Embed Size (px)
Citation preview
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 1 of 14
PHY 4604 Problem Set #2 Solutions Problem 1 (10 points): The time-dependent Schrödinger’s equation is
),()(),(2
),(2
22
txxVx
txmt
txi Ψ+∂Ψ∂
−=∂
Ψ∂ hh ,
where we take the potential energy, V, to be only a function of x (not of time). We look for solutions of the form
)()(),( txtx Φ=Ψ ψ . The time-dependent equation separates into two equations (separation constant E).
)()()()(2 2
22
xExxVxdxd
mψψψ
=+−h and )()( tE
dttdi Φ=
Φh .
Thus, h/)( iEtet −=Φ . (a) (3 points) For normalizable solutions, show that the separation constant E must be real. Solution: Following Griffith’s problem 2.1 we set Γ+= iEE r , where Er and Γ are real. Thus,
hhh /// )()(),( tiEtiEt reexextx −Γ− ==Ψ ψψ and
∫∫+∞
∞−
∗Γ+∞
∞−
∗ ==ΨΨ dxxxedxtxtx t )()(1),(),( /2 ψψh .
The second term is independent of time and hence Γ = 0 and E is real. One can prove E is real by noting that if V(x) is real then
( )∫∫+∞
∞−
∗+∞
∞−
∗ = dxxxHdxxHx opop )()()()( ψψψψ ,
where
)(22
)(2
222
xVxd
dm
Vm
pH op
opxop +−=+=
h .
The time-independent Schrödinger equation is )()( xExHop ψψ = which yields
∫∫+∞
∞−
∗+∞
∞−
∗ = dxxxEdxxHx op )()()()( ψψψψ
( ) ( ) ∫∫∫+∞
∞−
∗∗+∞
∞−
∗+∞
∞−
∗ == dxxxEdxxxEdxxxHop )()()()()()( ψψψψψψ
Thus, E = E* which means that E is real. (b) (3 points) Show that E corresponds to the total energy and that )()(),( txtx Φ=Ψ ψ corresponds to a state with definite energy (i.e. ΔE = 0). Solution:
EdxxxEdxxHxHE opop ==>=>=<< ∫∫+∞
∞−
∗+∞
∞−
∗ )()()()( ψψψψ
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 2 of 14
22
222
)()(
)())()(()())((
EdxxxE
dxxHHxdxxHxHE opopopop
==
=>=>=<<
∫
∫∫∞+
∞−
∗
+∞
∞−
∗+∞
∞−
∗
ψψ
ψψψψ
Thus, (ΔE)2 = <E2> - <E>2 = 0. Also note that since h/)(),( iEtextx −=Ψ ψ we have
EdxxxEdxtxt
itxdxtxEtxEE opop ==Ψ⎟⎠⎞
⎜⎝⎛
∂∂
Ψ=ΨΨ>=>=<< ∫∫∫+∞
∞−
∗+∞
∞−
∗+∞
∞−
∗ )()(),(),(),(),( ψψh .
(c) (4 points) Show that the time-independent wave function, )(xψ , can always be taken to be a real function. (Hint: take V to be real and see Griffith’s problem 2.1) Solution: If )(xψ satisfies )()( xExHop ψψ = we can take the complex conjugate of both sides
and get )()( xExHop∗∗∗∗ = ψψ , but Hop
* = Hop and E* = E and hence )()( xExHop∗∗ = ψψ . Thus,
( ) ( ))()()()( xxExxHop∗∗ +=+ ψψψψ and we see that ( ))()( xx ∗+ψψ is a solution to the time-
independent equation and ( ))()( xx ∗+ψψ is real. Problem 2 (50 points): Consider an infinite square well defined by V(x) = 0 for 0 < x < a, and V(x) = ∞ otherwise. (a) (4 points) Show that the stationary states are given by
h/)(),( tiEnn
nextx −=Ψ ψ with )/sin(2)( axna
xn πψ =
and 2
222
2manEn
hπ= , and n is a positive integer.
Solution: For the region outside of 0 < x < a 0)( =xψ and inside the region
)()(2 2
22
xExdxd
mψψ
=−h or )()( 2
2
2
xkxdxd ψψ
−= with mkE
2
22h=
The most general solution is of the form )cos()sin()( kxBkxAx +=ψ .
The boundary condition at x = 0 gives 0)0( == Bψ and the boundary condition at x = a gives 0)sin()( == kaAaψ which implies that πnka = with n = 1, 2, 3,… Thus,
)/sin()( axnAxn πψ = with 2
222
2manEn
hπ= . The normalization is arrived at by requiring that
24)2sin(
2)(sin)/(sin1)()(
2
0
2
0
22
0
22 aAyynaAdyy
naAdxaxnAdxxx
nna
nn =⎟⎠⎞
⎜⎝⎛ −==== ∫∫∫
+∞
∞−
∗ππ
πππψψ
Thus, aA /2= . These states are called stationary because the probability density and all the expectation values are independent of time. (b) (2 points) Show that the states, )(xnψ , form an orthonormal set. Namely, show that
mnnm dxxx δψψ =∫+∞
∞−
∗ )()( .
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 3 of 14
Solution:
mn
a
nm dymymydxaxnaxma
dxxx δπ
ππψψπ
=== ∫∫∫+∞
∞−
∗
00
)sin()sin(2)/sin()/sin(2)()(
(c) (8 points) Calculate <x>, <x2>, <px>, and <px2> for the nth stationary state.
Answer: 2ax n=>< , ⎟⎟
⎠
⎞⎜⎜⎝
⎛−=>< 2
22
)(21
31
πnax n , 0=>< nxp , 2
2222
anp nx
hπ=><
Solution:
28)2cos(
4)2sin(
4)(2
)(sin2)/(sin2)())((
0
2
2
0
22
0
2
ayyyyn
a
dyyyna
adxaxnx
adxxxxx
n
na
nopnn
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
⎟⎠⎞
⎜⎝⎛===>< ∫∫∫
+∞
∞−
∗
π
π
π
ππψψ
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
⎟⎠⎞
⎜⎝⎛===>< ∫∫∫
+∞
∞−
∗
22
3
3
2
0
23
3
2
0
223
0
2222
)(21
31
46)(
)(2
4)2cos()2sin(
81
46)(2
)(sin2)/(sin2)())((
πππ
ππ
ππψψ
π
π
nann
nayyyyy
na
dyyyna
adxaxnx
adxxxxx
n
na
nopnn
There are two ways to calculate <px>n. The easiest is to take the time derivative of <x> as follows:
0=><
=><dtxdmp n
nx .
The more difficult way is to evaluate
02
)(sin2
)cos()sin(2)/cos()/sin(2
)()()())((
0
2
00
=⎟⎟⎠
⎞⎜⎜⎝
⎛−=
−=−=
⎟⎠⎞
⎜⎝⎛
∂∂
−==><
∫∫
∫∫+∞
∞−
∗+∞
∞−
∗
π
π
πππ
ψψψψ
n
na
nnnopxnnx
ya
i
dyyya
idxaxnaxna
na
i
dxxx
ixdxxpxp
h
hh
h
There are also two ways to calculate <px2>n . The easy way is to use the fact that since in this
problem V = 0 we have
mp
H opxop 2
)( 2
= and nnnop EH ψψ = so that nnnopx mEp ψψ 2)( 2 = and hence
2
222
2
22222
22)()(2)())((
an
manmdxxxmEdxxpxp nnnnopxnnx
hh ππψψψψ =⎟⎟⎠
⎞⎜⎜⎝
⎛===>< ∫∫
+∞
∞−
∗+∞
∞−
∗ .
The more difficult way is to evaluate
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 4 of 14
2
222
0
2
0
22
0
22
2
2
2222
4)2sin(
22
)(sin2)/(sin2
)()()())((
anyy
an
a
dyya
na
dxaxna
na
dxxx
xdxxpxp
n
na
nnnopxnnx
hh
hh
h
ππ
πππ
ψψψψ
π
π
=⎟⎠⎞
⎜⎝⎛ −⎟⎠⎞
⎜⎝⎛=
⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
−==><
∫∫
∫∫+∞
∞−
∗+∞
∞−
∗
(d) (6 points) Compute Δx = σx and Δpx = xpσ for the nth stationary state. Is the product ΔxΔpx
consistent with the uncertainty principle? Which state comes closest to the uncertainty limit?
Answer: 2)(2
31
2)(
πnax n −=Δ ,
anp nxhπ
=Δ )( , 23
)(2
)(2
−=ΔΔπnpx nx
h
Solution: For From (c) we see that
22
2
2222
)(2
31
2)(21
121
4)(21
31)(
πππ na
naa
naxxx nnn −=−=−⎟⎟
⎠
⎞⎜⎜⎝
⎛−=><−><=Δ
an
anppp nxnxnx
hh ππ==><−><=Δ 2
22222)(
and
22
3)(
2)(2
31
2)(
2
2hhh
≥−=−=ΔΔπ
ππ n
na
anpx nx .
The smallest ΔxΔpx is the ground state n = 1 where
2)136.1(2
32)(
2
1hh
=−=ΔΔ =π
nxpx
(e) (8 points) Suppose that particle in this infinite square well has an initial wave function which is an even mixture of the first two stationary states:
[ ])()()0,( 21 xxAx ψψ +=Ψ . What is the normalization A? If you measure the energy of this particle, what are the possible values 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 ½ and energy E2 with probability ½, 2
22
45
maE hπ>=<
Solution: The normalization is arrived at by requiring that
[ ][ ]
( ) 222122111
2
21212
2)()()()()()()()(
)()()()(1)0,()0,(
AdxxxxxxxxxA
dxxxxxAdxxx
=+++=
++==ΨΨ
∫
∫∫∞+
∞−
∗∗∗∗
+∞
∞−
∗∗+∞
∞−
∗
ψψψψψψψψ
ψψψψ
and 2/1=A . Thus, hh /
22/
1121 )()(),( tiEtiE excexctx −− +=Ψ ψψ ,
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 5 of 14
where c1 = c2 = 2/1 which means that you get energy E1 with probability ½ and energy E2 with probability ½. The expectation value of E is
2
22
2
22
2
22
221
121
45
24
221
mamamaEEE hhh πππ
=⎟⎟⎠
⎞⎜⎜⎝
⎛+=+>=<
(f) (10 points) Suppose that particle in this infinite square well has the initial wave function
⎩⎨⎧
−=Ψ
)()0,(
xaAAx
x axa
ax≤≤
≤≤2/
2/0
Sketch )0,(xΨ and determine the normalization A. Find ),( txΨ . What is the probability that a measurement of the energy of this state will yield the ground state energy E1? What is the expectation value of the energy for the state ),( txΨ ?
Answer: 3/32 aA = , ∑∞
=
−−−=ΨL
h
,5,3,1
/2
2/)1(2 )/sin(1)1(264),(
n
tiEn neaxnna
tx ππ
,
the probability of measuring E1 is 9855.09641 ≈=
πP , 2
26ma
E h>=< .
Ψ(x,0)
0
1
2
0.0 0.5 1.0
x/a
Solution: First we normalize the distribution as follows:
232/
0
22
2/
222/
0
22
122
)()0,()0,()(1
AadxxA
dxxaAdxxAdxxxdxx
a
a
a
a
==
−+=ΨΨ==
∫
∫∫∫∫+∞
∞−
∗+∞
∞−
ρ
Thus, 33 /32/12 aaA == . We know that
∑∞
=
−=Ψ1
/)(),(n
tiEnn
nexctx hψ with ∫+∞
∞−
∗ Ψ= dxxxc nn )0,()(ψ .
Thus,
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 6 of 14
[ ] [ ]
[ ] ( ))2/cos()2/()2/sin(2)1(1
)sin(2)1(1)/sin(2)1(1
)/sin(2)/sin(2
)/sin()(2)/sin(2
2
2/
0
22/
0
2/
0
2/
0
2/
2/
0
ππππ
ππ
πππ
ππ
π
nnnan
aA
dyyyan
aAdxaxnxa
A
dxaxnnxa
Adxaxnxa
A
dxaxnxaa
Adxaxnxa
Ac
n
nn
an
aa
a
a
a
n
−⎟⎠⎞
⎜⎝⎛−−=
⎟⎠⎞
⎜⎝⎛−−=−−=
=−+=
−+=
∫∫
∫∫
∫∫
For even n cn = 0 and for odd n we have
222/)1(
2
2/32/)1( 64)1(2322)1(
ππ nana
ac nn
n−− −=⎟
⎠⎞
⎜⎝⎛−=
and
∑∞
=
−−−=ΨL
h
,5,3,1
/2
2/)1(2 )/sin(1)1(264),(
n
tiEn neaxnna
tx ππ
where 2
222
2manEn
hπ= .
The probability of measuring E1 is
9855.096644
2
22
1 ≈=⎟⎟⎠
⎞⎜⎜⎝
⎛==
ππncP .
The expectation value of the energy is
2
2
22
2
222222
2
,5,3,1222
2
,5,3,1
2222
221
2
6)2(4871
51
31
1148148
264
mamamanma
mn
nEcE
n
nnnn
hhL
hh
h
L
L
==⎟⎠⎞
⎜⎝⎛ ++++==
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=>=<
∑
∑∑∞
=
∞
=
∞
=
ξπππ
ππ
where I used
871
51
311
2
222π
=++++ L .
(g) (12 points) Suppose a particle of mass m in this infinite square is in the ground state (i.e. n = 1). Now suppose that the well suddenly expands to twice its original size (the right wall moves from a to 2a) leaving the wave function (momentarily) undisturbed. If the energy of the particle is now measured, what is the most probable value and what is the probability of getting this value? What is the next most probable value, and what is its probability?
Answer: Most probable energy is 2
22
2 2maE hπ
= with probability P2 = ½. The next most probable
energy is 2
22
1 8maE hπ
= , with probability 36025.0932
21 ≈=π
P .
Solution: The new allowed energies are
2
222
)2(2 amnEn
hπ=
and the new wave functions are
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 7 of 14
⎟⎠⎞
⎜⎝⎛= x
an
axn 2
sin22)( πψ with ⎟
⎠⎞
⎜⎝⎛=Ψ x
aax πsin2)0,(
We know that
∑∞
=
−=Ψ1
/)(),(n
tiEnn
nexctx hψ with ∫+∞
∞−
∗ Ψ= dxxxc nn )0,()(ψ .
For n ≠ 2 we get,
( )( )41sin24
11
111
2sin
22
12
12
sin
12
12
sin
22
12
12
sin
12
12
sin
22
12
cos12
cos2
2sin2
sin2
22
21
21
0
00
−+
=⎥⎦
⎤⎢⎣
⎡+
−−⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛ +
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +
−⎟⎠⎞
⎜⎝⎛ −
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛ +
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +
−⎟⎠⎞
⎜⎝⎛ −
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −
=
=⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +−⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −=⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛= ∫∫
nnnn
n
n
n
n
an
axn
an
axn
a
dxaxn
axn
adxx
ax
an
ac
n
a
aa
n
ππ
ππ
π
π
π
π
π
π
π
π
ππππ
For n = 2 we get,
22
4)2sin(
22)(sin2)/(sin2
00
2
0
22 =⎟
⎠⎞
⎜⎝⎛ −=== ∫∫
ππ
πππ yydyydxax
ac
a
.
Thus,
⎪⎩
⎪⎨
⎧
==
===
−
L
L
,8,6,4,5,3,1
2
0222 )4(
3221
2
nn
ncP
nnn π
The most probable energy is 2
22
2 2maE hπ
= (same as before the change) with probability P2 = ½.
The next most probable energy is 2
22
1 8maE hπ
= , with probability
36025.0932
2)4(322
11 222 ≈===− ππ n
cP .
Problem 3 (50 points): Consider the one dimensional harmonic oscillator with potential energy
2
21)( kxxV = . The Hamiltonian is given by
( )222
)(21
2xmp
mV
mp
H xx ω+=+=
where mk /=ω . We express the Hamiltonian in terms of the operators (px)op and (x)op, where
xip opx ∂
∂−= h)( and xx op =)( and 2
222 )()()(
xppp opxopxopx ∂
∂−== h .
Namely,
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 8 of 14
( )opopxop xmpm
H )()(21 2222 ω+= ,
and we define the two operators opa )( ± as follows
( )opxopop pixmm
a )(2
1)( mh
ωω
≡± .
(a) (2 points) Prove that the commutator of (px)op and (x)op is given by hixp opopx −=])(,)[( , where [Aop,Bop] = AopBop – BopAop. Solution: The commutator is an operator and to figure out what it is equal to we must operate on a function as follows:
( )ψψψψψψ
ψψ
hhhhhh ix
xix
xixxi
xxix
xi
xpxxpxxp opxopopopxopopx
−=∂∂
+∂∂
−∂∂
−=∂∂
+∂∂
−=
−=
)()(
)()()()()()(])(,)[(.
Thus, hixp opopx −=])(,)[( . (b) (2 points) Prove that [(a-)op,(a+)op] = 1. Solution: We see that
( )
1],)[(
])(,)[(],)[(])(,[],[2
1
])(,)([2
1])(,)[(
22
==
++−=
−+=+−
opopx
opxopxopopxopxopopop
opxopopxopopop
xpi
ppxpimpximxxmm
pixmpixmm
aa
h
h
h
ωωωω
ωωω
(c) (2 points) Prove that ( ) ( )21
21 )()()()( −=+= +−−+ opopopopop aaaaH ωω hh .
Solution: We see that
( )( )
( )
211
21
21
2)(1],)[()(
21
)()(2
1)()(
222
2222
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−+=−+=
+−=−+
op
opopx
opopxopxop
opxopopxopopop
H
xmm
pxpimpxm
m
pixmpixmm
aa
ω
ωωω
ωωω
ωωω
h
hhh
h
and hence ( )2
1)()( += −+ opopop aaH ωh Also,
( ) ( )( )2
1
21
21
)()(
])(,)[()()()()(
−=
++=+=
+−
−++−−+
opop
opopopopopopop
aa
aaaaaaH
ω
ωω
h
hh
(d) (4 points) Show that opopop aaH )(])(,[ ±± ±= ωh . Solution: We see that
opopopopopopop
opopopopopopopop
aaaaaaa
aaaaaaaH
)()]()(,)[(])(,)[()(
])(,)()[(])(,)()[(])(,[ 21
±−±+±−+
±−+±−+±
±=+=
=+=
ωωω
ωω
hhh
hh
where I used [AB,C]=A[B,C]+[A,C]B.
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 9 of 14
(e) (4 points) Show that if the state Eψ satisfies Schrödinger’s equation with energy E (i.e.
EEop EH ψψ = ), then the state Eopa ψψ )( ++ = satisfies Schrödinger’s equation with energy )( ωh+E and the state Eopa ψψ )( −− = satisfies Schrödinger’s equation with energy )( ωh−E .
Solution: From (d) we see that ( )
( ) ±±±
±±±±
±=±
+==
ψωψω
ψψψ
)()()(
])(,[)()(
hh EaHa
aHHaaHH
Eopopop
EopopopopEopopop
(f) (4 points) Show that if the state Eψ satisfies Schrödinger’s equation with energy E (i.e.
EEop EH ψψ = ), then ωh21≥E . (Hint: use the fact that the norm of all allowed stated is positive
definite) Solution: The norm of any state must be positive definite and we know that Eopa ψψ )( −− = is either zero or a stationary state and hence
0)()( ≥∫+∞
∞−−
∗− dxxx ψψ
and
( )
( ) ( ) ( )211
211
211 )()()()(
)()())(()()()()()()(
−=−=−=
==
∫∫
∫∫∫∞+
∞−
∗∞+
∞−
∗
+∞
∞−−+
∗+∞
∞−−
∗−
+∞
∞−−
∗−
EdxxxEdxxHx
dxxaaxdxxaxadxxx
EEEopE
EopopEEopEop
ωωω ψψψψ
ψψψψψψ
hhh
Thus, ( ) 02
11 ≥−Eωh and ωh21≥E .
Note that I used the fact that
( ) ∫∫+∞
∞−−
∗+∞
∞−−
∗± = dxxaxdxxxa opEEop )())(()()()( ψψψψ m
which comes from
( ) ( )( )
( ) ∫∫
∫∫
∫∫
∞+
∞−−
∗∞+
∞−−
∗
∞+
∞−−
∗∞+
∞−−
∗
+∞
∞−−
∗+∞
∞−−
∗±
=±=
⎟⎠⎞
⎜⎝⎛
∂∂
±=⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛
∂∂
=
=
dxxaxdxxpixmxm
dxxx
xmxm
dxxxx
xmm
dxxxpixmm
dxxxa
opEopxopE
EE
EopxopEop
)())(()()()(2
1
)()(2
1)()(2
1
)()()(2
1)()()(
ψψψωψω
ψωψω
ψψωω
ψψωω
ψψ
mh
hh
hmh
mh
where I integrated by parts and dropped the I integrated by parts and dropped the boundary term. (g) (4 points) The ground state, 0ψ , is defined to be the stationary state with the lowest energy (i.e. 000 ψψ EHop = ). Show that ωh2
10 =E and
2
24/1
0 )(xm
emx h
h
ω
πωψ
−⎟⎠⎞
⎜⎝⎛=
Solution: We know that 0)( 0 =− ψopa and thus
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 10 of 14
( )
ωψψω
ψψωψψ
hh
h
21
00
021
000
)()(2
)()()()()()(
==
+=>=>=<<
∫
∫∫∞+
∞−
∗
+∞
∞−−+
∗+∞
∞−
∗
dxxx
dxxaaxdxxHxHE opopopop
and also
( ) 000 21)(
210)( ψω
ωψω
ωψ ⎟
⎠⎞
⎜⎝⎛
∂∂
+=+==− xxm
mpixm
ma opxopop h
hh
Thus 0)(0 =xψ satisfies the following differential equation,
)()(0
0 xxmx
x ψωψh
=∂
∂ ,
with solution 2
20 )(
xm
Aex h
ω
ψ−
= . The normalization A is determined by requiring that
ωπψψ
ω
mAdxeAdxxx
xmh
h 2200
2
1)()( === ∫∫+∞
∞−
−+∞
∞−
∗
and hence 4/1
⎟⎠⎞
⎜⎝⎛=
hπωmA .
(h) (4 points) Prove that 11)( ++ += nnop na ψψ and 1)( −− = nnop na ψψ , were nψ is the nth
excited state and show that 0)(!
1 ψψ nopn a
n += .
Solution: From (f) we have
( ) ∫∫+∞
∞−
∗+∞
∞−
∗± = dxxaxdxxxa nopmnmop )())(()()()( ψψψψ m
From (g) we know that 1)( ++ = nnnop ca ψψ and 1)( −− = nnnop da ψψ where cn and dn are constants.
In addition we know that ( ) nopopnnnnop aanEH ψωωψψψ 2
121 )()()( +=+== −+hh
and hence nnopop naa ψψ =−+ )()( .
Similarly, ( ) nopopnnnnop aanEH ψωωψψψ 2
121 )()()( −=+== +−hh
and hence nnopop naa ψψ )1()()( +=+− .
Thus,
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 11 of 14
( )
)1()()()1()()())((
)()()()()()( 211
2
+=+==
==
∫∫
∫∫∞+
∞−
∗∞+
∞−+−
∗
+∞
∞−+
∗+
+∞
∞−+
∗+
ndxxxndxxaax
cdxxxcdxxaxa
nnnopopn
nnnnnopnop
ψψψψ
ψψψψ
so that 1+= ncn . Similarly,
( )
ndxxxndxxaax
ddxxxddxxaxa
nnnopopn
nnnnnopnop
===
==
∫∫
∫∫∞+
∞−
∗∞+
∞−−+
∗
+∞
∞−−
∗−
+∞
∞−−
∗−
)()()()())((
)()()()()()( 211
2
ψψψψ
ψψψψ
so that ndn = . We see that
01 )( ψψ opa+= , 02
12 )(2
1)(2
1 ψψψ opop aa ++ == , 03
23 )(123
1)(3
1 ψψψ opop aa ++ ⋅⋅==
and 0)(!
1 ψψ nopn a
n += .
(i) (8 points) Find <x>, <x2>, <px>, and <px2> for the nth stationary state.
Answer: 0=>< nx , ωm
nx nh
⎟⎠⎞
⎜⎝⎛ +=><
212 , 0=>< nxp , ωmnp nx h⎟
⎠⎞
⎜⎝⎛ +=><
212
Solution: We see that
( )opopop aam
x )()(2
)( −+ +=ωh ( )opopopx aamwip )()(
2)( −+ −=
h .
Thus,
( )
0)()(2
)()(12
)()()()(2
)())((
11 =++=
+==><
∫∫
∫∫∞+
∞−
∗−
∞+
∞−+
∗
+∞
∞−−+
∗+∞
∞−
∗
dxxxnm
dxxxnm
dxxaaxm
dxxxxx
nnnn
nopopnnopnn
ψψω
ψψω
ψψω
ψψ
hh
h
where I used
11)( ++ += nnop na ψψ and 1)( −− = nnop na ψψ and mnnm dxxx δψψ =∫+∞
∞−
∗ )()( .
We can calculate <px>n from <x>n as follows:
0=><
=><dtxdmp n
nx
or we can calculate it the long way
( )
0)()(2
)()(12
)()()()(2
)())((
11 =−+=
−==><
∫∫
∫∫∞+
∞−
∗−
∞+
∞−+
∗
+∞
∞−−+
∗+∞
∞−
∗
dxxxnmidxxxnmi
dxxaaxmidxxpxp
nnnn
nopopnnopxnnx
ψψωψψω
ψψωψψ
hh
h
Now
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 12 of 14
( )
( )
ωψψψψ
ω
ψψω
ψψω
ψψ
mndxxxndxxxn
m
dxxaaaaaaxm
dxxaaxm
dxxxxx
nnnn
nopopopopopopn
nopopnnopnn
hh
h
h
⎟⎠⎞
⎜⎝⎛ +=⎥
⎦
⎤⎢⎣
⎡++++=
+++=
+==><
∫∫
∫
∫∫
∞+
∞−
∗∞+
∞−
∗
∞+
∞−−+−−++
∗
+∞
∞−−+
∗+∞
∞−
∗
210)()()1()()(0
2
)()()()()()()()(2
)()()()(2
)())((
22
222
where I used ( ) 21
2 211)()( ++++ ++=+= nnopnop nnnaa ψψψ
( ) nnnopnopop nnnnaaa ψψψψ === −+−+ 1)()()(
( ) nnnopnopop nnnnaaa ψψψψ )1(111)()()( 1 +=++=+= +−+−
( ) 212 1)()( −−−− −== nnopnop nnnaa ψψψ
Also,
( )
( )
ωψψψψω
ψψω
ψψωψψ
mndxxxndxxxnm
dxxaaaaaaxm
dxxaaxmdxxpxp
nnnn
nopopopopopopn
nopopnnopxnnx
hh
h
h
⎟⎠⎞
⎜⎝⎛ +=⎥
⎦
⎤⎢⎣
⎡++++=
+−−−
=
−−
==><
∫∫
∫
∫∫
∞+
∞−
∗∞+
∞−
∗
∞+
∞−−+−−++
∗
+∞
∞−−+
∗+∞
∞−
∗
210)()()1()()(0
2
)()()()()()()()(2
)()()()(2
)())((
22
222
.
(j) (4 points) Find Δx = σx and Δpx = xpσ for the nth stationary state and check that the
uncertainty principle is satisfied.
Answer: ωm
nx nh
⎟⎠⎞
⎜⎝⎛ +=Δ
21)( , ωmnp nx h⎟
⎠⎞
⎜⎝⎛ +=Δ
21)( ,
2)12()( h
+=ΔΔ npx nx
Solution: From (h) we see that
ωmnxxx nnn
h⎟⎠⎞
⎜⎝⎛ +=><−><=Δ
21)( 22
ωmnppp nxnxnx h⎟⎠⎞
⎜⎝⎛ +=><−><=Δ
21)( 22
Thus,
22)12(
21
21)( hh
hh
≥+=⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +=ΔΔ nmn
mnpx nx ω
ω.
(k) (4 point) Compute <T> and <V> for the nth stationary state, where T is the kinetic energy and V is the potential energy. What is sum <T> + <V>?
Answer: ωh⎟⎠⎞
⎜⎝⎛ +=><
21
21 nT n , ωh⎟
⎠⎞
⎜⎝⎛ +=><
21
21 nV n , nn EnVT =⎟
⎠⎞
⎜⎝⎛ +=><+>< ωh
21)(
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 13 of 14
Solution: From (h) we see that
ωh⎟⎠⎞
⎜⎝⎛ +=
><=><
21
21
2
2
nm
pT nxn
ωω h⎟⎠⎞
⎜⎝⎛ +=><=><
21
21
21 22 nxmV nn
Note that <T>n = <V> n and that nnn EnVT =⎟⎠⎞
⎜⎝⎛ +=><+>< ωh
21 .
(l) (8 points) Suppose that a particle in the harmonic oscillator potential starts out in the state [ ])(4)(3)0,( 10 xxAx ψψ +=Ψ .
What is <x> and <px> for this state? If you measure the energy of this particle, what are the possible values 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: E0 with probability 9/25 and E1 with probability 16/25,
)cos(225
24 tm
x ωωh
>=< , )sin(225
24 tmpx ωωh−>=< , and ωh50
57>=< E .
Solution: The normalization A is determined by requiring that
[ ][ ]
( ) 211011000
2
10102
25)()(16)()(12)()(12)()(9
)(4)(3)(4)(31)0,()0,(
AdxxxxxxxxxA
dxxxxxAdxxx
=+++=
++==ΨΨ
∫
∫∫∞+
∞−
∗∗∗∗
+∞
∞−
∗∗+∞
∞−
∗
ψψψψψψψψ
ψψψψ
Thus, A = 1/5 and
2/315
42/05
3
/15
4/05
3/11
/00
)()(
)()()()(),( 1010
titi
tiEtiEtiEtiE
exex
exexexcexctxωω ψψ
ψψψψ−−
−−−−
+=
+=+=Ψ hhhh
.
which means that you get energy E0 with probability 9/25 and energy E1 with probability 16/25. The expectation value of E is
ωωω hhh 5057
23
2516
21
259
12516
0259 =+=+>=< EEE .
The expectation value of x at time t is
[ ]( )[ ]
[ ] [ ] [ ] [ ]
( ) )cos(225
24225
12
)()()(2
)()()(2
)()()()()()(2
),())(,(
2/05
32/315
42/315
42/05
3
2/315
42/05
32/315
42/05
3
tm
eem
dxexaexm
dxexaexm
dxexexaaexexm
dxtxxtxx
titi
tiop
titiop
ti
titiopop
titi
op
ωωω
ψψω
ψψω
ψψψψω
ωω
ωωωω
ωωωω
hh
hh
h
=+=
+=
+++=
ΨΨ>=<
+−
∞+
∞−
−+
+∗∞+
∞−
−−
+∗
∞+
∞−
−−−+
+∗+∗
+∞
∞−
∗
∫∫
∫
∫
The expectation value of px at time t is
PHY4604 Fall 2007 Problem Set 2 Solutions
Department of Physics Page 14 of 14
[ ]( )[ ]
[ ] [ ] [ ] [ ]
( ) )sin(225
24225
12
)()()(2
)()()(2
)()()()()()(2
),())(,(
2/05
32/315
42315
42/05
3
2/315
42/05
32/315
42/05
3
tmeemi
dxexaexm
dxexaexmi
dxexexaaexexmi
dxtxptxp
titi
tiop
titiop
ti
titiopop
titi
opxx
ωωω
ψψω
ψψω
ψψψψω
ωω
ωωωω
ωωωω
hh
hh
h
−=−−=
−−=
+−+=
ΨΨ>=<
+−
∞+
∞−
−+
+∗∞+
∞−
−−
+∗
∞+
∞−
−−−+
+∗+∗
+∞
∞−
∗
∫∫
∫
∫
Note that
>=<−=−==><
xptmtm
mtdtd
mm
dtxdm )sin(
22524)sin(
22524)cos(
22524 ωωω
ωωω
ωhhh .
