CH 126 HW 1, Solutions

1. (a) Normalization implies N2 〈ψ|ψ〉 = 1 where N is the normalization constant and |ψ〉

is the eigenvector.

Plugging this condition in yields: 1 = N2 〈ψ|ψ〉 = N2(1 + 4 + 1). So N = 1/√

6 where

the normalized eigenvector |ψ′〉 = N |ψ〉

(b) |ψ〉 = 1/√

6(|a〉+ 2i |b〉 − |c〉)

So

〈H〉 = 〈ψ|H|ψ〉 = 1/6(0.5 + 40 + 11)hω = 1/6(43/2)hω = (103/12)hω

〈A〉 = 〈ψ|A|ψ〉 = 1/6(−2i− 2i− 1) = 1/6(−4i− 1)

(c)

〈x|H|y〉 = hω

0.5 0 0

0 10 0

0 0 11

〈x|A|y〉 = hω

〈a|A|a〉 〈a|A|b〉 〈a|A|c〉

〈b|A|a〉 〈b|A|b〉 〈b|A|c〉

〈c|A|a〉 〈c|A|b〉 〈c|A|c〉

=

0 0 1

1 0 0

0 1 0

Now|ψ〉 = 1/√

6

1

2i

−1

and 〈ψ| = 1/√

6[1 −2i −1

]Plugging in we get the same answer as (b):

〈ψ|H|ψ〉 = (103/12)hω

〈ψ|A|ψ〉 = 1/6(−4i− 1)

(d) If the system is in state |ψ(t = 0)〉 = 131/2

(|a〉 + |b〉 − i |c〉) at time t=0, then we add a

phase factor of e−iET/h on each eigenstate, where E is the energy of the state, to find the

state of the system at time t=T:

|ψ(t = T )〉 = 131/2

(e−i0.5ωT |a〉+ e−i10ωT |b〉 − ie−i11ωT |c〉)

(e) Given |ψ(t = 0)〉 = 131/2

1

1

−i

and

1

|ψ(t = T )〉 = 131/2

e−i0.5ωT

e−i10ωT

−ie−i11ωT )

we have

ρ(t = 0) = 1/3

1 1 i

1 1 i

−i −i 1

ρ(t = T ) = 1/3

1 ei9.5ωT iei10.5ωT

e−i9.5ωT 1 ieiωT

−ie−i10.5ωT −ie−iωT 1

2

(1)

Problem 2

(a) First write out the Schrödinger equation with the proper terms added in the Hamiltonian

ψψε ExexKdxd

m=⎟⎟

⎠

⎞⎜⎜⎝

⎛−+− 2

2

22

22h

We can write this in the same form as the harmonic oscillator equation if we complete the square for x and make some variable substitutions.

ψψ

εε

ψεψε

ψψεε

ψψεεε

'22

2'

222

22

222

22

22

2

22

2

22

22

2

22

222

2

22

EzKdxd

m

KeKEE

Kexz

KeKE

KexK

dxd

m

EKe

KexK

dxd

m

EKe

Kex

KexK

dxd

m

=⎟⎟⎠

⎞⎜⎜⎝

⎛+−

⎟⎠⎞

⎜⎝⎛+=−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+=⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −+−

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ −+−

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛+−+−

h

h

h

h

Since this has the same mathematical form as the harmonics oscillator differential equation, it will have the same solutions, etc., so the energy eigenstates are

( )Ke

mKnE

mKnE

221

21'

2ε−⎟

⎠⎞

⎜⎝⎛ +=

⎟⎠⎞

⎜⎝⎛ +=

h

h

This problem is very similar to a classical simple harmonic oscillator (vertically oriented mass and spring) in a gravitational field. The gravitational field serves to change the equilibrium position of the mass, though the amplitude and frequency of the oscillations do not change. This is easy to see, since the spring constant does not change. Similarly, the potential energy of the quantum harmonic oscillator shifts, though the wave functions have the same form and the frequency and amplitude are unaffected. To explicitly show this, the classical mechanics are explored below.

(2)

No Gravity: Gravity:

0)()(

0)()(

)(

202

20

2

2

=+

=

=+

==−=

txtxdtd

mk

tkxtxdtdm

txdtdmmakxF

ω

ω

0)()(

0)()(

)(

20

202

20

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛++

=

=−+

==+−=

ωω

ω

gtxtxdtd

mk

mgtkxtxdtdm

txdtdmmamgkxF

let

dxdu

gxu

=

−= 20ω

Solutions have the form:

πω

ϕω

2

)cos()(

0

0

=

+=

f

tAtx

20

0

0

0

)cos()(

2

)cos()(

ωϕω

πω

ϕω

gtAtx

f

tAtu

−+=

=

+=

(b) To show that the first order correction in energy vanish, just note that the perturbation Hamiltonian H’=-eEx is odd and remember that )0(

nψ are even functions . And then because of symmetry

0)0()0()1( == nnn xeEE ψψ

Alternatively, write x as combination of raising and lowering operators (as in (c), see below) and show that the term vanishes due to orthonormality. The second order corrections are given by

( )

( ) ( )∑

∑

≠

≠

−=

−=

mnn

mn mn

mnn

mnmxn

eEE

EE

xeEE

ω

ψψ

h

22)2(

)0()0(

2)0()0(2)2(

Using the raising and lowering operators, the matrix element in the above can be found

( )1,1,†

†

12 +− ⋅−+⋅=+

+=

nmnm nnm

maan

maanmxn

δδωh

Plugging into the second order expression:

( )( )

( )( ) ( )

( )( )

( )2

2)2(

2

2)2(

1,1,2

2)2(

2

1,1,2)2(

2

11

12

12

12

ω

ω

δδω

ω

δδω

meEE

nnmeEE

mnnn

meEE

mn

nnm

eEE

n

n

mn

nmnmn

mn

nmnm

n

−=

⎟⎠⎞

⎜⎝⎛

−+

+⋅=

−⋅++⋅

=

−

⋅++⋅

=

∑

∑

≠

+−

≠

+−

h

h

which is identical to our result above. Because this is identical to the analytical method in (a), we can expect any higher order corrections to be zero.

3. (a) H ′ = −µEe−t2/τ2 and P10 = |H10/ih|2∫∞−∞ e

iω10tf(t)dt

H10 = H ′

f(t) = e−t2/τ2

So

P10 = |H10/ih|2∫∞−∞ e

iω10te−t2/τ2dt = E2µ2πτ2e−ω

2τ2/2

h2

(b) The prefactor is π(100kV/cm ∗ 1D)2/h2 = 3.14 ∗ 1023s−2

3500 cm−1 = 105 THz

Looking at the plot, the population is maximized near 1.3 ∗ 10−14 s = 13 fs. This makes

sense, considering the the vibrational frequency of the OH-stetch is close to this value:

1/(105 THz) = 9 fs.

2.×10-14 4.×10-14 6.×10-14 8.×10-14 1.×10-13tau (s)

5.×10-6

0.00001

0.000015

0.00002

Probability

4. (a) See plot on next page...The MP2 scan shows a clear dissociation energy of about

0.2 Hartrees, while the HF scan is diverging at large bond length. Considering the actual

dissociation energy of 4.75 eV = 0.17 Hartrees it seems that MP2 yields the more reliable

result.

(b) The fit value is somewhat dependent on the range you use but should be close to the

literature value of 4401 cm−1. Near the bottom of the well I obtain values of:

HF/3-21G: k = Angstroms−2*hartrees = 610 N/m

MP2/3-21G: k = 1.5 Angstroms−2*hartrees = 650 N/m

The reduced mass of H2 is µ = 1/2 amu and ν = 1/(2π)√k/µ, so we get:

HF/3-21G: ν = 136THz = 4490 cm−1

3

MP2/3-21G: ν = 140THz = 4620 cm−1

4