5.61 F17 Solutions to Problem Set 6

5.61 Fall 2017 Problem Set #6 Solutions 1. Harmonic Oscillator Subjected to Perturbation by an Electric Field This problem is related to the example discussed in Lecture #19 of a harmonic oscillator perturbed by an oscillating electric field. An electron is connected by a harmonic spring to a fixed point at x = 0. It is subject to a field–free potential energy 1 V (x) = kx2 . 2 The energy levels and eigenstates are those of a harmonic oscillator where ω = [k/me ]1/2 Ev = }ω(v + 1/2) ˆ† )v ψv=0 (x). ψv (x) = (v!)−1/2 (a Now a constant electric field, E0 , is applied and V (x) becomes 1 V (x) = kx2 + E0 ex (e > 0 by definition). 2 Note on dipole interactions and signs: The interaction energy of a charge q located at position x in a uniform DC electric field E0 is aways H = −µE0 = −E0 qx. Note the negative sign! This means that when a dipole, µ ~ = q~x, points along the same direction as an electric field, there is a favorable interaction (i.e. negative interaction energy). For an electron, q = qe− ≡ −e, where e is the elementary charge and is strictly positive, making the electron’s charge negative. Therefore, an electron in a field in the +x direction has an interaction expressed as ~ 0 = −E0 qe− x = −E0 (−e)x = +E0 ex. H = −µ ~ ·E As the electron’s position x increases, its interaction energy with the field increases (assuming E0 > 0, i.e. the field points in the +x direction). This makes physical sense: we know from 8.02 that an electron likes to go away from the direction that the field points (and positive charges like to go toward the direction of the field). 5.61 Fall 2017 Problem Set #6 Solutions Page 1 You are going to approach this problem two ways: (i) by a simple and exact way first, and then (ii) by perturbation theory. A. Solve for xmin , V (xmin ), and V (x0 ) where x0 = x − xmin for this harmonic oscillator in a constant electric field. Is the system still a harmonic oscillator? What is ω for this oscillator? Solution: The total potential, including the interaction with the electric field is V (x) = mω 2 2 x + E0 ex. 2 (1.1) We find its minimum to be dV = mω 2 x + E0 e = 0 dx E0 e ⇒ xmin = , mω 2 mω 2 E02 e2 E02 e2 V (xmin ) = − 2 m2 ω 2 mω 2 2 2 E e = 0 2. 2mω (1.2) (1.3) (1.4) (1.5) Defining the displacement from the minimum x0 = x − xmin , we arrive at     E0 e 2 E0 e mω 2 0 0 0 x − + E0 e x − V (x ) = 2 mω 2 mω 2 E 2 e2 mω 2 0 2 = x − 0 2. 2 2mω (1.6) (1.7) Thus, we see that the system is still harmonic! All we have done is to shift the minimum position and minimum energy, but the potential is still quadratic. The harmonic frequency ω remains unchanged. B. Write an expression for the energy levels as a function of the strength of the electric field. Solution: Since the potential now is a harmonic oscillator with frequency ω and a constant offset, we can easily write down the energy levels: E02 e2 . 2mω 2 (1.8) Problem Set #6 Solutions Page 2 Ev = }ω(v + 1/2) − 5.61 Fall 2017 Try not to mix up the fact that we use the letter E for both energy and electric field (an unfortunate, but common convention!). Note that the electric field strength only shifts the offset of each level. The spacing between levels (}ω) remains unchanged. C. One definition of the polarizability, α, is the second derivative of the energy with respect to the electric field d2 Ev αv = − . dE02 What is the value of αv ? Is it v–dependent? Solution: The polarizability for the v th level is given by d2 Ev dE02   d2 E02 e2 = − 2 }ω(v + 1/2) − 2mω 2 dE0 e2 . =+ mω 2 αv = − (1.9) (1.10) (1.11) This value is independent of v, so each level has the same polarizability. D. Another definition of the polarizability is µ(E0 ) − µ(E = 0) = αE0 where µ is the electric dipole moment. Using this definition of α, what is µ(E0 )? Solution: A second (and in this circumstance equivalent) definition of α is µv (E0 ) − µv (E = 0) = αv E0 , (1.12) which let’s us solve for µv (E0 ), the dipole moment of state v as a function of field strength. Substituting our known expressions from above, we get 5.61 Fall 2017   e2 µv (E0 ) − −e hxiv,Ek=0 = E0 mω 2 µv (E0 ) − 0 = −exmin (1.14) µv (E0 ) = −exmin . (1.15) Problem Set #6 Solutions (1.13) Page 3 This makes physical sense. The electric field pulls the oscillator to a new equilibrium position xmin . The dipole moment of the state is always qx. For the oscillator, the expectation value of x is the minimum position for any state, yielding our result −exmin . E. Now let’s approach this problem by perturbation theory. The zero-order energies and wavefunctions are those of the harmonic oscillator at E0 = 0. The perturbation term is b (1) = E0 ex̂ H where x̂ is the usual harmonic oscillator displacement coordinate. If  x̂ = } 2µω 1/2 (â + ↠), write a general formula for all of the non-zero Z xv0 ,v ≡ dxψv?0 x̂ψv integrals. Solution: We now treat the problem as a perturbation of a harmonic oscillator with perturbation term b (1) = E0 ex̂, H where  x̂ = } 2mω 1/2 (1.16) (â + ↠) Non-zero integrals that involve x̂ can be readily evaluated: Z xv0 ,v = dxψv?0 x̂ψv  =  = } 2mω } 2mω 1/2 Z 1/2 Z (1.18) dxψv?0 (â + ↠)ψv (1.19)   dxψv?0 v 1/2 ψv−1 + (v + 1)1/2 ψv+1 (1.20)  ⇒ xv+1,v =  xv−1,v = 5.61 Fall 2017 (1.17) } 2mω } 2mω 1/2 1/2 (v + 1)1/2 (1.21) v 1/2 (1.22) Problem Set #6 Solutions Page 4 (1) b 0 and then F. Using the value you found for xv0 ,v write all of the E0 –dependent values for H v ,v compute the energy levels of the harmonic oscillator perturbed by an electric field, where Ev = Ev(0) + Ev(1) + Ev(2) and the perturbed wavefunctions are ψv = ψv(0) + ψv(1) . Solution: We can now evaluate integrals of the perturbation term: b (1) 0 H v 0 ,v = E0 exv ,v 
1/2 }  (v + 1)1/2 v 0 = v + 1  E0 e 2mω
1/2 1/2 } = E0 e 2mω v v0 = v − 1   0 else (1.23) (1.24) Using these values, we evaluate the first and second order corrections to the energies Ev = Ev(0) + Ev(1) + Ev(2) . (1.25) The first order term is a diagonal (v 0 = v) integral of the perturbation term b (1) Ev(1) = H v 0 ,v =0 (1.26) for all v. (1.27) We see that there is strictly no first order shift to the energies. Now we move onto the second order correction Ev(2) = = b (1) H v 0 ,v v0 Ev − Ev0 (0) b (1) H v−1,v (0) E02 e2 }v 2mω (1.28) (0) 2 (0) Ev − Ev−1 = 2 X b (1) H v+1,v + (0) 2 (0) (1.29) Ev − Ev+1 E02 e2 }(v+1) 2mω + }ω −}ω 2 2 E e = 0 2 (v − v − 1) 2mω E 2 e2 = 0 2. 2mω (1.30) (1.31) (1.32) The perturbed energies are thus Ev = Ev(0) + Ev(1) + Ev(2) (1.34) Problem Set #6 Solutions Page 5 = }ω(v + 1/2) − 5.61 Fall 2017 (1.33) E02 e2 . 2mω 2 Note that the second-order perturbation theory expression matches the exact result from part B. (1) Let’s now compute ψv , the correction to the zero-order wavefunctions. We begin with the definition of the first order correction ψv(1) = = b (1) H v 0 ,v X (0) v 0 6=v Ev − b (1) H v−1,v (0) Ev −  (0) (0) Ev0 ψv0 . (0) (0) Ev−1 ψv−1 + (1.35) b (1) H v+1,v (0) Ev − (0) Ev+1 (0) ψv+1 ! 1/2 } v 1/2 (0) (v + 1)1/2 (0) = E0 e ψ + ψv+1 2mω }ω v−1 −}ω  2 2 1/2   E0 e 1/2 (0) 1/2 (0) v ψ − (v + 1) ψ = v−1 v+1 . 2m}ω 3 (1.36) (1.37) (1.38) d2 Ev compute the polarizability, αv . Is the polarizability v–dependent? Does αv agree dE02 with the value you obtained in part C? G. Using Solution: Re-evaluating the polarizability, we obtain d2 Ev dE02 e2 = , mω 2 α= (1.39) (1.40) exactly the result we obtained earlier. Everything is thus far consistent. H. Using the n o (1) ψv , compute µv using Z µv = e dxψv? x̂ψv where the ψv here are the perturbed ψv . Is µ v–dependent? Should it be v–dependent? Does it agree with the result you obtained in part D? Solution: Using the perturbed wavefunctions, let’s calculate the induced dipole moment of each 5.61 Fall 2017 Problem Set #6 Solutions Page 6 state µv Z µv = q hxiv = −e  = −e     dx ψv(0)? + ψv(1)? x̂ ψv(0) + ψv(1) } 2mω 1/2 Z (1.41)     dx ψv(0)? + ψv(1)? â + ↠ψv(0) + ψv(1) 1/2  2 2 1/2 } E0 e = −e (2v − 2(v + 1)) 2mω 2m}ω 3 e2 = E0 = −exmin . mω 2 (1.42)  (1.43) (1.44) Once again, the perturbation theory result agrees with the exact result. ¬ 2. Some Short Answer Questions A. Evaluate the following integrals for ψJM eigenfunctions of J2 and Jz . (i) R  4 ∗ b+ ψ2,−2 dτ ψ22 J Solution: Z Z = ∗ ψ22  Jˆ+ 4 Z ψ2,−2 dτ = ∗ ψ22 p 2(2 + 1) − (−2)(−2 + 1)(Jˆ+ )3 ψ2,−1 dτ p p ∗ ψ22 2(2 + 1) − (−2)(−2 + 1) 2(2 + 1) − (−1)(−1 + 1)(Jˆ+ )2 ψ2,0 dτ Z p p ∗ = ψ22 2(2 + 1) − (−2)(−2 + 1) 2(2 + 1) − (−1)(−1 + 1) p × 2(2 + 1) − (0)(0 + 1)(Jˆ+ )ψ2,1 dτ Z p p ∗ = ψ22 2(2 + 1) − (−2)(−2 + 1) 2(2 + 1) − (−1)(−1 + 1) p p × 2(2 + 1) − (0)(0 + 1) 2(2 + 1) − (1)(1 + 1)ψ22 dτ √ √ √ √ Z ∗ = 4 × 6 × 6 × 4 ψ22 ψ22 dτ = 24 (ii) R (2.1) (2.2) (2.3) (2.4) (2.5) (2.6) (2.7) (2.8)  4 ∗ b+ ψ2,−1 dτ ψ33 J Solution: 5.61 Fall 2017 Problem Set #6 Solutions Page 7 3 Jˆ+ ψ2,−1 = Cψ2,2 (2.9) for some constant C. One more application of Jˆ+ annihilates this state Jˆ+ ψ2,2 = 0 so (iii) Z R (2.10) ∗ ˆ4 ψ3,3 J+ ψ2,−1 dτ = 0. (2.11) 4 Jˆ+ ψ3,−3 = Cψ3,1 (2.12)  4 ∗ b+ ψ3,−3 dτ ψ33 J Solution: for some constant C. This is orthogonal to ψ3,3 so Z ∗ ˆ4 ψ3,3 J+ ψ3,−3 dτ = 0. b = 1 kx2 + B. Which of the following operators commutes with H 2 (2.13) p̂2 2µ ? (i) a† Solution: No. b , ↠] = [}ω(N b + 1/2), ↠] = [}ω[N b , ↠] = }ω = [H 6 0. (2.14) (ii) a† a Solution: b , which surely commutes with H b = }ω(N b + 1/2). Yes. This operator is N 5.61 Fall 2017 Problem Set #6 Solutions Page 8 (iii) aa† Solution: b + 1, which surely commutes with H. b Yes. This operator is N (iv) aaaa† a† a† Solution: b Yes. Let this operator be called A. b ψv = }ω(v + 1/2)(v + 1)(v + 2)(v + 3)ψv b Âψv = (v + 1)(v + 2)(v + 3)H H (2.15) and b ψv = Â}ω(v + 1/2)ψv = }ω(v + 1/2)(v + 1)(v + 2)(v + 3)ψv AˆH (2.16) which implies that b Aˆ = AˆH. b H (2.17) b x + iL b y Hermitian? b+ = L C. Is L Solution: No b † − iL b† = L b† = L b x − iL by = L b − 6= L b+. L x y + (2.18) b + ? Explain. D. Is 2−1/2 [ψLML + ψLML −1 ] an eigenfunction of L Solution: b + produces a linear combination of ML + 1 and ML states, which cannot be the given state No. L because it contains no ML − 1 state. 3. Anharmonic Oscillator The potential energy curves for most stretching vibrations have a form similar to a Morse potential (x is displacement from equilibrium, written as Q in lecture). VM (x) = D[1 − e−βx ]2 = D[1 − 2e−βx + e−2βx ]. 5.61 Fall 2017 Problem Set #6 Solutions Page 9 Expand in a power series VM (x) = D[β 2 x2 − β 3 x3 + 7 4 4 β x + . . . ]. 12 In contrast, most bending vibrations have an approximately quartic form 1 VQ (x) = kx2 + ax4 . 2 Here is some useful information:   } 3/2 3 x ˆ = (a ˆ+a ˆ† )3 2µω   } 2 4 x ˆ = (a ˆ+a ˆ† )4 2µω ω = (k/µ)1/2 (k/µ)1/2 2πc † 3 3 b + 1)a ba (a ˆ+a ˆ ) =a ˆ + 3(N ˆ + 3N ˆ† + a ˆ†3 b − 2] + [6N b 2 + 6N b + 3] + a b + 6) + a (a ˆ+a ˆ † )4 = a ˆ4 + a ˆ2 [4N ˆ†2 (4N ˆ†4 b =a N ˆ† â. ω ˜= The power series expansion of the vibrational energy levels is Ev = hc[ω̃(v + 1/2) − ω̃x̃(v + 1/2)2 + ω̃ỹ(v + 1/2)3 ]. A. For a Morse potential, use perturbation theory to obtain the relationships between (D, β) and ˜ Treat the (ˆ a+a ˆ† )3 term through second-order perturbation theory and the (ˆ a+a ˆ † )4 (ω̃, ω˜ ˜x, ω ˜y). term only through first-order perturbation theory. [HINT: you will find that ω̃y˜ = 0.] Solution: We can interpret a Morse potential as a perturbation of a perfect harmonic oscillator. These perturbations are the higher order terms in the power pseries expansion of the potential. Considering only the terms of order 4 or less and defining ω0 = 2Dβ 2 /m, we have p2 mω02 2 + x 2m 2 7D 4 4 = −Dβ 3 x3 + β x . 12 H (0) = (3.1) H (1) (3.2) The zero-order energies are that for a harmonic oscillator with frequency ω0 Ev(0) = }ω0 (v + 1/2) 5.61 Fall 2017 Problem Set #6 Solutions (3.3) Page 10 To calculate the first-order corrections it’s necessary to rewrite the perturbation term H (1) in terms of raising and lowering operators. 7D 4 4 H (1) = −Dβ 3 x3 + β x 12 3/2 2   } 7D 4 } † 3 3 (â + â ) + (â + ↠)4 = −Dβ β 2mω0 12 2mω0 =− }3/2 D1/4 β 3/2 7}2 β 2 † 3 (â + â ) + (â + ↠)4 96m 29/4 m3/4 (3.4) (3.5) (3.6) (1) The first-order corrections are the integrals Hv,v . In the equation above the cubic term has selection rules ∆v = ±3, ±1, and so it will not contribute to the first-order corrections. Expanding the quartic term and keeping only those parts that have a selection rule ∆v = 0 (i.e. those terms which have two â’s and two ↠’s), we arrive at Z 7}2 β 2 b 2 + 6N b + 3]ψv Ev(1) = Hv(1) = dxψv? [6N (3.7) ,v 96m 7}2 β 2 = (6v 2 + 6v + 3) (3.8) 96m 7}2 β 2 2 = (v + v + 1/2) (3.9) 16m 7}2 β 2 7}2 β 2 = (v + 1/2)2 + . (3.10) 16m 64m The second-order term is a bit more complicated. We will make the simplification of only considering the second-order contributions from the cubic term in the potential (this is reasonable because in real systems quartic terms are generally an order of magnitude smaller than cubic terms). The second order correction to the energies is (1) 2 Ev(2) = X Hv0 ,v v0 = 6 v Ev − Ev0 (0) (0) (3.11) where (1) Hv0 ,v Z }3/2 D1/4 β 3/2 =− dxψv0 (â + ↠)3 ψv 29/4 m3/4 Z }3/2 D1/4 β 3/2 b + 1)â + 3N b ↠+ b 3]ψv . =− dxψv0 [â3 + 3(N 29/4 m3/4 (3.12) (3.13) The selection rules for this integral are ∆v = ±1, ±3. Evaluating it in the second-order correction 5.61 Fall 2017 Problem Set #6 Solutions Page 11 sum yields 4 terms: Ev(2)   }3 D1/2 β 3 1 (v 3 − 3v 2 + 2v) 3v 3 3(v 3 − 3v 2 + 3v + 1) (v 3 + 6v 2 + 11v + 6) + = 9/2 3/2 + + }ω0 3 1 −3 −1 2 m (3.14) −}2 β 2 [−12v 2 − 12v − 5] 32m 3}2 β 2 2 = [v + v + 5/12] 8m 3}2 β 2 }2 β 2 = (v + 1/2)2 + . 8m 6m = (3.15) (3.16) (3.17) Okay! Now we’ve done the hard work, let’s make the final connections. First let’s add up all of the perturbation corrections to the energy levels: Ev = Ev(0) + Ev(1) + Ev(2)  2 2   2 2  7} β 7}2 β 2 3} β }2 β 2 = [}ω0 (v + 1/2)] + (v + 1/2)2 + + (v + 1/2)2 + 16m 64m 8m 6m 2 2 2 2 53 } β 13 } β = + }ω0 (v + 1/2) + (v + 1/2)2 192 m 16 m (3.18) (3.19) (3.20) Compare this to the “dumb” (i.e. a priori un-insightful) power series expansion of the energy levels (which we would measure experimentally), Ev = hc[ω̃(v + 1/2) − ω̃x̃(v + 1/2)2 + ω̃ỹ(v + 1/2)3 + . . . ] (3.21) Matching powers of (v + 1/2), we can determine the following relations between the experimentally determined molecular constants (ω̃, ω̃x̃ . . . ) and the potential curve parameters (the information we actually care about). hcω̃ = }ω0 −hcω̃x̃ = 13 }2 β 2 16 m hcω̃x̃ = 0 (3.22) (3.23) (3.24) ˜x, ω ˜y) ˜ and (k, b) B. (Optional Problem) For a quartic potential, find the relationship between (ω̃, ω˜ † 4 by treating (a ˆ+a ˆ ) through second-order perturbation theory. Solution: We approach the quartic bending potential in the same manner as above. First we find the perturbation theory expressions for the energy levels and manipulate them to be a power series in 5.61 Fall 2017 Problem Set #6 Solutions Page 12 (v + 1/2), and then compare this to the experimental power series expansion of the vibrational energy levels. Our Hamiltonian H = H (0) + H (1) , where with ω = p p2 mω 2 2 + x 2m 2 2  } 4 (â + ↠)4 = bx = b 2mω H (0) = (3.25) H (1) (3.26) k/m. Our zero order energies are those of H (0) (a harmonic oscillator) Ev(0) = }ω(v + 1/2). (3.27) The first order corrections will use the same expression as part 3.A (Eqs. 3.8–3.10). Ev(1) 2 } (6v 2 + 6v + 3) =b 2mω     3 } 2 3 } 2 2 = b (v + 1/2) + b . 2 mω 8 mω  (3.28) (3.29) The second order correction will require a sum over states. The selection rules for the terms in this sum are (excluding ∆v = 0) ∆v = ±2, ±4, leading to four non–zero terms Ev(2)  4  (v)(v − 1)(v − 2)(v − 3) (v − 1)(v)(4v − 2) + 4}ω 2}ω  (v + 2)(v + 1)(4v + 6) (v + 4)(v + 3)(v + 2)(v + 1) + + −2}ω −4}ω   2 3 9 13 b } = − 4 5 v3 + v2 + v + 3 4m ω 2 2   2 3 b } 11 3 3 2 = − 4 5 (v + 1/2) + 3(v + 1/2) + (v + 1/2) + 4m ω 4 4 2 =b } 2mω Summing our zero-order energies with the first- and second-order corrections yields     3 b2 }3 11 b2 }3 3 b}2 Ev = − + }ω − (v + 1/2) 8 m2 ω 2 16 m4 ω 5 16 m4 ω 5   3 b}2 3 b2 }3 + − (v + 1/2)2 2 m2 ω 2 4 m4 ω 5   1 b2 }3 (v + 1/2)3 − 4 m4 ω 5 (3.30) (3.31) (3.32) (3.33) (3.34) (3.35) (3.36) As before, the molecular constants from the empirical energy level power series in (v + 1/2) are 5.61 Fall 2017 Problem Set #6 Solutions Page 13 related to the potential parameters by matching powers: 11 b2 }3 16 m4 ω 5 2 3 b2 }3 3 b} −hcω̃x ˜= − 2 m2 ω 2 4 m 4 ω 5 1 b2 }3 hcω̃y˜ = − . 4 m4 ω 5 hcω ˜ = }ω − (3.37) (3.38) (3.39) Note that when we treat the quartic term up through second-order perturbation theory it contributes to the linear, quadratic, and cubic terms in the (v + 1/2) power series. The first-order correction only contributes to the quadratic term (and both contribute to the constant offset). 5.61 Fall 2017 Problem Set #6 Solutions Page 14 4. Semi–Classical Treatment of H–Atom–Like Systems The radial part of the H-atom Hamiltonian includes the effective potential 1 `(` + 1) V` (r) = − + . r 2r2 The eigenfunctions and eigenvalues of this radial problem are obtained from a 2nd order differential equation for χn,` (r) = rRn,` (r)  2 2  −} d − V` (r) χn,` (r) = En,` χn,` (r). 2µH dr2 You are going to understand this problem using a semi-classical approximation, without actually solving the differential equation. A. For a 1-dimensional problem (OK to choose ` = 0), where n = 0 is the quantum number for the lowest energy level, what is the relationship between n and the number of internal nodes? Solution: If ` = 0, n = number of nodes B. The de Broglie relationship, λ = h/p, may be generalized to a system where p is dependent on r, via the classical mechanical definition of p(r) p(r) = [2m(En` − V` (r))]1/2 = h/λ(r). Thus the semi-classical wavelength is λ(r) = h[2m(En,` − V` (r))]−1/2 . The following integral equation enables you to compute the number of semi-classical wavelengths between the turning points of the V` (r) at energy En,` : Z 1 r+ (En,` ) pEn,` (r)dr = # of wavelengths. h r− (En,` ) Justify this equation by reference to an infinite box of width a. Solution: En = n2 h2 8ma2 Between turning points, V = 0 r p= 5.61 Fall 2017 p 2mEn = 2mn2 h2 nh = 2 8ma 2a Problem Set #6 Solutions Page 15 1 nh h 2a Z a/2 dr = −a/2 nha n = = # of wavelengths 2ha 2 C. If the lowest energy level has quantum number n = 0 and ψn=0 (r) has zero internal nodes, then how many internal nodes does the ψn=4 (r) function have? 4 nodes How many wavelengths fit between the turning points of V` (r) at En=4 ? Solution: There are four nodes, therefore the wavefunction must cross the r–axis 4 times, leading to 52 wavelengths. Since n actually starts with 0 in this case, n = 4, corresponds to the 5th energy level. Therefore, the number of wavelengths matches with the formula derived in part B. Below is a qualitative picture (node spacing, magnitude of wavefunction, and the wavefunction as r → 0 and r → ∞ are not depicted in this cartoon). D. There is a semi-classical quantization rule (corrected for tunneling of the wavefunction into the classically forbidden E < V` (r) regions) Z r+ (En ) h pE (r)dr = (n + 1/2) 2 r− (En ) that tells you (i) The energy of the nth level (by iterating E until the quantization condition is satisfied); (ii) how many energy levels lie at or below whatever value of E you choose; and (iii) what is the density of states, dn dE E , or n(E + dE) − n(E − dE) . 2dE Use the semi-classical quantization condition to find the energy levels of a harmonic oscillator. (You have to evaluate a do-able integral.) 5.61 Fall 2017 Problem Set #6 Solutions Page 16 Solution: p 2m(En − V (x)) 1 1 V (x) = kx2 = mω 2 x2 2 2 p(x) = (4.1) (4.2) Plugging Eq. (4.2) into Eq. (4.1) s p(x) =   1 2 2 2m En − mω x . 2 Turning points are when En = V (x) = 12 mω 2 x2 r x+ = x− = − q x+ (En ) Z r 2En ; mω 2 Z pEn (x)dx = x− (En ) q − 2Em2 mω q 2En mω 2 Z = − mω 2 x2 Let y 2 = −→ dy = 2En Z x+ (En ) pEn (x)dx = x− (En ) 2En mω 2 s q s 2En mω 2   1 2 2 2m En − mω x dx 2 s  2mEn 2En mω 2 mω 2 dx 2En 2En ω Z 1 p 1 − y 2 dy = −1 Note: The above integral was evaluated by recognizing that circle, and then simply dividing the area of a unit circle by 2. Z x+ (En ) x− (En )  mω 2 x2 1− dx 2En p En π . ω 1 − y 2 is the top half of the unit   h 1 En π pEn (x)dx = = n+ ω 2 2   1 En = }ω n + . 2 These are the same energy levels that we have previously calculated using operator algebra for the harmonic oscillator. E. What does the generalized de Broglie definition of λ(r) tell you about the locations of nodes for ` = 0 of the H atom? 5.61 Fall 2017 Problem Set #6 Solutions Page 17 (i) Compute the r value of the innermost internal node in χn,` (r) for ` = 0, n = 10 and n = 20, where −hc< H E10s = , < H = 109679 cm−1 . 102 This will show you that the innermost nodes for all n & 6 members of any n` Rydberg series occur at approximately the same value of r. Solution: λ(r) = p h 2m(En` − V` (r) Using atomic units, let } = me = 4π0 = e− = 1. Replace λ with 2r and V` (r) with − 1r : 2r = q 2π 2 En` + 1 r
(2π)2
2 En` + 1r   1 2 = (2π)2 8r En` + r 4r2 = 8En` r2 + 8r − (2π)2 = 0 p −8 ± 64 + 32En` (2π)2 r= 16En` 1 −hc< H Ens = = 2 2 nq 2n −1 ± r= 1 n2 1 + π 2 n12 1 n2  1 when n ≥ 6, therefore we can take a Taylor series expansion of our radical r= −1 ± 1 + 21 π 2 n12 1 n2
= π2 a0 2 Note that, because we were able to take a Taylor expansion, the radical due to the large energy denominator n does not appear in our final answer for the position of the first node. We can therefore conclude that the position of the first node is approximately independent of n, when n ≥ 6. (ii) For 10s, sketch the locations of all of the internal nodes. Which nodes are closest together and which are the furthest apart? 5.61 Fall 2017 Problem Set #6 Solutions Page 18 Solution: 10s will have 9 total nodes. We have already calculated the position of the first node. To determine the position of the second node, we must set λ = r. r=q 2π 2 En` + 1 r
(2π)2
2 En` + 1r   1 2 2r En` + = (2π)2 r r2 = 2En` r2 + 2r − (2π)2 = 0 q −1 ± 1 + 8π 2 n12 r= 1 n2 rsecond rthird node node = = −1 ± 1 + 42 π 2 n12
1 n2 −1 ± 1 + 92 π 2 n12 1 n2
= 2π 2 a0 9 = π 2 a0 2 Clearly, node spacing increases with node number. ww w w 5.61 Fall 2017 w w w w Problem Set #6 Solutions w Page 19 The next sequence of questions (parts F through J) are optional. They will lead you to estimate the amplitudes of the lobes of |ψn` (r)|2 between each pair of adjacent nodes. F. The classical mechanical oscillator period for each n` level may be estimated from τn,` = h . En+1/2,` − En−1/2,` Estimate the oscillation period for the 10s state. Solution:  τn` = h 10.52 9.52 + −hc< H hc< H  = 500 = 152 f s. c< H G. Derive the time required for a classical electron to travel from the (n − 1)th to the (n + 1)th internal node with reference to the equations below: r of the nth node is denoted as r[n]
h
λ r[n] = p r[n]

 ν r[n] = p r[n] me
λ r[n]
. ∆t[n] = ν r[n] Solution: ∆t[n] = λ(r[n] ) hme h = = 2 v(r[n] ) p(r[n] ) 2 (En` − V` (r)) H. The probability of finding a classical particle moving at positive velocity between the (n − 1)th and (n + 1)th nodes is ∆t[n] . τn,` λ∗n,` (r)λn,` (r) is an oscillatory function, oscillating between 0 and twice the node–to–node proba


∗ bility. Estimate ψn,` r[n] ψn,` r[n] and ψn,` r[n] . Solution: 5.61 Fall 2017 Problem Set #6 Solutions Page 20 En+ 1 ` − En− 1 `

∆t[n] ∗ 2 2 ψn` r[n] ψn` r[n] ∝ = 2τn` 2(En` − V` (r)) I. How would you use the results that you have derived here to estimate the expectation value of rk for any value of k [HINT: a sum of each node-to-node region] D E Z ∞ ∗ ψn,` rk ψn,` dr? rk = −∞ Solution No solution given. J. For non-hydrogenic atoms, it is possible to replace the Rydberg equation (integer-n) by En∗ ,` = − hc< n∗2 ` n∗` = n − δ` where δ` is the “quantum defect.” n∗` can be determined empirically from E∞ − En∗ ` = hc< n∗2 `  hc< n = E∞ − En∗ ` ∗ 1/2 and [En∗ ` − En∗ `−1 ] ≈ 2hc<
3 . n∗` Are these semi-classical equations sufficient to approximate the expected values of all properties of all Rydberg states of all atoms? Solution No solution given. 5.61 Fall 2017 Problem Set #6 Solutions Page 21