Download Problem set 4 Engel P7

Problem set 4 Engel P7.1, P7.6-10.
Problem 7.1 The force constant for a H18F molecule is 966 N m-1.
a)
Calculate the zero-point vibrational energy for this molecule for a harmonic potential.
The following relation gives the zero-point vibrational energy
E=1/2 hν
where ν is the quantum vibrational frequency. We can calculate ν from the relation
1
2π
⎛k⎞
⎜⎜ ⎟⎟
⎝µ⎠
1/ 2
where m is the reduced mass and k is the force constant, 966 N m-1.
We can obtain the reduced mass from the relation
µ=
m1 m 2
m1 + m 2
We must be careful here to stick with mks units.
m1=mass of 1H/atom in kilograms = 1 g/mol × 0.001 kg/g ÷ 6.022 ×1023
(atom/mol)=0.166×10-26 kg/(atom 1H)
Similarly, m2=2.99×10-26
The reduced mass is then
µ = 1.57 × 10-27 kg, a little smaller than the mass of 1H (hence reduced mass).
Substituting µ in the expression for the vibrational frequency, we obtain
ν=1.25×1014 s-1
We now have what we need to calculate the zero-point energy
E=1/2 hν = 0.5 × 6.626 × 10-34 J-s × 1.25×1014 s-1 = 4.14 × 10-20 J.
11/14/2005 2:03 PM
-1-
PChem05ProbSet4HO.doc
Problem 7.6 Evaluate the average vibrational amplitude, <x>, of the quantum harmonic
oscillator about its equilibrium value
Common sense and symmetry tells us that the answer must be zero for any state of the quantum
harmonic oscillator. The argument goes as follows: We want to evaluate integrals of the form
∞
∞
−∞
−∞
∫ ΨxΨdx = ∫ xΨΨdx
We’ve switched the order of multiplication in the integral to focus attention on the product ψψ.
The HO wave functions are either symmetric or antisymmetric. Thus, the product ψψ must be
symmetric for all wave functions. We multiply ψψ by an antisymmetric function, x. We
conclude that the product xψψ must be antisymmetric. Looking at the integral again, we see that
it must have the value zero.
Problem 7.7 Evaluate the average linear momentum <px>=< − ih
d
for various states of the
dx
quantum harmonic oscillator.
Again, common sense tells us the answer must be zero, although we have to do a little more
work to show it. The common sense part here is that if the particle had a net momentum other
than zero the whole assembly would be moving along through space. It’s not. The oscillator is
fixed in place! To be rigorous, we will need to consider integrals of the general form
∞
d
∫ Ψ dx Ψdx
−∞
We will not need to worry about the constant, as the idea is to show the integral is zero.
a)
Ground state. The ground state for the HO has the general form (α/π)1/4 exp(-αx2/2).
When we differentiate this function with respect to x, we recover the same function (to within a
constant) but multiplied by x. Thus the integral (again to within a constant) becomes
∞
∫ xΨΨdx
−∞
just as we had for the average amplitude. And by the same argument about the symmetry of the
functions in the integrand, this integral is zero.
Now let’s generalize. Inspection of the other wave functions reveals that all the Hermite
polynomials are either symmetric (even) or antisymmetric (odd) functions. When differentiated,
each term of the polynomial “loses” a power of x. If the exponent of the term was originally
even, it becomes odd after differentiation. If the exponent of the term was odd, it becomes even.
Thus differentiation of the Hermite polynomials converts even functions into odd functions, and
odd functions into even functions.
11/14/2005 2:03 PM
-2-
PChem05ProbSet4HO.doc
We also know from our treatment of ground state, above, that when the exponential term in the
complete HO wave functions is differentiated, the leading exponential is converted from an even
function to an odd one, i.e., from (α/π)1/4 exp(-αx2/2) to -αx times the original even function.
Putting all these arguments together for two cases
Case 1. Hermite polynomial is even
The derivative term must be odd because we have
d/dx (exponential (even) * Hermite (even))=[d/dx exponential] (odd) * Hermite (even) +
exponential(even)*[d/dx Hermite] (odd). Overall, the derivative is an odd function.
To form the expectation value of the momentum we multiply this derivative (odd) by ψ, which
by hypothesis is even. Thus the integrand is the product of an even function and an odd
derivative, which is odd. The integral is zero.
Case 2. Hermite is odd
Here we find the derivative must be an even function.
d/dx (exponential (even) * Hermite (odd)) = [d/dx exponential] (odd) * Hermite (odd)
+exponential(even)*[d/dx Hermite] (even)
Here, to form the expectation value of the momentum we multiply this derivative (even) by ψ,
which by hypothesis is odd. Thus, once again, the integrand is the product of an even function
and an odd derivative, which is odd. The integral is zero.
Problem 7.8
Non-zero integral this time. We want <x2> =<ψ x2 ψ> for the following integrals
∞
1)
∫
∞
(α/π)1/2 exp(-αx2) x2 dx = (α/π)1/2
−∞
∫
exp(-αx2) x2 dx
−∞
∞
2)
∫
Ground state nvib=0
∞
(4α3/π)1/2 x2 exp(-αx2) x2 dx = (4α3/π)1/2
−∞
∫
exp(-αx2) x4 dx
1st excited state nvib=1
−∞
∞
3)
∫
(α/[4π])1/2 (2αx2 - 1)2 exp(-αx2) x2 dx =
−∞
∞
(α/π)1/2 [4 α2
∫
−∞
∞
exp(-αx2) x6 -4α
∫
∞
exp(-αx2) x4 dx +
−∞
∫
exp(-αx2) x2dx] 2nd excited
−∞
state nvib=2
These integrals can be evaluated from the standard formula
11/14/2005 2:03 PM
-3-
PChem05ProbSet4HO.doc
∞
∫
−∞
∞
x2nexp(-αx2)dx = 2 ∫ x2nexp(-αx2)dx =
−0
=2×
1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ (2n − 1) π
α
2 n +1 α n
using the values n=1, n=2, and n=3. Note well that the “n” that appears in this equation is NOT
the quantum number of the harmonic oscillator.
For n=1 (ground state - vibrational quantum number nvib=0), we have
2×
1
2 α
2
π 1 π
=
α 2 α3
Multiplying by the constant in (1) above, we obtain <x2>
1
= 1/2 hν/k
2α
For n=2 (first excited state - vibrational quantum number nvib=1), we have
2×
1⋅ 3
23α 2
π 3 π
=
α 4 α5
Multiplying by the constant in (2) above, we obtain <x2> =
3
= 3/2 hν/k
2α
For n=3 (second excited state - vibrational quantum number nvib=2), we have for the integral in
x6
2×
1 ⋅ 3 ⋅ 5 π 15 π
=
24α 3 α 8 α 7
The three terms in (3) are as follows
⎛α ⎞
⎟
⎜
⎝ 4π ⎠
1
2
⎡ 2 15 π
3 π
1 π ⎤ 5
−
4
α
+
=
= <x2> = 5/2 hν/k
⎢4α
7
5
3 ⎥
8 α
4 α
2 α ⎥⎦ 2α
⎢⎣
Hmmm. There seems to be a pattern!
Problem 7.9
Evaluate the average of the square of the linear momentum of the quantum harmonic oscillator
< p x2 > for the ground state and the first two excited states
11/14/2005 2:03 PM
-4-
PChem05ProbSet4HO.doc
We want Ψ − h 2
d2
Ψ . The second derivative is a big nuisance. Although the work to be
dx 2
done is straightforward and you can solve the problem by brute force, here’s some trickery that
makes use of other things we already know. The average value < p x2 > is related to the average
value of the kinetic energy, T:
< p x2 >/2m= <T>
Also, by conservation of energy
<T> +<V> = E
where <V> is the average value of the potential energy. Re-writing, we have
< p x2 > = 2m [E - <V>]
Here’s the trick: we know that V=1/2 kx2 so <V>=<1/2 kx2>=1/2 k<x2>, where k is the spring
constant. This observation saves a lot of work because we already know <x2> from problem 7.8
and we know that for the harmonic oscillator
E=(nvib+1/2)hν with nvib=0,1,2 where ν=
1
2π
k
m
So our problem is solved without taking derivatives or doing any integrals!!!!
Before we plug in, it will be convenient to express ν in terms of α. From the text, we have
α=
2π km
h
so
ν=
k
hα
Now we can plug in.
For nvib=0
⎛1
⎝2
k ⎞ 1 mk
1 1 ⎞
⎛k
=1/2 mhν=
⎟=
⎟ = m⎜ −
2 2α ⎠
⎝ α 2α ⎠ 2 α
⎛3
⎝2
1 3 ⎞
⎛ 3k 3k ⎞ 3 mk
=3/2 mhν
−
⎟=
⎟ = m⎜
2 2α ⎠
⎝ α 2α ⎠ 2 α
< p x2 > = 2m⎜ hν − k
For nvib=1
< p x2 > = 2m⎜ hν − k
11/14/2005 2:03 PM
-5-
PChem05ProbSet4HO.doc
And by extrapolation, for For nvib=2
< p x2 > = 5/2 mhν
Problem 7.10
Using the results for problems 7.6 through 7.9, calculate the uncertainties in the position and
momentum,
2
σ p2 = p 2 − p
and σ x2 = x 2 − x
2
for the ground state and the first two excited states of
the quantum harmonic oscillator. Compare your results with the predictions of the Heisenberg
uncertainty pinricple.
We will do the calculation for the ground state and leave the next two states to you.
2
σ p2 = p 2 − p
σ x2 = x 2 − x
2
=
=
1
1
2
h mk -0 = h mk
2
2
1
1
1
1
-02=
= h
2α
2α
2
mk
It turns out that the magnitude of σx is of the same order as the range of motion, calculated
classically, for x.
multiplying, we get
1
4
σ 2pσ x2 = h 2
And taking square roots,
σ pσ x ≅
h
2
as expected.
11/14/2005 2:03 PM
-6-
PChem05ProbSet4HO.doc