Post on 25-Jan-2022
Lecture 16: Probability Current II and 1D
Scattering
Phy851 Fall 2009
Continuity Equation
• This is the standard continuity equation, valid forany kind of fluid
• For energy eigenstates (stationary states), weneed:
• This gives:
• Must have spatially uniform current in steadystate (of course j can be zero)
dt
txdPxjxj
),()()( =+−− εε
dt
txdxjxj
ερεε
2),()()( =+−−
dt
txdxjxj ),(
2
)()( ρε
εε=
+−−
),(),( txdt
dtxj
dx
dρ=−
)0,(),(0),( xtxtxdt
dρρρ =⇒=
€
ddx
j(x, t) = 0 ⇒ j(x, t) = j0
€
ddtj(x, t) = 0⇒ j(x, t) = j(x,0)
Derivation of the probability current:
Current of a plane wave
• For a plane wave we have:
• The corresponding probability current is:
• So for a plane wave, we find:
ikxaex =)(ψ
€
j = −i h
2mψ∗ ′ ψ −ψ∗' ψ( )
density
velocity
00),( vtxjrr
ρ=
( )m
kaikik
m
ai
hh 22
)(2
=−−−=
This result is fairly intuitive
Quantum Interference terms
• Consider a superposition of plane waves:
• The probability density is:
• The probability current density is:
• Note that the interference term in j(x,t)vanishes for k2 = -k1– This is always the case for Energy
Eigenstates Currents are then purelyadditive
– There is still interference in the probability density due to the presence of leftand right currents, just not in the probabilitycurrent.
xikxik eaeax 2121)( +=ψ
( )..)( )(21
2
2
2
112 cceaaaax xkki +++= −∗ρ
Interference Term
Interference Term
€
j(x) = a12 hk1m
+ a22 hk2m
+ a1∗a2e
i(k2−k1 )x + c.c( ) h(k1 + k2)2m
Return to the Step Potential
• The probability current density is:
• Spatially uniform current requires:– So the probability conservation law is:
E
x
V0
I II
E
21
21
kk
kkr
+
−=
21
12
kk
kt
+=
xkixkiI reex 11)( −+=ψ xki
II tex 2)( =ψ
m
kr
m
kr
m
kxjI
12121 )1()(hhh
−=−=
m
ktxjII
22)(
h=
)()( xjxj III =
m
kt
m
kr 2212)1(
hh=−
inout jj =11
222=+
k
ktr
Rearrange terms toget more intuitive
result
• Transmission and reflection probabilities arederived from conservation law: jin = jout
• For step potential, this gives:
Continued
( )( ) ( )
124
221
2221
21
221
212
21 =+
++=
+
+−=+
kk
kkkk
kk
kkkkTR
21
21
kk
kkr
+
−=
21
12
kk
kt
+=
11
222=+
k
ktr
€
T := jout (x > 0)jin
€
R := jout (x < 0)jin
Any constant we mighthave put in front of the
incident wave wouldcancel out here
2
21
21
1
2
24
kk
kk
k
ktT
+==
2
1
1
2
rk
krR ==
Probability current:Summary/conclusions
• The proper way to compute probability inscattering is via probability current.
• The probability for a particle to scatter into acertain channel is the ratio of the outgoing currentin that channel to the total incoming current.
• In 1D scattering at fixed energy, we can treat theleft-traveling and right-traveling components of thecurrent 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 amplitudesquared 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 thepossibility for left and right travelling incomingwaves:
• Boundary condition equations:
E
xVI
I II
xkixkiI beaex 11)( −+=ψ
xkixkiII decex 22)( −+=ψ
( )h
III
VEmEViUVEUk
−−+−=
2)()(1
VII
( )h
IIIIII
VEmEViUVEUk
−−+−=
2)()(2
dcbaIII +=+⇒= )0()0( ψψ
x=0
( ) ( )dckbakIII −=−⇒′=′ 21)0()0( ψψ
Scattering Matrix
• The scattering matrix gives the outgoingamplitudes in terms of the incomingamplitudes:
=
d
aS
c
b
dcba +=+ dacb +−=−
−
−=
−
d
a
kkkkc
b
21
1
21
1111
−
−+=
d
a
kkk
k
kkc
b
211
2
21
11
1
11
−
−
+=
d
a
kkk
kkk
kkc
b
121
221
21 2
21
−
−
+=
121
221
21 2
21kkk
kkk
kkS
Boundaryconditionequations
Solve for theoutgoing
amplitudes
Calculate theinverse
Multiplymatrices to
get:
Move outgoing amplitudes tol.h.s. and incoming to r.h.s.
Definition ofscattering matrix, S:
Result:
dkakckbk 2121 +=+
−=
−
d
a
kkc
b
kk 2121
1111
( ) ( )dckbak −=− 21
b.c. eqs inmatrixform
Example: wave incident from left
• Consider case a=1, b=r, c=t, and d=0:
• We find:
• For left and right incoming waves:
−
−
+=
121
221
21 2
21kkk
kkk
kkS
−
−
+=
0
1
2
21
121
221
21 kkk
kkk
kkt
r
21
21
kk
kkr
+
−=
=
d
aS
c
b
out in
−
−
+=
R
L
c
c
kkk
kkk
kkt
r
121
221
21 2
21
21
12
kk
kt
+=
( )21
221 2
kk
ckckkr RL
+
+−=
21
121 )(2
kk
ckkckt RL
+
−+=
22
12 |||| kckcj RLin +=
( ){ }( )22
122
21
21222
222
2112
||||
Re4||4||||
kckckk
cckkkckckk
j
krR
RL
LRRL
in ++
−++−==
∗∗
RT −=1