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PHY3063 Spring 2006 R. D. Field

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PHY 3063 Final Exam Name_________________

Problem 1 (35 points): Consider an particle with mass m

confined within an infinite square well defined by

V(x) = 0 for0 < x < L,

V(x) = + otherwise.

(a) (5 points): Using Schrdingers equation calculate the

allowed stationary state eigenfunctions n(x), where the completewavefunctions are given by

h/)(),(

tiE

nnnextx

= . Normalizethe eigenfunctions so that the probability of finding the particle somewhere in the box is one.

(b) (5 points): Show that the allowed energy levels of the system are, En = E0 n2,

where )2/(222

0 mLE h= is the ground state energy and n = 1, 2, 3, . Why is n = 0 excluded asa possible energy level?

(c) (10 points): Consider the operator, O = (x)op(px)op (i.e. the product of the position operatortimes the momentum operator). Is the operatorO hermitian? Calculate the expectation value of

the operatorO for the nth

stationary state (i.e. calculate ).

(d) (15 points): Suppose the particle in this infinite square well has an initial wave function at

t = 0 given by

)/(sin)0,(2 LxAx = .

What is the normalization A? If you measure the energy of this particle, what is the probabilitythat you will measure the ground state energy E0?

V = +infinityV = +infinity

Infinite Square Well

0 L x

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PHY 3063 Final Exam Name_________________Problem 1 Scratch Paper

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Problem 3 (30 points): Suppose we have two vector operators

opJ)( 1r

and opJ )( 2r

with 0])(,)[( 21 =opop JJrr

and each of the vectors obey the same SU(2) lie algebra

opkijkopjopi JiJJ )(])(,)[( 111 = and opkijkopjopi JiJJ )(])(,)[( 222 = .

The states |j1m1> 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.(a) (5 points):Show that

opzopzopopopopopop

opopopopopopopopop

JJJJJJJJ

JJJJJJJJJJJ

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

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

212121

2

2

2

1

21

2

2

2

12121

2

++++=

++=++==++

rrrrrrrr

(b) (5 points): Evaluate the following in SU(2).3 2 = 4 3 = 5 3 =5 4 = 2 3 4 =

(c) (20 points):Now consider the case where j1 = 1 and j2 = (i.e. 3 2) and define the states as

follows:

>=>

>=>

>=>

11||

10||

11||

111

110

111

Y

Y

Y

and>=>

>=>

21

21

2

21

21

2

||

||

Now consider the two superposition states

21102111 ||3

2||

3

1| >>+>>>+ YY and 21102111 ||

3

1||

3

2| >>>>> YY .

Calculate the following and express your answer in terms of |>:

(1) >|zJ (2) >|2

1J (3) >|2

2J

(4) >|2 21 zzJJ (5) >+++ |)( 2121 JJJJ (6) >|2J

Are the states >| eigenstates of the J and Jz and if so what are their eigenvalues?

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PHY 3063 Final Exam Name_________________Problem 3 Scratch Paper

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Final Exam Page 7 of 7 April 25, 2006

Useful Math

Trigonometric Relations:

)sin()sin(cossin2

)cos()cos(sinsin2

)cos()cos(coscos2

sinsincoscos)cos(

sincoscossin)sin(

BABABA

BABABA

BABABA

BABABA

BABABA

++=

+=

++==

=

m

Indefinite Integrals:

4

2cos2sin

8

1

46cos

8

2cos

4

2sin

4cos

4

2sin

2cos

sincoscos

4

2cos2sin

8

1

46sin

8

2cos

4

2sin

4sin

32

4sin

4

2sin

8

3sin

cos3

cossin

4

2sin

2

sin

cossinsin

2

sincossin

2322

22

2

2322

22

4

33

2

2

xxx

xxxdxx

xxxxxdxx

xxxdx

xxxxdxx

xxx

xxxdxx

xxxxxdxx

xxxxdx

xx

xdx

xxxdx

xxxxdxx

xxdxx

+

+=

++=

+=

+=

=

=

+=

=

=

=

=

Definite Integrals:

mn

mn

nxdxmx

nxdxmx

2coscos

2sinsin

0

0

=

=