EE1 2006: Solution to homework assignment 6

Problem 1:

(a) Show that the functions ψ0(x) = N0e−αx2/2 and ψ1(x) = N1xe

−αx2/2 are eigenfunctions of theHamiltonian operator for the harmonic oscillator with spring constant k where α =

√km/h.

(b) What are the eigenvalues, E0 and E1?

(c) Find the normalization constants, N0 and N1.

(d) Sketch the probability distribution for the location of the particle as a function of x for each one of thetwo states.

(e) Show that the two functions are orthogonal (as any two eigenfunctions of the Hamiltonian must be,unless they correspond to the same eigenvalue). (Hint: you can use symmetry arguments).

Solution:

Need to operate with the Hamiltonian operator on ψ0(x) = N0e−αx2/2 and show that the outcome is a

constant, the eigenvalue, times ψ0(x). For the harmonic oscillator, the Hamiltonian operator is

H = − h2

2md2

dx2+ V (x)

So,

Hψ0(x) = N0

[− h2

2md

dx

(−2αx

2

)e−αx2/2 +

k

2x2e−αx2/2

]

= N0

[− h2

2m

(2α2x2

2−α)e−αx2/2 +

k

2x2e−αx2/2

]In order for ψ0(x) to be an eigenfunction, the x2 term needs to vanish. This means that α has to have avalue such that

h2α2

2m=k

2That is

α =

√km

h

The result of applying the Hamiltonian operator then becomes

Hψ0(x) = N0

[− h2

2m(−α)e−αx2/2

]

(b) The result in part (a) shows that the eigenvalue corresponding to ψ0(x) is

E0 = − h2

2m(−α) =

h2

2m

(√km

h

)=h

2

√k

m

(c) Find the normalization constant, N0: Must have∫ ∞−∞

|ψ0(x)|2 dx = 1

Plugging in the form of the function gives,

N20

∫ ∞−∞

e−αx2dx = 1

The integral can be found in tables of integrals∫ ∞−∞

e−αx2dx =

√π

α

which means that

N20

√π

α= 1

that is

N0 =(απ

)1/4

.

(d) See figure 5-7 in the textbook by McQuarrie.

(e) Show that ψ0(x) and ψ1(x) are orthogonal, that is∫ ∞−∞

ψ0(x)ψ1(x) dx = N0N1

∫ ∞−∞

xe−αx2dx.

This can be seen most easily from a symmetry argument. The ground state wavefunction ψ0(x) is symmetricabout x = 0, that is ψ0(−x) = ψ0(x). It is an even function. The first excited state wavefunction isantisymmetric, that is ψ1(−x) = −ψ1(x). It is an odd function. The product of an even function and an oddfunction is an odd function. The integral over an odd function is necessarily zero if the range of integrationis symmetric about x = 0.

Problem 2:

What is the probability that a quantum mechanical harmonic oscillator is within the classically forbiddenregion when it is in the first excited state, n = 1? Compare your result with the corresponding probabilityfor the ground state, which was shown to be 16%. Explain the trend.

Solution:

First find the classical turning for the oscillations of the classical harmonic oscillator when its energyis equal to the first excited state energy of the quantum mechanical harmonic oscillator. At the classicalturning point, the potential energy equals the total energy

kA21

2=(

12

+ 1)h

√k

m.

This gives

A1 =

√3h√km

=

√3α.

The probability of finding the oscillator in the classically forbidden region is the probability of x > A1 orx < −A1. This is given by the appropriate integral over the probability distribution function

P =∫ −√3α

−∞|ψ1(x)|2dx +

∫ ∞√

3α

|ψ1(x)|2dx

= 2∫ ∞√

3α

|ψ1(x)|2dx = 2∫ ∞√

3α

√α

4π4α x2 e−αx2

dx

= 4

√α3

π

∫ ∞√

3α

x2 e−αx2dx

The integral can be related to the error function erf(z) = 2√π

∫ z

0e−t2dt or, if preferred, the complementary

error function erfc(z) = 1 − erf(z) (use whichever is more convenient to find). Make the substitution ofvariables t =

√α x, so dx = (1/

√α)dt and the probability becomes

P =4√π

∫ ∞√

3

t2 e−t2dt .

To relate this integral to the error function, do integration by parts∫ ∞√

3

t2 e−t2dt = −12

∫ ∞√

3

t · (−2t) e−t2dt

= −12

([t · e−t2

]∞√

3−∫ ∞√

3

1 · e−t2dt

)= −1

2

(−√

3e−3 − erfc(√

3))

=12

(√3e−3 + 1 − erf(

√3))

The probability is

P =4√π

12

(√3e−3 + 1 − erf(

√3))

= 2√

3πe−3 + erfc(

√3) = 2

√3πe−3 + 1 − erf(

√3)

Using tables (for example the Mathematical Handbook by Schaum) to evaluate erf(√

3) or erfc(√

3) gives

P = 0.112 = 11.2% .

This is smaller than the comparable probability for the ground state of the quantum harmonic oscillator,16% (see section 5-10 in the text book by McQuarrie, and lectures notes). The reason is that the potentialis rising more steeply, i.e. it is ’harder’, at the classical turning point for E1 than at the classical turningpoint for E0. The steeper the potential wall is, the smaller the tail of the wavefunction into the classicallyforbidden region.

Problem 3:

A simple function that is frequently used to describe the potential energy of rare gas dimers (such as Ar2)is the Lennard-Jones potential

U(r) = 4ε((σ

r

)12

−(σr

)6)

where r is the distance between the two atoms and the two parameters, ε and σ depend on which atoms areinvolved. The shape of the curve is qualitatively similar to the Morse potential curve, but is more appropriatefor van der Waals interactions (while Morse is more appropriate for molecules with a covalent bond betweenatoms). In each one of the questions (a-e) below you should give an expression that contains the parametersof the potential function, ε and σ, and possibly also the mass of the two atoms, m1 and m2.

(a) What is the distance rb at which the potential energy is smallest (the ”bond length”) and find the valueof the potential energy at that distance (the binding energy of the dimer).

(b) Expand U(r) in a Taylor series about r = rb up to second order and give an expression for the forceconstant, k, of the harmonic oscillator approximation to U(r).

(c) Assuming the Harmonic Oscillator approximation is good enough for the ground state, what is thedissociation energy of a dimer described by a Lennard-Jones potential (that is, what is the minimum energyrequired to break the dimer a part)?

(d) A dimer initially in the ground state can get excited to the first excited state as it collides with a surface(for example, the walls of the container). What is the energy required to excite the dimer from the ground

vibrational state to the first vibrational excited state assuming the harmonic oscillator approximation isvalid?

(e) What is the first anharmonic correction term to the energy of the dimer (that is, next term beyond theharmonic oscillator term in the Taylor expansion of the Lennard-Jones potential function)?

Solution:

(a) At the minimum, the first derivative is zero. Find dUdr by differentiating the expression for the Lennard-

Jones potential, and then set it to zero to find rb.

dU

dr= 4ε

(−12

σ12

r13+ 6

σ6

r7

)=

24εr

(−2

(σr

)12

+(σr

)6)

This must be zero when r = rb, so rb must satisfy

2(σr

)12

=(σr

)6

That isrb = 21/6 σ

The potential energy at the minimum is U(rb)

U(rb) = 4ε(( σ

21/6 σ

)12

−( σ

21/6 σ

)6)

= 4ε(

14− 1

2

)= −ε

(b) Let x = r − rb and V (x) = U(r). Need to expand V (x) up to and including the second order term in aTaylor series

V (x) = V (0) + xdV

dx(0) +

x2

2d2V

dx2(0)

= −ε + 0 +12kx2

where k = d2Vdx2 (0). The first derivative term vanishes because x = 0 corresponds to a minimum of the V (x)

function. This is a potential energy function for a harmonic oscillator with a spring constant k. In oder toget an expression for the spring constant in terms of the potential parameters ε and σ and the atomic massesm1 and m2, find the second derivative

k =d2V

dx2(0) =

d2U

dr2(rb) = 4ε

(12 · 13

(σ12

rb14

)− 6 · 7

(σ6

r8b

))

=4εr2b

(12 · 13

(σ

rb

)12

− 6 · 7(σ

rb

)6)

=4ε

(21/6σ)2

(12 · 13

(12

)2

− 6 · 7(

12

))

=4ε

21/3σ2(39 − 21) =

72ε21/3σ2

=36ε

22/3σ2

(c) The dissociation energy is the energy needed to bring the dimer up from the ground state to the stateof two separated atoms.

Ed = E2Ar − EAr2

The energy of two separated atoms is E2Ar = limr→∞ U(r) = 0 and the energy of the ground state is

EAr2 = −ε+ h2

√kµ where k is given by the expression obtained in (b) and µ is the reduced mass of the dimer

µ = m1m2/(m1 +m2). It is important to add to −ε the zero point energy of the harmonic oscillator, whichis the energy of the ground state with respect to the potential energy minimum.

(d) The energy required to excite a harmonic oscillator from a stationary state to the next higher state is

∆E = hω = h

√k

µ

where the expression for k is given in part (b) and the expression for µ is given in part (c).

(e) The first anharmonic correction term to the harmonic oscillator approximation to the Lennard-Jonespotential is the third order term in the Taylor expansion, the next term after the second order expansiongiven in part (b). This is

δE =x3

3!d3U

dr3(rb) .

The expression for the third derivative can be obtained by differentiating the second derivative given in part(b).

d3U

dr3(rb) = 4ε

(−12 · 13 · 14

(σ12

rb15

)+ 6 · 7 · 8

(σ6

r9b

)).

Now plug in rb = 21/6 and simplify.