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PHY 4604 Exam 1 Solutions

Monday October 3, 2005 (Total Points = 100)

Problem 1 (10 points): Circle true or false for following (1 point each).

(a) (True or False) One of the breakthroughs that lead to quantum mechanics was the idea of

associating differential operators with the dynamical variables.

(b) (True or False) The wave function (x,t) must vanish in a region of infinite potential.(c) (True or False) It is possible for a free particle to have a definite energy.

(d) (True or False) In quantum mechanics particles can enter the classically forbidden region

where V0 > E (i.e. KE < 0).

(e) (True or False) The operator(Aop-A

op) is hermitian.

Note is we let O = Aop-Aop then O

= A

op- Aop = - (Aop-A

op )= -O.

(f) (True or False) IfAop and Bop are hermitian then Aop-Bop is also hermitian.

Note is we let O = Aop-Bop then O

= A

op-Bop = Aop-Bop= O.

(g) (True or False) IfPop is the parity operator, Pop(x) = (-x), then Pop2 = 1.(h) (True or False) Solutions of Schrdingers equation of the form )()(),( txtx = correspond to states with definite energy E.

(i) (True or False) Solutions of Schrdingers equation of the form )()(),( txtx =

correspond to states in which the probability density 2|),(|),( txtx = is independent of time.

(j) (True or False) Schrdingers equation is valid for all velocities even when v c.

Problem 2 (30 points): Consider an infinite square well defined by

V(x) = 0 forL/2 < x < +L/2,

V(x) = + otherwise.

We look for stationary states of the formh/)(),( tiEnn

nextx= . Parity is a

good quantum number in this problem since V(-x) = V(x) and hence the

stationary state solutions are either even or odd under parity as follows:

)()()( )()()( xxxPnnn

++++++ ==

)()()( )()()( xxxPnnn

==

(a) (5 points). Calculate the (normalized) even parity wave functions )()( xn

++ and their

corresponding energies, +nE .

V = +infinityV = +infinity

Infinite Square Well

-L/2 +L/2 x

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Answer: )/)12cos((2

)()( LxnL

xn =++

+ ,2

2

2222

)12(22

== ++

+ nmLm

kE n

n

hh, where n

+= 1, 2, 3,

Solution: We look for solutions of the time-independent Schrdinger equation

)()()()(

22

22

xExxV

dx

xd

m

=+h

or )())((2)(

22

2

xxVEm

dx

xd

=

h

with h/)(),( iEtextx = . In the region L/2 < x < L/2 for E > 0 and V(x) = 0 we have

)()(2)( 2

22

2

xkxmE

dx

xd

==

hwith

2

2

h

mEk= and

m

kE

2

22h

=

The most general solution is

ikxikx BeAex + +=)( .The even parity solution have (-x) = (x) and hence A = B and hence

)cos(2)()()( kxAeeAx ikxikx =+= ++ The boundary condition at x = L/2 is

)2/cos(20)2/()( kLAL ==+ , which implies that kL/2 = (n+-1/2),where n

+= 1, 2, 3, Thus,

)/)12cos((2)()( LxnAxn =++

+ .

The normalization is determined by requiring that

12

4

)2sin(

2)12(

4)(cos

)12(

4

)/)12((cos4|)(|

2

)(

)(

2)(

)(

22

2/

2/

22

2/

2/

2)(

21

21

21

21

==

+

=

=

=

+

+

+

++

+

+

+

+

LA

yy

n

LAdyy

n

LA

dxLxnAdxx

n

n

n

n

L

L

L

L

n

.

Hence LA 2/1= and

)/)12cos((2

)()( LxnL

xn =++

+ .

The even parity energy levels are given by

2

2

2222

)12(22

== ++

+ nmLm

kE n

n

hh, where n

+= 1, 2, 3,

(b) (5 points). Calculate the (normalized) odd parity wave functions )()( xn

and their

corresponding energies, nE .

Answer: )/2sin(2

)()( LxnL

xn

= ,2

2

2222

)2(22

==

nmLm

kE n

n

hh, where n

-= 1, 2, 3,

Solution: The odd parity solution have (-x) = -(x) and hence A = -B and hence)sin(2)()()( kxAeeAx ikxikx == + ,

where I dropped the i to make it real. The boundary condition at x = L/2 is

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)2/sin(20)2/()( kLAL == , which implies that kL/2 = n-,where n

-= 1, 2, 3, Thus,

)/2sin(2)()( LxnAxn

= .The normalization is determined by requiring that

12

4

)2sin(

22

4)(sin

2

4

)/2(sin4|)(|

2

)(

)(

2)(

)(

22

2/

2/

22

2/

2/

2)(

21

21

21

21

==

==

=

+

+

+

+

LA

yy

n

LAdyy

n

LA

dxLxnAdxx

n

n

n

n

L

L

L

L

n

.

Hence LA 2/1= and

)/2sin(2

)()( Lxn

L

xn

= .

The even parity energy levels are given by

2

2

2222

)2(22

==

nmLm

kE n

n

hh, where n

-= 1, 2, 3,

(c) (3 points). What is the state of lowest energy (i.e. ground state) and what is its energy, E0. Isit a parity even or parity odd state.

Answer:2

22

02mL

Eh

= (even parity)

Solution: The state of lowest energy is the parity even n+

= 1 state with energy

2

22

0 2mLE

h

= and the wave function is given by )/cos(

2

)(

)(

1 LxLx =

+

.

(d) (3 points). What is the state with the 2nd

lowest energy (i.e. 1st

excited state) and what is itsenergy, E1. Is it a parity even or parity odd state?

Answer: 2

22

1

2

mLE

h= (odd parity)

Solution: The next lowest energy is the parity odd n-= 1 state with energy

2

22

1

2

mLE

h= and the wave function is given by )/2sin(

2)()(1 Lx

Lx = .

(e) (14 points). Suppose that a particle in this infinite square well has the initial wave function at

t = 0 given by

=A

x0

)0,(4/4/

4/||

LxL

Lx

>

where A is a constant. Determine the normalization A. If a measurement ofthe energy of this state is made at a later time t, what is the probability that

the measurement will yield the ground state energy, E0? What is the

probability that it will yield the 1st

excited state energy, E1?

V = +infiniV = +infinity

Infinite Square Well

-L/2 +L/2

(x,0)

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Answer:L

A2

= , 81.08

20=

P , 01 =P

Solution: The normalization is determined by requiring that

12/|)0,(| 24/

4/

2

4/

4/

2 ===

LAdxAdxx

L

L

L

L

.

HenceL

A2

= . The overlap of(x,0) with )/cos(2)()(1 LxL

x =+ is given by

22

2

22)sin(

2)cos(

2

)/cos(2

)/cos()0,(2

4/

4/

4/

4/

4/

4/

2/

2/

1

=

=

=

=

==

+

+

+

+

ydyyL

L

dxLxL

dxLxxL

cL

L

L

L

The probability of measuring the ground state energy E0 is

81.08||2

2

10 ==

cP

(x,0) has even parity and hence it has no overlap with the odd parity state)/2sin(

2)()(1 Lx

Lx = .

We can check by doing the integral,

0)cos(2)sin(2

)/2sin(2

)/2sin()0,(2

4/4/

4/

4/

4/

4/

2/

2/

1

===

==

+

+

+

+

ydyyLL

dxLxL

dxLxxL

cL

L

L

L

The probability of measuring the 1st

excited state energy E1 is

0|| 211 ==cP .

Problem 3 (30 points): Suppose that particles with energy E > V0 enter from the

left and travel to the right and encounter both a delta-function potential and a step-

function potential at x = 0 as follows:

+=

)(

0)(

0 xVxV

0

0

0 and V(x) = 0 we have

)()(2)( 2

22

2

xkxmE

dx

xd

==

hwith

2

2

h

mEk= and

m

kE

2

22h

=

The most general solution is

ikx

L

ikx

LL eBeAx+ +=)( ,

In the region x > 0 (right region) we have

iqx

R

iqx

RR eBeAx+ +=)( , where rkVEmq ==

2

0 )(2

h

Since there are no particles entering from the right in the right region we set BR= 0. Theboundary conditions at x = 0 are

)0()0( RL = which implies that (1) AL + BL = AR.

Also,

)0(2)()(

2 R

x

L

x

R m

dx

xd

dx

xd

h=

=+=

which implies

RLLR A

m

ikBikAiqA 22

h

=+ or (2) ( ) RRLL AiRrAkm

ik

q

BA +=

+= 22

h

Adding (1) + (2) yields

( ) RL AiRrA ++= 12 which implies ( )iRrA

A LR ++=

1

2

And subtracting (1)-(2) gives

( ) RL AiRrB = 12 which implies ( )( )

( )iRrAiRr

AiRrB LRL ++

==1

11

21

.

The reflection probability is

22

22

2

2

)1(

)1(

||

||

Rr

Rr

A

B

PLm

k

Lmk

R ++

+

== h

h

.

(b) (5 points) Calculate the quantum mechanical transmission probability, PT, and express youranswer in terms ofr and R.

Answer: 22)1(

4

Rr

rPT ++

=

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Solution: The transmission probability is

222

2

)1(

4

||

||

Rr

r

A

AP

Lmk

Rm

q

T ++==

h

h

.

(c) (5 points) Show that PT + PR= 1.Solution: Adding PRand PT gives

1)1(

)1(

)1(

4)1(

)1(

4

)1(

)1(22

22

22

22

2222

22

=++++

=++

++=

+++

+++

=+Rr

Rr

Rr

Rrr

Rr

r

Rr

RrPP TR .

(d) (5 points) What is the transmission probability, PT1, for the case R = 0 (i.e. no deltafunction)? What is PT1 for the case V0/E = 3/4 (with R = 0)?

Answer: 889.09

8

)1(

42/121

+

= =rT rr

P

Solution: For R = 0 the transmission probability becomes

889.09

8

)2/11(

2

)1(

422/121 =+ += =rT r

rP .

Note that V0/E = 3/4 implies that r = .

(e) (5 points) What is the transmission probability, PT2, for the case V0 = 0 (i.e. no step)?What is PT2 for the case R = 2 (with V0 = 0)?

Answer:2

1

4

4222

+

= =RT RP

Solution: For r = 0 the transmission probability becomes

2

1

44

4

4

4222

=+

+

= =RTR

P .

Note that V0 = 0 implies that r = 1.(f) (5 points) What is the overalltransmission probability, PT, for the case V0/E = 3/4 andR = 2? Does PT = PT1PT2?

Answer: 32.025

8==TP , TTT PPP = 44.0

9

421 .

Solution: For r = and R = 2 the transmission probability becomes

32.025

8

44/9

2

)1(

42,2/122

=+

++

= == RrT Rrr

P .

We that

TTT PPP =

= 44.094

2

1

9

821 .

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Problem 4 (30 points): Consider bound states of a delta-function potential at x = 0 combinedwith an infinite step at x = L as follows:

+

=

)()(

xxV

Lx

Lx

L (region 3) we have

0)(3 =x ,

Since V = +. The boundary conditions at x = 0 are

)0()0( 21 = which implies that (1) A = C + D.Also,

)0(2)()(

12

12

h

m

dx

xd

dx

xd

xx

==+=

which implies

Am

ADC2

2

h

= or(2) A

mDC

=

2

21

h

The boundary condition at x = L is

0)(2 =L which implies that 0=++ LL

DeCe

.Thus,

LCeD2+= and Le

AmC

22 1

21

++

=h

and

V(x)

-(x)

V = +infi

L

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Am

C

=2

1h

and A

mD

=2

h

.

Thus.

Lemm

222

1+

=

hh

or

m

e L2

2 1h

=

Let Ly 2= and we have

cye y = 1 ,where )2/(2 mLc h= . For If )/(2 mLh= then c = 0.5 the

energy is arrived at by solving yey

211= . Let f(y) = e-y

and g(y) = 1 y/2 and plot the two functions. Using mycalculator I get y = 1.594 and

2

2

2

22

0 317.08 mLmL

yE

hh== .

The wave function isxAex +=)(1 and

L

xLxxLx

e

eeeAeeeCx

2

22

21

)()(+

++++

== .

(b) (15 points) Find the allowed bound state energies if

)4/(2 mLh= and sketch the wave functions. How manybound states are they?

Answer: There are no bound states.

Solution: If )4/(2 mLh= then c = 2 the energy is arrived

at by solving yey

21=

. Let f(y) = e-y

and g(y) = 1 - 2yand plot the two functions. Now there is no solution andhence there is no bound state. Note that y = 0 is not an

allowed solution since it implies that = 0 which impliesthat

Ax =)(1 and Ax =)(2 ,

and hence (x) is not renormalizable.

(x)

-1.0 0.0 1.0

x/L

-1

0

1

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

f(y)

g(y) c = 1/2

-1

0

1

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

f(y)

g(y)

c = 2