3063 exam2 solutions sp06
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8/9/2019 3063 Exam2 Solutions Sp06
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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
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PHY3063 Spring 2006 R. D. Field
Exam 2 Solutions Page 2 of 7 March 9, 2006
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].
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PHY3063 Spring 2006 R. D. Field
Exam 2 Solutions Page 3 of 7 March 9, 2006
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 =>
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8/9/2019 3063 Exam2 Solutions Sp06
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PHY3063 Spring 2006 R. D. Field
Exam 2 Solutions Page 4 of 7 March 9, 2006
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
=
=
==
=>==< .
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PHY3063 Spring 2006 R. D. Field
Exam 2 Solutions Page 5 of 7 March 9, 2006
(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 >====
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