8.04 Spring 2013, Problem Set 9 Solution

Problem Set 9 8.04 Spring 2013 Solutions Thursday, May 2 Problem 1. (10 points) Coulomb Potential Superposition States (a) (2 points) Our wavefunction is given by √ ψ = C φ100 + 4iφ210 − 2 2φ22−1 (1) To normalize, we compute (φ|φ) and set it to 1: ∞ ∞ ∞ −∞ −∞ −∞ (ψ|ψ) = √ C ∗ φ100 + 4iφ210 − 2 2φ22−1 ∗ × (2a) √ C φ100 + 4iφ210 − 2 2φ22−1 d3Nr ∞ ∞ (2b) ∞ φ∗100 φ100 d3Nr + = |C|2 −∞ −∞ (2c) −∞ ∞ ∞ ∞ ∞ ∞ φ∗210 φ210 d3Nr + 8 +16 −∞ −∞ −∞ � ∞ ∗ φ22−1 φ22−1 d3Nr −∞ −∞ (2d) −∞ = |C|2 [(φ100 |φ100 ) + 16(φ210 |φ210 ) + 8(φ22−1 |φ22−1 )] = |C|2 (1 + 16 + 8) = 25|C|2 , (2e) (2f) where we have used the fact that the energy eigenfunctions are normalized. Note that “cross-terms” do not appear because the energy eigenstates are orthogonal. Choosing C to be real gives 1 C= . (3) 5 (b) (2 points) Finding the expectation value of Eˆ requires very similar manipulations, except this time we need to “sandwich” the energy operator in between the two copies of ψ: ˆ Ê = (ψ|Eψ). (4) Note that the cross-terms still do not appear, because the energy eigenstates satisfy the energy eigenvalue equation i.e. Êψnlm = En ψnlm , (5) ˆ 311 ) = E3 (ψ100 |ψ311 ) = 0. (ψ100 |Eψ (6) so for example 1 Armed with this, we can immediately say Eˆ ˆ = (ψ|Eψ) (7a) ˆ 100 ) + 16(φ210 |Eφ ˆ 210 ) + 8(φ22−1 |Eφ ˆ 22−1 ) = |C|2 (φ100 |Eφ (7b) = |C|2 [E1 (φ100 |φ100 ) + 16E2 (φ210 |φ210 ) + 8E2 (φ22−1 |φ22−1 )] 1 = (E1 + 24E2 ) . 51 (7c) (7d) (c) (2 points) The algebra required to find L̂2 is very similar to what we have already done: L̂2 = (ψ|L̂2 ψ) = |C|2 [(φ100 |L̂2 φ100 ) + 16(φ210 |L̂2 φ210 ) + 8(φ22−1 |L̂2 φ22−1 )] = |C|2 [(0h2 )(φ100 |φ100 ) + 16(2h2 )(φ210 |φ210 ) + 8(6h2 )(φ22−1 |φ22−1 )] 80h2 = . 25 (8a) (8b) (8c) (8d) (d) (2 points) Finally, for L̂z we have L̂z = (ψ|L̂z ψ) = |C|2 [(φ100 |L̂z φ100 ) + 16(φ210 |L̂z φ210 ) + 8(φ22−1 |L̂z φ22−1 )] = |C|2 [(0h)(φ100 |φ100 ) + 16(0h)(φ210 |φ210 ) + 8(−h)(φ22−1 |φ22−1 )] 8h = − . 25 (9a) (9b) (9c) (9d) (e) (2 points) Each energy eigenstate evolves in time by a phase: ψnlm (Nr, t) = ψnlm (Nr, 0)e−iEn t/h . (10) Note that the energy (and therefore the time evolution) depends only on the principal quantum number n. Using the superposition principle, we have √ ψ(Nr, t) = C(φ100 (Nr)e−iE1 t/h + 4iφ210 (Nr)e−iE2 t/h − 2 2φ22−1 (Nr)e−iE2 t/h ), (11) where En = −13.6 eV/n2 since we are dealing with hydrogen. 2 Problem Set 9 8.04 Spring 2013 Solutions Thursday, May 2 Problem 2. (20 points) Where does a weakly bound electron live? (a) (10 points) The general hydrogen eigenfunctions are given by: 2 nr0 ψnlm (r, θ, φ) = 3 (n − l − 1)! 2n(n + l)! 1/2 − nrr e 2r nr0 0 l L2l+1 n−l−1 2r nr0 Ylm (θ, φ), (12) where Lkp are the associated Laguerre polynomials. For the case when l = n − 1 the Laguerre polynomial in the above expression (12) becomes L20n−1 , but in general Lk0 = 1, so in this case the hydrogen eigenfunctions simplify to: ψn,n−1,m (r, θ, φ) = 3 2 nr0 1 2n(2n − 1)! 1/2 − nrr e 0 2r nr0 n−1 Yn−1,m (θ, φ). (13) From eq. (13) we determine that: Z r = ψn,n−1,m (r, θ, φ)∗ rψn,n−1,m (r, θ, φ)d3 r = 2 nr0 3 1 2n(2n − 1)! Z ∞ r 3 \0 2r nr0 2 nr0 = nr0 n + 1 2 2n−2 e 2r − nr 0 Z dr Yn−1,m (θ, φ)∗ Yn−1,m (θ, φ)dΩ ' '\ 1 −4 (2n+1)! , (14) and � 2� r = = Z 2 nr0 ψn,n−1,m (r, θ, φ)∗ r2 ψn,n−1,m (r, θ, φ)d3 r 3 1 2n(2n − 1)! Z ∞ r \0 4 2r nr0 2 nr0 = (nr0 )2 n + 1 2 (n + 1). 2n−2 e −5 (2n+2)! 2r − nr 0 Z dr Yn−1,m (θ, φ)∗ Yn−1,m (θ, φ)dΩ ' '\ 1 (15) (b) (6 points) From eq. (15) we get: 3 � h r 2 i = n 2 r0 1+ 3 1 + 2 −→ n2 r0 . 2n 2n n→∞ (16) By definition the standard deviation (uncertainty) in r is: Δr = hr 2 i − h r i 2 =    2 1 1 nr0 (nr0 ) 2 n + (n + 1) − (nr0 ) 2 n + = √ , 2 2 2 (17) so the relative uncertainty in position becomes: Δr nr0 =√ hri 2nr0 n + 1 2 =√ 1 2 n+ 1 2 −→ 0. n→∞ (18) In the expression for energy there are no powers of n which we can neglect for high n, so the energy retains its exact expression. The point of this exercise is to demonstrate the correspondence principle: for high quantum numbers the quantum systems start to behave classically. This kind of highly excited atoms are called Rydberg atoms. (c) (4 points) As shown in the previous part (b) the radius of the bound statute increases as n2 r0 . This behavior is specific to the long range potentials (actually infinite in the case of Coulomb potential). For example in the case of a potential well (finite range) the bound state is limited more or less to the width of the well. 4 Problem Set 9 8.04 Spring 2013 Solutions Thursday, May 2 Problem 3. (10 points) Mighty fine structure Recall the relativistic correction to the kinetic energy that we have seen in class: KE = where we assume that |p| : mc p pN4 pN2 m2 c4 + pN2 c2 − mc2 ≈ − , 2m 8m3 c2 << 1. The total energy operator then becomes Ê = p̂2 p̂4 e2 − − . r 2m 8m3 c2 (19) We can estimate the correction to the energy by evaluating1   pN4 δEnlm = (φnlm | − 3 2 φnlm ), 8m c (21) where φnlm are the hydrogen eigenstates that we worked out in class. Note that (0) (0) (φnlm |p̂4 φnlm ) = (φnlm p̂2 |p̂2 φnlm ) = (2m)2 (φnlm (Enlm − V̂ )|(Enlm − V̂ )φnlm ) = 1 We can write (19) as ˆ ˆ = Ê0 + δ E, E ˆ0 is the Hydrogen Hamiltonian, and its eigenstates as where E E = E0 + δE, where E0 are the Hydrogen energy eigenvalues that we worked out in class. Similarly, we can also write ψ = φ + δφ, ˆ Note that we expect all the quantities with the δ to be small. where ψ is an exact energy eigenstate of E. The Schödinger equation for eigenstates then becomes (Ê0 + δÊ)(φ + δφ) = (E0 + δE)(φ + δφ). Dotting with (φ| and expanding, we obtain (φ|Ê0 φ) + (φ|δÊ0 φ) + (φ|Ê0 δφ) + (φ|δÊ0 δφ) = E0 + δE0 + E0 (φ|δφ) + δE0 (φ|δφ). (20) Note that the last terms in both members are much smaller than the others, as they contain two δ’s, so we can drop them. Moreover, from Ê0 φ = E0 φ, ˆ0 is a Hermitian operator, we can write and from the fact that E (φ|Ê0 δφ) = (Ê0† φ|δφ) = (Ê0 φ|δφ) = E0 (φ|δφ), and therefore (20) gives (φ|δÊ0 φ) = δE0 . 5 (0) = (2m)2 (φnlm |(Enlm − Vˆ )2 φnlm ), where we used  p̂2 +V 2m  (0) φnlm = Enlm φnlm . Therefore (0) (0) (φnlm |p̂4 φnlm ) = (2m)2 (φnlm |(Enlm )2 φnlm )−2(2m)2 (φnlm |(Enlm Vˆ )φnlm )+(2m)2 (φnlm |V̂ 2 φnlm ). (22) The first term in the above equation is just (0) (0) (0) (φnlm |(Enlm )2 φnlm ) = (Enlm )2 (φnlm |φnlm ) = (Enlm )2 , and the second term is (0) (0) −2(φnlm |(Enlm Vˆ )φnlm ) = 2Enlm (φnlm | e2 φnlm ). r To evaluate the above expression, notice that  2  e2 e d pN e2 =− − , r 2 de 2m r (23) and recall from class that  (φnlm | pN2 e2 − 2m r  φnlm ) = − me4 . 2h2 n2 Using this together with (23), we obtain (φnlm | e2 me4 |φnlm ) = 2 2 . r hn We obtain the third term in (22) with a similar trick, but we need to do a little more work. Recall from class that for eigenstates of L̂2 , we can write the energy of the hydrogen atom as p̂r2 h2 l(l + 1) e2 + − , 2m 2m r2 r whose associated eigenvalues can be written in terms of l by using jmax = l + 1 − n. In this notation, we saw that the eigenvalues read − me4 . 2h2 (jmax + l + 1)2 (24) Now observe that 1 2m 1 d = 2 2 h 2l + 1 dl r  h2 l(l + 1) 2m r2  1 d = 2l + 1 dl 6  p̂2r h2 l(l + 1) e2 + − 2m 2m r2 r  , and the expression in the bracket is precisely the energy operator of the hydrogen atom, for which we know the expectation value on φnlm . Therefore, using (24)   1 2m 1 d me4 m2 e 4 − 2 = . (φnlm | 2 φnlm ) = 2 r h 2l + 1 dl 2h (jmax + l + 1)2 h4 (l + 21 )n3 Summing up everything, we obtain  2 8  me me4 me4 m2 e 8 3m4 e8 4m4 e8 4 2 − 2 + = − + , (φnlm |p̂ φnlm ) = (2m) 2h2 n2 h2 n2 h4 (l + 12 )n3 h4 n4 4h4 n4 h4 (l + 12 )n3 and the total energy is Enlm me4 3me8 me8 En2 =− 2 2 + 2 4 4 − 4 2 = E + n 2h n 8c h n 2mc2 2h c (l + 12 )n3  3− 4n (l + 12 )  . We can write this in terms of the so-called fine structure constant α,   1 2 4 3 1 Enlm = En + mc α − , 2 4n4 n3 (l + 12 ) where e2 . hc This dimensionless number is a fundamental constant of Physics, it appears every time we keep into account any quantum corrections in electrodynamics. In particular, it accounts for the corrections to the structures of the atomic spectra, and this is ultimately the reason why it is called “fine structure” constant.2 Note that to�obtain this � � 1 �result we have used a series of trick. It is also possible to evaluate the 1 integrals of r and r2 directly, although the calculation is very lengthy. A full treatment can be found on Bethe and Salpeter, “Quantum mechanics of one- and two-electron atoms”, 1977, p.17. α= 2 For more about this, see R. Shankar, “Principles of Quantum Mechanics”. 7 Below is the diagram of the energy levels. On the left there are the unperturbed levels, on the right the corrected ones. The values of l for the n = 3 group goes from down to top (e.g. the lowest n = 3 level is l = −1). 2 -0.2 4 6 8 n=3 l=-1,-2,2,-1,0 n=2 l=0 l=1 l=-1 -0.4 -0.6 -0.8 l=0 -1.0 n=1 The numerical values of the energy levels in eV are showed in the following table: n l Before 1 0 -13.6057 2 -1 -3.40142 2 0 2 +1 3 -2 -1.51174 3 -1 3 0 3 +1 3 +2 After -13.6066 -3.40121 -3.40157 -3.40145 -1.51172 -1.51168 -1.51179 -1.51175 -1.51175 As for each pair (n, l) the energy is different, the degeneracy is equal to the number of independent eigenstates with quantum numbers (n, l). Since the only remaining quantum label is m, and m = −l, . . . , l, the degeneracy is 2l + 1. 8 Problem Set 9 8.04 Spring 2013 Solutions Thursday, May 2 Problem 4. (20 points) Short Answers (a) (3 points) If the atom is in its (n, l, m) = (1, 0, 0) state, its time evolution is given by Ψ(Nr, t) = Ψ100 (Nr)e−iE1 t/h . (25) Let’s see what happens when we try to work out something like x̂ : Z ∞ Z ∞ Z ∞ ∗ 3 ∗ −iE1 t/h 3 +iE1 t/h x̂ = Ψ x̂Ψd Nr = Ψ100 e xΨ100 e d Nr = Ψ∗100 xΨ100 d3Nr. (26) −∞ −∞ −∞ The time-dependence cancels out! Given that dynamical quantities like x̂ do not vary in time, it would probably be best to say that the electron is not in motion3 . (b) (3 points) Rough answer: as we saw in the previous part, the electron is not “in motion”, and thus not accelerating. Charges that are not in motion do not radiate. More precise, but less satisfying answer: a hydrogen atom in its ground state does not radiate because if it did, the electron would have to drop to an even lower energy state. No such energy state exists. It is quantum mechanically forbidden, in the sense that it is impossible to construct a wavefunction consistent with a lower energy than the ground state. (c) (2 points) For precisely the same reason as in part (a), the electron is not “in motion”. (d) (3 points) Rough answer: same as rough answer in part (b). More precise, but less satisfying answer: if our system is in an energy eigenstate, then it will remain in that eigenstate forever with its time evolution given by a multi­ plicative phase factor. It will not switch to a different eigenstate. This may mean that our model of “hydrogen” is too simplistic, since real atoms do not act in this way. The applicability of our model, however, is a separate debate. For our simple model, the Schrödinger equation tells us unambiguously that an atom in the (2, 1, 1) excited state should remain there forever. (e) (3 points) The angular momentum of Hydrogen in the state (n, l, m) is given by L̂2 φnlm = h2 l(l + 1)φnlm . By the conservation of angular momentum, if hydrogen drops from the (2, 1, 1) state to the (1, 0, 0) state, it will emit a photon of angular momentum L2l=1 − L2l=0 = 2h2 = 87.12 × 10−32 (eV · s)2 . 3 (27) Even though from a rigorous standpoint it’s not clear how one should define “in motion” from a quantum mechanical perspective. In quantum mechanics we simply have a wavefunction that extends over all space and evolves in time in a way that is specified by Schrödinger’s equation. 9 (f) (3 points) The energy of a photon is given by E= hc . λ (28) Plugging the energy that we found in the previous part into this formula, we get ◦ λ = 1216 A, (29) which is a particularly important wavelength in astronomy! This agrees with the result from the Rydberg-Ritz formula. (g) (3 points) Stationary states are, well, stationary. So they clearly can’t radiate. We can of course build states in which the electron does move – superpositions! However, (a) they are necessarily periodic with period given by 1/ the difference in energy of the terms in the superposition, and (b) the expectation value of the energy is necessarily time independent. So that also doesn’t allow for radiation and decay. If our atom really radiates, it must be true that the expectation value of the energy decays in time. However, we want the total probability that the electron is in some state to still be 1 – we want the energy to decrease, but we do not want the electron to disappear!! So we want the probability to look something like, (Eexcited ) = e−Γt , P (Eground ) = 1 − e−Γt . Meanwhile, we do not want the laws of physics do be time dependent, so the en­ ergy operator should be time-dependent. Putting all that together, we want a timeindependent energy operator which leads to decaying exponentials in the time evolution and a loss of the total energy. This is not compatible with unitary evolution by a her­ mitian energy operator. So how do we explain the decay using quantum? Well, we have not introduced the coupling between the EM field and the atom. Properly, the quantum system we should study is the quantum atom interacting with the quantum EM field. This is obviously a harder problem. The lack of hermiticity we saw above is a consequence of treating the EM field classically and allowing it to absorb energy. Precisely how this works leads to quantum field theory and the theory of light-atom interactions; some very good discussions on this are available in the various quantum and light-atom books of Claude Cohen-Tanoudji. 10 Problem Set 9 8.04 Spring 2013 Solutions Thursday, May 2 Problem 5. (40 points) Lab Exercise: Multiple Wells (a) (2 points) E1 = 1.52eV , E2 = 5.96eV , E3 = 12.84eV . The corresponding three wavefunctions are shown below: (b) i. (2 points) We have 6 levels in total, they are non-degenerate and we can note from the graph that they are grouped in three sets. Each set, or band, has two energy levels, thus two states. The mean energies of the three bands are: E¯1 = 1.55eV , E¯2 = 6.085eV , E¯3 = 13.095eV . These mean energies are higher than the energy levels that we found in part (a). Note that the differences E¯n − En increase with n. ii. (2 points) The following table reports the number of nodes that lie within the wells + the number of nodes that lie within the barriers: States within the same band are roughly linear combinations of states from the single wells with the same energy level. Note that the node theorem is respected, as the number of nodes is precisely equal to the excitation number. iii. (2 points) We have ΔE1 = 0.02eV , ΔE2 = 0.11eV , ΔE3 = 0.45eV , thus we observe that the energy width increases with the energy of the band. 11 level 1st band 1 0+0 2 0+1 2nd band 2+0 2+1 3rd band 4+0 4+1 iv. (2 points) If we decrease the separation between wells we see that the width of bands increases. This is expectable: decreasing the separation the wells feel more the presence of each other, so we expect the energy levels to look less similar to the single square well. v. (2 points) Given one band, the state which looks more periodic is the antisym­ metric combination, i.e. the state with higher energy. Its approximate wavelength is roughly a divisor of twice the width of the well. Looking at the plot, for the three bands we have then λ1 = 0.9nm, λ2 = 0.45nm, λ3 = 0.225nm. The corre­ sponding energies of a free particle of mass me are given by Ei = h2 1 , 2me λi2 thus, E1 = 1.86eV , E2 = 7.43eV , E3 = 16.72eV . Note that these values are comparable to the highest excited levels of the three bands. (c) (4 points) With 10 wells, we expect to have again 3 bands, with 10 energy levels per band, and E¯1 � E1 , E¯2 � E2 , E¯3 > E3 . For the zeroes, we expect the obvious extension of the corresponding table we obtained before, for example in the first band we expect to have no nodes lying within the wells and a progressive number of nodes lying within the barriers from 0 to 9. The bands should be wider than the previous case. We also expect λi and Ei to be the same as in the case of two wells. (d) (8 points) Our predictions of part (c) are confirmed, and we obtain the following values for the energies: level 1st band 1 1.14 2 1.17 3 1.24 4 1.31 5 1.4 6 1.52 7 1.63 8 1.72 9 1.82 10 1.87 Ē 1.48 ΔE 0.73 and for the nodes: 12 2nd band 4.73 4.87 5.14 5.41 5.73 6.15 6.53 6.88 7.25 7.49 6.02 2.76 3rd band 11.17 11.48 11.99 12.54 13.2 13.96 14.69 15.42 16.21 16.79 13.76 5.62 level 1st band 1 0+0 2 0+1 3 0+2 4 0+3 5 0+4 6 0+5 7 0+6 8 0+7 9 0+8 10 0+9 2nd band 10+0 10+1 10+2 10+3 10+4 10+5 10+6 10+7 10+8 10+9 3rd band 20+0 20+1 20+2 20+3 20+4 20+5 20+6 20+7 20+8 20+9 Moreover, λi and Ei are the same as those obtained in part (b). (e) (8 points) The plot is given in the following picture. The dashed lines represent the free-particle energies, and the dots are the energy lev­ els of the ten-well potential. Note that the energy levels group into three parabolic branches. Moreover, the free-particle energies perfectly coincide with the highest en­ ergy levels of each band. 13 If we set the separation to infinity the wells do not feel the presence of each other, and the widths of the bands would shrink to zero, giving 10 degenerate energy levels per band, as is shown in the plot below. 12 10 8 6 4 2 5 10 15 20 25 30 If we send the separation to zero, the potential looks like the finite square well in part (a) with width 10 times longer. The energy eigenvalues will thus be those of a single large square well, without degeneracy. Note that they are distributed along a single parabolic branch. It turns out that the parabola it generates is very similar to the free-particle energy-momentum relation. 20 15 10 5 5 10 15 20 25 30 (f) (8 points) As the number of wells N increases, the energy levels within a band tend to form a continuum, and in each band we see that E = E0 + εn2 , where E0 , ε are some constants, and n = 1, ..., N . Note that since the energy dependence is quadratic in n, the electron can be seen as a free particle in a box (the larger N , the bigger the box), with an effective mass related to the aperture of the parabola that corresponds to the given band. Moreover, the free-particle energies perfectly coincide with the highest energy levels of each band. Below is a plot of the energy bands. 14 15 10 5 0 5 10 15 15 20 25 30