224 21 A Quantum Box 1 2 3 4 -3 -2 -1 0 321 L z / h E / hω Fig. 21.3. Allowed quantum numbers for the couple L z ,E correspond to the degeneracy of an energy level of ˆ H 0 found previously. This justifies the fact that {ˆn l , ˆn 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 ˆ L z corresponding to the two eigenvalues ±¯h. A first method for finding these eigenvectors consists in calculating the action of ˆ L z on the vectors of the basis {|n x ,n y }.Using the expression of ˆ L z in terms of ˆa x , ˆa 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:  0 −i¯h i¯h 0  . 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 ˆa † l in order to obtain the eigenvector of energy 2¯hω and angular momentum −¯h.Of course, we recover the previous result. 21.6 Solutions 225 Section 21.3: Quantum Box in a Magnetic Field 21.3.1. By expanding ˆ H B , one finds: ˆ H B = ˆp 2 x +ˆp 2 y 2µ +  µω 2 2 + q 2 B 2 8  (ˆx 2 +ˆy 2 )+ ω c ˆ L z 2 . 21.3.2. If we set Ω =  ω 2 + ω 2 c /4, we can rewrite ˆ H B = ˆ H (Ω) 0 + ω c ˆ L z /2, where ˆ H (Ω) 0 is the Hamiltonian of a two-dimensional oscillator of frequency Ω: ˆ H (Ω) 0 = ˆp 2 x +ˆp 2 y 2µ + µΩ 2 2 (ˆx 2 +ˆy 2 ) . 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 H (Ω) 0 and ˆ L z , which we continue to note {|n l ,n r }, the eigenvalues being ¯hΩ(n l + n r +1)andm¯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 − ω c 2  n 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  ω 2 ω c  ω. We therefore have ˜ E n l ,n r  ¯hω c n r if n r =0, and ˜ E n l ,0  ¯hω 2 ω c n l . For a non-vanishing n r , the energy level increases linearly with B, the slope being proportional to n r .Forn 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. 226 21 A Quantum Box 012345 0 1 2 ω c /ω E / h Ω Fig. 21.4. Variation of the energy levels ˜ E n l ,n r 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 =0andn 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 =2Ω, i.e. ω c = ω/ √ 2. 21.3.4. The value of the field B which corresponds to ω c = ω/ √ 2is 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 ˆ H 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. Section 21.4: Experimental Verification 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 |u 0 ↔ |u −  and |u 0 ↔|u + . The corresponding frequencies ν ± are such that ν ± = Ω 2π ± ω c 4π . 21.6 Solutions 227 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 × 10 11 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 = π 2 ¯h 2 n 2 /(2µD 2 ), 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 =3π 2 ¯h 2 /(2µD 2 ). (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 ⊗ |χ 0 . 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 3π 2 ¯h 2 2µD 2  ¯hω , or equivalently D  π 0 . (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  π 0 , 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. Section 21.5: Anisotropy of a Quantum Box 21.5.1. For vanishing B, the Hamiltonian is ˆ H x + ˆ H y with ˆ H x = ˆp 2 x 2µ + 1 2 µω 2 (1 + )ˆx 2 , ˆ H y = ˆp 2 y 2µ + 1 2 µω 2 (1 − )ˆx 2 . 228 21 A Quantum Box One can find a common eigenbasis for ˆ H x and ˆ H 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| ˆ L z |0, 0 + µω 2 2 0, 0|ˆx 2 − ˆy 2 |0, 0 . The state |0, 0 is an eigenstate of ˆ L z with eigenvalue 0. The first term in this sum therefore vanishes. By symmetry, we have 0, 0|ˆx 2 |0, 0 = 0, 0|ˆy 2 |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 ˆ L z in the basis under consideration in question 2.6. We must calculate the matrix elements of ˆx 2 and ˆy 2 . In order to do that, the simplest is to use the expressions of ˆx and ˆy in terms of creation and annihilation operators. One has: ˆx 2 = ¯h 2µω (ˆa x +ˆa † x )(ˆa x +ˆa † x ) , which leads to 1, 0|ˆx 2 |1, 0 = ¯h 2µω 1, 0|ˆa † x ˆa x +ˆa x ˆa † x |1, 0 = ¯h 2µω (1 + 2) = 3¯h 2µω , 1, 0|ˆx 2 |0, 1 = 0, 1|ˆx 2 |1, 0 =0, 0, 1|ˆx 2 |0, 1 = ¯h 2µω 0, 1|ˆa x ˆa † 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 [ ˆ H 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 ± ¯h 2   2 ω 2 + ω 2 c . 21.7 Comments 229 (c) Setting tan 2α = ω c /(ω), the above matrix is written as: [ ˆ H B, ]=2¯hω + ¯h 2   2 ω 2 + ω 2 c  cos 2α −i sin 2α i sin 2α −cos 2α  , whose eigenvectors are |u − (B,) =  i sin α cos α  |u + (B,) =  cos α i sin 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 ,were- 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 − )isoftheorderof0.06. 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). 22 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 (C n H n+2 ) − ,whichcan be considered as deriving from polyethylene molecules, such as hexatriene CH 2 =CH-CH=CH-CH=CH 2 , 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: (CH 2 ···CH ···CH ···CH ···CH 2 ) − , (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<0orx>L n =0 for 0≤ x ≤ L n . (22.2) 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 E 0 andofthe first excited state E 1 of the set of n + 1 electrons? We recall that  n k=1 k 2 = n(n + 1)(2n +1)/6. 232 22 Colored Molecular Ions 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: λ C = h/(m e c)=2.426 ×10 −2 ˚ A. 22.1.4. Experimentally, one observes that the ions n =9,n =11andn =13 absorb blue light (λ 9 ∼ 4700 ˚ A), yellow light (λ 11 ∼ 6000 ˚ A) and red light (λ 13 ∼ 7300 ˚ A), respectively. Is the previous model in agreement with this observation? Are the ions n ≤ 7orn ≥ 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: (CH 2 ···CH ···N ···CH ···CH 2 ) − . (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 = −V 0 for |x − L n 2 |≤α/2 , where α/d  1andV 0 > 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 λ N n are systematically shorter (blue-shifted) if n =4p+1, and systematically longer (red-shifted) if n =4p+ 3, than those λ 0 n of the cor- responding hydrocarbons (22.1). Explain this phenomenon and show that λ N n and λ 0 n are related by: λ 0 n λ N n =1−(−1) n+1 2 γ n n +2 , where γ is a parameter to be determined. 22.2.3. The nitrogenous ion n = 11 absorbs red light (λ N 11 ∼ 6700 ˚ A). Check that the ion n = 9 absorbs violet light (λ N 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, 22.3 Solutions 233 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 Section 22.1: Hydrocarbon Ions 22.1.1. The energy levels are ε k = π 2 ¯h 2 k 2 2mL 2 n k =1, 2, . 22.1.2. The ground state energy of the n + 1 electrons is E 0 = π 2 ¯h 2 mL 2 n (n+1)/2  k=1 k 2 = π 2 ¯h 2 24 mL 2 n (n +1)(n +2)(n +3). The energy of the first excited state is E 1 = E 0 + π 2 ¯h 2 8 mL 2 n  (n +3) 2 − (n +1) 2  = E 0 + π 2 ¯h 2 2 mL 2 n (n +2). 22.1.3. One has hν = E 1 − E 0 = π 2 ¯h 2 (n +2)/(2 mL 2 n ). Since λ = c/ν,we obtain an absorption wavelength λ n = 8 d 2 λ C n 2 (n +2) . 22.1.4. From the general form λ n = 646.33 n 2 /(n + 2), we obtain λ 9 = 4760 ˚ A, λ 11 = 6020 ˚ A, λ 13 = 7280 ˚ A, in good agreement with experiment. For smaller n, the wavelengths λ 7 = 3520 ˚ Aandλ 5 = 2310 ˚ Aareinthe ultraviolet part of the spectrum. The ions n ≤ 7 do not absorb visible light and are thus not colored. For n ≥ 15, the wavelengths λ 15 = 8550 ˚ Aandλ 17 = 9830 ˚ Aareinthe 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. Section 22.2: Nitrogenous Ions 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)| 2 dx = −V 0  L n +α/2 L n −α/2 |ψ k (x)| 2 dx. 234 22 Colored Molecular Ions Setting y = x − L n /2, one obtains δε k = − 2V 0 L n  +α/2 −α/2 sin 2  kπ 2 + kπy nd  dy. There are two cases: • k even: δε k = − 2V 0 L n  +α/2 −α/2 sin 2  kπy nd  dy, i.e. δε k = O((α/d) 3 ) . The perturbation is negligible. • k odd: δε k = − 2V 0 L n  +α/2 −α/2 cos 2  kπy nd  dy. To first order in α/d,wehaveδε k = −2V 0 α/nd < 0. The exact formulas are: δε k = − V 0 L n  α − (−1) k L n kπ 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 E 1 −E 0 of 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 = − 2V 0 α nd < 0 . • (n +1)/2 odd, i.e. n =4p +1,δε (n+3)/2 =0, δE = −δε (n+1)/2 = 2V 0 α nd > 0 . We can summarize these results in the compact form E 1 − E 0 + δE = π 2 ¯h 2 2md 2 n +2 n 2  1 − (−1) n+1 2 γ n n +2  , . 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, 22.3 Solutions 233 one observes the reverse effect, as. valid, the transverse extension ∆L of the quantum box must be large compared to  0 . The condition obtained in the previous question D  π 0 , put together with  0  ∆L, imposes that the box. 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