5.61 F17 Solutions to Problem Set 5

5.61 Fall 2017 Problem Set #5 Solutions 1. Phase Ambiguity b operators to generate all Harmonic Oscillator wavefunctions and When one uses â, ↠and N calculate all integrals, it is easy to forget what the explicit functional forms are for all of the ψv (x). In particular, is the innermost   (near x− ) or outermost (near x+ ) lobe of the ψv † −1/2 b always positive? Use â = 2 x˜ − i b p˜ to show that the outermost lobe of all ψv (x) is always positive, given that ˆ† )v ψ0 (x) ψv (x) = [v!]−1/2 (a and that ψ0 (x) is a positive Gaussian. Apply x̂ and −ipˆ to the region of ψ0 (x) near x+ (E0 ) to discover whether the region of ψ1 (x) near x+ (E1 ) is positive or negative. Solution: We know that ˆ† )v ψ0 (x) ψv (x) = [v!]−1/2 (a (1.1) b̃ − ib p̃)v ψ0 (x) ψv (x) = [v!]−1/2 (2−1/2 [x (1.2) And plugging in for ↠gives us or, after putting the momentum operator into position space   v ∂ −1/2 −1/2 b x˜ − } ψ0 (x). 2 ψv (x) = [v!] ∂x (1.3) For large values of x, the b x˜ operator should give a large, positive value. The derivative of a positive gaussian in this region should be negative and very small (and will contribute positively due to the negative sign in the equation). And the ψ0 itself is defined as a positive gaussian and is therefore positive over all space. Therefore, because all terms are positive for very large x values, the far right lobe should be positive. A Note about Phase Ambiguity b operators to generate all Harmonic Oscillator wavefuncWhen one uses â, ↠and N tions and calculate all integrals, it is easy to forget what the explicit functional forms are for all of the ψv (x). In particular, is the innermost  (near  x− ) or outermost (near † −1/2 b̃ b x+ ) lobe of the ψv always positive? Use â = 2 x − ip˜ to show that the outermost lobe of all ψv (x) is always positive, given that ˆ† )v ψ0 (x) ψv (x) = [v!]−1/2 (a and that ψ0 (x) is a positive Gaussian. Apply x̂ and k − ipˆ to the region of ψ0 (x) near x+ (E0 ) to discover whether the region of ψ1 (x) near x+ (E1 ) is positive or negative. 2. Anharmonic Oscillator The potential energy curves for most stretching vibrations have a form similar to a Morse potential VM (x) = D[1 − e−βx ]2 = D[1 − 2e−βx + e−2βx ]. Expand in a power series   7 4 4 2 2 3 3 VM (x) = D β x − β x + β x + . . . . 12 In contrast, most bending vibrations have an approximately quartic form 1 VQ (x) = kx2 + bx4 . 2 Here is some useful information:  3/2 } 3 ˆ+a ˆ† )3 (a xˆ = 2µω  2 } 4 ˆ+a ˆ† )4 xˆ = (a 2µω ω = (k/µ)1/2 [radians/second] 1/2 (k/µ) [cm−1 if c = 3.0 × 1010 cm/second] 2πc b + 1)a ba ˆ+a ˆ† )3 = a ˆ3 + 3(N ˆ + 3N ˆ† + a ˆ†3 (a b − 2] + [6N b 2 + 6N b + 3] + a b + 6) + a ˆ+a ˆ† )4 = a ˆ4 + a ˆ2 [4N ˆ†2 (4N ˆ†4 (a ω e= b =a ˆ† â. N The power series expansion of the vibrational energy levels is   Ev = hc ω e (v + 1/2) − ω e x̃(v + 1/2)2 + ω e ỹ(v + 1/2)3 . Hint: The goal of this problem is to relate information about the potential surface (i.e. D and β) to information about the energy level pattern we can obtain experimentally (i.e. ω̃, ω̃x̃, etc.). We make these connections via perturbation theory. A. For a Morse potential, use perturbation theory to obtain the relationships between (D, β) ˆ + ↠)3 term through second–order perturbation theory and and (ω e, ω e x̃, ω e ỹ). Treat the (a † 4 ˆ + â ) term only through first order perturbation theory. the (a [HINT: you will find that ω e y˜ = 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 expansion of the potential. Considering 5.61 Fall 2017 Problem Set #5 Solutions Page 2 only the terms of order 4 or less and defining ω0 = p 2Dβ 2 /m, we have p2 mω02 2 + x 2m 2 7D 4 4 H (1) = −Dβ 3 x3 + β x. 12 The zero-order energies are those for a harmonic oscillator with frequency ω0 H (0) = Ev(0) = }ω0 (v + 1/2) (2.1) (2.2) (2.3) 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 β x H (1) = −Dβ 3 x3 + 12  3/2  2 } 7D 4 } 3 † 3 ˆ+a ˆ) + ˆ+a ˆ † )4 = −Dβ (a β (a 12 2mω0 2mω0 3/2 1/4 3/2 2 2 } D β 7} β ˆ+a ˆ † )3 + ˆ+a ˆ† )4 = − 9/4 3/4 (a (a 96m 2 m (2.4) (2.5) (2.6) (1) The first-order corrections are the integrals Hv,v . In the equation above, the cubic term has selection rules Δv = ±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 Δv = 0 (i.e. those terms which have two â’s and two ↠’s), we arrive at Z 7}2 β 2 (1) (1) b 2 + 6N b + 3]ψv Ev = Hv,v = dxψv? [6N (2.7) 96m 7}2 β 2 = (6v 2 + 6v + 3) (2.8) 96m 7}2 β 2 2 = (v + v + 1/2) (2.9) 16m 7}2 β 2 7}2 β 2 = (v + 1/2)2 + . (2.10) 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) Ev(2) = 2 X Hv0 ,v v 0 =v 6 Ev − Ev0 (0) (0) (2.11) where (1) Hv0 ,v 5.61 Fall 2017 Z }3/2 D1/4 β 3/2 ˆ+a ˆ† )3 ψv = − 9/4 3/4 dxψv0 (a 2 m Z }3/2 D1/4 β 3/2 b + 1)a ba ˆ3 + 3(N ˆ + 3N ˆ† + a ˆ†3 ]ψv . = − 9/4 3/4 dxψv0 [a 2 m Problem Set #5 Solutions (2.12) (2.13) Page 3 The selection rules for this integral are Δv = ±1, ±3. Evaluating it in the second-order correction sum yields 4 terms:   }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) (2) + + + Ev = 9/2 3/2 3 1 −1 −3 2 m }ω0 (2.14) −}2 β 2 = [−12v 2 − 12v − 5] (2.15) 32m 3}2 β 2 2 = [v + v + 5/12] (2.16) 8m }2 β 2 3}2 β 2 = (v + 1/2)2 + . (2.17) 8m 6m 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 2 2 = [}ω0 (v + 1/2)] + (v + 1/2) + + (v + 1/2) + 16m 64m 8m 6m 2 2 2 2 13 } β 53 } β = + }ω0 (v + 1/2) + (v + 1/2)2 192 m 16 m (2.18) (2.19) (2.20) 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 + . . . ] (2.21) Matching powers of (v + 1/2), we can determine the following relations between the experimentally determined molecular constants (ω̃, ω˜ ˜x . . . ) and the potential curve parameters (the information we actually care about). hcω ˜ = }ω0 13 }2 β 2 −hcω̃x˜ = 16 m hcω̃y˜ = 0 (2.22) (2.23) (2.24) B. Optional Problem For a quartic potential, find the relationship between (ω e, ω e x̃, ω e ỹ) and (k, b) by treating † 4 ˆ + â ) through second–order perturbation theory. (a 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 5.61 Fall 2017 Problem Set #5 Solutions Page 4 series in (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 p2 mω 2 2 H (0) = + x 2m 2  2 } (1) 4 ˆ+a ˆ† )4 H = bx = b (a 2mω (2.25) (2.26) p with ω = k/m. Our zero order energies are those of H (0) (a harmonic oscillator) Ev(0) = }ω(v + 1/2). The first order corrections will use the same expression as part 2.A.  2 } (1) (6v 2 + 6v + 3) Ev = b 2mω  2  2 b } b } 2 . = (v + 1/2) + 24 mω 96 mω (2.27) (2.28) (2.29) The second order correction will require a sum over states. The selection rules for the terms in this sum are (excluding Δv = 0) Δv = ±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 b} 9 13 = − 4 5 v3 + v2 + v + 3 2 2 4m ω   2 3 b} 11 3 3 2 = − 4 5 (v + 1/2) + 3(v + 1/2) + (v + 1/2) + 4m ω 4 4 =b 2 } 2mω (2.30) (2.31) (2.32) (2.33) Summing our zero-order energies with the first- and second-order corrections yields     1 b}2 3 b2 }3 11 b2 }3 Ev = − + }ω − (v + 1/2) (2.34) 16 m2 ω 2 16 m4 ω 5 16 m4 ω 5   1 b 2 }2 3 b2 }3 + − (v + 1/2) (2.35) 24 m2 ω 3 4 m4 ω 5  2 3  1 b} + (v + 1/2)3 (2.36) 4 m4 ω 5 As before, the molecular constants from the empirical energy level power series in (v + 1/2) 5.61 Fall 2017 Problem Set #5 Solutions Page 5 are related to the potential parameters by matching powers: 11 b2 }3 16 m4 ω 5 3 b2 }3 1 b}2 −hcω̃x˜ = − 24 m2 ω 2 4 m4 ω 5 1 b2 }3 hcω̃y˜ = . 4 m4 ω 5 hcω ˜ = }ω − (2.37) (2.38) (2.39) 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 firstorder correction only contributes to the quadratic term (and all contribute to the constant offset). 3. Perturbation Theory for Harmonic Oscillator Tunneling Through a δ–function Barrier V (x) = (k/2)x2 + Cδ(x) (3.1) where C > 0 for a barrier. δ(x) is a special, infinitely narrow, infinitely tall function centered at x = 0. It has the convenient property that Z ∞ δ(x)ψv (x)dx = ψv (0) (3.2) −∞ where ψv (0) is the value at x = 0 of the v th eigenfunction for the harmonic oscillator. Note that, for all v = odd, Z ∞ δ(x)ψodd (x)dx = 0 (3.3) −∞ A. (i) The {ψv } are normalized in the sense Z ∞ |ψv |2 dx = 1 (3.4) −∞ What are the units of ψ(x)? Solution: The normalization for the zero-order wavefunctions is given as Z ∞ |ψv |2 dx = 1. (3.5) −∞ The right-hand side is dimensionless, so |ψv |2 dx must be dimensionless as well, implying that |ψv |2 has units of reciprocal length [`−1 ], so ψv has units of [`−1/2 ]. 5.61 Fall 2017 Problem Set #5 Solutions Page 6 (ii) From Eq. (3.2), what are the units of δ(x)? Solution: The δ function is defined by Z ∞ δ(x)ψv (x)dx = ψv (0). (3.6) −∞ Therefore, δ(x)ψv (x)dx must have the same dimensions as ψv (x) itself. That means that δ(x)dx is dimensionless, so δ(x) has dimensions of `−1 . (iii) V (x) has units of energy. From Eq. (3.1), what are the units of the constant, C? Solution: The perturbed harmonic oscillator potential is given as V (x) = (k/2)x2 + Cδ(x). (3.7) Cδ(x) must have units of energy [m`2 t−2 ]. We know that δ(x) has units of `−1 , so C must have units of [m`3 t−2 ]. B. In order to employ perturbation theory, you need to know the values of all integrals of b (1) H b (1) ≡ Cδ(x) H Z (3.8) +∞ b (1) ψv (x)dx = Cψv0 (0)ψv (0) ψv0 (x)H (3.9) −∞ b (0) ψv (x) = }ω(v + 1/2)ψv (x). H (1) (3.10) (2) Write general formulas for Ev and Ev (do not yet attempt to evaluate ψv (0) for all even–v). Use the definitions in Eqs. (3.11) and (3.12). (1) Ev(1) = Hvv Ev(2) = (3.11)  2 (1) Hvv0 X (0) v 0 =v 6 (0) Ev − Ev0 Solution: All ψv (x) with v odd are anti-symmetric functions with a node at x = 0, so ( 0 v or v 0 odd (1) Hv0 ,v = Cψv0 (0)ψv (0) v and v 0 even 5.61 Fall 2017 Problem Set #5 Solutions (3.12) (3.13) Page 7 The corrections to the energy levels are ( (1) Ev(1) = Hvv = Ev(2) = 0 Cψv (0)2 2 (0) ( = (3.14)  (1) Hv0 ,v X v 0 =v 6 v odd v even (3.15) (0) Ev − Ev0 0 Pv0 even v 0 6=v (Cψv0 (0)ψv (0))2 }ω(v−v 0 ) v odd v even (3.16) C. The semi-classical amplitude of ψ(x) is proportional to [vclassical (x)]−1/2 where vclassical (x) is the classical mechanical velocity at x vclassical (x) = pclassical (x)/µ = h 2}ω(v+1/2) µ 1 [2µ(Ev − V (x))]1/2 . µ (3.17) i1/2 At x = 0, vclassical (0) = . The proportionality constant for ψ(x) is obtained from the ratio of the time it takes to move from x to x + dx to the time it takes to go from x− (Ev ) to x+ (Ev ), which is 1/2 of the period of oscillation. dx/vclassical (0) τv /2 2dx 2ωdx = = 2πvclassical (0) vclassical (0)(h/}ω)  1/2 (ω/π) for even-v ψv (0) ≈ vclassical (0) ψ(0)2 dx = (1) (2) Use this semi-classical evaluation of ψv (0) to estimate the dependence of Hvv and Hvv0 on the vibrational quantum numbers, v and v 0 . Solution: A semi-classical argument estimates the amplitude of the even–v wavefunctions at the origin to be  1/2 (ω/π) ψv (0) ≈ (3.18) , where vclassical (0)  1/2 2}ω(v + 1/2) (3.19) vclassical (0) = µ  1/4 ωµ . (3.20) ⇒ ψv (0) ≈ hπ(v + 1/2) Note that the amplitude at x = 0 decreases as vclassical increases, as expected. 5.61 Fall 2017 Problem Set #5 Solutions Page 8 We can now estimate the value of the integrals of the perturbation term 1/2 ωµ = Cψv (0) ≈ C hπ(v + 1/2)  1/2 ωµ = Cψv0 (0)ψv (0) ≈ C hπ(v + 1/2)1/2 (v 0 + 1/2)1/2 (1) Hvv (1) Hv0 ,v  2 (3.21) (3.22) We note that the diagonal terms (v 0 = v), are ∝ v −1/2 and the off-diagonal terms (v 0 6= v) are ∝ (v 0 v)−1/4 . D. Make the assumption that all terms in the sum over v 0 (Eq. 3.12) except the v, v + 2 (0) (1) (2) and v, v − 2 terms are negligibly small. Determine Ev = Ev + Ev + Ev and comment on the qualitative form of the vibrational energy level diagram. Are the odd–v lev(0) els shifted at all from their Ev values? Are the even–v levels shifted up or down rela(0) tive to Ev ? How does the size of the shift depend on the vibrational quantum number? Solution: Making the assumption that only the v 0 = v − 2, v, v + 2 terms are non-negligible, we can calculate the perturbed energy levels of the v–even states. Ev = Ev(0) + Ev(1) + Ev(2)  ≈ }ω(v + 1/2) + C ωµ hπ(v + 1/2)1/2 1/2  (3.23)  1 1 1 + + 1/4 1/4 (v + 1/2) (v − 3/2) (v + 5/2)1/4 (3.24) (Note: the 1/(v − 3/2)1/4 term does not occur for v = 0). The selection rules on the perturbation term do not affect the v-odd levels at all. The vodd levels remain unperturbed. The v-even levels are all shifted up. The size of the shift decreases approximately as v −1/2 , decreasing slowly with increasing v. E. Estimate E1 − E0 and E3 − E2 . Is the effect of the δ–function barrier on the level pattern increasing or decreasing with v? Solution: Since the shifts in the energy levels decreases as ∼v −1/2 , the shifts in the level differences will go down by ∼v −3/2 . Therefore we expect the E1 − E0 difference to be larger than the E3 − E2 difference. 5.61 Fall 2017 Problem Set #5 Solutions Page 9 F. Sketch (freehand) the superposition state, Ψ(x, t = 0) = 2−1/2 [ψ0 (x) + ψ1 (x)]. Predict the qualitative behavior of Ψ? (x, t)Ψ(x, t). Solution: The time-dependence of this two-level superposition state’s probability distribution Ψ? (x, t)Ψ(x, t) will exhibit oscillations at the frequency corresponding to the two-level energy difference ω − (E1 − E0 )/}. As this motion causes the wavepacket to move back and forth through the barrier, this oscillation frequency is called the tunneling frequency. G. Compute hx̂it for the coherent superposition state in part F. Recall that Z ˆ+a ˆ† )ψv dx. xv+1,v = (some known constants) ψv+1 (a Solution: To compute hxit , we need expressions for the perturbed wavefunctions in terms of the zeroorder harmonic oscillator wavefunctions (1) ψv = ψv(0) + X Hv0 ,v v 0 =v 6 Ev − Ev0 (0) (0) (3.25) (0) Odd-v wavefunctions are unperturbed, so ψ1 = ψ1 , but we do need to calculate the energy of the perturbed ground state (1) ψ0 = (0) ψ0 (0) + = ψ0 + X Hv0 ,0 (0) v 0 6=0 E0 C h − (0) Ev 0 ωµ }π(1/2)1/2 (5/2)1/2 −2}ω (0) (0) = ψ0 + C2 ψ2 + . . . 5.61 Fall 2017 (0) ψv 0 (3.26) i (0) ψ2 + . . . Problem Set #5 Solutions (3.27) (3.28) Page 10 We can now calculate the expectation value of x s Z   1 } (0)? (0)? (0)? hx̃it = dx [ψ0 + C2 ψ2 + . . . ]e+iE0 t/} + ψ1 e+iE1 t/} 2 2µω   ˆ+a ˆ† ) [ψ0(0) + C2 ψ2(0) + . . . ]e−iE0 t/} + ψ1(0) e−iE1 t/} (a s Z h 1 } (0)? (0) (0)? (0) = dx C2 21/2 ψ1 ψ1 e+i(E1 −E0 )t/} + ψ0 ψ0 e−i(E1 −E0 )t/} 2 2µω i (0)? (0) +i(E1 −E0 )t/} 1/2 (0)? (0) +i(E0 −E1 )t/} + ψ1 ψ1 e + C2 2 ψ2 ψ2 e + orthogonal terms s } = (1 + C2 21/2 ) cos(ω10 t), 2µω (3.29) (3.30) (3.31) (3.32) (3.33) where ω10 = (E1 − E0 )t/} (3.34) The value of ω10 will be ≈ ω plus additional perturbation corrections as determined in part 3D. Note that the motion remains sinusoidal (why is this always true for a two-level superposition state?). H. Discuss what you expect for the qualitative behavior of hx̂it for the v = 0, 1 superposition vs. that of the v = 2, 3 superposition state. How will the right↔left tunneling rate depend on the value of C? Solution: The tunneling rate for a superposition of two adjacent states will be proportional to the energy difference between them. As we saw in parts 3.D and 3.E, the difference in energy of the v = 0 and v = 1 level is smaller than the difference in energy between the v = 2 and v = 3 states (because lower–v states are pushed up more). Therefore the v = 0, 1 superposition will have a slower tunneling rate. Increasing the barrier height by increasing C will decrease the tunneling rates. It pushes up each even-v state closer to the unperturbed odd-v state above it. For example, the v = 0, 1 states will be closer together in energy and a superposition state of them will tunnel at a slower rate. 5.61 Fall 2017 Problem Set #5 Solutions Page 11 4. Perturbation Theory for a Particle in a modified infinite box b (0) = pˆ2 /2m + V (0) (x) H V (0) (x) = ∞ V (0) (x) = 0 x < 0, x > a 0≤x≤a b (1) = V 0 (x) H V 0 (x) = 0 a−b a+b ,x > 2 2 a−b a+b 0 2 2 x< V 0 (x) = −V0 where a > 0, b > 0, and a > b. A. Draw V (0) (x) + V 0 (x). Solution: V (x)6 x=0 0 −V0 (0) 0 x=a (a − b)/2 (a + b)/2 a (0) B. What are ψn (x) and En ? Solution: (0) (0) ψn and En are the familiar PIB wavefunctions and energy levels: r  nπ  2 (0) sin x ψn = a a n2 h2 En(0) = 8ma2 5.61 Fall 2017 Problem Set #5 Solutions (4.1) (4.2) Page 12 C. What is the selection rule for non-zero integrals Z (1) b (1) ψ (0) ? Hnm = dxψn(0) H m Solution: The perturbation integrals are H(1) nm Z = Z = b (1) ψ (0) dxψn(0) H m (4.3) (0) dxψn(0) V 0 (x)ψm (4.4) Without evaluating any further, we can determine the selection rules for these integrals. We require that the integrand is symmetric (or has a symmetric part). Since V 0 (x) is a (0) symmetric function relative to the center of the box, this requires that the product of ψn (0) and ψm must be symmetric. This requires that n and m are either both even or both odd. D. Use sin A sin B = and 1 [cos(A − B) − cos(A + B)] 2 Z dx cos Cx = (0) (1) 1 sin Cx C (2) to compute En = En + En + En for n = 0, 1, 2, and 3 and limiting the second-order perturbation sums to n ≤ 5. Solution: We will now explicitly calculate the perturbation integrals in order to evaluate the first and second-order corrections to the energy levels. 5.61 Fall 2017 Problem Set #5 Solutions Page 13 H(1) nm Z = Z (0) dxψn(0) V 0 (x)ψm a+b 2 = a−b 2 (4.5) (0) dxψn(0) (−V0 )ψm a+b 2 (4.6)  nπ  r 2  mπ  2 = −V0 dx sin x sin x a−b a a a a 2 Z a+b  nπ   mπ  2 2 = −V0 dx sin x sin x a a a a−b 2      Z a+b 2 −V0 (n − m)π (n + m)π = dx cos x − cos x a−b a a a 2   ⎤ a+b ⎡  (n−m)π  2 (n+m)π x sin x a a −V0 ⎣ sin ⎦ = − π n−m n+m Z r (4.7) (4.8) (4.9) (4.10) a−b 2 The first-order corrections are equal to these integrals when n = m. For n = 1, 2, 3, they are   sin bπ/a (1) (4.11) E1 = −V0 b/a + π   sin 2bπ/a (1) (4.12) E2 = −V0 b/a − 2π   sin 3bπ/a (1) (4.13) E3 = −V0 b/a + 3π The second-order corrections are equal to the sum (1) En(2) = 2 Hmn X (0) (4.14) (0) m6=n En − Em Limiting the infinite sum to m ≤ 5 and remembering our selection rule for the perturbation integrals, we obtain (1) (2) E1 = −8h2 /8ma2 (1) (2) E2 = (2) 5.61 Fall 2017 (1) + 2 H51 −24h2 /8ma2 + ... (4.15) 2 H42 −12h2 /8ma2 (1) E3 = 2 H31 + ... 2 H13 +8h2 /8ma2 (4.16) (1) + 2 H53 −16h2 /8ma2 Problem Set #5 Solutions + ..., (4.17) Page 14 (1) where Hmn 2 is evaluated from Eq. (4.10). E. Now reverse the sign of V0 and compare the energies of the n = 0, 1, 2, 3 levels for V0 > 0 vs. V0 < 0. Solution: The effect of switching the sign of V0 will change the perturbation from being a “well in the box” to a “hill in the box”. We expect the energies to be perturbed upward. This can easily be seen by the form of the first-order corrections in Eqs. (4.11)-(4.13). Switching the sign of the perturbation will simply switch the sign of the first order corrections. 5.61 Fall 2017 Problem Set #5 Solutions Page 15