †

r in order to obtain the eigenvector of energy 2¯hω and angular momentum +¯ h, (ii) the operator ˆ a †

l in order to obtain the eigenvector of energy 2¯hω and angular momentum −¯h Of

course, we recover the previous result

Section 21.3: Quantum Box in a Magnetic Field

21.3.1 By expanding ˆH B, one finds:

ˆ

H B= pˆ

2

x+ ˆp2

y

2µ +

µω2

2 +

q2B2

8

(ˆx2+ ˆy2) +ω c

ˆ

L z

2 .

21.3.2 If we set Ω =

ω2+ ω2

c /4, we can rewrite ˆ H B = ˆH0(Ω) + ω c Lˆz /2,

where ˆH0(Ω) is the Hamiltonian of a two-dimensional oscillator of frequency

Ω:

ˆ

H0(Ω)= pˆ

2

x+ ˆp2

y

2µ +

µΩ2

2 (ˆx

2+ ˆy2) One can then repeat the method of the previous section, by replacing ω by Ω in

the definition of the operators ˆa x, ˆa y, One constructs an eigenbasis common

to H0(Ω) and ˆL z, which we continue to note{|n l , n r
}, the eigenvalues being

¯

hΩ(n l + n r + 1) and m¯ h Each vector |n l , n r
is also an eigenvector of ˆ H B, corresponding to the energy

E n l ,n r = ¯hΩ(n l + n r+ 1) + ¯hω c (n r − n l )/2

= ¯h



Ω + ω c

2



n r+ ¯h



Ω − ω2cn l+ ¯hΩ

21.3.3 (a) Two limiting regimes of the magnetic field can be considered,

corresponding to the limits ω c  ω (very weak magnetic field) and ω c  ω

(very strong magnetic field) In the first case, we have in first order in B:

˜

E n l ,n r  ¯hω(n l + n r) + ¯hω c (n r − n l )/2 , which corresponds to a linear variation in B of the N + 1 levels arising from the level E N in the absence of the field The slope (¯hqB/(2µ) (n r − n l) is

different for each level, which means that there is no degeneracy if B does not

vanish

For strong fields, one finds

Ω + ω c

2  ω c , Ω − ω c

2  ω2

ω c  ω

We therefore have

˜

E n l ,n r  ¯hω c n r if n r = 0 , and E˜n l ,0  ¯hω2

ω c n l .

For a non-vanishing n r , the energy level increases linearly with B, the slope being proportional to n r For n r= 0, the energy ˜E tends to zero as 1/B.

(b) The energy levels ˜E n l ,n r corresponding to N = 0, 1, 2 are represented on

Fig 21.4

0 1 2 3 4 5 0

1 2

ωc/ω

E / h Ω

Fig 21.4 Variation of the energy levels ˜E n l ,nr arising from N = 0, 1, 2 as a function

of the magnetic field B

(c) We notice on Fig 21.4 that the levels n l = 2, n r = 0 and n l = 0, n r= 1

cross each other The corresponding value of the field B is given by the solution

of the equation

Ω + ω c

2 = 2



Ω − ω c

2



,

or 3 ω c = 2 Ω, i.e ω c = ω/ √

2

21.3.4 The value of the field B which corresponds to ω c = ω/ √

2 is

µω/(q √

2) 26 T.

21.3.5 If we assume that the field B is smaller than 21 T, the three first

energy levels of ˆH B correspond to n l = n r = 0 (ground state of energy

¯

hΩ), n l = 1, n r = 0 (energy 2¯hΩ − ¯hω c /2), and n l = 0, n r = 1 (energy 2¯hΩ + ¯ hω c /2) These three states are eigenstates of ˆ L z with the eigenvalues

0, −¯h and ¯h respectively.

21.4.1 In the absence of a magnetic field, we saw in question 1.2 that only

the level n x = n y = 0 is appreciably populated for T = 10 K As the magnetic

field increases, the energy splitting between the ground state and the first

excited state diminishes but it stays much smaller than k B T if ω c is less

than ω/ √

2 For ω c = ω/ √

2, the splitting is ¯hω/ √

2 For that value, the ratio between the populations of the first excited state and the ground state is

r = exp(−49) = 3.7 × 10 −22.

Since only the ground state is populated, all the detectable absorption lines will occur from transition starting from this state

21.4.2 The first two absorption peaks correspond to the transitions|u0
↔

|u −
and |u0
↔ |u+
The corresponding frequencies ν ± are such that

ν ± =

Ω

2π ± ω c

4π .

The domain in B explored on the experimental figure of the text corresponds

to values of ω c which are small compared to ω We can therefore use the

weak-field expansion of question 3.3 in order to simplify this expression

ν ± =

ω

2π ± ω c

4π .

We therefore expect that the frequencies ν ± will vary linearly with B, the

slopes being±q/(4πµ), and that the two straight lines will cross for a vanishing

field at the frequency ω/(2π).

This linear variation of ν ± does appear on the figure for higher values of

B and the measured slope is close to the expected value (2 × 1011 Hz T−1).

However, the experimentally observed behavior for a very weak field does not correspond to our theoretical prediction Instead of two lines of same frequency

for B = 0, there is a finite difference ν+− ν −

21.4.3 Role of the z Dimension

(a) The energy levels of an infinite square well of size D are given by E n=

π2¯h2n2/(2µD2), with n positive integer, the corresponding eigenstates are the functions χ n (z) ∝ sin(nπz/D) The splitting between the ground state and

the first excited state is ∆E = 3π2¯h2/(2µD2)

(b) In order to consider that the motion along z is “frozen”, the energy

splitting ∆E between the two first levels of the square well must be very large

compared to ¯hω If this condition is satisfied, the accessible states for the

electron confined in the quantum box (in a reasonable domain of temperatures and exciting frequencies) will be simply combinations of the vectors|n x , n y
⊗

|χ0
It is then legitimate to neglect the dynamics of the electron along z.

If this condition is not satisfied, absorption lines can appear for frequencies near those presented on the experimental figure They will correspond to the

excitation of the motion along the z axis The condition that the z motion be

“frozen” is

3π2¯h2

2µD2  ¯hω , or equivalently D  π0. (21.7)

(c) In order for the harmonic approximation of the transverse motion to

be valid, the transverse extension ∆L of the quantum box must be large compared to 0 The condition obtained in the previous question D  π0,

put together with 0  ∆L, imposes that the box must have a very flat

geometrical shape: the height D along z must be very small compared to its transverse extension in xy We conclude that the vertical scale of the picture

21.1 is very dilated

21.5.1 For vanishing B, the Hamiltonian is ˆ H x+ ˆH y with

ˆ

H x= pˆ

2

x

2µ +

1

2µω

2(1 + )ˆ x2 , Hˆy= pˆ

2

y

2µ+

1

2µω

2(1− )ˆx2 .

One can find a common eigenbasis for ˆH xand ˆH y, corresponding to products

of Hermite functions in the variable x

µω(1 + )/¯ h by Hermite functions in

the variable y

µω(1 − )/¯h The corresponding eigenvalues are

¯

hω √

1 +  (n x + 1/2)+ ¯ hω √

1−  (n y + 1/2)  ¯hω (n x + n y+ 1) +¯ hω

2 (n x −n y)

where n x , n y are non-negative integers

21.5.2 To first order in B and , the shift of the ground state energy is given

by the matrix element

∆E 0,0=0, 0| ˆ W |0, 0
= ω c

2 0, 0|ˆL z |0, 0
+ µω2

2 0, 0|ˆx2− ˆy2|0, 0

The state|0, 0
is an eigenstate of ˆL zwith eigenvalue 0 The first term in this sum therefore vanishes By symmetry, we have 0, 0|ˆx2|0, 0
= 0, 0|ˆy2|0, 0
,

which means that the second term also vanishes, to first order in  and B.

21.5.3 (a) We have already determined the matrix ˆL z in the basis under consideration in question 2.6 We must calculate the matrix elements of ˆx2

and ˆy2 In order to do that, the simplest is to use the expressions of ˆx and ˆ y

in terms of creation and annihilation operators One has:

ˆ

x2= ¯h

2µω(ˆa x+ ˆa

†

x)(ˆa x+ ˆa †

x ) ,

which leads to

1, 0|ˆx2|1, 0
= ¯h

2µω 1, 0|ˆa †

x aˆx+ ˆa x aˆ†

x |1, 0

= ¯h

2µω(1 + 2) =

3¯h

2µω ,

1, 0|ˆx2|0, 1
= 0, 1|ˆx2|1, 0
= 0 ,

0, 1|ˆx2|0, 1
= ¯h

2µω 0, 1|ˆa x aˆ†

x |0, 1

= ¯h

2µω ,

where we have set |0, 1
≡ |n x = 0, n y = 1
, etc for simplicity We obtain a

similar result by exchanging the roles of x and y The restriction of ˆ H B, in the subspace of interest is therefore

[ ˆH B,] = 2¯hω +¯h

2

 ω −iω c

iω c − ω

.

(b) The energy eigenvalues are obtained by diagonalizing this 2× 2 matrix

E ± (B, ) = 2¯ hω ±¯h22ω2+ ω2

c

(c) Setting tan 2α = ω c /( ω), the above matrix is written as:

[ ˆH B,] = 2¯hω + ¯h

2



2ω2+ ω2

c

cos 2α −i sin 2α

i sin 2α − cos 2α

,

whose eigenvectors are

|u − (B, )
= i sin α cos α

|u+(B, )
= i sin α cos α

.

21.5.4 (a) The variation of E ± (B, ) − E 0,0 with B reproduces well the experimental observations For large values of B such that  ω  ω c, we

re-cover the linear variation with B of the two transition frequencies When B tends to zero (ω c   ω), one finds two different Bohr frequencies

correspond-ing respectively to the two transitions n x = n y = 0→ n x = 0, n y = 1 and

n x = n y= 0→ n x = 1, n y = 0

(b) When B tends to zero, one finds experimentally that the limit of (ν+−

ν − )/(ν++ ν − ) is of the order of 0.06 The theoretical prediction for this ratio

is /2 We therefore conclude that   0.12.

21.7 Comments

Quantum boxes of semiconductors, a simple model of which has been exam-ined here, are the subject of many investigations both academic (Coulomb correlations) and applied (optronics) Here we have only considered electronic excitations, but collective modes in a lattice (phonons) also play an important role in the dynamics of quantum boxes It has been shown recently that the two types of excitations are strongly coupled This is in contrast with the usual situation encountered in semiconductors, for which the coupling between the electrons and the phonons is weak

The data presented here come from S Hameau et al., Phys Rev Lett 83,

4152 (1999)

Colored Molecular Ions

Some pigments are made of linear molecular ions, along which electrons move freely We derive here the energy levels of such an electronic system and we show how this energy scheme explains the observed color of the pigments Consider molecular ions of the chemical formula (CnHn+2)−, which can

be considered as deriving from polyethylene molecules, such as hexatriene

CH2=CH-CH=CH-CH=CH2, with an even number of carbon atoms, by

re-moving a CH+ group In an ion of this type, the bonds rearrange themselves

and lead to a linear structure of the following type:

(CH2· · · CH · · · CH · · · CH · · · CH2)− , (22.1)

with an odd number n of equally spaced carbon atoms separated by d = 1.4 ˚A

In this structure, one can consider that the n + 1 electrons of the double bonds

of the original polyethylene molecule move independently of one another in a

one-dimensional infinite potential well of length L n = nd:

V (x) = + ∞ for x < 0 or x > L n

Actually, one should write L n = (n − 1)d + 2b where b represents the edge

effects Experimentally, the choice b = d/2 appears to be appropriate.

22.1 Hydrocarbon Ions

22.1.1 What are the energy levels ε k of an electron in this potential?

22.1.2 Owing to the Pauli principle, at most two electrons can occupy the

same energy level What are the energies of the ground state E0 and of the

first excited state E1 of the set of n + 1 electrons?

We recall thatn

k=1 k2= n(n + 1)(2n + 1)/6.

22.1.3 What is the wavelength λ n of the light absorbed in a transition be-tween the ground state and the first excited state? One can introduce the

Compton wavelength of the electron: λC= h/(mec) = 2.426 × 10 −2 ˚A.

22.1.4 Experimentally, one observes that the ions n = 9, n = 11 and n = 13

absorb blue light (λ9 ∼ 4700 ˚ A), yellow light (λ11 ∼ 6000 ˚A) and red light

(λ13 ∼ 7300 ˚A), respectively Is the previous model in agreement with this

observation? Are the ions n ≤ 7 or n ≥ 15 colored?

22.2 Nitrogenous Ions

One can replace the central CH group by a nitrogen atom, in order to form ions of the type:

(CH2· · · CH · · · N · · · CH · · · CH2)− . (22.3)

The presence of the nitrogen atom does not change the distances between atoms but it changes the above square well potential The modification

con-sists in adding a small perturbation δV (x), attractive and localized around

the nitrogen atom:

δV (x) = 0 for |x − L n

2 | > α/2

=−V0 for |x − L n

2 | ≤ α/2 ,

where α/d  1 and V0> 0.

22.2.1 Using first order perturbation theory, give the variations δε k of the

energy level ε k of an electron in the well For convenience, give the result to

leading order in α/d.

22.2.2 Experimentally, one observes that, for the same value of n, the

spec-trum of the nitrogenous ions (22.3) is similar to that of the ions (22.1) but

that the wavelengths λN

n are systematically shorter (blue-shifted) if n = 4p+1, and systematically longer (red-shifted) if n = 4p + 3, than those λ0

nof the

cor-responding hydrocarbons (22.1) Explain this phenomenon and show that λ N n and λ0

n are related by:

λ0

n

λ N n = 1− (−1) n+12 γ n

n + 2 ,

where γ is a parameter to be determined.

22.2.3 The nitrogenous ion n = 11 absorbs red light (λN11∼ 6700 ˚A) Check

that the ion n = 9 absorbs violet light (λN

9 ∼ 4300 ˚A) What is the color of

the nitrogenous ion n = 13?

22.2.4 For sufficiently large n, if the nitrogen atom is placed not in the

central site but on either of the two sites adjacent to the center of the chain,

one observes the reverse effect, as compared to question 2.2 There is a red

shift for n = 4p + 1 and a blue shift for n = 4p + 3 Can you give a simple

explanation for this effect?

22.3 Solutions

22.1.1 The energy levels are

ε k= π

2¯h2k2

2mL2

n

k = 1, 2,

22.1.2 The ground state energy of the n + 1 electrons is

E0= π

2¯h2

mL2

n

(n+1)/2

k=1

k2= π

2¯h2

24 mL2

n

(n + 1)(n + 2)(n + 3)

The energy of the first excited state is

E1= E0+ π

2¯h2

8 mL2

n

"

(n + 3)2− (n + 1)2#

= E0+ π

2¯h2

2 mL2

n

(n + 2)

22.1.3 One has hν = E1− E0 = π2¯h2(n + 2)/(2 mL2

n ) Since λ = c/ν, we

obtain an absorption wavelength

λ n =8 d

2

λC

n2

(n + 2) .

22.1.4 From the general form λ n = 646.33 n2/(n + 2), we obtain λ9 =

4760 ˚A, λ11= 6020 ˚A, λ13= 7280 ˚A, in good agreement with experiment

For smaller n, the wavelengths λ7 = 3520 ˚A and λ5= 2310 ˚A are in the

ultraviolet part of the spectrum The ions n ≤ 7 do not absorb visible light

and are thus not colored

For n ≥ 15, the wavelengths λ15 = 8550 ˚A and λ17 = 9830 ˚A are in the infrared region These ions do not absorb visible light in transitions from the ground state to the first excited state They are nevertheless colored because

of transitions to higher excited states

22.2.1 The normalized wave functions are ψ k (x) = 

2/L n sin(kπx/L n) One has

δε k=



δV (x) |ψ k (x) |2dx = −V0

 L n +α/2

L −α/2 |ψ k (x) |2dx

Setting y = x − L n /2, one obtains

δε k =− 2V0

L n

 +α/2

−α/2

sin2 kπ

2 +

kπy nd

dy

There are two cases:

• k even:

δε k =− 2V0

L n

 +α/2

−α/2

sin2 kπy nd

dy , i.e δε k = O((α/d)3)

The perturbation is negligible

• k odd:

δε k =− 2V0

L n

 +α/2

−α/2

cos2 kπy nd

dy

To first order in α/d, we have δε k =−2V0α/nd < 0.

The exact formulas are:

δε k =− V0

L n



α − (−1) k L kπ n sin kπα

L n



.

The (single particle) energy levels corresponding to even values of k are prac-tically unaffected by the perturbation; only those with k odd are shifted This

is simple to understand For k even, the center of the chain is a node of the wave function, and the integral defining δε k is negligible For k odd, on the

contrary, the center is an antinode, we integrate over a maximum of the wave function, and the perturbation is maximum

22.2.2 The perturbation to the excitation energy E1− E0of question 1.2 is

δE = δε (n+3)/2 − δε (n+1)/2

• (n + 1)/2 even, i.e n = 4p + 3, δε (n+1)/2= 0,

δE = δε (n+3)/2=− 2V0α

nd < 0

• (n + 1)/2 odd, i.e n = 4p + 1, δε (n+3)/2= 0,

δE = −δε (n+1)/2= 2V0α

nd > 0

We can summarize these results in the compact form

E1− E0+ δE = π

2¯h2

2md2

n + 2

n2 1− (−1) n+12 γ n

n + 2

,