Scattering from
one dimensional square potential
barrier
Collision theory
In a collision problem, the energy is specified in advance and the behavior of the wave
function at great distances is found in terms of it. This asymptotic behavior is related to the
amount of scattering of the particle by the force field .
Application:
Collision studies
Reactive collisions
Fragmentations
Tunneling trough the potential barriers, Intersystem crossing
One dimensional square potential barrier
We make use of stationary solutions of the time-independant wave equation
that correspond to particular values of the incident momentum
0
V(x)
x<0, x>a
V(x)=
V0
V0 >0 0<x<a
E0 >0
2
h
−
∇ 2 Φ ( x ) = EΦ ( x )
2m
0
0
a
x
h2 2
−
∇ Φ ( x ) = EΦ ( x )
2m
h2 2
[−
∇ + V ( x )]Φ( x ) = EΦ( x )
2m
Quantum particle approaches from the region of negative x
and is reflected or transmitted by the barrier.
Asymptotic behavior
Particle possesses certain energy E0 >0 and momentum
p=hk/2π=(2mE)1/2
– In the regions x<0 the wave function represent particle moving to the right(incident
particle) and left (reflected particle); for x>a, wave function should represent a
particle moving to the right (transmitted particle).
φ ( x ) = Ae ikx + Be − ikx
x<=0
x>=a
φ ( x ) = Ce ikx
The solutions are
appropriate for the forcefree regions that are
external to any scattering
potential
– In the force region [0, a] particle has definite momentum and can be represented by
a one dimensional momentum eigenfunction
φ ( x ) ∝ e ipx / h
Moving in positive x direction
φ ( x ) ∝ e − ipx / h
Moving in negative x direction
Normalization
Physical meaning of the coefficients A, B, C can be inferred by substituting the wave functions
to the one-dimensional probability current density equation
∂
P ( x , t ) + ∇S ( x , t ) = 0
∂t
h
S ( x, t ) =
[Ψ * ∇Ψ − (∇Ψ*)Ψ ]
2im
def
This has the familiar form associated
with the conservation of flow of a fluid
of density P and current density S
Probability current density
L. Schiff, Quantum Mechanics
The rate of change of the number of particles in volume is due entirely to
the particle current flowing through the area bounding
S(x)=v(IAI2-IBI2)
S(x)=vICI2
v=p/m
for x<0
for x>0
where v= hk/2πm is the speed of a
particle with propagation number k
This can be interpreted as the net flux (positive to the right) in the two regions and A, B,
C are amplitudes of the incedent, reflected and transmitted wae functions
The absolute normalization is not important, since the interest is only
in obtaining the reflection and transmission coefficients wich are in the ratios
to the incident particle:
IBI2/IAI2 , ICI2/IAI2
Solutions inside the potential barrier
Character of the solution inside the potential barrier depends on
whether E is greater or less than V0.
For E>V0
one can obtain the propagation number α=[8π2 m(E-V0)/h]1/2
and the solution inside the barrier is:
φ ( x ) = Fe iαx + Ge − iαx
0 <= x <= a
The continuity of f and df/dx at x=0 and x=a required by the boundary
conditions provides four relations between the five coefficients.
Solution:
B
( k 2 − α 2 )(1 − e 2iαa )
=
A ( k + α )2 − ( k − α )2 e 2 iαa
C
4kαe 2 i (α − k )a )
=
A ( k + α )2 − ( k − α )2 e 2 iαa
The absolute squares of ratios are the scattering
(reflection and transmission) coefficients:
4k 2α 2
4 E ( E − V0 ) −1
B
−1
]
[
1
]
= [1 + 2
=
+
2 2
2
2
2
A
( k − α ) sin αa
V0 sin αa
2
( k 2 − α 2 )2 sin 2 αa −1
C
V0 sin 2 αa −1
] = [1 +
]
= [1 +
2 2
4 E ( E − V0 )
A
4k α
2
2
One can verify that summe of these coefficients gies 1 as would be expected
Properties of the solution:
For E---Æ V0
the particle energy approaches the top of the barrier:
2
C
mV0a 2 −1
→(
)
2
A
2h
For E>V0
the transmission coefficient oscillates between a steadily
increasing lower enelope and unity (interference phenomena).
There is perfect transmission when αa =π, 2π, ... .
Whenever the barrier contains an integer number of half wavelengths.
It is known as a transmission of light through thin refracting layers
For 0<E<V0,
the reflection and transmission coefficents are obtained
by replacing α by iβ=[8π2 m(V0-E)/h]1/2
C
V0 sinh 2 β a −1
= [1 +
]
A
4 E (V0 − E )
2
2
βa>>1
(V0 − E ) − 2 βa
16 E
e
2
V0
Tunneling through the barrier
The transmission coefficient of a square barrier as a function of
particle energy for 4π2 mV0a2/h
Scattering of the wave packet
The time dependant Schrödinger equation can be used to study
the motion of a wave packet:
The solution can be expanded in energy eigenfunctions and
the summation carried out explicitly. For frre particle is this possible
but it is impossible when a potential is present.
Other method is based on a numerical integration. The diiferential
equation in the time are approximating by a difference equation.
see also previous lectures
Gaussian wave packet
Scattering from a square
barrier <E>=1/2V0
Gaussian wave packet
Scattering from a square
barrier <E>=V0
Figures shows how the probability decays
slowely and bounces back and forth within
the barrier when the mean kinetic energy
of the particle is zero.
Gaussian wave packet
scattering from a square
well <E>=1/2V0
Gaussian wave packet
scattering from a square
well <E>=V0