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Lecture 21. Time-Dependent Spin Problem from Midterm. WKB Revisited. Problem: A spin ½ particle is initially in the +z state (the zS eigenstate with the

eigenvalue 2/h+ ). It is placed in a magnetic field that points in the x direction, until 0=zS .

Show that the particle is now in yS eigenstate.

Solution: First of all, let me write the Hamiltonian of the system: BSBSH xˆˆˆ γγ −=⋅−=

rr,

where mceh

−=γ . As a result, the time-dependent Schrödinger equation takes the form

Ψ−=∂Ψ∂

xSBtih ˆγ . (21.1)

As usual, we can solve the stationary problem first, i.e. find the eigenstates and eigenvalues of

the Hamiltonian

Ψ=Ψ− ESB xˆγ . (21.2)

Since the operators H and xS commute, the eigenstates +x , −x of the operator xS

(remember that ( ) ±± ±= xxSx 2/ˆ h ) are simultaneously the eigenstates of the operator H .

Therefore,

+++ =− xExB2hγ , −−− =

−− xExB

2hγ →

2h

mBE γ

=± .

As a result, a general solution of the equation (21.1) is

( ) −−

+− −+ +=Ψ xebxeat tiEtiE hh // , 122 =+ ba . (21.3)

According to the initial condition ( ) +=Ψ z0 . We know that −++ += xxz2

12

1 . Then,

( ) −−

+− −+ +=Ψ xexet tiEtiE hh //

21

21 . (21.4)

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Notice that we didn’t use any particular representation till this point. We were able to work out

the solution (21.4) in Dirac ket space!

At some moment 1tt = , the expectation 0=zS , i.e. ( ) ( ) 0ˆ11 =ΨΨ tSt z .

In zS matrix representation ( ) ( )( )

−

=

−

+

=Ψ −+ −−

tit

eet tiEtiE

ωω

sincos

11

21

11

21 // hh , where

02

2/ >=−=mcBe

Bh

γω ; ( ) ( )( ) ( )( ) 0

sincos

1001

2sincos

1

11 =

−

− ti

ttit

ωω

ωω h

→ ( ) ( ) 0sincos 12

12 =− tt ωω . As a result,

41πω =t , and at 1tt =

( )

−

=Ψi

t1

21

1 , (21.5)

which is the eigenstate of yS .

Problem: This problem is about tunneling of the electron through the barrier:

>

<<

<

=

bx

bxbxU

x

xV

,0

0,

0,0

)( 0 .

(a) Write down the exact wave function for the electron in the region 0<x in terms of the

incoming and reflected waves.

(b) Write down the exact wave function for the electron in the region bx > in terms of

the transmitted wave.

(c) It is obvious that bx = is the second turning point. Find the position of the first

turning point.

(d) Find the exponential factor of the transmission coefficient for the case 0UE <

(transmission coefficient in “pure” WKB approximation).

(e) Write down the WKB wave function for the electron in the region ax < , where a is

the first turning point. Write down the derivative of this function in WKB

approximation.

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(f) Write down the WKB wave function for the electron in the region bxa << . Write

down the derivative of this function in WKB approximation.

(g) Try to pick up the right connection formula for the WKB wave functions between the

regions bxa << and ax < .

(h) Discuss how you can derive the expression for the transmission coefficient (still in

WKB approximation) that takes into account rectangular wall at bx = . Note: you

don’t need to actually derive this expression.

Solution: (a) ( ) ( )xikBxikA 00exact exp −+=Ψ ,h

mEk 20 = ; 0<x .

(b) ( )xikF 0exact exp=Ψ , h

mEk 20 = ; bx > .

(c) 0UEba = .

(d) ( )∫−

b

adxx

eTκ2

~ , ( ) ( )h

ExVmx

−=

)(2κ ;

−−

2/3

0

0 1324

exp~UEmUb

Th

(e) ( )

( )( )

( )

−+

=Ψ ∫∫

x

a

x

a

dxxkixkBdxxki

xkA ''exp''expWKB , ( ) ( )( )

h

xVEmxk

−=

2;

ax < .

( ) ( ) ( ) ( )

−−

=

Ψ∫∫x

a

x

a

dxxkixkBidxxkixkAidx

d ''exp''expWKB .

(f) ( )

( )( )

( )

−+

=Ψ ∫∫

b

x

b

x

dxxx

HdxxxD ''exp''expWKB κ

κκ

κ, ( ) ( )( )

h

ExVmx

−=

2κ ;

bxa << .

( ) ( ) ( ) ( )

−+

−=

Ψ∫∫b

x

b

x

dxxxkHdxxxDdx

d ''exp''expWKB κκκ .

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Of course, we can use another form: ( )

( )( )

( )

−+

=Ψ ∫∫

x

a

x

a

dxxxNdxx

xM ''exp''expWKB κ

κκ

κ

as well.

(g) The appropriate connection formula takes the form (see the end of the lecture 5 and the

formula (3.2) at the third lecture):

<<

−

<

+

≅Ψ

∫

∫

.,')'(exp)(

;,4

')'(sin)(

2

)(WKB

bxadxxxN

axdxxkxkN

x x

a

a

x

κκ

π

The main assumption to derive this formula is to keep only decaying exponential function as

bx→ in the region bxa << .

(h) To take into account a rectangular wall at bx = we need to use boundary conditions

( ) ( )

( ) ( ),00

,00

−Ψ

=+Ψ

−Ψ=+Ψ

bdxdb

xdd

bb

instead of using the connection formula

<<

>

=Ψ∫

∫ +

.,)(

,,)(

)(')'(

4/')'(

bxaexD

bxexkD

x b

x

x

b

dxx

dxxki

κ

π

κ

As a result, the connection formula in (g) will take an absolutely different form.