Quantum Mechanics, Sixth Edition [6th Ed] (Instructor's Solution Manual) (Solutions)

158 Pages·2015·1.043 MB·English

by  Alastair I. M. Rae

Contents
AboutthisManual ........................................... 1
SolutionstoProblemsattheendofChapter1 ................. 3
SolutionstoProblemsattheendofChapter2 ................ 15
SolutionstoProblemsattheendofChapter3 ................ 25
SolutionstoProblemsattheendofChapter4 ................ 33
SolutionstoProblemsattheendofChapter5 ................ 39
SolutionstoProblemsattheendofChapter6 ................ 51
SolutionstoProblemsattheendofChapter7 ................ 63
SolutionstoProblemsattheendofChapter8 ................ 77
SolutionstoProblemsattheendofChapter9 ................ 85
SolutionstoProblemsattheendofChapter10 ...............97
SolutionstoProblemsattheendofChapter11 ..............111
SolutionstoProblemsattheendofChapter12 ..............123
SolutionstoProblemsattheendofChapter13 ..............135
SolutionstoProblemsattheendofChapter14 ..............111
SolutionstoProblemsattheendofChapter15 ..............123
SolutionstoProblemsattheendofChapter16 ..............135
393

About this manual
This manual contains solutions to the problems set at the end of each chapter of
“Quantum Mechanics”. It is divided into sections corresponding to the chapters in
thetextandtitledaccordingly.Bracketednumbersrefertoequationsinthemaintext.
1

1
CHAPTER
The Physics and
Mathematics of
Waves
1.1 UseEuler’sformulatofindapurelyrealexpressionforii.
Solution
ii=(cid:16)eiπ/2(cid:17)i=e−π/2
1.2 Showthat(1.3)canbewrittenas
x(t)=Acos(ωt+φ)
andderiveexpressionsforAandφintermsofBandC.Assumingtheoscillatorstarts
out at position x(0)= x with velocity v(0)=v , determine A and φ in terms of x
0 0 0
andv .Note:WecallAtheamplitudeandφthephaseoftheoscillation.
0
Solution
Replace the constants B and C in x(t) = Bcosωt+Csinωt with two different
constants A and φ which solve B = Acosφ and C = −Asinφ. This results in
x(t) = Acos(ωt+φ). Now x(0) = Acosφ = x and x˙(0) = −ωAsinφ = v so A =
0 0
(x2+v2/ω2)1/2andφ=−tan−1(v /x ω).
0 0 0 0
1.3 Aspringwithstiffnesskhangsverticallyfrompointontheceiling.Amassmis
attachedtothelowerendofthespringwithoutstretchingit,andthenisreleasedfrom
rest.Showthatwhenthegravitationalforcemgistakenintoaccount,themotionis
stillsinusoidalwithω=(k/m)1/2 butwithanequilibriumpositionshiftedtoalower
point.Findthenewequilibriumpositionintermsofm,k,andg.
3
4 (cid:4) QuantumMechanicsSixthEditionSolutionsManual
Solution
Let y measure the vertical position of the mass, with y=0 the unstretched string.
Thenmy¨=−ky−mg=−k(y+mg/k).Defining x≡y+mg/k,getmx¨=−kx,soonce
againω2=k/m.Theequilibriumpointisx=0ory=−mg/k.
1.4 Consider a system of a mass and spring, such as in Figure 1.1, but with an
additional force F = −bv proportional to velocity but acting in the direction
damp
opposite to the motion. Reformulate the equation of motion, and find the solution
for x(0)= x and v(0)=v . Use the ansatz x(t)=exp(iαt) to solve for α. You may
0 0
assumethatb2/m2islessthan4k/m.
Solution
The equation of motion is mx¨=−kx−bx˙ so −mα2 =−k−ibα. Defining ω2 ≡k/m
0
andβ≡b/2m,yieldstheequationα2−2iβα−ω2=.Solvingthis,
0
α= 1(cid:20)2iβ±(cid:16)−4β2+4ω2(cid:17)1/2(cid:21)=iβ±ω
2 0
where ω2 ≡ ω2−β2. (Note that β2/ω2 = (b2/m2)/(4k/m) < 1 so ω is real.) The
0 0
solution becomes x(t) = Ae−βtcos(ωt+φ) where Acosφ = x and −A(βcosφ+
0
ωsinφ) = v . These can be solved in principle, but the interesting solution is
0
when β(cid:28)ω, i.e. “lightly damped” motion. The result in oscillation which slowly
damps.
1.5 Two equal masses m move in one dimension and are each connected to fixed
wallsbyspringswithstiffnessk.Themassesarealsoconnectedtoeachotherbya
third,identicalspring,asshown:
k m k m k
Write the (differential) equations of motion for the positions x (t) and x (t) of the
1 2
two masses. Solve those equations with the ansatz x (t)= A exp(iαt) and x (t)=
1 1 2
A exp(iαt);youwilldiscovernontrivialsolutionsonlyfortwovaluesofω2.(Those
2
3m 2m
twovaluesarecalledeigenfrequkencies.)Whatkkindofmotioncorrespondstoeachof
thesetwoeigenfrequencies?
Solution
Label the two masses #1 and #2 from left to right. The force on m #1 is −kx +
1
k(x −x )=−2kx +kx ,andtheforceonm#2is−kx −k(x −x )=−2kx +kx ,
2 1 1 2 2 2 1 2 1
so,definingω2≡k/m,theequationsofmotionare
0
x¨ =−2ω2x +ω2x and x¨ =−2ω2x +ω2x
1 0 1 0 2 2 0 2 0 1
Nowinserttheansatzsolution.Afteralittlerearranging,youfind
(2ω2−ω2)A −ω2A =0 and −ω2A +(2ω2−ω2)A =0
0 1 0 2 0 1 0 2
ThePhysicsandMathematicsofWaves (cid:4) 5
ThesearetwohomogenousequationsforA andA .TheonlysolutionisA =A =0,
1 2 1 2
thatisnomotion,unlessthedeterminantvanishes:
ω4=(ω2−2ω2)2 so ω2=ω2 or ω2=3ω2
0 0 0 0
For ω2 = ω2, find A = Ak so thme two mkasses omscillatekin phase with the same
0 1 2
amplitude. For ω2 =3ω2, find A =−A so the two masses oscillate out of phase
0 1 2
withthesameamplitude.
1.6 Findtheeigenfrequenciesforthetwo-mass,two-springsystemshownhere:
3m 2m
k k
Solution
Labelmass3m#1andmass2m#2.Thentheequationsofmotionare
3mx¨ =−kx +k(x −x )=−2kx +kx and2mx¨ =−k(x −x )=kx −kx
1 1 2 1 1 2 2 2 1 1 2
Usingthestandarddefinitionsandansatz,
(3ω2−2ω2)A +ω2A =0 and ω2A +(2ω2−ω2)A =0
0 1 0 2 0 1 0 2
Next,setthedeterminantequaltozerotofind
(3ω2−2ω2)(2ω2−ω2)−ω4=6ω4−7ω2ω2+ω2=(6ω2−ω2)(ω2−ω2)=0
0 0 0 0 0 0 0
so the eigenfrequencies are ω2 =ω2, in which case the two masses oscillate with
0
equalamplitudebutoutofphase,andω2=ω2/6,inwhichcasetheoscillationsare
0
inphasewithA /A =2/3.
1 2
1.7 For the two-mass, three-spring system discussed in Problem 1.5, find expres-
sionsforx (t)andx (t)subjecttotheinitialconditionsx (0)=Aandx (0)=v (0)=
1 2 1 2 1
v (0)=0.Makeaplotofx (t)andx (t),andalsoplotthequantitiesx (t)+x (t)and
2 1 2 1 2
x (t)−x (t).Commentonyourobservations.
1 2
Solution
Nowweneedtowritethethegeneralsolutionforthemotionofthetwomasses:
√ √
x (t) = aeiω0t+be−iω0t+ce 3iω0t+de− 3iω0t
1
√ √
x (t) = aeiω0t+be−iω0t−ce 3iω0t−de− 3iω0t
2
Note that we have maintained the amplitude ratios and relative phases between the
different solutions for the particular eigenfrequencies. This is necessary in order
6 (cid:4) QuantumMechanicsSixthEditionSolutionsManual
to make sure that each of the four terms separately solves the coupled differential
equations.Nowwecanapplytheinitialconditions:
A = a+b+c+d
√
0 = a−b+ 3(c−d)
0 = a+b−c−d
√
0 = a−b− 3(c−d)
Addingfirstandthirdgivesa+b=A/2andaddingsecondandfourthgivesa−b=0,
soa=b=A/4.Subtractingthirdfromfirstgivesc+d=A/2andsubtractingfourth
fromsecondgivesc−d=0soc=d=A/4.Therefore
A(cid:104) √ (cid:105)
x (t) = cos(ω t)+cos( 3ω t)
1 0 0
2
A(cid:104) √ (cid:105)
x (t) = cos(ω t)−cos( 3ω t)
2 0 0
2
Belowleft,plotsofx (t)andx (t).Belowright,plotsofx (t)+x (t)andx (t)−x (t).
1 2 1 2 1 2
1.0 1.0
0.5 0.5
2 4 6 8 10 12 2 4 6 8 10 12
-0.5 -0.5
-1.0 -1.0
The wiggling motion is a superposition of different eigenfrequencies, but the sum
anddifferenceshowtheindividualisolatedeigenfrequencies.
1.8 Repeat Problem 1.7, but this time let the “coupling” spring between the two
masseshaveaspringconstantk =k/100.Showthattheoverallmotion“oscillates”
c
between cases were the first mass is in simple harmonic motion by itself, to one
wherethesecondmassisinsimpleharmonicmotion,andthenbackagain.Whatis
thefrequencyoftheselowfrequencyoscillationsbetweenthetwomasses?
Solution
First, go back to Problem 5. Now, the force on m #1 is −kx +k (x −x )=−(k+
1 c 2 1
k )x +kx ,andtheforceonm#2is−kx −k (x −x )=−(k+k )x +kx ,so,defining
c 1 2 2 c 2 1 c 2 1
ω2≡k/mandα2=2k /m=2(k /k)ω2,theequationsofmotionare
0 c c 0
x¨ =−(ω2+α2/2)x +(α2/2)x and x¨ =−(ω2+α2/2)x +(α2/2)x
1 0 1 2 2 0 2 1
Nowinserttheansatzsolution.Afteralittlerearranging,youfind
(ω2+α2/2−ω2)A −(α2/2)A =0 and −(α2/2)A +(ω2+α2/2−ω2)A =0
0 1 2 1 0 2
ThePhysicsandMathematicsofWaves (cid:4) 7
so ω2+α2/2−ω2 =±α2/2 and the solutions are ω2 =ω2 and ω2 =ω2+α2. It is
0 0 0
easytosee,asinProblem5,thatthesetwosolutionscorrespondtoequalamplitude
oscillationsinphaseandoutofphase,respectively.Atthispoint,themotionsofthe
twomassesworkoutjustasinProblem7,andwehave
A(cid:20) (cid:113) (cid:21)
x (t) = cos(ω t)+cos( ω2+α2t)
1 2 0 0
A(cid:20) (cid:113) (cid:21)
x (t) = cos(ω t)−cos( ω2+α2t)
2 2 0 0
Whenk=k ,α2=2ω2 andwegetthecorrectsolutiontoProblem7.Followingare
c 0
the same plots, but for k =k/10, that is, α2 =ω2/5 (which plot more nicely than
c 0
k =k/100):
c
1.0 1.0
0.5 0.5
50 100 150 50 100 150
-0.5 -0.5
-1.0 -1.0
Therightplotshowsthattheeigenfrequenciesareveryclosetoeachother,resulting
in the beat pattern shown in the left plot. With α2 (cid:28) ω2, and the trigonometric
0
identities
(cid:18)u+v(cid:19) (cid:18)u−v(cid:19)
cosu+cosv = 2cos cos
2 2
(cid:18)u+v(cid:19) (cid:18)u−v(cid:19)
cosu−cosv = −2sin sin
2 2
itisclearthattheplotsaretheproductofahighfrequencycomponent
1(cid:18) (cid:113) (cid:19)
ω + ω2+α2 ≈ω
2 0 0 0
withanenvelopewithlowfrequency
 
21(cid:18)(cid:113)ω20+α2−ω0(cid:19)≈ ω201+2αω22 −1= ω20kkc
0
That is, for the left plot above, there are ≈ 20 crests within one envelope wave-
length.
1.9 Derive the solution (1.13) to the wave equation (1.12) by going through the
following steps. Consider a change of variables from x and y to ξ= x−vt and η=
x+vt.Thenusethechainruletorewritethewaveequationintermsofξandη.You
shouldfindthat
∂2y
=0
∂ξ∂η
8 (cid:4) QuantumMechanicsSixthEditionSolutionsManual
Then argue that this means that y is a function of either ξ or η, but not both at the
same time. In other words, the solution is (1.13). If you are not familiar with the
chainruleforpartialdifferentiation,itmeansthatifwandzarefunctionsofxandy,
then
∂ ∂f ∂w ∂f ∂z
f(x,y)= +
∂x ∂w ∂x ∂z ∂x
andsimilarlyfor∂/∂y.Youcanassumethatyougetthesameresultregardlessofthe
orderinwhichthepartialderivativesaretaken.
Solution
Asdirected,applythechainruletothewaveequation
∂y ∂y ∂y
= +
∂x ∂ξ ∂η
∂2y ∂2y ∂2y ∂2y
= +2 +
∂x2 ∂ξ2 ∂ξ∂η ∂η2
∂y ∂y ∂y
= −v +v
∂t ∂ξ ∂η
∂2y ∂2y ∂2y ∂2y
= v2 −2v2 +v2
∂t2 ∂ξ2 ∂ξ∂η ∂η2
1 ∂2y ∂2y ∂2y
− = −4v2 =0
v2 ∂t2 ∂x2 ∂ξ∂η
leadingtotheexpressionwesought.Theobvioussolutiontothisdifferentialequation
is the sum of two “constants” in ξ and η, respectively, that is y(ξ,η) = f(ξ)+
g(η).
1.10 Provetheprincipleoflinearsuperpositionforthewaveequation(1.12).That
is,showthatify (x,t)andy (x,t)aresolutionsofthewaveequation,theny(x,t)=
1 2
ay (x,t)+by (x,t)isalsoasolution,whereaandbarearbitraryconstants.
1 2
Solution
Allyouhavetodoisplugitinanditfallsouteasily:
1 ∂2y ∂2y 1 ∂2y 1 ∂2y ∂2y ∂2y
− = a 1 +b 2 −a 1 −b 2
v2 ∂t2 ∂x2 v2 ∂t2 v2 ∂t2 ∂x2 ∂x2
(cid:34) (cid:35) (cid:34) (cid:35)
1 ∂2y ∂2y 1 ∂2y ∂2y
= a 1 − 1 +b 2 − 2
v2 ∂t2 ∂x2 v2 ∂t2 ∂x2
= 0+0=0
1.11 Astringwithlinearmassdensityµhangsmotionlessbetweentwofixedpoints
(x,y)=(±a,0) where y measures the vertical direction. The length of the string is
greaterthan2a,sothelowestpointisat(x.y)=(0,b).Derivethedifferentialequation