3063 exam2 solutions sp06

8/9/2019 3063 Exam2 Solutions Sp06

1/7

PHY3063 Spring 2006 R. D. Field

Exam 2 Solutions Page 1 of 7 March 9, 2006

PHY 3063 Exam 2 Solutions

Problem 1 (25 points): Consider an atom consisting of a proton and a muon. A proton has an

electric charge +e and rest mass energy Mpc2 940 MeV. A muon is an elementary particle with

charge e similar to an electron but with a mass that is 207 times the mass of an electron (mec2

0.511 MeV). Use Bohrs model of an atom to calculate the following:

(a) (8 points) The radius of the first Bohr orbit (i.e. the ground state) of the muonic atom (in

fm).

Answer: 284 fm

Solution: Bohrs model tell us that

fmAArm

mr

m

Zr

e

ee 2841084.2)529.0(186

1

186

1 3001 ===

oo

,

where I used

e

pe

e

pe

pem

Mm

m

Mm

Mm

Mm

mM186

/2071

1207

207

207

+

=

+

=

+

andoD

Ar e 529.00 =

.

(b) (8 points) The ground state energy of the muonic atom (in keV).

Answer: -2.53 keV

Solution: Bohrs model tell us that

keVeVeVEm

ZEe

53.2530,2)6.13(18602

1 ==== ,

where

eVcmE e 6.13)(2

1 220 == and

137

12 =c

Ke

h .

(c) (9 points) The ground state energy of the hydrogen atom can be estimated using the

uncertainty relation. For a hydrogen atom in the ground state the uncertaintyr in the positionof the electron is approximately the 1

stBohr radius, r0, consequently the elecrrons momentum

pr is uncertain by about r/h . Compute the ground state energy of the hydrogen atom (in eV),

r

Ke

m

pE

e

r

22

2= ,

by setting r = r0 and 0/ rpp rr h== , where r0 is the Bohr radius of the ground state.

Answer: -13.6 eVSolution: We see that

eVcmcmcm

cm

Ke

cmmr

Ke

rmr

Ke

m

pE

eee

eeeee

r

6.13)(2

1)()(

2

1

)/())/((222

222222

2

2

22

0

2

2

0

222

===

==

hh

hh


2/7



Problem 2 (25 points): The parity operator, Pop, is defined by

)()( xxPop = .

(a) (10 points) Prove that the parity operator is hermitian. Remember hermitian operators have

the property that Oop = Oop where

[ ]

= dxxOxdxxxO opop )()()()( 1*

21

*

2 .

Solution: We see that

[ ] +

+

= dxxxdxxxPop )()()()( 121*

2 ,

but

+

+

+

+

=== dxxxdyyydyyydxxx )()()()()()()()( 12121212 ,

where I changed variables with y = -x. Thus,

[ ] +

+

+

+

=== dxxPxdxxxdxxxdxxxP opop )()()()()()()()( 1212121*

2

and hence P

op = Pop which means that Pop is hermitian.

(b) (5 points) Prove that the eigenvalues of the parity operator are real.

Solution: Let Pop|> = |> where is a constant and |> is the normalized eigenket (i.e. = 1). We see that

= = = ( )*= ( )

*=

and therefore is real since it is equal to its complex conjugate.

(c) (5 points) Prove that the eigenvalues of the parity operator are +1 and -1.Solution: Let Pop|> = |> where is a real constant and |> is the normalized eigenket. Wesee that

= = = 2,

where I used Pop Pop = 1. Hence 2= 1, which means that = -1 or = +1.

(d) (5 points) Prove that if(x) is an eigenstate of both Pop and the Hamiltonian operatorHopthen Pop and Hop commute (i.e.[Pop,Hop] = 0).

Solution: Let Hop|> = h|> and Pop|> = |>. We know that h and are real constants. Wesee that

[Pop,Hop]|>= (PopHop- HopPop)|> = (h- h)|> = 0

and hence [Pop,Hop].


3/7



Problem 3 (50 points): Consider an electron with mass me

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 electron somewhere in the box is one.

Answer: )/sin(2

)( LxnL

xn =

Solution: For the region outside of 0 < x < L 0)( =x and inside the region

)()(

2 2

22

xExd

xd

me

=

hor )(

)( 22

2

xkxd

xd

= with

em

kE

2

22h

=

The most general solution is of the form

)cos()sin()( kxBkxAx += .The boundary condition at x = 0 gives 0)0( ==B and the boundary condition at x = L gives

0)sin()( == kLAL which implies that nkL = with n = 1, 2, 3, Thus,

)/sin()( LxnAxn = with 2222

2mL

nEn

h= . The normalization is arrived at by requiring that

24

)2sin(

2)(sin)/(sin1)()(

2

0

2

0

22

0

22 LAyy

n

LAdyy

n

LAdxLxnAdxxx

nnL

nn =

====

+

Thus, LA /2= . These states are called stationary because the probability density and all theexpectation values are independent of time.

(b) (5 points) Show that the wavefunctions n(x,t) correspond to states with definite energy (i.e.show that E = 0).Solution: We see that

nnnnn

nnopnn EdxtxtxEdxt

txtxidxtxHtxE ==

==>< +

+

+

),(),(),(

),(),(),( h

and

22

2

2222 ),(),(

),(),(),(),( nnnn

nnnopnn EdxtxtxEdx

t

txtxdxtxHtxE ==

==>< +

+

+

h .

Hence

0)()( 22 =>


4/7



Answer: )(2

2

222

cmL

nE e

en

=D

, MeVcmE e 52.2)(2

22

1 =

Solution: From part (a) we see that k = n/L and hence

)(

222

2

222

2

22222

cm

L

n

mL

n

m

kE e

en

===Dhh

.

The ground state corresponds to n = 1 and hence

MeVMeVcmcmL

E eLee

e52.2)511.0(

2)(

2)(

2

22

22

22

1 =

= =

D

D.

(d) (5 points) Show that the states, )(xn , form an orthonormal set. Namely, show that

mnnm dxxx =+

)()( .

Solution:

mn

L

nm dymymydxLxnLxmL

dxxx

=== +

00

)sin()sin(2

)/sin()/sin(2

)()(

(e) (5 points) Calculate for nth

state stationary state.

Answer:2

Lx n =><

Solution:

28)2cos(

4)2sin(

4)(2

)(sin2

)/(sin2

)())((

0

2

2

0

2

2

0

2

Lyyyyn

L

dyyyn

L

LdxLxnx

Ldxxxxx

n

nL

nopn

=

=

==>=<

+

(f) (5 points) Calculate for nth

state stationary state.

Answer: 0=>< nxp Solution:

02

)(sin2

)cos()sin(2

)/cos()/sin(2

)()()())((

0

2

00

=

=

==

=>==< .


5/7



(g) (10 points) Suppose the electron in this infinite square well has an initial wave function at t =

0 which is an equal mixture of the first two stationary states:

[ ])()()0,( 21 xxAx += .What is the normalization A? If you measure the energy of this particle, what are the possiblevalues you might get, and what is the probability of getting each of them? What is the

expectation value of the energy for this state?

Answer: 2/1=A , E1 with probability and energy E2 with probability , 2

22

4

5

maE

h>=<

Solution: The normalization is arrived at by requiring that

[ ][ ]

( ) 2221221112

2121

2

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

)()()()(1)0,()0,(

AdxxxxxxxxxA

dxxxxxAdxxx

=+++=

++==

+

+

+

and 2/1=A . Thus,hh /

22/

1121 )()(),( tiEtiE excexctx += ,

where c1 = c2 = 2/1 which means that you get energy E1 with probability and energy E2 with

probability . The expectation value of E is

2

22

2

22

2

22

221

121

4

5

2

4

22

1

mamamaEEE

hhh =

+=+>=<

(h) (10 points) Suppose as in part (d) the electron in this infinite square well has an initial wave

function at t = 0 which is an equal mixture of the first two stationary states:

[ ])()()0,( 21 xxAx += .What is the probability density 2|),(|),( txtx = for this state. Does it depend on time? What is

and for this state. Do they depend on time?

Answer:

+>=< )cos(

9

321

2 2t

Lx

, )sin(

3

8t

Lpx >====

3063 exam2 solutions sp06

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