Lecture 16: Probability Current II and 1D Scattering Phy851 Fall 2009

Lecture 16:
Probability Current II and 1D
Scattering
Phy851 Fall 2009
Continuity Equation
dP( x, t )
j( x − ε ) − j( x + ε ) =
dt
dρ ( x, t )2ε
j( x − ε ) − j( x + ε ) =
dt
j ( x − ε ) − j ( x + ε ) dρ ( x, t )
=
2ε
dt
d
d
−
j ( x, t ) = ρ ( x, t )
dx
dt
•
This is the standard continuity equation, valid for
any kind of fluid
•
For energy eigenstates (stationary states), we
need:
d
ρ ( x, t ) = 0 ⇒ ρ ( x, t ) = ρ ( x,0)
•
•
dt
d
j(x,t) = 0 ⇒ j(x,t) = j(x,0)
dt
d
j(x,t) = 0 ⇒ j(x,t) = j 0
This gives:
dx
€Must have spatially uniform current in steady
state (of course j can be zero)
€
Derivation of the probability current:
Current of a plane wave
• For a plane wave we have:
ψ ( x) = aeikx
• The corresponding probability current is:
h
∗
∗
′
j = −i
ψ
ψ
−
ψ
' ψ)
(
2m
ha
€
2
hk
(ik − (−ik ) )= a
= −i
2m
m
2
density
• So for a plane wave, we find:
r
r
j ( x, t ) = ρ 0 v0
This result is fairly intuitive
velocity
Quantum Interference terms
• Consider a superposition of plane waves:
ψ ( x) = a1eik1x + a2 eik 2 x
• The probability density is:
ρ ( x) = a1 + a2 + (a1∗ a2 ei ( k2 − k1 ) x + c.c.)
2
2
Interference Term
• The probability current density is:
hk1
h(k1 + k 2 )
2 hk 2
∗
i(k2 −k1 )x
j(x) = a1
+ a2
+ ( a1 a2e
+ c.c )
m
m
2m
2
Interference Term
• Note that the interference term in j(x,t)
vanishes for k2 = -k1
– This is always the case for Energy
Eigenstates  Currents are then purely
additive
– There is still interference in the
probability density due to the presence of left
and right currents, just not in the probability
current.
Return to the Step Potential
E
E
V0
I
II
x
ψ I ( x) = ei k1x + re − i k1x
k1 − k 2
r=
k1 + k 2
ψ II ( x) = te i k2 x
2k1
t=
k1 + k 2
• The probability current density is:
jI ( x) =
hk1
2 hk1
2 hk
−r
= (1 − r ) 1
m
m
m
2 hk 2
jII ( x) = t
m
• Spatially uniform current requires:
jI ( x) = jII ( x)
– So the probability conservation law is:
2
(1 − r )
Rearrange terms to
get more intuitive
result
hk1
2 hk 2
=t
m
m
k2
r +t
=1
k1
2
2
jout = jin
Continued
k2
r +t
=1
k1
2
2
2k1
t=
k1 + k 2
k1 − k 2
r=
k1 + k 2
• Transmission and reflection probabilities are
derived from conservation law: jin = jout
T :=
R :=
j out (x > 0)
j in
j out (x < 0)
j in
€
Any constant we might
have put in front of the
incident wave would
cancel out here
• For step potential, this gives:
€
2
T=
t k2
k1
=
2
4k1k 2
k1 + k 2
2
R=
2
(
k1 − k 2 ) + 4k1k 2
R +T =
(k1 + k2 )2
r k1
k1
=r
2
k12 + 2k1k 2 + k 22
=
=1
2
(k1 + k2 )
Probability current:
Summary/conclusions
•
The proper way to compute probability in
scattering is via probability current.
•
The probability for a particle to scatter into a
certain channel is the ratio of the outgoing current
in that channel to the total incoming current.
•
In 1D scattering at fixed energy, we can treat the
left-traveling and right-traveling components of the
current as independent
– because there are no interference terms in the current
density for +k and –k currents.
– Allows us to group components into ‘incoming’ and
‘outgoing’ currents
•
For a plane-wave, the current is the amplitude
squared times the velocity.
Important Shortcut:
• If you are asked to compute R and T, for a 1d
scattering problem, you can:
– Compute R=|r|2
– Then use T=1-R (conservation law)
– i.e. you don’t need to compute t or worry about
current unless specifically asked to do so.
– Wrong: T=|t|2 then R = 1-T would not work
Scattering From a Potential Step revisited
•
Lets solve the step potential again, but with the
possibility for left and right travelling incoming
waves:
E
I
ψ I ( x) = ae
i k1 x
+ be
II
VII
− i k1 x
ψ II ( x) = cei k2 x + de − i k2 x
VI
x
x=0
k1 = (U ( E − VI ) + iU (VI − E ) )
k 2 = (U ( E − VII ) + iU (VII − E ) )
•
2m E − VI
h
2m E − VII
h
Boundary condition equations:
ψ I (0) = ψ II (0) ⇒ a + b = c + d
ψ I′ (0) = ψ II′ (0) ⇒ k1 (a − b ) = k 2 (c − d )
Scattering Matrix
• The scattering matrix gives the outgoing
amplitudes in terms of the incoming
amplitudes:
Definition of
scattering matrix, S:
a+b = c+d
b
a
  = S  
c
d 
b − c = −a + d
k1 (a − b ) = k 2 (c − d )
Boundary
condition
equations
Solve for the
outgoing
amplitudes
Calculate the
inverse
Multiply
matrices to
get:
Result:
k1b + k 2 c = k1a + k 2 d
Move outgoing amplitudes to
l.h.s. and incoming to r.h.s.
1

 k1
− 1 b   − 1 1  a 
  = 
 
k 2  c   k1 k 2  d 
b.c. eqs in
matrix
form
−1
 b   1 − 1  − 1 1  a 
 
 
  = 
 c   k1 k 2   k1 k 2  d 
b
1  k 2 1 − 1 1  a 


 
  =
 c  k1 + k 2  − k1 1 k1 k 2  d 
b
1  k1 − k 2

  =
 c  k1 + k 2  2k1
1  k1 − k 2

S=
k1 + k 2  2k1
2k 2  a 
 
k 2 − k1  d 
2k 2 

k 2 − k1 
Example: wave incident from left
• Consider case a=1, b=r, c=t, and d=0:
b
a
  = S  
c
d 
out
•
1  k1 − k 2

S=
k1 + k 2  2k1
2k 2 

k 2 − k1 
in
r
1  k1 − k 2

  =
 t  k1 + k 2  2k1
We find: r = k1 − k 2
k1 + k 2
2k 2  1 
 
k 2 − k1  0 
2k1
t=
k1 + k 2
• For left and right incoming waves:
r
1  k1 − k 2

  =
 t  k1 + k 2  2k1
r=
(k1 − k2 )cL + 2k2cR
k1 + k 2
2k 2  cL 
 
k 2 − k1  cR 
t=
2k1cL + (k 2 − k1 )cR
k1 + k 2
jin =| cL |2 k1 + | cR |2 k 2
| r |2 k1
R=
jin
2
=
2
{
k1 − k 2 | cL |2 +4 k 2 | cR |2 +4 Re k 2∗ (k1 − k 2 )cR∗ cL
2
(
k1 + k 2 | cL |2 k1 + | cR |2 k 2
T = 1− R
)
}