Chemistry 342
Spring, 2005
The Harmonic Oscillator
1. Classical description.
A particle of mass m is subject to a force Fx, which is proportional to its displacement
from the origin (Hooke’s Law).
m

dV( x )
dx
 Fx
k
  kx
x=0
where k is the force constant. If we take the zero of the potential energy V to be at
the origin x = 0 and integrate,
V(x)
V(x) 

x
0
x
dV   k  xdx
0

1
2
kx 2
x
Note that this potential energy function differs from that in the particle-in-the-box
problem in that the walls do not rise steeply to infinity at some particular point in
space (x = 0 and x = L), but instead approach infinity much more slowly.
From Newton’s second law,
F  ma
 m
thus,
d2x
dt 2
 
k
x
m
d2x
dt 2
  kx
Chemistry 342
Spring, 2005
This second-order differential equation is just like that for the free particle, so
solutions must be of the form
1
1




2
2
k
k







x ( t )  A sin   t  B cos   t 
 m  
 m  




where A and B are constants of integration. If we assume that x = 0 at t = 0, then
B = 0 and
1


2
k



x ( t )  x 0 sin   t  , where x 0  A, the max . displ . amplitude .
 m  


Since this can also be written as
x( t )  x 0 sin( ωt )  x 0 sin( 2πt ),
we see that the position of the particle oscillates in a sinusoidal manner with
frequency
1
 
1  k 2
 
2π  m 
, so
k  4π 2 m  2 .
The energy of the classical oscillator is
E  T  V 
1
1
mv 2  kx 2
2
2
Since
1
1


2
k
k
 
 2
v( t ) 
   x 0 cos   t 
 m  
m


1 k 2
1
E

m  x 0 cos 2    k x 02 sin 2 
2 m
2
dx
dt
E

1 2
kx 0 , and is notquantized .
2

Chemistry 342
Spring, 2005
2. Quantum mechanical description.
Following our prescription, we begin by writing down the classical energy
expression for the oscillator
E  TV 
1
1
mv 2  kx 2
2
2

p 2x 1 2
 kx
2m 2
and then convert this to the quantum mechanical analog, the Hamiltonian operator Ĥ ,
by replacing each of the dynamical variables (px and x) by their operator equivalents,
px
 p̂ x
  i

x
, x  x̂  x
This yields
Ĥ  
2 2
2m x 2

1 2
kx
2
We then use this form of Ĥ in the time-independent Schrödinger equation
Ĥψ  Eψ
not zero. New feature.
yielding

2 2
ψ( x) 
2m x 2
1 2
kx ψ(x)  Eψ(x).
2
(A)
Now, we want to solve this equation; i.e., to find the set of functions ψ(x) which,
when operated on by the operator Ĥ , yield a constant (E) times the function itself.
The wavefunctions should also be finite, single-valued, and continuous throughout
the range from x → -∞ to x → ∞.
As in the case of the free particle, Eq. (A) can be solved by expanding ψ in a power
series, substituting this series into (A), and solving for the coefficients. In this case, it
is a bit involved so we will only outline the procedure here. Details may be found in
Pauling and Wilson (pp. 37-73) or Eyring, Walter, and Kimball (pp. 75-79).
First, we transform Eq. (A) into a more useful form by introducing some new
variables. To be consistent with Atkins, we choose
Chemistry 342
Spring, 2005
y 
αx , α 
4π 2 m
h

mω

Rewriting (A) as
2 2
1


ψ( x )   E  kx 2  ψ( x )  0
2
2m x
2


we first multiply through by 2m/  2 α , yielding
α 1
2
mkx 2 
 2mE
ψ
(
x
)



 ψ( x )  0.
x 2
 2α 
  2α
We then define
ε 
2mE
 2α
Recognizing that mk/  2 = α 2 , we have
α 1
2
ψ( x) 
x 2
and, since α 1  2 / x 2
ε
 αx 2  ψ(x )  0
  2 / y 2 ,
2
ψ( y) 
y 2
ε
 y 2  ψ( y)  0
(B)
We now proceed to solve this equation.
When y becomes very large, Eq. (B) reduces to
 2 ψ( y)
y 2
 y 2 ψ( y)  0.
In the limit as y → ± ∞, this equation has the asymptotic solution
ψ( y)  c e  y
2
/2
,
where c is a constant.
The solution with the positive exponential does not behave properly, as ψ(y) → ∞ for
y → ± ∞. We want ψ(y) → 0 in this limit. So we choose the solution with the
negative exponential,
Chemistry 342
Spring, 2005
ψ( y)  c e  y
2
for y → ± ∞.
/2
But we are primarily interested in solving Eq. (B) for small, or at least finite, y. To
do this, we assume a solution of the form
ψ( y)  c H( y) e  y
2
/2
(C)
where H(y) is a power series
H( y)  a 0
 a1 y  a 2 y 2
 
To find the values of the coefficients a 0 , a1 , , we substitute (C) into (B), which
yields (after some algebra, worked out in Levine)
d 2 H ( y)
dy 2
 2y
dH( y)
dy
 (ε  1) H( y)  0.
(D)
This equation is very similar to a famous differential equation known as Hermite’s
equation
d 2 H ( y)
dy 2
 2y
dH( y)
dy
 2 v H ( y)  0
(E)
In fact, (D) and (E) are identical if
(ε  1)  2v
The solutions H(y) to Eq. (E), known as the Hermite polynomials, are of the form
H v ( y) 

k 0
(1) k v! (2 y) v2 k
( v  2k)! k!
where v is an integer (v = 0, 1, 2, …) and k is an index, running from k = 0 to v/2 if v
is even, and from k = 0 to k = (v-1)/2 if v is odd. The Hermite polynomials may also
be generated using the function
H v ( y)  (1) v e y 2
dv
2
(e  y ).
v
dy
At this point, we have solved the 1D harmonic oscillator problem.
Chemistry 342
Spring, 2005
3. Eigenvalues and eigenfunctions.
The energies (eigenvalues) of the one-dimensional harmonic oscillator may be found
from the relations
(ε  1)  2v,
ε 
2mE
,
 2α
α 
mω

Combining these, we obtain
Ev
1
1


  v   ω   v   hν,
2
2


v  0, 1, 2, 
Unlike the corresponding classical result, we find that the quantum mechanical
energy is quantized, in units of ω , where ω is the classical frequency ω2 = k/m. v is
called the vibrational quantum number. We also find that the lowest state, with v = 0,
does not have zero energy but instead has E = ω /2, the so-called zero point energy.
We can summarize these results in the form of an energy level diagram
3
Ev
 7ω / 2
Ev
 5ω / 2
Ev
 3ω / 2
Ev
 ω / 2
ω
2
ω
1
ω
v0
Eigenfunctions.
The wavefunctions of the one-dimensional harmonic oscillator are of the form
ψ v ( y)  c H ( y) e  y
2
/2
1

2
where c (or Nv) is a normalizing constant,  1α v  . The functions ψv(y) look
 π 2 v! 
 2

complicated, but in reality they are not, at least for small v. To see this, let us
examine the first few Hermite polynomials…
Chemistry 342
Spring, 2005
H v  0 ( y)  1



H v 1 ( y)  2 y


2
H v  2 ( y)  4 y  2 

H v 3 ( y)  8 y 3  12 y 
Note that for v even, only even powers
of y appear whereas for v odd, only odd
powers of y appear!
The corresponding wavefunctions are
1
 α 1 2
ψ v 0 ( y)   12 
π 
 2
e
y2
2
1
 α 12  2
ψ v 1 ( y)   1 
 2π 
 2
2y  e
y2
2
1
 α 1 2
ψ v  2 ( y)   21 
 8π 
 2
(4 y 2  2)  e
y2
2
, etc.
These are plotted on the next pages, together with their corresponding probability
distributions.
Note that these functions (and their magnitudes squared) are very similar to the
corresponding functions for the particle-in-the-box problem, being respectively even
or odd with respect to reflection about y = 0. There is one important difference,
however, and this is that the HO functions do not go to zero at the “walls” of the
potential. Thus, there is a finite probability ~ (2v + 1) that the particle will be found
in “classically forbidden” regions. This is the origin of the QM tunneling effect.
Expectation values.
Using these functions, we can calculate the expectation value of any dynamical
variable we wish, according to the recipe
 ψ Â ψ dT
 ψ ψ dT
*
a  A 
*
.
Suppose, for example, that we wish to calculate the average position, <x>, in some
particular state. (Clearly, it is zero, but can we prove it?). We have (since the ψ’s are
normalized)
Chemistry 342
Spring, 2005
1
y 
α
x 
1


1  α2  
y2
 1
  H v ( y) ŷ H v ( y) e dy
α 2 v 
 π 2 v! 
Now, you may substitute in the explicit expressions for Hν(y) if you wish, and
integrate, but it is easier in the present case to make use of the following useful
relation (called a recursion relation)
ŷ H v ( y)  v H v 1 ( y) 
1
H v 1 ( y)
2
Then, the integral becomes



 



1

 2
H v ( y)v H v 1 ( y)  H v 1 ( y) e  y dy
2


Separating the pieces, we have



 


H v ( y) v H v 1 ( y) e  y dy


1
2
  H v ( y) H v 1 ( y) e  y dy

2

y2
y2
2
 v  H v ( y)  e
 H v 1 ( y) e 2 dy

1 
y2
y2
  H v ( y) e 2  H v 1 ( y) e 2 dy


2
2
But Hv(y) and Hv±1(y) are, respectively, either even and odd or odd and even
functions of y. Since the limits on the integral are ± ∞, the integrals are zero.
Therefore,
<x> = 0.
One can proceed in a similar fashion when calculating <x2>, <x3>, …etc. Also, one
can make use of another relation
d H v ( y)
dy
 2v H v1
when calculating expectation values of other dynamical variables that involve, in their
operator form, derivatives with respect to displacements (e.g., the momentum
operator).
Chemistry 342
Spring, 2005
Fitts, D. D., Principles of Quantum Mechanics as Applied to Chemistry and Chemical
Physics, Cambridge University Press, 1999.
Chemistry 342
Spring, 2005
Fitts, D. D., Principles of Quantum Mechanics as Applied to Chemistry and Chemical
Physics, Cambridge University Press, 1999.