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Modern Physics
lecture 4
The Schroedinger Equation
As particles are described by a wave
function, we need a wave equation for
matter waves
 The general form for a wave travelling in
one dimension is

d2
1 d2
y  x , t   2 2 y  x, t 
2
dx
v dt
The Schroedinger Equation


For a matter wave moving on the x-axis we can
write the full wave function as
( x, t )   ( x)e it
For our purpose we will look only at the spatial
part of this wave function which satisfies the wave
equation
d2
2m



( E  U )
2
2
dx


This is the Schrodinger Equation (E is kinetic and
U is potential energy)
A particle in a box
1. Classically
A particle in a box
Classically if we put a particle in a box, and
it is confined to move in one direction. The
particle will reflect off the walls and rattle
backward and forward in the box
 Any time that we open the box we find that
the particle has equal probability of being
somewhere in the box

String Harmonics



Recall the harmonics
of a string held at both
ends
Waves exist only
when the wavelength
is an integer multiple
of half wavelengths
The wavelength of the
standing wave of a
string is quantised
Ln

2
2L
 
n
n=3
n=2
n=1
L
A particle in a box

The wave function for a standing wave is
given by
yx  A sin( kx)

Substituting the quantised formula for the
wavelength we arrive at
 nx 
y x   A sin 

 L 
A Particle in a Box


The box is constructed from walls that are
infinitely high potential energy barriers and in
between the potential energy is zero
We can re-write the S.E. for the region between
the walls to be
d2
2mE
2






k

2
2
dx


This has a solution
 x  Asin kx

Which is the same solution as was found by comparing matter
waves to waves in a string
A Particle in a Box
2. Quantum mechanics
Potential energy:
 for x  (, 0)  ( L, )
U ( x)  
0 for x  (0, L)
2
2
(0)  ( L)  0
Schroedinger equation
inside the box:
2 d 2 

 E
2
2m dx
A particle in a box
Quantum mechanically we find the same
solution for particles moving inside a box
 The wave function is described by

 nx 
 x   A sin 

 L 

The probability of finding a particle in a
position inside the box when it is opened
will therefore be given by
 x 
2
 nx 
 A sin 

 L 
2
A particle in a box: solution
n 
2
n
sin(
x)
L
L
A standing wave
A particle in a box

Because the deBroglie wavelength is
quantised the momentum is also quantised
h
px  h /  
2L / n

Therefore kinetic energy is also quantised
2
1 2 p 2 nh / 2 L 
E  mv 

n
2
2m
2m
A particle in a box: solution
E
 
2 2
2
n
2
2mL
with
n = 1, 2, 3, ...
The
lowest allowed
energy corresponds to n=1
therefore it is not zero
The
quantum mechanical
lowest energy state is
called zero point energy
Quantum Mechanics
Energy levels are important because if the
particle is charged (such as an electron) a
photon will be emitted as the particle
changes energy levels
 It can also absorb a photon raising the
energy level
 This gives us the opportunity of probing
quantum effects using spectroscopy

A particle in a finite well



If the walls of the box are
finite
If the energy of the
particle is lower than the
potential then classically
the particle is bound
(stuck) in the box (region
II)
In quantum mechanics
sometimes the particle can
be found in regions I and
III
I
II
III
E
U
L
0
x
A particle in a finite well

In regions I and III the Schrodinger
equation is
d2
2m
2


(
U

E
)


C

2
2
dx


This has the solution
  AeCx  Be Cx
A particle in a finite well

Thus we arrive at the following
interpretation of the solutions
 I  Ae
Cx
 III  Be Cx
Harmonic oscillator
Potential energy
1
U ( x )  m 2 x 2
2
SE :
2 d 2  m 2 x 2


  E
2
2m dx
2
solution
1
En  (n  ) 
2
gdzie n  1, 2, 3, ....
Harmonic oscillator