Physical Chemistry -The Harmonic Oscilator_lecture12-15

5.61 Fall 2007 Lectures #12-15 page 1



THE HARMONIC OSCILLATOR


• Nearly any system near equilibrium can be approximated as a H.O.

• One of a handful of problems that can be solved exactly in quantum
mechanics

examples


m1 m2 B (magnetic
field)
A diatomic molecule µ
(spin
magnetic
moment)
E (electric
field)




Classical H.O.



m
k



X0 X


Hooke’s Law: (
f = −k X − X 0 ≡ −kx)
(restoring force)
d2x d2x ⎛ k ⎞
f = ma = m 2 = −kx ⇒ + x=0
dt dt 2 ⎜⎝ m⎠⎟

,5.61 Fall 2007 Lectures #12-15 page 2



Solve diff. eq.: General solutions are sin and cos functions


k
() ( )
x t = Asin ω t + B cos ω t ( ) ω=
m
or can also write as


() (
x t = C sin ω t + φ )
where A and B or C and φ are determined by the initial conditions.


e.g. ()
x 0 = x0 ()
v 0 =0
spring is stretched to position x 0 and released at time t = 0.

Then

() ()
x 0 = A sin 0 + B cos 0 = x0 () ⇒ B = x0
dx
()
v 0 =
dt
()
= ω cos 0 − ω sin 0 = 0() ⇒ A=0
x=0



So ()
x t = x0 cos ω t ( )
k
Mass and spring oscillate with frequency: ω =
m
and maximum displacement x0 from equilibrium when cos(ωt)= ±1


Energy of H.O.
Kinetic energy ≡ K

2
1 1 ⎛ dx ⎞ 1 1
( ) ( )
2
K = mv 2 = m ⎜ ⎟ = m ⎡⎣ −ω x0 sin ω t ⎤⎦ = kx02 sin 2 ω t
2 2 ⎝ dt ⎠ 2 2

Potential energy ≡ U

dU 1 1 2
()
f x =−
dx
⇒ U = − ∫ f x dx =() ∫ ( kx )dx =
2
kx 2
=
2
( )
kx0 cos 2 ω t

, 5.61 Fall 2007 Lectures #12-15 page 3



Total energy = K + U = E

1 2 1 2
E=
2
( ) ( )
kx0 ⎡⎣sin 2 ω t + cos 2 ω t ⎤⎦ E= kx
2 0

x (t )
x 0(t )


0 t

-x0(t)

U K
1 2
kx E
2 0




0 t

Most real systems near equilibrium can be approximated as H.O.

e.g. Diatomic molecular bond A B


X
U


X
X0 A + B separated atoms




equilibrium bond length