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PHY4604 Fall 2006 Final Exam Solutions

Department of Physics Page 1 of 25

PHY 4604 Final Exam Solutions

Wednesday December 6, 2006 (Total Points = 100)

Problem 1 (25 points): Consider the following one-dimensional potential:

+= 0

0

)( VxV

Lx

Lx

Lx

0 enter from the left in

region 1 and travel to the right and encounter the potential V(x). Let2

2

2

2

mLE

h=

andE

Vr 01+= .

(1) (5 points) Calculate the probability that particles entering from the left in region 1 will be

reflected back (i.e. calculate the ratio of the probability flux in region 1 traveling to the left tothe incident flux in region 1 traveling to the right, 11 / jjPR

rs= ). Express your answer in terms of

and r.Answer:PR= 1.

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

)()()()(

2 2

22

xExxVdx

xd

m

=+

hor )())((

2)(22

2

xxVEm

dx

xd

=

h

withh/)(),( iEtextx = . In the region x < 0 (region 1) for E > 0 and V(x) = 0 we have

)()(2)( 2

22

2

xkxmE

dx

xd

==

h

withL

mEk

==

2

2

h

and 2

2

222

22

mLm

kE

hh==

The most general solution is

ikxikx BeAex + +=)(1 ,

In the region 0 < x < L (region 2) we have

iqxiqx DeCex + +=)(2 ,

with rkVEm

q =+

=2

0 )(2

h, where

E

V

k

qr 01+== . In the region x > L (region 3) we have

0)(3 =x ,

The boundary conditions at x = L implies that

0)(2 =L and hence 0=++ iqLiqL DeCe or

iqLCeD 2+= .

Thus,

)()( 22iqxiqLiqx eeeCx ++ = ,

At x = 0 the wave function must be continuous with a continuous derivative which yields

V(x)

-V0

V = +in

L

Region 1 Region 2 Reg


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)0()0( 21 = and )1(2iqLeCBA +=+

0

2

0

1 )()(

==

=xx dx

xd

dx

xd and )1(

2iqLeCiqikBikA ++=

Thus,

)1( 2iqLeCBA +=+ and )1( 2iqLeCrBA ++=

And

))1()1((2 2iqLerrCA +++= and ))1()1((2 2iqLerrCB +++=

Thus,

iqLerr

AC

2)1()1(

2+++

= and Aerr

errB

iqL

iqL

2

2

)1()1(

)1()1(+

+

++++

= .

The relfection probability is given by

1)1)(1()1)(1()1()1(

)1)(1()1)(1()1()1(

)1()1(

)1()1(

)1()1(

)1()1(

)1()1(

)1()1(

||

||

2222

2222

2

2

2

22

2

2

2

2

=++++++++++++

=

++++

++++

=++++

==

+

+

+

+

+

+

iqLiqL

iqLiqL

iqL

iqL

iqL

iqL

iqL

iqL

mk

m

k

R

errerrrr

errerrrr

err

err

err

err

err

err

A

BP h

h

(2) (5 points) Calculate the probability that particles traveling to the right in region 2 will be

reflected back (i.e. calculate the ratio of the probability flux in region 2 traveling to the left to

flux in region 2 traveling to the right, 22 / jjRrs

= ). Express your answer in terms of and r.Answer:R = 1.

Solution: We see that

1||

|| 222

2

=== + iqL

m

q

mq

eC

DR

h

h

.

(3) (5 points) Calculate the ratio of the probability flux in region 2 traveling to the right to the

incident flux in region 1 traveling to the right, 12 / jjPrr

= ). Express your answer in terms ofand r. What are the maximum and minimum values of P() (express your answer in terms of r)?Evaluate the maximum and minimum values of P() for the case V0 = 3E. Explain what is goingon.

Answer:

)2cos()1(1

222

rrr

rP

++= ,

2

112min

= =rr

P , 22

max ==r

rP



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)2cos()1(1

2

)2cos()1(1

2

)2cos(2)1)(1()1()1(

4

)1)(1()1)(1()1()1(

4

)1()1(

2

||

||

22

2222

2222

2

22

2

rrr

r

qLrr

r

qLrrrr

r

errerrrr

r

errr

A

CP

iqLiqL

iqL

mk

m

q

++=

++=

++++=

++++++=

=++

==

+

+h

h

The minimum value of P ocurs when cos(2r) = 1 and hence

2

11

)1(1

2222min

=++

= =rrrr

rP ,

where I used

213

0

0

+== EVE

Vr .

The maximum value of P ocurs when cos(2r) = -1 and hence

2)1(1

2222max

=+

= =rrrr

rP .

(B)Now consider the case E < 0 and look for possible bound states. Let 22

2

2

mLE

h= and

2

2

2

0

2

mL

Vh

= .

(1) (5 points) How many bound states are there if = /4? What is the ground state energy (i.e.what is for the ground state)?Answer:No bound states for < /2.Solution: We look for solutions of the time-independent Schrdinger equation

)()()()(

2 2

22

xExxVdx

xd

m

=+

hor )())((

2)(22

2

xxVEm

dx

xd

=

h

withh/)(),( iEtextx = . In the region x < 0 (region 1) for E < 0 and V(x) = 0 we have

)()(2)( 2

22

2

xxmE

dx

xd

==

h

with

L

mE =

=

2

2

h

and 22

222

22

mLm

Ehh

==

The most general solution is

xx BeAex + +=)(1 ,

However, =

x

xe and hence we must set B = 0. In the region 0 < x < L (region 2) we

have

iqxiqx DeCex + +=)(2 ,


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with2

0)(2

h

VEmq

+= , and

2

022 2

h

mVq =+ . Also, 2

2

0

222 2)()( ==+

h

VmLLqL and

222)( =+qL . In the region x > L (region 3) we have

0)(3 =x ,

The boundary conditions at x = L implies that

0)(2 =L and hence 0=++ iqLiqL DeCe or

iqLCeD 2+= .

Thus,

)()( 22iqxiqLiqx eeeCx ++ = ,

At x = 0 the wave function must be continuous with a continuous derivative which yields

)0()0( 21 = and )1(2iqLeCA +=

0

2

0

1 )()(

==

=xx dx

xd

dx

xd and )1(

2iqLeCiqA ++=

Dividing the two equations yields

)cot()(

)(

)1(

)1(2

2

qLqee

eeiq

e

eiq

iqLiqL

iqLiqL

iqL

iqL

=+

==+

=+

+

+

+

If we let y = qL then we see that

y

y

qL

L

qy

22

)cot(

===

,

where 222 =+y and )(22

22

2

22

2

2

ymLmL

E == hh

. The evengy levels are determined by

letting )cot()( yyf = andy

yyg

22

)(=

and looking for the places where f(y) = g(y).

-1

0

1

2

3

4

5

0.0 1.6 3.1 4.7 6.3 7.9 9.4

f(y)g(y)

y

/2 3/2 2

= /4

We see that there are no solution fory > . Thus, there are no bound states if < /2.(2) (5 points) How many bound states are there if = 11/4? What is the ground state energy(i.e. what is for the ground state)?


5/25



Solution: There are three bound states.2

02

2

02

mL

Eh

= with 0 8.17.

Solution: For = 11/4 we have three bound states. The ground state has the smallest y, whichis y 2.81 and hence

74.662)17.8(2))81.2()4/11((2)(2 2

22

2

222

2

22

min

2

2

2

0mLmLmLymLE

hhhh

==

-1

0

1

2

3

4

5

0.0 1.6 3.1 4.7 6.3 7.9 9.4

f(y)g(y)

y

/2 3/2 2

= 11/4

Problem 2 (25 points): The Hamiltonian of a charged particle with intrinsic spin Sr

at rest in a

magnetic field is BSBHrrrr

== , wherem

q= (q is the electric charge, m is the mass).

For spin 1 we have zSySxSS zyx ++=r

with

=

010

101

010

2

hxS

=

010

101

010

2

hiSy

=

100

000

001

hzS

(A) (2 points) Show that

=++=

100

010

001

2 22222 hzyx SSSS .


=

=

101

020

101

2010

101

010

010

101

010

2

222 hh

xS

=

=

101

020

101

2010

101

010

010

101

010

2

222 hh

yS

=

=

100

000

001

100

000

001

100

000

001222

hhzS

Hence,


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=++=

100

010

001

2 22222 hzyx SSSS .

(B) (6 points) The W+

boson is a spin 1 elementary particle with charge q = +e and mass MW.

Suppose a W+

is at rest in a uniform magnetic field which points in the z-direction, zBB 0=

r

.What are the energy levels and the corresponding eigenkets of the system? What is the ground

state energy and ground state eigenket? Express your answers in terms of 0B= .

Answer: The ground state (i.e. lowest) energy is h=0E with eigenket

>=>=

0

0

1

11|| 0 .

The 1st

excited state has energy 01 =E with eigenket

>=>=

0

1

0

10|| 1 .

The 2nd

excited state has energy h+=2E with eigenket

>=>=

1

0

0

11|| 2 .

Solution: The eigenvalues of Sz are determined from

0

00

00

00

=

h

h

and hence 0)()( =+ hh

Thus, there are three eigenvalues hh = ,0, . The eigenket of Sz corresponding to = 0 isdetermined from

=

=

0

0

0

0

100

000

001

c

a

c

b

a

hh which implies that a = c = 0 and

>=

0

1

0

10|

The eigenkets of Sz corresponding to h= is determined from

=

=

c

b

a

c

a

c

b

a

hhh 0

100

000

001

which implies that a = a and -c = c and hence

>=

0

0

1

11| and

>=

1

0

0

11| and where >>= mmmSz 1|1| h .

The Hamiltonian is given by zz SSBBSH === 0rr

. Hence there are three energy levels

h=0E , 01 =E , h+=2E . The ground state (i.e. lowest) energy is h=0E with eigenket


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>=>=

0

0

1

11|| 0 .

The 1st

excited state has energy 01 =E with eigenket

>=>=

0

10

10|| 1 .

The 2nd

excited state has energy h+=2E with eigenket

>=>=

1

0

0

11|| 2 .

(C) Suppose that at t = 0 the W+ boson is in the state

>=1

1

1

31)0(| .

(1) (2 points) What is >)(| t ? Express your answer in terms of 0B= .

Answer:

>=

+

ti

ti

e

e

t

13

1)(|

Solution: We know thathhh /

22

/

11

/

00210 |||)(| tiEtiE

tiEececect

>+>+>>=

where

3

1

1

1

1

3

1)001()0(|00 =

>==< c

3

1

1

1

1

3

1)010()0(|11 =

>==< c

3

1

1

1

1

3

1)100()0(|22 =

>==< c

Hence,


8/25



=

+

+

=

>+>+>>=

+

+

+

ti

ti

titi

titi

e

e

ee

eccect

13

1

1

0

0

3

1

0

1

0

3

1

0

0

1

3

1

|||)(| 221100

where 0B= .

(2) (2 points) If I measure the energy of the state >)(| t , what values might I get and what isthe probability of getting these values.

Answer: We will measure h=0E with probability 31

0 =P and we will measure 01 =E with

probability31

1 =P and we will measure h+=2E with probability 31

2 =P .

Solution: We will measure E0

with probability3

12

00|| == cP and we will measure E

1with

probability312

11 || == cP and we will measure E2 with probability 312

22 || == cP .

(3) (2 points) What is the average energy, , for the state >)(| t ? Express your answer interms of 0B= .Answer: = 0.Solution: We can calculate in two ways. It is given by

0)()0()( 031

31

031

221100 =+++=++>=< BBEPEPEPE hh .

( )

( ) 0)11(3

013

1100

000

001

13

)(||)()(||)(

=

=

=

>=>=)(| t ? Express youranswer in terms of 0B= .

Answer: h32=E .

Solution: We can calculate in two ways. It is given by 2322

0322

0312

312

0312

22

2

11

2

00

2 )()()()0()( hhhh ==+++=++>=< BBBEPEPEPE And it is also given by


9/25



( )

( )3

)(2)11(

3

)(01

3

)(

1

100

000

001

13

)(

)(||)()(||)(

222

2

2222

hhh

h

=+=

=

>=>=)(| t ? Express your answer interms of 0B= .

Answer: )cos(23

4

)( ttSx

h

>=< , )sin(23

4

)( ttSy

h

>=< , 0)( >=< tSz Solution: We see that

( )

( ) ( )

)cos(23

4

231

1

123

1

010

101

010

123

)(||)()(

0

t

eeeeeeee

e

e

eetSttS

titititititititi

ti

ti

titi

xx

h

hh

h

=

+++=

+=

>=>==>=


10/25



( )

( ) ( ) 011301

3

1

100

000

001

13

)(||)()(

0

0

==

=

>=>=)(| t ? Express your answer interms of 0B= .

Answer: )(cos233

)( 2 ttSx =h

, )(sin233

)( 2 ttSy =h

, h3

2)( = tSz .


( )

( ) ( )

( ) ( )1)(cos23

1)cos()(3

)cos(22)cos(26

216

1

101

020

101

16

)(||)()(

222

22

222

0

+=++=

++=

+

+

=

>=>===>=


11/25



( )

( ) ( ))(sin239

)(sin13

)(sin1)(sin23

)()())((

22

2

32

2

2

382

2222

tt

tttStStS yyy

==

+=>==>== are the eigenkets of opJ )(2

1 and opzJ )( 1 and the states |j2m2> are the eigenkets

of opJ )(2

2 and opzJ )( 2 as follows:

>>=

>+>=

111111

111111

2

1

||)(

|)1(|)(

mjmmjJ

mjjjmjJ

opz

op

>>=

>+>=

222222

222222

2

2

||)(

|)1(|)(

mjmmjJ

mjjjmjJ

opz

op

Also we know that

>+>=

>+>=

1|)1()1(|)(

1|)1()1(|)(

222222222

111111111

mjmmjjmjJ

mjmmjjmjJ

op

op

where opyopxop JiJJ )()()( 111 =

and opyopxop JiJJ )()()( 222 =

.Now consider the vector

sum of the two operators,

opopop JJJ )()()( 21rrr

+= or opiopiopi JJJ )()()( 21 += for i = 1,2, 3.


12/25



(A) (1 point):Show that

opzopzopopopopopop

opopopopopopopopop

JJJJJJJJ

JJJJJJJJJJJ

)()(2)()()()()()(

)()(2)()()()()()()(

212121

2

2

2

1

21

2

2

2

12121

2

++++=

++=++==++

rrrrrrrr

Solution: First we note that

)(

)(

1121

1

1121

1

+

+

=+=

JJJ

JJJ

iy

x

and)(

)(

2221

2

2221

2

+

+

=+=

JJJ

JJJ

iy

x

Hence,

zz

zz

zzyyxx

JJJJJJJJ

JJJJJJJJJJJJ

JJJJJJJJJJJJJ

212121

2

2

2

1

21221121

2211212

2

2

1

212121

2

2

2

121

2

2

2

1

2

2

2))(())((

2222

++++=

+++++=

++++=++=

++

++++

rr

(B)Now consider the case where j1 = 1 and j2 = 1 (i.e. 3 3) and define the states as follows:

1111

1110

1111

11||

10||

11||

>=>>=>

>=>

Y

Y

Y

and

2211

2210

2211

11||

10||

11||

>=>>=>

>=>

Y

Y

Y

Consider the three superposition states

211111210110211111 ||3

1||

3

1||

3

1| >>+>>>>> YYYYYYa

211111211111 ||2

1||

2

1| >>>>> YYYYb

211111210110211111 ||

6

1||

3

2||

6

1| >>+>>+>>> YYYYYYc

Calculate the following:

(1) (1 point) >=< aa |

(2) (1 point) >=< bb |

(3) (1 point) >=< cc |

(4) (1 point) >=< ba |

(5) (1 point) >=< ca |

(6) (1 point) >=< cb |

Calculate the following and express your answer in terms of|a>,|b>, and |c> :(1) (1 point) >=+ aJJ |)(

2

2

2

1

(2) (1 point) >=+ bJJ |)(2

2

2

1

(3) (1 point) >=+ cJJ |)(2

2

2

1

(4) (1 point) >=azJ |


13/25



(5) (1 point) >=bzJ |

(6) (1 point) >=czJ |

(7) (1 point) >=++ ++ azzJJJJJJ |)2( 212121

(8) (1 point) >=++ ++ bzzJJJJJJ |)2( 212121

(9) (1 point) >=++ ++ czzJJJJJJ |)2( 212121

(10) (2 points) >=aJ |2

(11) (2 points) >=bJ |2

(12) (2 points) >=cJ |2

(3 points) Are the states |a>, |b>, and |c> eigenstates of the J2 and Jz and if so what are theireigenvalues?

Answer:

(1) >>=+ aaJJ |4|)(2

2

2

1

(2) >>=+ bbJJ |4|)( 2221

(3) >>=+ ccJJ |4|)(2

2

2

1

(4) >>= aazJ |0|

(5) >>= bbzJ |0|

(6) >>= cczJ |0|

(7) >>=++ ++ aazzJJJJJJ |4|)2( 212121

(8) >>=++ ++ bbzzJJJJJJ |2|)2( 212121

(9) >+>=++ ++ cczzJJJJJJ |2|)2( 212121

(10) >>= aaJ |0|2

(11) >>= bbJ |2|2

(12) >>= ccJ |6|2

The state |a> is an eigenstate ofJ2 and Jz withj = 0 and mj = 0 (i.e.|a>=|00>). The state |b>is an eigenstate ofJ

2and Jz withj = 1 and mj = 0 (i.e.|b>=|10>). The state |c> is an eigenstate

ofJ2

and Jz withj = 2 and mj = 0 (i.e.|c>=|20>).Solution: We know that

11111

2

1111

2

1

11011

2

1110

2

1

11111

2

1111

2

1

|211|)11(111||

|210|)11(110||

|211|)11(111||

>=>+=>=>

>=>+=>=>

>=>+=>=>

YJYJ

YJYJ

YJYJ

11111

2

1111

2

1

11011

2

1110

2

1

11111

2

1111

2

1

|211|)11(111||

|210|)11(110||

|211|)11(111||

>=>+=>=>

>=>+=>=>

>=>+=>=>

YJYJ

YJYJ

YJYJ

1111111111

1111111

1111111111

|111|111||

010|010||

|111|111||

>=>=>=>

=>=>=>

>=>=>=>

YJYJ

JYJ

YJYJ

zz

zz

zz

1111111111

1111111

1111111111

|111|111||

010|010||

|111|111||

>=>=>=>

=>=>=>

>=>=>=>

YJYJ

JYJ

YJYJ

zz

zz

zz


14/25



1101111111

1111111101

111111

|210|)11(1)11(111||

|211|)10(0)11(110||

011||

>=>+++=>=>

>=>++=>=>

=>=>

+

+

++

++

YJYJ

YJYJ

JYJ

011||

|211|)10(0)11(110||

|210|)11(1)11(111||

111111

1111111101

1101111111

=>=>

>=>+=>=>

>=>+=>=>

JYJ

YJYJ

YJYJ

and hence

>>=+=

>>+>>>>+>=+

aa

a JJJJ

|4|)22(

11|11|3

100|00|

3

111|11|

3

1)(|)( 212121

2

2

2

1

2

2

2

1

>>=+=

>>>>+>=+

bb

b JJJJ

|4|)22(

11|11|2

111|11|

2

1)(|)( 2121

2

2

2

1

2

2

2

1

>>=+=

>>+>>+>>+>=+

cc

c JJJJ

|4|)22(

11|11|6

100|00|

3

211|11|

6

1)(|)( 212121

2

2

2

1

2

2

2

1

>=

>>++>>+>>=

>>+>>>>+>=

a

zzaz JJJ

|0

11|11|

3

1)11(00|00|

3

1)00(11|11|

3

1)11(

11|11|3

100|00|

3

111|11|

3

1)(|

212121

21212121

>=

>>+>>=

>>>>+>=

b

zzbz JJJ

|0

11|11|2

1)11(11|11|

2

1)11(

11|11|2

111|11|

2

1)(|

2121

212121

>=

>>++>>++>>=

>>+>>+>>+>=

c

zzcz JJJ

|0

11|11|3

1)11(00|00|

3

2)00(11|11|

6

1)11(

11|11|6

100|00|

3

211|11|

6

1)(|

212121

21212121


15/25



>=

>>+>>++>>=

>>+>>>>+

>>>>+

>>+>>=

>>+>>>>+

>>+>>>>+

>>+>>>>=

>>+>>>>++=

>++

+

+

++

++

a

zz

zz

azz

JJ

JJ

JJ

JJJJJJ

JJJJJJ

|4

11|11|3

1)22(10|10|

3

1)22(11|11|

3

1)22(

11|11|

3

1)1)(1)(2(10|10|

3

1)0)(0)(2(11|11|

3

1)1)(1)(2(

11|11|3

12210|10|

3

122

10|10|3

12211|11|

3

122

11|11|3

110|10|

3

111|11|

3

1)2(

11|11|3

110|10|

3

111|11|

3

1)(

11|11|3

110|10|

3

111|11|

3

1)(

11|11|3

110|10|

3

111|11|

3

1)2(

|)2(

212121

212121

2121

2121

21212121

21212121

21212121

212121212121

212121

and hence

>>=>=++++>= ++ aaazza JJJJJJJJJ |0|)44(|)2(| 2121212

2

2

1

2 .

Thus, the state |a> is an eigenstate ofJ2

and Jz withj = 0 and mj = 0 (i.e.|a>=|00>).

>=

>>>>++>>=

>>>>+

>>+

>>=

>>>>+

>>>>+

>>>>=

>>>>++=

>++

+

+

++

++

b

zz

zz

bzz

JJ

JJ

JJ

JJJJJJ

JJJJJJ

|2

11|11|3

1)2(10|10|

2

1)22(11|11|

2

1)2(

11|11|2

1)1)(1)(2(11|11|

2

1)1)(1)(2(

10|10|2

122

10|10|2

122

11|11|2

111|11|

2

1)2(

11|11|2

111|11|

2

1)(

11|11|2

111|11|2

1)(

11|11|2

111|11|

2

1)2(

|)2(

212121

2121

21

21

212121

212121

212121

2121212121

212121

and hence

>>=>=++++>= ++ bbbzzb JJJJJJJJJ |2|)24(|)2(| 2121212

2

2

1

2 .


16/25



Thus, the state |b> is an eigenstate ofJ2

and Jz withj = 1 and mj = 0 (i.e.|b>=|10>).

>=

>>+>>++>>=

>>+>>+>>+

>>+>>+

>>+>>=

>>+>>+>>+

>>+>>+>>+

>>+>>+>>=

>>+>>+>>++=

>++

+

+

++

++

c

zz

zz

czz

JJ

JJ

JJ

JJJJJJ

JJJJJJ

|2

11|11|6

1)24(10|10|

3

2)11(11|11|

6

1)24(

11|11|61)1)(1)(2(10|10|

32)0)(0)(2(11|11|

61)1)(1)(2(

11|11|3

22210|10|

6

122

10|10|6

12211|11|

3

222

11|11|6

110|10|

3

211|11|

6

1)2(

11|11|6

110|10|

3

211|11|

6

1)(

11|11|6

110|10|3

211|11|6

1)(

11|11|6

110|10|

3

211|11|

6

1)2(

|)2(

212121

212121

2121

2121

21212121

21212121

21212121

212121212121

212121

and hence

>>=+>=++++>= ++ ccczzc JJJJJJJJJ |6|)24(|)2(| 2121212

2

2

1

2 .

Thus, the state |c> is an eigenstate ofJ2

and Jz withj = 2 and mj = 0 (i.e.|c>=|20>).

Problem 4 (25 points): Consider the case of two non-interactingparticlesboth with mass m in a one-dimensional infinite square well given by

V(x) = 0 for0 < x < L, and V(x) = . For one particle we know that the

stationary states of Schrdingers equation are given byh/

)(),( tiEnnnextx

= and 02EnEn = , and n is a positive integer and

2

22

02mL

Eh

= and where )/sin(2

)( LxnL

xn = . For two (non-interacting)

particles we look for a solution of the formhh /

21

/

2121 )()(),(),,(iEtiEt exxexxtxx == with

Em

p

m

p xx =+2

)(

2

)( 222

1 .

(A) (1 point): Show that )()(),( 2121 xxxx = , is a solution to the two particle non-

interaction Schrdinger equations, where and are positive integers and (x) and (x) arethe one particle stationary state solutions. Show that the allowed energy levels are given by

0

22 )( EE += .


0 L

Two Particles in a Box

(x1)(x2)


17/25



hh /

21

/

2121 )()(),(),,(iEtiEt exxexxtxx == with E

m

p

m

p xx =+2

)(

2

)( 222

1 .

Thus,

),(),(

2

),(

2

212

2

21

22

2

1

21

22

xxE

dx

xxd

mdx

xxd

m

=hh

)()()(

)(2

)()(

2212

2

2

2

1

2

2

1

1

2

2

2

xxEdx

xdx

mdx

xdx

m

=

hh

Edx

xd

xmdx

xd

xm=

2

2

2

2

2

2

2

1

1

2

1

2 )(

)(

1

2

)(

)(

1

2

hh

Hence, E = E1 + E2 and

)()(

2112

1

1

22

xEdx

xd

m

=

h )(

)(

2222

2

2

22

xEdx

xd

m

=

h

The total energy is therefore given by

2

2222

212

)()()(mL

EEE +=+=h

2

222

12

)(mL

E h=

2

222

22

)(mL

E h=

where = 1, 2, 3, ... and = 1, 2, 3, ... and)/sin()/sin(

2)()(),( 212121 LxLx

Lxxxx ==

)/(sin)/(sin4

)()()()(),( 22

1

2

2212

*

1

*

21 LxLxL

xxxxxx ==

This solution corresponds to the case where the two particles are distinguishable. Define the

following three probabilities. Let LLP be the probability of finding both particles in the left 1/3

of the box as follows:

=3/

0

3/

0

21

2

21 |),(|L L

LL dxdxxxP .

Let LRP be the probability of finding one particle in the left 1/3 of the box and one particle in the

right 1/3 of the box as follows:

=3/

0 3/2

21

2

21 |),(|L L

L

LR dxdxxxP .

Let CCP be the probability of finding both particles in the center 1/3 of the box as follows:

=

3/2

3/

3/2

3/21

2

21 |),(|

L

L

L

L

CC

dxdxxxP .

Calculate the following observables for the ground state and the first excited state for the

distinguishable case. (Show your work and fill in the table)

Points Observable Ground State 1st

Excited State

2E

2 LLP


18/25



2 LRP

2 CCP

Answer: Distinguishable

Observable Ground State = 1 = 1 1st Excited State = 1 = 2E 2E0 05E

LLP ( ) 0.03824

3

3

12

2

1

=

LP ( )( ) 0.07868

3

3

1

4

3

3

121

+

=

LL PP

LRP ( )( ) 0.03824

3

3

12

11

=

RL PP ( )( ) 0.07868

3

3

1

4

3

3

121

+

=

RL PP

CCP ( ) 0.37092

3

3

12

2

1

+=

CP ( )( ) 0.1191

4

3

3

1

2

3

3

121

+=

CC PP

Solution: Let me first work on some integrals that I will need

4

3

3

1

2

)3/2sin(

3

1

4

)3/2sin(

6

2

4

)2sin(

2

2)(sin

2)/(sin

2 3/

0

3/

0

23/

0

2

1

==

=

=

==

yyL

Ldxy

L

LdxLx

LP

LL

4

3

3

2

2

)3/4sin(

3

2

4

)3/4sin(

6

22

4

)2sin(

2

2)(sin

2)/(sin

23/2

0

3/2

0

2

3/2

0

2

1

+==

=

=

==

yyL

Ldxy

L

LdxLx

LP

L

T

2

3

3

1111 +==LTC PPP

L

LL

L

R PdxLxL

dxLxL

P 1

3/

0

2

3/2

2

1 )/(sin2

)/(sin2

===

Note that 1111 =++RCL PPP .

8

3

3

1

4

)3/4sin(

3

1

4

)3/4sin(

3

1

4

)2sin(

22

2)(sin

2

2)/2(sin

23/2

0

3/2

0

2

3/

0

2

2

+==

=

=

== yyL

Ldxy

L

LdxLx

LP

L

L

8

3

3

2

4

)3/8sin(

3

2

4

)3/8sin(

6

41

4

)2sin(

22

2)(sin

2

2)/2(sin

23/4

0

3/4

0

2

3/2

0

2

2

==

=

=

== yyL

Ldxy

L

LdxLx

LP

L

T


19/25



4

3

3

1222 ==LTC PPP

L

LL

L

R PdxLxL

dxLxL

P 2

3/

0

2

3/2

2

2 )/2(sin2

)/2(sin2

===

Note that 1222 =++ RCL PPP .

3

1

23

2

4

)2sin(

23

2)(sin

3

2)/3(sin

2

00

2

3/

0

2

3 =

=

=

==

yyL

Ldxy

L

LdxLx

LP

L

L

( )3

2

3

2

4

)2sin(

23

2)(sin

3

2)/3(sin

22

0

2

0

2

3/2

0

2

3 =

=

=

==

yyL

Ldxy

L

LdxLx

LP

L

T

3

1333 ==LTC PPP

L

LL

L

R PdxLxL

dxLxL

P 3

3/

0

2

3/2

2

3 )/3(sin2

)/3(sin2

===

Note that 1333 =++RCL PPP . These results are summarized in the following table.

n=1 n=2 n=3

L

nP 1955.04

3

3

1

4022.0

8

3

3

1+

3

1

C

nP 6090.02

3

3

1+

1955.0

4

3

3

1

3

1

R

nP 1955.04

3

3

1

4022.0

8

3

3

1+

3

1

sum 1 1 1

We will also need the following integrals.

( )

( ) ( )

2

3)sin(

1)sin(

3

1)/cos(

1)/3cos(

1

)/cos()/3cos(1

)/2sin()/sin(2

3/

00

3/

0

3/

0

3/

0

3/

0

12

==

===

yydxLxL

dxLxL

dxLxLxL

dxLxLxL

I

LL

LL

L

( )

( ) ( ) 0)sin(1

)sin(3

1)/cos(

1)/3cos(

1

)/cos()/3cos(1

)/2sin()/sin(2

3/2

3/

23/2

3/

3/2

3/

3/2

3/

3/2

3/

12

==

===

yydxLxL

dxLxL

dxLxLxL

dxLxLxL

I

L

L

L

L

L

L

L

L

C

( )

( ) ( )

2

3)sin(

1)sin(

3

1)/cos(

1)/3cos(

1

)/cos()/3cos(1

)/2sin()/sin(2

3/2

3

2

3/23/2

3/23/2

12

+==

===

yydxLxL

dxLxL

dxLxLxL

dxLxLxL

I

L

L

L

L

L

L

L

L

R


20/25



( )

( ) ( )

8

33

4

3

8

3

)sin(21)sin(

41)/2cos(1)/4cos(1

)/2cos()/4cos(1

)/3sin()/sin(2

3/20

3/40

3/

0

3/

0

3/

0

3/

0

13

==

=

===

yydxLxL

dxLxL

dxLxLxL

dxLxLxL

I

LL

LL

L

( )

( ) ( )

4

33

2

3

4

3

2

3

2

3

2

1

2

3

2

3

4

1

)sin(2

1)sin(

4

1)/2cos(

1)/4cos(

1

)/2cos()/4cos(1

)/3sin()/sin(2

3/4

3/2

3/8

3/4

3/2

3/

3/2

3/

3/2

3/

3/2

3/

13

=+=

+=

=

===

yydxLxL

dxLxL

dxLxLxL

dxLxLxL

I

L

L

L

L

L

L

L

L

C

( )

( ) ( )

8

33

4

3

8

3

)sin(2

1)sin(

4

1)/2cos(

1)/4cos(

1

)/2cos()/4cos(1

)/3sin()/sin(2

2

3/4

4

3/8

3/23/2

3/23/2

13

==

=

===

yydxLxL

dxLxL

dxLxLxL

dxLxLxL

I

L

L

L

L

L

L

L

L

R

The results are summerized in the following table.

12 13

L 2757.0

2

3

2067.0

8

33

C 0 4135.0

4

33+

R 2757.0

2

3++

2067.0

8

33

sum 0 0

Now for distinguishable particles we have)/sin()/sin(

2)()(),( 212121 LxLx

Lxxxx ==

)/(sin)/(sin4

|),(|),( 22

1

2

2

2

2121 LxLxL

xxxxD ==

The ground state is the state with = 1 and = 1, Thus, 00 2EED = and


21/25



( )2

2

1

3/

0

22

2

3/

0

11

2

3/

0

3/

0

212

2

1

2

211

4

3

3

1)/(sin

2)/(sin

2

)/(sin)/(sin4

)(

==

=

L

LL

L L

D

LL

PdxLxL

dxLxL

dxdxLxLxL

P

and

( )2

2

1

3/2

3/

22

2

3/2

3/

11

2

3/2

3/

3/2

3/

212

2

1

2

211

2

3

3

1)/(sin

2)/(sin

2

)/(sin)/(sin4

)(

+==

=

C

L

L

L

L

L

L

L

L

D

CC

PdxLxL

dxLxL

dxdxLxLxL

P

and

( )2

2

111

3/2

22

23/2

0

11

2

3/2

0 3/2

212

2

1

2

211

4

3

3

1)/(sin

2)/(sin

2

)/(sin)/(sin4

)(

===

=

LRLL

L

L

L L

L

D

LR

PPPdxLxL

dxLxL

dxdxLxLxL

P

The 1st

excited state has = 1 and = 2, Thus, 01 5EED = and

( )( )

+

==

=

8

3

3

1

4

3

3

1)/2(sin

2)/(sin

2

)/2(sin)/(sin4

)(

21

3/

0

22

2

3/

0

11

2

3/

0

3/

0

212

2

1

2

212

LL

LL

L L

D

LL

PPdxLxL

dxLxL

dxdxLxLxL

P

and

( )( )

+==

=

4

3

3

1

2

3

3

1)/2(sin

2)/(sin

2

)/2(sin)/(sin4

)(

21

3/2

3/

22

2

3/2

3/

11

2

3/2

3/

3/2

3/

212

2

1

2

212

CC

L

L

L

L

L

L

L

L

D

CC

PPdxLxL

dxLxL

dxdxLxLxL

P

and

+

==

=

83

31

43

31)/2(sin2)/(sin2

)/2(sin)/(sin4

)(

21

3/2

222

3/2

0

112

3/2

0 3/2

212

2

1

2

212

RL

L

L

L

L L

L

D

LR

PPdxLxL

dxLxL

dxdxLxLxL

P

(B) For two identical bosons (i.e. particles with integral spins in the same spin state) we must use

the symmetric wavefunction

( )),(),(2

1),( 122121 xxxxxx

S

+= ( symmetric under 12)


22/25



( )),(),(2

1),( 122121 xxxxxx

S

+= ( = symmetric under 12)

Calculate the following observables for the ground state and the first excited state for theidentical boson case. (Show your work and fill in the table)


Excited State

2 E

2 LLP

2 LRP

2 CCP

Answer: Bosons

Observable Ground State = 1 = 1 1st Excited State = 1 = 2E 2E0 05E

LLP ( ) 0.038243312

2

1

=

LP ( )( ) ( )0.15460.07600.0786

2

3

8

3

3

1

4

3

3

12

2

1221

+

+

+

=+ LLL

IPP

LRP ( )( ) 0.03824

3

3

12

11

=

RLPP

( )( ) ( )( )

0.00260.0760-0.0786

2

3

8

3

3

1

4

3

3

12

121221

+

=+

RLRLIIPP

CCP ( ) 0.37092

3

3

12

2

1

+=

CP ( )( ) ( )

0.1191

4

3

3

1

2

3

3

121221

+=+

CCC IPP

Solution: In this case we have

[ ])/sin()/sin()/sin()/sin(2

),( 212121 LxLxLxLxL

xxS += ()

)/sin()/sin(2

),( 2121 LxLxL

xxS =

and

),(),(),( 21int

2121 xxxxxxclassicalBE

+= ()

where

( )),(),(Re),( 212121int xxxxxx ()

and),(),( 2121 xxxx

classicalBE

= .where

( ))/(sin)/(sin)/(sin)/(sin2),( 22122212221 LxLxLxLxLxxclassical +=

( ))/sin()/sin()/sin()/sin(4

),(2211221

int LxLxLxLxL

xx =


23/25



and

)),(),((),( 122121

21 xxxxxxDDclassical

+= .

The ground state is the state with = 1 and = 1, Thus, 00 2EEBE = , and

D

LL

BE

LL PP )()( 1111 = , DCC

BE

CC PP )()( 1111 = , DLR

BE

LR PP )()( 1111 = .

The 1st

excited state has = 1 and = 2, Thus, 01 5EED = and

( )( ) ( )2

2

1221

3/

0

3/

0

2121

int

12

3/

0

3/

0

212112

3/

0

3/

0

21211212

2

3

8

3

3

1

4

3

3

1

),(),(),()(

+

+

=+=

+==

LLL

L LL L

classical

L L

BE

BE

LL

IPP

dxdxxxdxdxxxdxdxxxP

where I used

( )2123/

0

222

3/

0

111

3/

0

3/

0

2122112

3/

0

3/

0

2121

int

12

)/2sin()/sin(2

)/2sin()/sin(2

)/2sin()/sin()/2sin()/sin(4),(

L

LL

L LL L

IdxLxLxL

dxLxLxL

dxdxLxLxLxLxL

dxdxxx

==

=

Also,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )2

12122112121221

2121

3/

0 3/2

2121

int

12

3/

0 3/2

212112

3/

0 3/2

21211212

2

3

8

3

3

1

4

3

3

1

),(),(),()(

+

=+=++=

+==

RLRLRLRLRL

L L

L

L L

L

classical

L L

L

BE

BE

LR

IIPPIIPPPP

dxdxxxdxdxxxdxdxxxP

where I used

( )( )RLL

L

L

L L

L

L L

L

IIdxLxLxL

dxLxLxL

dxdxLxLxLxLxL

dxdxxx

1212

3/2

222

3/

0

111

3/

0 3/2

2122112

3/

0 3/2

2121

int

12

)/2sin()/sin(2

)/2sin()/sin(2

)/2sin()/sin()/2sin()/sin(4

),(

==

=

Also,

( )( ) ( )( ) ( ) ( )( ) ( )

+=+=++=

+==

4

3

3

1

2

3

3

1

),(),(),()(

2

1221

2

121221

2121

3/2

3/

3/2

3/

2121

int

12

3/2

3/

3/2

3/

212112

3/2

3/

3/2

3/

21211212

CCCCCCCC

L

L

L

L

L

L

L

L

classical

L

L

L

L

BE

BE

CC

IPPIPPPP

dxdxxxdxdxxxdxdxxxP

where I used


24/25



( )2123/2

3/

222

3/2

3/

111

3/

3/

3/2

3/

2122112

3/2

3/

3/2

3/

2121

int

12

)/2sin()/sin(2

)/2sin()/sin(2

)/2sin()/sin()/2sin()/sin(4

),(

C

L

L

L

L

L

L

L

L

L

L

L

L

IdxLxLxL

dxLxLxL

dxdxLxLxLxLxL

dxdxxx

==

=

(C) For two identical fermions (i.e. particles with half-integral spins in the same spin state) wemust use the symmetric wavefunction

( )),(),(2

1),( 122121 xxxxxx

A

= (antisymmetric under 12)

Calculate the following observables for the ground state and the first excited state for the

identical fermion case. (Show your work and fill in the table)


Excited State

2E

2 LLP

2 LRP

2 CCP

Answer: Fermions

Observable Ground State = 1 = 2 1st Excited State = 1 = 3E 05E 010E

LLP ( )( ) ( )

0.00260.0760-0.0786

2

3

8

3

3

1

4

3

3

12

2

1221

+

=

LLL IPP ( )( ) ( )

0.02240.0427-0.0652

8

33

3

1

4

3

3

12

2

1331

=

LLL IPP

LRP ( )( ) ( )( )

0.15460.07600.0786

2

3

8

3

3

1

4

3

3

12

121221

+

+

+

=

RLRL IIPP ( )( ) ( )( )

0.02240.0427-0.0652

8

33

3

1

4

3

3

12

131331

=

RLRL IIPP

CCP ( )( ) ( )

0.1191

4

3

3

1

2

3

3

121221

+=+

CCC IPP ( )( ) ( )

0.03200.1710-0.2030

8

33

3

1

2

3

3

12

2

1331

+=

CCC IPP

Solution: In this case we have

[ ])/sin()/sin()/sin()/sin(2),( 212121 LxLxLxLxL

xxA =

and

),(),(),( 21int

2121 xxxxxxclassicalFD

=

where

( )),(),(Re),( 212121int xxxxxx ().


25/25


The ground has = 1 and = 2, Thus, 00 5EEFD = and

( )( ) ( )2

2

1221

3/

0

3/

0

2121

int

12

3/

0

3/

0

212112

3/

0

3/

0

21211212

2

3

8

3

3

1

4

3

3

1

),(),(),()(

+

==

==

LLL

L LL L

classical

L L

FD

FD

LL

IPP

dxdxxxdxdxxxdxdxxxP

Also,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )2

12122112121221

2121

3/

0 3/2

2121

int

12

3/

0 3/2

212112

3/

0 3/2

21211212

2

3

8

3

3

1

4

3

3

1

),(),(),()(

+

+

==+=

==

RLRLRLRLRL

L L

L

L L

L

classical

L L

L

FD

FD

LR

IIPPIIPPPP

dxdxxxdxdxxxdxdxxxP

Also,

( )( ) ( )( ) ( ) ( )( ) ( )

+=+=+=

==

4

3

3

1

2

3

3

1

),(),(),()(

2

1221

2

121221

2121

3/2

3/

3/2

3/

2121int12

3/2

3/

3/2

3/

212112

3/2

3/

3/2

3/

21211212

CCCCCCCC

L

L

L

L

L

L

L

L

classical

L

L

L

L

FDFD

CC

IPPIPPPP

dxdxxxdxdxxxdxdxxxP

The 1st

excited state has = 1 and = 3, Thus, 00 10EEFD = and

( )( ) ( )

2

21331

3/

0

3/

0

2121

int

13

3/

0

3/

0

212113

3/

0

3/

0

21211313

833

31

43

31

),(),(),()(

==

==

LLL

L LL L

classical

L L

FD

FD

LL

IPP

dxdxxxdxdxxxdxdxxxP

Also,

( )( ) ( )( ) ( )( ) ( )( ) ( )( )2

13133113131321

3121

3/

0 3/2

2121

int

13

3/

0 3/2

212113

3/

0 3/2

21211313

8

33

3

1

4

3

3

1

),(),(),()(

==+=

==

RLRLRLRLRL

L L

L

L L

L

classical

L L

L

FD

FD

LR

IIPPIIPPPP

dxdxxxdxdxxxdxdxxxP

Also,

( )( ) ( )( ) ( ) ( )( ) ( )2

2

1331

2

131321

3121

3/2

3/

3/2

3/

2121

int

13

3/2

3/

3/2

3/

212113

3/2

3/

3/2

3/

21211313

8

33

3

1

2

3

3

1

),(),(),()(

+==+=

==

CCCCCCCC

L

L

L

L

L

L

L

L

classical

L

L

L

L

FD

FD

CC

IPPIPPPP

dxdxxxdxdxxxdxdxxxP

4604 exam3 solutions fa06

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