Download L6 QM scattering

Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 3 for Electrical Engineering
Lecturers: Daniel Rohrlich, Ron Folman
Teaching Assistants: Daniel Ariad, Barukh Dolgin
Week 6. Quantum mechanics – probability current • 1D scattering
and tunneling • simple 1D potentials • parity • general 1D potential •
2D and 3D square wells and degeneracy
Sources: Merzbacher (2nd edition) Chap. 6;
Merzbacher (3rd edition) Chap. 6;
Tipler and Llewellyn, Chap. 6 Sects. 1-3,6;
7 ‫ יחידה‬,‫פרקים בפיסיקה מודרנית‬
Probability current
When we solved the finite square well potential for bound
states (E < V0 ), we threw away the outside term ek′|x| because
it is not normalizable.
Recall k  2mE /  , k '  2m | V0  E | /  .
V(x)
V0
outside
inside
−L/2
0
outside
L/2
x
Probability current
But for free states (E > V0 ) the solutions outside
2 d 2
( E  V0 ) ψ( x)  
ψ( x )
2
2m dx
, | x| L/2 ,
are linear combinations of eik′x and e−ik′x ; and they are not
normalizable either. Should we throw them away?
V(x)
V0
outside
inside
−L/2
0
outside
L/2
x
Probability current
We know why the wave eik′x is not normalizable: its momentum
is p  k ' with zero uncertainty (Δp = 0), so the uncertainty Δx in
its position must be infinite – the particle can be anywhere along
the x-axis with equal probability.
V(x)
V0
outside
inside
−L/2
0
outside
L/2
x
Probability current
We know why the wave eik′x is not normalizable: its momentum
is p  k ' with zero uncertainty (Δp = 0), so the uncertainty Δx in
its position must be infinite – the particle can be anywhere along
the x-axis with equal probability. Should we throw away the wave
eik′x?
V(x)
V0
outside
inside
−L/2
0
outside
L/2
x
Probability current
If we ask what fraction of an incoming beam ψ(x) = Aeikx is
reflected and what fraction is transmitted, then the normalization
doesn’t matter.
Probability current
If we ask what fraction of an incoming beam ψ(x) = Aeikx is
reflected and what fraction is transmitted, then the normalization
doesn’t matter.
The flux density is ρv where ρ = |ψ(x)|2 = |A|2 and v = k / m.
Probability current
Consider the time derivative of the probability density ρ(x,t):

 *
 ( x, t )  [ ( x, t ) ( x, t )]
t
t
 *

*
  ( x, t )  ( x, t )   ( x , t )  ( x , t )
t
t
2

i 
2 *

*

  ( x, t ) 2  ( x, t )   ( x , t ) 2  ( x , t ) 
2m 
x
x

i  
 *


*

 ( x, t )  ( x , t )   ( x, t )  ( x , t )  ,

2m x 
x
x

where we have applied Schrödinger’s equation and its complex
conjugate. (What happened to the potential term?)
Probability current
We found

i  
 *


*
 ( x, t )  

(
x
,
t
)

(
x
,
t
)


(
x
,
t
)

(
x
,
t
)
,


t
2m x 
x
x

which implies the 1D “continuity equation”


 ( x, t )   J x ( x, t ) ,
t
x
if we define
i 
 *


*
J x ( x, t ) 

(
x
,
t
)

(
x
,
t
)


(
x
,
t
)

(
x
,
t
)
.


2m 
x
x

Probability current
Let’s check: If Ψ(x,t) = Aeikx-iωt , then
i 
 *


*

(
x
,
t
)

(
x
,
t
)


(
x
,
t
)

(
x
,
t
)

2m 
x
x

i
k
2
*

A A(ik  ik ) 
A
,
2m
m
J x ( x, t ) 


which is just what we found from our earlier calculation.
So we could look for an outside solution of the form
ψ( x)  Aeik ' x  Be ik ' x , x   L / 2 ,
 Feik ' x
, x  L/2 ,
2
in which the incoming flux k ' A /m splits into a reflected flux
2
2
k ' B /m and a transmitted flux k ' F /m .
V(x)
V0
outside
inside
Ae ik ' x  Be ik ' x Ceikx De ikx
−L/2
0
L/2
outside
Fe ik ' x
x
1D scattering and tunneling
But let’s consider a simpler scattering problem:
1D scattering and tunneling
But let’s consider a simpler scattering problem:
V(x)
V0
0
x
We take E > V0 and
ψ( x)  Aeikx  Be ikx
, x0 ,
 Ceik ' x
, x0 ,
in which the incoming flux k A 2/m splits into a reflected flux
2
2
k B /m and a transmitted flux k ' C /m . The reflection
2
probability is R = B / A and the transmission probability is
V(x)
T = C / A 2 k' / k .
V0
Aeikx  Be ikx
Ceik ' x
0
x
Continuity of ψ at x = 0: A + B = C .
Continuity of dψ/dx at x = 0: ik(A – B) = ik′C .
Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′)
and B/A = (k–k′)/(k +k′) .
V(x)
V0
Aeikx  Be ikx
Ceik ' x
0
x
Continuity of ψ at x = 0: A + B = C .
Continuity of dψ/dx at x = 0: ik(A – B) = ik′C .
Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′)
and B/A = (k–k′)/(k +k′) . Does R + T = 1?
V(x)
V0
Aeikx  Be ikx
Ceik ' x
0
x
Continuity of ψ at x = 0: A + B = C .
Continuity of dψ/dx at x = 0: ik(A – B) = ik′C .
Now 2A = (1+k′/k) C and 2B = (1–k′/k) C; hence C/A = 2k/(k +k′)
and B/A = (k–k′)/(k +k′) . Does R + T = 1?
2
R
2
B
(k  k ' )
4kk'
k' C


1


1

 1 T
2
2
A
k A
(k  k ' )
(k  k ' )
2
V(x)
V0
Aeikx  Be ikx
Ceik ' x
0
x
Isn’t something very odd – “quantumly” odd – going on here?
V(x)
V0
0
x
Isn’t something very odd – “quantumly” odd – going on here?
A classical particle with E > V0 would never reflect back from the
potential step…but here there is reflection unless k = k′ , i.e.
unless V0 = 0.
V(x)
V0
0
x
Isn’t something very odd – “quantumly” odd – going on here?
A classical particle with E > V0 would never reflect back from the
potential step…but here there is reflection unless k = k′ , i.e.
unless V0 = 0.
In fact, even if we replace V0 by –V0 so that k >> k′, there is still
reflection at x = 0!
V(x)
0
−V0
x
The case E < V0 :
ψ( x)  Aeikx  Be ikx
 Cek ' x
, x0 ,
, x0 .
Now R= 1, T = 0: no transmission.
How deep do particles penetrate the potential step?
V(x)
V0
Aeikx  Be ikx
Ce  k ' x
0
x
Summary of scattering from the potential step:
1.0
0.5
Snapshots of the probability density, for an incident wave packet:
Another scattering problem: the “square” potential barrier:
V(x)
V0
0
L
x
Another scattering problem: the “square” potential barrier:
ψ( x)  Aeikx  Be ikx
, x0
;
 Ce k ' x  De k ' x
, 0 x L ;
 Feikx
, L x
;
As before, k  2mE /  , k '  2m | V0  E | /  .
V(x)
V0
Aeikx  Be ikx
Ce k ' x De  k ' x
0
L
Feikx
x
Boundary conditions (ψ and dψ/dx continuous) at x = 0:
A B  C  D ;
ik  A  B   k ' C  D  ;
Hence
 k' 
 k' 
 k' 
 k' 
2 A  1  C  1   D ; 2B  1  C  1   D .
 ik 
 ik 
 ik 
 ik 
V(x)
V0
Aeikx  Be ikx
Ce k ' x De  k ' x
0
L
Feikx
x
Boundary conditions (ψ and dψ/dx continuous) at x = L:

Cek 'L  De k 'L  FeikL ;

k ' Ce k 'L  De k 'L  ikFeikL ;
Hence
2Ce
k 'L
 ik  ikL
 1   Fe
;
 k' 
2De
k ' L
 ik  ikL
 1   Fe
 k' 
.
V(x)
V0
Aeikx  Be ikx
Ce k ' x De  k ' x
0
L
Feikx
x
If we combine these equations we can show
F
e ikL

A cosh k ' L  i  k '  k  sinh k ' L


2 k
Hence
F
T
A
2
k'


1
k
'
k


 cosh2 k ' L     sinh 2 k ' L 
4  k k' 


2
1
V(x)
V0
Aeikx  Be ikx
Ce k ' x De  k ' x
0
L
Feikx
x
k 'L
2
For k′L >> 1, both cosh k′L and sinh k′L approach e
and the expression for T simplifies:
F
T
A
2
16 e

2k ' L
k ' / k  k / k '2
V(x)
V0
Aeikx  Be ikx
Ce k ' x De  k ' x
0
L
Feikx
x
Simulation of a Gaussian wave packet with kinetic energy 500 eV,
incident on a square potential barrier with height 600 eV and
thickness 25 pm. About 17% of the wave packet tunnels through
the barrier.
To see the probability density:
http://www.youtube.com/watch?v=4-PO-RHQsFA&NR=1
To see the real (black) and imaginary (red) components of the
wave function:
http://www.youtube.com/watch?v=_3wFXHwRP4s
Three applications of quantum tunneling:
1. Light crosses a “total internal reflection barrier”.
2. The NH3 molecule:
Distance of N atom from plane of 3 H atoms,
at various energy levels Ei
Three applications of quantum tunneling:
1. Light crosses a “total internal reflection barrier”.
2. The NH3 molecule:
Distance of N atom from plane of 3 H atoms,
at various energy levels Ei
3. α–decay of a U238 nucleus:
Simple 1D potentials – summary
Simple 1D potentials – summary
Parity
When the potential term V(x) of the 1D Schrödinger equation is
symmetric, i.e. V(x) = V(−x) , then the Schrödinger equation itself
is symmetric and the solutions must have the same physical
symmetry. But what is physical, ψ(x) or |ψ(x)|2 ?
Parity
When the potential term V(x) of the 1D Schrödinger equation is
symmetric, i.e. V(x) = V(−x) , then the Schrödinger equation itself
is symmetric and the solutions must have the same physical
symmetry. But what is physical, ψ(x) or |ψ(x)|2 ?
Only |ψ(x)|2 (and the probability current) are physical. Hence
ψ(x) can be a solution whether ψ(−x) = ψ(x) or ψ(−x) = −ψ(x).
But we give these cases different names: in the first case ψ(x) has
even parity; in the second ψ(x) has odd parity.
Parity
When the potential term V(x) of the 1D Schrödinger equation is
symmetric, i.e. V(x) = V(−x) , then the Schrödinger equation itself
is symmetric and the solutions must have the same physical
symmetry. But what is physical, ψ(x) or |ψ(x)|2 ?
Only |ψ(x)|2 (and the probability current) are physical. Hence
ψ(x) can be a solution whether ψ(−x) = ψ(x) or ψ(−x) = −ψ(x).
But we give these cases different names: in the first case ψ(x) has
even parity; in the second ψ(x) has odd parity.
If two solutions are degenerate (have the same energy) then a
linear combination of them need not have definite parity. But if
the energies are nondegenerate (one solution for each energy)
then the solutions will have definite parity (even or odd).
General 1D potential
Now let’s consider a 1D potential without any special symmetry:
V(x)
x
What can we say about the solutions of the time-independent
Schrödinger equation?
General 1D potential
d 2ψ
 2 V ( x)  E  ψ
dx

2m
2
V(x)
E
x
If ψ(x) > 0 and E > V(x) , then ψ(x) will turn down until it
becomes negative. If ψ(x) < 0 and E > V(x), then ψ will turn up
until it becomes positive. The larger E − V(x) , the more up-anddown curves in ψ(x) and the more nodes (zeros) in ψ(x).
General 1D potential
d 2ψ
 2 V ( x)  E  ψ
dx

2m
2
ψ(x) V(x)
E0
x
Hence the lowest-energy solution will have no node.
General 1D potential
d 2ψ
 2 V ( x)  E  ψ
dx

2m
2
ψ(x) V(x)
E1
x
The next-lowest-energy solution will have one node.
General 1D potential
d 2ψ
 2 V ( x)  E  ψ
dx

2m
2
ψ(x) V(x)
E5
x
As the energy increases, so does the number of nodes.
2D and 3D square wells and degeneracy
The time-independent Schrödinger equation for a particle in two
dimensions:
2 
 2  2



  V ( x, y )  ψ ( x, y )
Eψ( x, y )  

 2m  x 2 y 2 

Suppose V(x,y) is an infinite square well potential, vanishing for
|x| < Lx/2 and |y| < Ly/2, and infinite otherwise. What are the five
lowest energy levels? Are the energies nondegenerate? If not,
what are their degeneracies? How do we normalize the solutions
in 2D? What are the normalized solutions? Does it matter
whether or not Lx and Ly are equal?
What are the corresponding questions and answers in 3D?
The time-independent Schrödinger equation for a particle in three
dimensions is
 2 2

Eψ(r)  
  V (r) ψ(r)
 2m

where r = (x,y,z). How do we define the probability current J(r,t)
in 3D so that the continuity condition,

 (r, t )    J (r, t )  0
t
holds?
,