Download EE1 2006: Solution to homework assignment 6 Problem 1: (a) Show

EE1 2006: Solution to homework assignment 6
Problem 1:
2
2
(a) Show that the functions ψ0 (x) = N0 e−αx /2 and ψ1 (x) = N1 xe−αx /2 are√eigenfunctions of the
Hamiltonian operator for the harmonic oscillator with spring constant k where α = km/h̄.
(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 the
two 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:
2
Need to operate with the Hamiltonian operator on ψ0 (x) = N0 e−αx /2 and show that the outcome is a
constant, the eigenvalue, times ψ0 (x). For the harmonic oscillator, the Hamiltonian operator is
Ĥ = −
h̄2 d2
+ V (x)
2m dx2
So,
h̄2 d −2αx −αx2 /2 k 2 −αx2 /2
Ĥψ0 (x) = N0 −
e
+ x e
2m dx
2
2
h̄2 2α2 x2
k 2 −αx2 /2
−αx2 /2
= N0 −
−α e
+ x e
2m
2
2
In order for ψ0 (x) to be an eigenfunction, the x2 term needs to vanish. This means that α has to have a
value such that
k
h̄2 α2
=
2m
2
That is
√
km
α=
h̄
The result of applying the Hamiltonian operator then becomes
2
h̄2
Ĥψ0 (x) = N0 −
(−α)e−αx /2
2m
(b) The result in part (a) shows that the eigenvalue corresponding to ψ0 (x) is
!
r
√
h̄2
h̄2
km
h̄ k
E0 = −
(−α) =
=
2m
2m
h̄
2 m
(c) Find the normalization constant, N0 : Must have
Z ∞
|ψ0 (x)|2 dx = 1
−∞
Plugging in the form of the function gives,
N02
Z
∞
−∞
2
e−αx dx = 1
The integral can be found in tables of integrals
Z
∞
r
−αx2
e
dx =
−∞
which means that
N02
r
that is
N0 =
π
α
π
=1
α
α 1/4
.
π
(d) See figure 5-7 in the textbook by McQuarrie.
(e) Show that ψ0 (x) and ψ1 (x) are orthogonal, that is
Z
∞
∞
Z
2
xe−αx dx.
ψ0 (x)ψ1 (x) dx = N0 N1
−∞
−∞
This can be seen most easily from a symmetry argument. The ground state wavefunction ψ0 (x) is symmetric
about x = 0, that is ψ0 (−x) = ψ0 (x). It is an even function. The first excited state wavefunction is
antisymmetric, that is ψ1 (−x) = −ψ1 (x). It is an odd function. The product of an even function and an odd
function is an odd function. The integral over an odd function is necessarily zero if the range of integration
is symmetric about x = 0.
Problem 2:
What is the probability that a quantum mechanical harmonic oscillator is within the classically forbidden
region when it is in the first excited state, n = 1? Compare your result with the corresponding probability
for 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 energy
is equal to the first excited state energy of the quantum mechanical harmonic oscillator. At the classical
turning point, the potential energy equals the total energy
kA21
=
2
r
1
k
+ 1 h̄
.
2
m
This gives
s
A1 =
r
3h̄
√
=
km
3
.
α
The probability of finding the oscillator in the classically forbidden region is the probability of x > A1 or
x < −A1 . This is given by the appropriate integral over the probability distribution function
√
− 3α
Z
Z
2
|ψ1 (x)| dx +
P =
−∞
Z
= 2
∞
√
|ψ1 (x)|2 dx = 2
3α
r
=4
α3
π
Z
∞
√
Z
∞
√
∞
√
|ψ1 (x)|2 dx
3α
r
3α
2
2
α
4α x2 e−αx dx
4π
x2 e−αx dx
3α
Rz
2
The integral can be related to the error function erf (z) = √2π 0 e−t dt or, if preferred, the complementary
error function√erf c(z) = 1 − erf
√(z) (use whichever is more convenient to find). Make the substitution of
variables t = α x, so dx = (1/ α)dt and the probability becomes
4
P = √
π
Z
∞
√
2
t2 e−t dt .
3
To relate this integral to the error function, do integration by parts
Z
∞
√
2
t2 e−t dt = −
3
∞
1
2
Z
1
2
h
=−
√
2
t · (−2t) e−t dt
3
2
t · e−t
i∞
√
3
Z
−
∞
√
2
1 · e−t dt
3
√ 1 √ −3
=−
− 3e − erf c( 3)
2
√ 1 √ −3
=
3e + 1 − erf ( 3)
2
The probability is
√
√
√ √
√
4 1 √ −3
3 −3
3 −3
P =√
3e + 1 − erf ( 3) = 2
e + erf c( 3) = 2
e + 1 − erf ( 3)
π
π
π 2
√
√
Using tables (for example the Mathematical Handbook by Schaum) to evaluate erf ( 3) or erf c( 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 potential
is rising more steeply, i.e. it is ’harder’, at the classical turning point for E1 than at the classical turning
point for E0 . The steeper the potential wall is, the smaller the tail of the wavefunction into the classically
forbidden 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
σ 12 σ 6
−
U (r) = 4
r
r
where r is the distance between the two atoms and the two parameters, and σ depend on which atoms are
involved. The shape of the curve is qualitatively similar to the Morse potential curve, but is more appropriate
for van der Waals interactions (while Morse is more appropriate for molecules with a covalent bond between
atoms). In each one of the questions (a-e) below you should give an expression that contains the parameters
of the potential function, 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 value
of 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 force
constant, k, of the harmonic oscillator approximation to U (r).
(c) Assuming the Harmonic Oscillator approximation is good enough for the ground state, what is the
dissociation energy of a dimer described by a Lennard-Jones potential (that is, what is the minimum energy
required 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 is
valid?
(e) What is the first anharmonic correction term to the energy of the dimer (that is, next term beyond the
harmonic oscillator term in the Taylor expansion of the Lennard-Jones potential function)?
Solution:
(a) At the minimum, the first derivative is zero. Find
Jones potential, and then set it to zero to find rb .
dU
dr
by differentiating the expression for the Lennard-
σ 6 σ 12
dU
σ6
24
σ 12
=
+
= 4 −12 13 + 6 7
−2
dr
r
r
r
r
r
This must be zero when r = rb , so rb must satisfy
2
σ 12
r
That is
=
σ 6
r
rb = 21/6 σ
The potential energy at the minimum is U (rb )
U (rb ) = 4
12
σ
21/6 σ
−
6 σ
21/6 σ
= 4
1 1
−
4 2
= −
(b) Let x = r − rb and V (x) = U (r). Need to expand V (x) up to and including the second order term in a
Taylor series
dV
x2 d2 V
V (x) = V (0) + x
(0) +
(0)
dx
2 dx2
1
= − + 0 + kx2
2
2
where k = ddxV2 (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 to
get an expression for the spring constant in terms of the potential parameters and σ and the atomic masses
m1 and m2 , find the second derivative
12 6 σ
d2 U
σ
d2 V
(0) =
(rb ) = 4 12 · 13
k =
−6·7
dx2
dr2
rb 14
rb8
4
= 2
rb
12 · 13
4
=
1/6
(2 σ)2
=
4
21/3 σ 2
σ
rb
12
−6·7
σ
rb
6 !
!
2
1
1
12 · 13
−6·7
2
2
(39 − 21) =
72
21/3 σ 2
=
36
22/3 σ 2
(c) The dissociation energy is the energy needed to bring the dimer up from the ground state to the state
of two separated atoms.
Ed = E2Ar − EAr2
The energy of q
two separated atoms is E2Ar = limr→∞ U (r) = 0 and the energy of the ground state is
h̄
EAr2 = − + 2 µk where k is given by the expression obtained in (b) and µ is the reduced mass of the dimer
µ = m1 m2 /(m1 + m2 ). It is important to add to − the zero point energy of the harmonic oscillator, which
is 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
s
∆E = h̄ω = h̄
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-Jones
potential is the third order term in the Taylor expansion, the next term after the second order expansion
given in part (b). This is
x3 d3 U
δE =
(rb ) .
3! dr3
The expression for the third derivative can be obtained by differentiating the second derivative given in part
(b).
12 6 d3 U
σ
σ
(rb ) = 4 −12 · 13 · 14
+6·7·8
.
3
15
dr
rb
rb9
Now plug in rb = 21/6 and simplify.