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