Download Lecture 9 - Scattering and tunneling for a delta

Lecture 9
The delta-function potential
Scattering and tunnelling
Objectives
Solve the Schrödinger equation for a delta function potential.
Classical turning points
E
x1
Position, x (arb. units)
A bound state
x2
Potential, V(x) (arb. units)
Potential, V(x) (arb. units)
Bound and scattering states
Classical turning point
E
Position, x (arb. units)
x1
A scattering state
Classically, if a particle is in a potential energy well it will oscillate between the two
turning points x1 and x2 where V  x 1 =V  x 2 = E , the energy of the particle. The
particle is in a bound state.
Bound and scattering states II
Potential, V(x) (arb. units)
E
Potential, V(x) (arb. units)
If the particle energy exceeds V(x) on either or both sides then it cannot get trapped
in a potential well and it is said to be in a scattering state.
Classical turning points
E
x1
Position, x (arb. units)
A scattering state
x2
Position, x (arb. units)
A classically bound state but a quantum
scattering state
Bound and scattering states III
Depending on the shape of the potential we may find only bound states, only
scattering states or a mixture of the two.
In quantum mechanics there is always a non-zero probability that a particle can pass
through a (finite) potential barrier so the following generalisations can be made:
EV −∞ and EV ∞ → bound state
EV −∞ or EV ∞ → scattering state
If we assume that the potential goes to zero at infinity then we can write:
E 0 → bound state, E0 → scattering state.
For the infinite square well we find only bound states are solutions to the Schrödinger
equation. The solutions have discrete energy levels.
For the free particle we have only scattering states as solutions. The solutions are
labelled by a continuous variable k.
The Dirac delta function
The Dirac delta function is defined as follows:
  x =0 if x≠0
  x =∞ if x=0
and
∞
∫   x dx=1
−∞
It is an infinitesimally narrow spike located at the origin, whose area is 1.
Similarly, the function   x−a is a spike located at x=a .
We can write
∞
∫
−∞
∞
f  x  x−a dx= f a ∫   x−adx= f a .
−∞
The delta function is zero everywhere except where x=a , when multiplied by a
function f  x it has the same effect as multiplying by f a  .
The delta function potential
We will consider a potential of the form v  x=−   x  , where  is some constant.
This potential has a delta function well located at x=0 . The time independent
2
2
ℏ d 
−   x = E  .
Schrödinger equation becomes −
2 m dx 2
If E 0 then we will have bound states and if E 0 we will have scattering states.
First consider bound states ( E0 ) in the region where x0 . Here V  x=0 so we
2
d 
2mE
−2 m E
2
=
=−

=

have
with
.
2
2
ℏ
dx
ℏ
− x
x
A general solution to this Schrödinger equation is   x= A e  B e . But we know
that  is positive (because E is negative), so the A term becomes infinite as
x −∞ . So we have   x= B e  x for x0 .
Boundary conditions
If we look at the area where x0 we find a general solution of the form ,
  x= F e− x G e  x which reduces to   x= F e− x as G e  x ∞ as x  ∞ .
We now have two solutions for   x for x0 and x0 . We can find the solution
at x=0 by noting the following boundary conditions:
 is always continuous.
d  /dx is continuous, except where V  x is infinite.
The continuity of  tells us that F = B at x=0 .
∞
From normalisation we can find F and B:
∫ F 2 e−2  x dx=0.5 ,
0
F =  .
The wavefunction around x=0

We need to consider the region containing
the delta function potential. We will try to
solve the Schrödinger equation by integrating
from  to − , either side of x=0 :
½
 (x)
½ -x
 e
½ x
 e
0



ℏ2
d2
−
dx∫ V  x   x dx= E ∫   x  dx
∫
2 m − dx 2
−
−
0
Position, x (arb. units)
Bound state wavefunction for the
delta function potential
The right hand side becomes zero as   0
so we have:


d
2m
= 2 ∫ V  x   xdx
dx − ℏ −
[ ]
Bound state solutions
 
d
2 m
=− 2   0 .
dx
ℏ
d
− x
∣ =−  
We know that   x=   e
for x0 so
dx +
d
x
∣ = √  .
and   x=   e for x0 so
dx d
=−2    and at x=0  0=   .
So 
dx
2m
m
Putting it all together gives −2   =− 2   , i.e. = 2 .
ℏ
ℏ
2 2
2
ℏ 
m
=− 2 .
The corresponding energy is E =−
2m
2ℏ
m  −m∣x∣/ ℏ



x=
e
The wavefunction can be written as
.
ℏ
The result of the integration is 
 
2
There is only one bound state for any given value of .
Scattering states
E0
Scattering states have
so the Schrödinger equation
2
d 
2mE
2 m E
2
k
=
=−

=−k

with
(k is real and positive).
2
2
ℏ
dx
ℏ
becomes
ikx
−ikx
The general solution for x0 is   x= Ae  Be
. At this point we cannot
eliminate either the A or B term.
ikx
−ikx
For x0 we have   x= Fe G e
and at x=0
can write AB= F G .
  x is continuous so we
Differentiating   x we find that
d
d
=ik  F e ikx −G e−ikx  for x0 , so
∣+=ik  F −G  and
dx
dx
d
d
=ik  A e ikx− B e−ikx  for x0 , so
∣ =ik  A− B  .
dx
dx d
=ik  F −G− A B .
So 
dx
 
Scattering states II
Going back to our previous integration of the Schrödinger equation, we have the
d
2 m
=− 2   0 . For the scattering states  0= A B and we
results that 
dx
ℏ
2 m
m
ik

F
−G−
AB
=−

AB
=
have
. Let
, divide by ik and collect the A
2
2
ℏ
ℏ k
and B terms on one side and we get F −G= A12i − B 1−2i  .
 
What do the terms in the Schrödinger equation solution represent?
Ae
Be
V(x)
ikx
-ikx
ikx
Fe
-ikx
Ge
0
x
0
Aeikx represents a wave incident from the
left and Feikx is a wave travelling from the
potential well to the right.
Suppose Ge-ikx is zero, then Be-ikx
represents a wave reflected from the
potential well, moving to the left.
Reflection and transmission
We put G=0 and solve the two equations AB= F G and .
F −G= A12i − B 1−2i  .
The results are B=
i
1
A .
A and F =
1−i 
1−i 
We can define reflection and transmission coefficients, R and T, which are the
probabilities of reflection and transmission from the potential well.
2
2
i −i 

∣F∣
∣B∣2
1
T
=
=
and
, and RT =1 .
R= 2 =
=
2
2
2
∣A∣ 1
∣A∣ 1−i 1i  1
In terms of energy we can write:
R=
1
1
T
=
and
.
1  2 ℏ 2 E / m  2 
1  m  2 / 2 ℏ 2 E 
T and R versus 
If we plot T and R versus  we find that when =0
tends towards, but does not equal, zero.
T =1 , but as  increases T
Tunnelling
Notice that if the sign of  is reversed, i.e. we have a potential barrier instead of a
well, the reflection and transmission coefficients are unchanged.
Even if the potential  is much larger than E there is a non-zero probability that the
particle can cross the potential barrier. This is the effect known as tunnelling.
It can also be seen that if the particle energy is greater than  there is still a chance
that the particle will be reflected from the potential barrier.
Neither of these phenomena exist in classical mechanics.
The wavefunctions we have derived are not normalisable and so do not represent
physically realisable states. To solve this problem we must use the approach taken
for free particles and add together a combination of stationary states.
Scattering matrices
We can construct solutions to the Schrödinger equation to the left and right of the
potential well by making use of the superposition theorem.
ikx
−ikx
To the left of the well we have   x= Ae  Be
and to the right we have
2 m E
ikx
−ikx
k
=
  x= Fe G e
with
.
ℏ
In the region containing the potential well we can write   x=C f  x D g  x 
where f  x and g  x  are two linearly independent solutions.
Going back to our image of reflection and transmission from the potential well we
see that B can be written in terms of A and G i.e. B=S 11 AS 12 G . Similarly,
S
S 12
F =S 21 A S 22 G with S = 11
. S is known as the scattering matrix.
S 21 S 22


We can find the amplitudes of the reflected and transmitted waves using
B =S A
.
F
G
 
Scattering matrices II
If there is no wave incident from the right then G=0 and we can write
2
2
∣B∣
∣F∣
2
Rl = 2 =∣S 11∣ and T l = 2 =∣S 21∣2 .
∣A∣
∣A∣
Similarly, when A=0 and the wave is incident from the right we have
2
2
∣F∣
∣B∣
2
Rr = 2 =∣S 22∣ and T r = 2 =∣S 12∣2 .
∣G∣
∣G∣

i
1−i 
S
=
So, using the results from before have
1
1−i 
1
1−i 
i
1−i 

and =
m
ℏ2 k
.
We can use the scattering matrix to find the energy of the bound states. To do this we
make the substitution k →i  and look for terms in the S matrix which blow up.
Bound states from S matrix
For our delta potential we want to use the S matrix to find the energy of the bound
state(s).
For a bound state A=G =0 . Let’s look at the S12 term first: S 12 =
2
1
.
1−i 
ℏ k
1

S
=
=
Substitute for
: 12
.
1−i m  /ℏ 2 k ℏ 2 k −i m 
2
2
iℏ 
ℏ 
→ 2
Put k=i  : S 12 = 2
.
i ℏ −i m  ℏ −m 
m
This blows up when ℏ 2 =m  , so = 2 .
ℏ
2m E
√−2 m E
=i  , so =
We know that k= √
ℏ
ℏ
2
m   −2 m E
m
=
Therefore
, i.e. E =− 2 , which is the same result as before.
2
ℏ
ℏ
2ℏ
The scanning tunnelling microscope (STM)
The tunnelling current is kept constant via a feedback loop and the vertical deflection
of the tip is used to measure the surface topography. Atomic resolution can be
achieved.
STM images I
Surface of Ni (110)
Cu surface
Images courtesy of IBM
STM images II
Fe atoms on Cu. The ring has a radius of
7.13 nm.
Fe atoms on Cu (111)
SQUID magnetometer
Cooper-pairs of electrons tunnel through
the Josephson junction, which is an
insulator.
The junction can switch ten times faster
than a semiconductor.
SQUID devices can measure the magnetic fields from living organisms.
The SQUID is one of the most sensitive measurement devices known.
Conclusions
A delta function potential contains one bound state.
The probability of an incident wave being able to pass through a delta function
potential barrier is non-zero, even if the energy of the wave is less than the height of
the barrier.
Classical mechanics cannot predict the existence of tunnelling, but the effect is real
and has many applications.