Problem Set 4 Solutions Distributed: February 26, 2016 Due: March 4, 2016

McQuarrie Problems 5-9 Overlay the two plots using Excel or Mathematica. See additional comments below.

The final result of Example 5-3 defines the force constant in terms of the parameters 2 −19 in the Morse potential: k = 2Deβ . Given De = 7.31 × 10 J/molecule and β = × 10 −1 ( ) ( ) 1.81 10 m (for HCl, k is ) 2 − 1 N m k = 2D β2 = 2 7.31 × 10 19 J/molecule 1.81 × 1010 × e m J ( ) ( ) ( ) N molecule m 2 kJ = 479 × 6.022 × 1023 molecule · m mol 109 nm 1000 N m kJ = 2.88 × 105 mol nm2 Vibrational potentials of HCl VHxL, kJmol

400

Harmonic 300 Morse

200

100

x=r-re, nm -0.05 0.00 0.05 0.10 0.15 0.20 0.25 0.30 Vibrational potentials of HCl, small displacements VHxL, kJmol

15 Harmonic

Morse 10

5

x=r-re, nm -0.005 0.000 0.005 0.010 The Morse potential captures the more realistic steep rise as x → 0 and leveling off to the bond dissociation energy as x → ∞. For small displacements, the harmonic potential and Morse potential are very similar.

Boekelheide CH342: Quantum Chemistry Problem Set 4 Solutions 2

5-13 − There’s a strong line in the IR spectrum of H127I at 2309 cm 1. Given the intensity of e the line, this is probably the fundamental( ) vibrational frequency ω in wavenumbers. 1/2 e 1 k Using equation 5.39, ω = 2πc µ .

127 g The reduced mass µ from the masses of I m 127I = 126.904473 mol and mH = g 1.00782503207 mol from Wikipedia

m 127ImH g µ = = 0.999884347 mol m 127I + mH

g which is 1.000 mol with 4 sig figs. Rearranging equation 5.39, we can solve for k: ( ( )( )) ( ) − 2 k = (2πcωe)2 µ = 2π 2.9979 × 108 m 2309 cm 1 1.000 g s ( )( mol) ( ) ( ) g m2 kg 100 cm 2 mol N m =1.891 × 1025 × cm2 mol s2 1000 g m 6.022 × 1023 kg m2 s2 ( ) ( ) ( ) N molecule m 2 kJ k = 314.1 × 6.022 × 1023 m mol 109 nm 1000 N m kJ k = 1.892 × 105 mol nm2

The period τ has units of time per cycle, and the frequency has units of cycle per time, so the period of oscillation will be: ( ) 1 1 × −12 cm s × m τ = = e = 1.444 10 νobs ωc m ( 100 cm) 15 − 10 fs τ = 1.444 × 10 14 s × s τ = 14.44 fs

5-14 k = 319 N 35 35 Given m for Cl Cl, Calculate the fundamental( ) vibrational frequency and 1/2 1 k g the zero-point energy. Using equation 5.37, νobs = 2π µ with m35Cl = 34.96885268 mol

Boekelheide CH342: Quantum Chemistry Problem Set 4 Solutions 3

g from Wikipedia, so µ = 17.48 mol ( ) 1/2 1 319 N ν = m obs 2π g 17.48 mol ( ) (( )( ) ( ))1/2 1/2 kg m mol N 1 1000 g 2 = 0.680 × 6.0221 × 1023 s g m mol kg N 1 = 1.66 × 1013 s ωe = ν /c obs ( ) 1 m = 55600 × m 100 cm 1 = 556 . cm

1 The zero-point energy for a harmonic oscillator is E0 = 2 hν ( ) ( ) 1 1 −34 13 1 −21 E0 = hν = 6.6261 × 10 J · s 1.66 × 10 = 5.53 × 10 J 2 2 s

5-15

Given that the fundamental (from v = 0 to v = 1) and first overtone (from v = 0 − − to v = 2) for 12C16O occur at 2143.0 cm 1 and 4260.0 cm 1, and using McQuarrie’s equation 5.43, − ωobs = ωev ωexev(v + 1)

I get two equations:

−1 2143.0 cm = ωe − 2ωexe −1 4260.0 cm = 2ωe − 6ωexe.

Multiplying the first by −2 and adding both, I get

−1 −26.0 cm = −2ωexe,

−1 and ωexe = 13.0 cm . Plugging this into the first equation, ( ) −1 −1 2143.0 cm = ωe − 2 13.0 cm ,

−1 and ωe = 2169.0 cm . Note that this is about 1% larger than the fundamental frequency. 5-21 Only do n = 1. It’ll be easier to show this without including any normalization constants in

Boekelheide CH342: Quantum Chemistry Problem Set 4 Solutions 4 front of the wavefunction. Justify why it is okay to ignore the constant if you take that approach.

From table 5.4, the unnormalized wavefunction for a harmonic oscillator in the first 2 −α x excited state is ψ1(x) = xe 2 . To show that this equation satisfies the Schrödinger equation, we have to show that it is an eigenfunction of the Hamiltonian operator. ( ) b b b Hψ1(x) = T + V ψ1(x) ( ) ℏ2 2 d 1 2 = − + kx ψ1(x) 2µ dx2 2 ℏ2 2 d ψ 1 2 = − + kx ψ1(x) (⋆) 2µ dx2 2

We’ll do this in pieces. First, the differential part in the first term, using the product rule: ( ) 2 d ψ d d − x2 α 2 2 = xe dx dx d(x ) 2 2 d −α x 2 −α x = e 2 − αx e 2 dx ( ) 2 2 2 −α x −α x 3 −α2 x = −αxe 2 − 2αxe 2 − αx e 2

2 2 2 −α x 2 3 −α x −α x = −3αxe 2 + α x e 2 pulling out ψ1(x) = xe 2 ( ) 2 2 2 −α x = −3α + α x xe 2 √ ( ) 2 2 kµ = −3α + α x ψ1(x) using α = , equation 5.45 ( √ ) ℏ 2 d ψ kµ kµ 2 = −3 + x ψ1(x) dx2 ℏ ℏ2

Plugging this result in for the kinetic operator term in (⋆), ( √ ) ℏ2 2 ℏ2 d ψ kµ kµ 2 − = − −3 + x ψ1(x) 2µ dx2 2µ ℏ ℏ2 ( √ ) 3ℏ k 1 = − kx2 ψ (x) 2 µ 2 1

Plugging this result into (⋆), ( √ ) 3ℏ k 1 1 Hψb (x) = − kx2 ψ (x) + kx2ψ (x) 1 2 µ 2 1 2 1 √ 3ℏ k = ψ (x). 2 µ 1

Boekelheide CH342: Quantum Chemistry Problem Set 4 Solutions 5

√ k ℏ h If we rearrange equation 5.33, µ = 2πν and = 2π . Plugging this in, we get,

b 3 3 Hψ1(x) = hνψ1(x) ψ1(x) is preserved, and the eigenvalue is hν . 2 2

We could neglect the normalization constant because for a linear operator like the b b Hamiltonian, H (aϕ) = aHϕ for some constant a. 5-26 Summarize how the results of these types of calculations might be used. ⟨ ⟩ ∫ √ 2 2 ∞ 2 5 ℏ −α x x = x ψ (x)x ψ (x) = √ ψ (x) = N H ( αx) e 2 Showing that −∞d 2 2 2 µk , using 2 2 2 , where N2 and H2 are the normalization constant and Hermit polynomials for the sec- ond excited state√ given on page√ 225. To evaluate the integral, I’ll start by changing variables using αx = ξ and αdx = dξ: ⟨ ⟩ ∫ 2 ∞ 2 2 2 ξ 1 − ξ ξ − ξ = dξ √ N2H2 (ξ) e 2 N2H2 (ξ) e 2 α −∞∫ α α ⟨ ⟩ ∞ 2 √1 2 2 −ξ2 ξ = dξ N2 ξ H2 (ξ) H2 (ξ) e α −∞

2 I want to look only at the Hermite term ξ H2 (ξ), and use the recursion definition of 1 the Hermite polynomials, equation 5.55, Hn (ξ) = 2 Hn+1 (ξ) + nHn−1 (ξ), ( ) 1 ξ2H (ξ) = ξ H (ξ) + 2H (ξ) 2 2 3 1 1 = ξH3 (ξ) + 2ξH1 (ξ) 2 ( ) ( ) 1 1 1 = H (ξ) + 3H (ξ) + 2ξ H (ξ) + 2H (ξ) 2 2 4 2 2 2 0 1 5 = H (ξ) + H (ξ) + 4H (ξ) 4 4 2 2 0

If I plug this back in, and recognize that the the Hermite polynomials for which n ≠ 2 will integrate to 0, the average becomes ∫ ( ) ⟨ ⟩ ∞ 2 √1 2 5 −ξ2 ξ = dξ N2 H2 (ξ) H2 (ξ) e α −∞ 2 ∫ ∞ √5 2 2 −ξ2 = dξ N2 H2 (ξ) e Going back to x 2 α −∞∫ ⟨ ⟩ ∞ √ (√ ) 2 5√ 2 2 −αx2 x = dx α N2 H2 αx e 2α ∫α −∞ ∞ (√ ) 5 2 2 −αx2 = dx N2 H2 αx e The integral is 1 2α −∞ ⟨ ⟩ 5ℏ √ x2 = √ I used α = kµ/ℏ 2 kµ

Boekelheide CH342: Quantum Chemistry Problem Set 4 Solutions 6

This is the final result. It can be used to calculate the intrinsic error in a mea- √surement of position for a quantum harmonic oscillator in the second excited state, ⟨x2⟩. This limits the precision with which we can know where a quantum object is. 5-27 Summarize how the results of these types of calculations might be used.

To do the same thing as 5-26 for momentum, we want to define another recursion d relationship for the Hermite polynomials. We wanna get dξ Hn (ξ) starting with the − n ξ2 dn −ξ2 derivative definition of the Hermite polynomials, Hn (ξ) = ( 1) e dξn e , [ ] n d d n ξ2 d −ξ2 Hn (ξ) = (−1) e e dξ dξ dξn n n+1 2 d 2 2 d 2 = 2ξ(−1)neξ e−ξ + (−1)neξ e−ξ dξn dξn+1 n n+1 n+1 2 d 2 (−1) 2 d 2 = 2ξ(−1)neξ e−ξ + eξ e−ξ dξn (−1) dξn+1 n n+1 2 d 2 2 d 2 = 2ξ(−1)neξ e−ξ − (−1)n+1eξ e−ξ dξn dξn+1

= 2ξHn (ξ) − Hn+1 (ξ)

= 2nHn−1 (ξ) From the recursion relationship.

2 Pb2 = −ℏ2 d Using this result and dx2 , we can simplify the second derivative of the wavefunction for a general excited state [ ] 2 2 ( ) 2 d d √ − x α 2 2 ψn(x) = 2 NnHn αx e dx dx [ ] (√ ) 2 d d −α x = Nn Hn αx e 2 dx d[x ] √ (√ ) 2 (√ ) 2 d −α x −α x = Nn 2n αHn−1 αx e 2 − αxHn αx e 2 d[x (√ ) 2 (√ ) 2 −α x 3 −α x = Nn 4n(n − 1)αHn−2 αx e 2 − 2nα 2 xHn−1 αx e 2

(√ ) 2 (√ ) 2 −α x 3 −α x − αH αx e 2 − 2nα 2 xH − αx e 2 n ] n 1 (√ ) 2 2 2 −α x +α x Hn αx e 2 [ (√ ) 2 (√ ) 2 −α x 3 −α x = Nn 4n(n − 1)αHn−2 αx e 2 − 4nα 2 xHn−1 αx e 2 ] (√ ) 2 (√ ) 2 −α x 2 2 −α x −αHn αx e 2 + α x Hn αx e 2

Boekelheide CH342: Quantum Chemistry Problem Set 4 Solutions 7

I need to work with the second term in the final result to get rid of the x

(√ ) 2 ( √ (√ )) 2 3 −α x −α x 4nα 2 xHn−1 αx e 2 = 2nα 2 αxHn−1 αx e 2 ( (√ ) (√ )) 2 −α x = 2nα 2(n − 1)Hn−2 αx + Hn αx e 2 (√ ) 2 (√ ) 2 −α x −α x = 4n(n − 1)αHn−2 αx e 2 + 2nαHn αx e 2

Plugging this back in to the second derivative, the first term cancels, and we get [ ] 2 (√ ) 2 (√ ) 2 (√ ) 2 d −α x −α x 2 2 −α x ψn(x) = Nn −2nαHn αx e 2 − αHn αx e 2 + α x Hn αx e 2 dx2 2 2 = −2nαψn(x) − αψn(x) + α x ψn(x) ( ) 2 2 = −2nα − α + α x ψn(x). ⟨ ⟩ ∫ 2 ∞ b2 Now I can put this into the integrals for the second excited state, p = −∞dx ψ2(x)P ψ2(x), 2 ( ) d ψ (x) = −5α + α2x2 ψ (x) with dx2 2 2 ∫ ⟨ ⟩ ∞ 2 b2 p = dx ψ2(x)P ψ2(x) −∞∫ ∞ 2 −ℏ2 d = dx ψ2(x) 2 ψ2(x) ∫−∞ dx ∞ ( ) 2 2 2 = −ℏ dx ψ2(x) −5α + α x ψ2(x) −∞ ∫ ∞ ∫ ∞ 2 2 2 2 = 5ℏ α dx ψ2(x)ψ2(x) −ℏ α dx ψ2(x)x ψ2(x) | −∞ {z } | −∞ {z } 1 ⟨x2⟩= 5 ∫ 2α ∞ 5 = 5ℏ2α dx ψ (x)ψ (x) − ℏ2α2 2 2 2α | −∞ {z } 1 5 5 √ = ℏ2α = ℏ µk 2 2

This is the final result, from which we can calculate the uncertainty in p.

Note that by combining the results from 5-26 and 5-27, the Uncertainty principle holds, because if we increase mass or the force constant, although we can know x with more certainty, we don’t know p. 5-39 ν = 2.63 × 103 −1 79 ¯l = 141 obs cm for H√Br, and pm. The RMSD bond length⟨ in⟩ the ⟨x2⟩ x = l − ¯l x2 = ground( state) will be given by , where⟨ ⟩ . Using equation 5.60, √ℏ v + 1 x2 = √ℏ kµ 2 , so in the ground state, 2 kµ . I can also use equation 5.39,

Boekelheide CH342: Quantum Chemistry Problem Set 4 Solutions 8

k = (2πcωe)2 µ. Combining these, ⟨ ⟩ ℏ x2 = , 4πcωµe

g g where µ = 0.9951 mol to 4 sig figs, using m79Br = 78.9183371 mol from Wikipedia. Plugging in values, ⟨ ⟩ ℏ x2 = 4πcωµe ( ) ( ) 2 ( ) 2 ( ) kg m cm J mol s 1000 g m 2 1 = 1.07 × 10−47 × s 6.0221 × 1023 g m kg 100 cm J mol ( ) 12 − 10 pm = 6.44 × 10 23 m2 × m = 64.4 pm2 √ So the RMSD displacement is ⟨x2⟩ = 8.03 pm , or 5.69% of the overall bond length. Interesting. 5-41

Using the expression at the end of problem 5-40, the fraction of molecules in the −hν/k T −1 ground state is f0 = 1 − e B . Given that ωe = 2650 cm for HBr, we can 13 −1 calculate ν = ωec = 7.944 × 10 s . At 300 K,

−hν/k T f0 = 1 − e B − − − 6.626×10 34 7.944×10 34 1 − ( J s)( s ) − J 1.381×10 23 (300 K) = 1 − e ( K ) = 1 − 3.023 × 10−6 = 100%

To the precision of the measurement of ωe, all the molecules are in the ground state at 300 K At 2000 K,

f0 = 1 − 0.149 = 85.1%

Fewer of the molecules are in the ground state at 2000 K, but still a surprising ma- jority.

Boekelheide CH342: Quantum Chemistry