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