Download Lecture 5: Harmonic oscillator, Morse Oscillator, 1D Rigid Rotor

Lecture 5: Harmonic oscillator, Morse
Oscillator, 1D Rigid Rotor
It turns out that the boundary condition of the wavefunction going to
zero at infinity is sufficient to quantize the value of energy that are allowed.
1
Ev = v +
~ω
v = 0, 1, 2, ...
2
p
where ω = k/m is the angular frequency of the oscillator. Thus the energy
levels are equally spaced starting from 1/2~ω.
The corresponding wavefunctions are given by
r
x
~2
−y 2 /2
2
ψv (x) = Nv Hv (y)e
y=
α =
α
km
In the above solution, Hv (y) is a Hermite Polynomial function and Nv is the
normalization constant. The Hermite polynomials that satisfy a differential
equation
Hv′′ − 2yHv′ + 2vHv = 0
and they can be derived using the power series method. The first few polynomials are given by
H0 (y) = 1
H1 (y) = 2y
2
H2 (y) = 4y − 2 H3 (y) = 8y3 − 12y
Notice that the odd polynomials contain only powers of y and are odd functions and the even polynomials are even functions. This is a general feature of
the Hermite polynomials and some other polynomials that we will encounter.
Consider v = 0. We have
1
E0 = ~ω
2
1
the ground state energy or the zero point energy. The wavefunction is given
by
2
2
ψ0 (x) = N0 e−x /2α
which is simply a Gaussian function. The normalization of the Gaussian
integrals can be carried out using
Z+∞
√
2
e−x dx = π
−∞
√
and it turns out that N02 = 1/α π.
The wavefunction for the first excited state v = 1 is given by
ψ1 (x) = N1
2x −x2 /2α2
e
α
This function is odd and has a node at x = 0. We plot the first few wavefunctions and the squares of the wavefunctions, i.e. the probability densities
as a function of x.
The quantum harmonic oscillator shows a finite probability in classically
forbidden regions as described below. Consider the v = 0 state wherein the
total energy is 1/2~ω. We see that the total energy E is equal to the potential
energy V when
1
1
~ω = kx2m
2
2
which leads to xm = ±α, the maximum allowed displacement. Classically, an
oscillator of energy E0 will oscillate harmonically between x = α and x = −α.
However, the quantum mechanical oscillator has a nonzero probability of
being in the region beyond x = ±α. This phenomenon is referred to as
tunneling.
The Harmonic oscillator is a model for studying vibrations of molecules. In
fact, the vibrational energy is modeled using the Harmonic oscillator model
and can closely be matched with Infra-red spectroscopy experiments. However, it is not really sufficient to describe the potential energy of bonds, since
2
U(x)
x
Figure 1: Energy Levels of a Harmonic Oscillator shown in the potential
energy well
ψ
ψ0
ψ
ψ2
1
x
x
Figure 2: The first few wavefunctions of a harmonic osciallator
we would like to have a “breaking” of bonds when the bond is stretched. In
this regard, another model potential is used called the Morse potential given
by
2
V (x) = De 1 − e−ax
3
where the constant a is given by
a=
k
2De
1/2
We just mention this potential here for completion. It can be solved exactly
and the energy levels are closely related to the harmonic oscillator for lower
values of v but get closer as v gets larger.
R−Re
0
De
Figure 3: The potential energy of a Morse oscillator showing the energy levels
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1
The Rigid Rotor Model
We now go on to describe rotational motion of a particle. We first consider
the 1D rigid rotor which corresponds to a particle freely moving on a ring or
a rod freely spinning about an axis perpendicular to its length (Z-axis). The
Hamiltonian operator for this system is given by
Ĥ =
L̂2z
2I
where Lz refers to the Z-component of the angular momentum and I refers to
the momoent of inertia. Note that I = µR2 where µ is the reduced mass and
R is the radius of the orbit for circular motion. We can write the operator
for Z-component of the angular momentum as
∂
∂
L̂z = x̂p̂y − ŷ p̂x = −i~ x
−y
∂y
∂x
This is most conveniently expressed in spherical polar coordinates using the
transformations
x = r sin θ cos φ y = r sin θ sin φ z = r cos θ
Using elementary methods in partial derivatives, we can show that
L̂z = −i~
∂
∂φ
which leads to the expression for the Hamiltonian in terms of φ as
Ĥ = −
~2 ∂ 2
2I ∂φ2
which leads to the Schrödinger equation
−
~2 ∂ 2 ψ(φ)
= Eψ(φ)
2I ∂φ2
which is solved by
ψ(φ) = Ae
iml φ
ml = ±
5
2IE
~2
1/2
Invoking the periodicity condition ψ(φ) = ψ(φ + 2π), we get eiml 2π = 1 or
ml = 0, ±1, ±2, .... and the energy values are given by
Eml =
m2l ~2
2I
and the normalized wavefunctions are given by
1
ψ(φ) = √ eiml φ
2π
Notice that except the ground states, all states are doubly degenerate. Also,
notice that the energy eigenstates are also eigenfunctions of the angular momentum operator with eigenvalue Lz = ml ~.
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