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1、ErrataInstructor’sSolutionsManualIntroductiontoQuantumMechanicsAuthor:DavidGriffithsDate:June14,2001•Page3,Prob.1.6(b):lasttwolinesshouldread1π2π21=A+0+a=a+.2λλλ2λ22221211σ=x−x=a+−a=;σ=√.2λ2λ2λ•Page8,Prob.2.6(b):inthefirstbox,theargumentofthesecondsineshouldincludeanx.√•Page9,Pr
2、ob.2.9:inthelastline,c=815/π3.1•Page10,Prob.2.11:inthethirdlinetheproofassumesthatg(whichinourcasewillbea−ψ)doesnotactuallyblowupat±∞fasterthanf(inourcaseψ)goestozero.Idon’tknowhowtofixthisdefectwithoutappealingtotheanalyticapproach,wherewefind(Eq.2.60)thatψ(x)2goesasymptoticallylikee−
3、mωx/2,andhencesotoodoesaψ.−•Page10,Prob.2.12:Becauseoftheiand−iinsertedinEqs.2.52and2.53respectively(seeCorrections#2—June1997),theexpressionforcattheendof(a)shouldincludeafactorofi.Also,addattheendof(a):“(The√signsareconventional.)”Inpart(b),everyωshouldcarryafactorofi(i.e.inserti
4、threetimesinthefirstline,inthreetimesinthenextline,and(−i)nintheboxedanswer).•Page11,Prob.2.14(a):forthesamereason,inthethirdline,removetheiintheexpressionforψ1.•Page19,Prob.2.30:or“...tanz≈z=(z0/z)2−1=(1/z)z02−z2.Now(Eqs.2.130and2.137)z2−z2=κ2a2,soz2=κa.Butz2−z2=00z41⇒z≈z,soκa≈z2..
5、.”00•Page22,Prob.2.36:removebox,andcontinueasfollows:Ψ(x,t)=√13ψ(x)e−iE1t/−ψ(x)e−iE3t/,13102122E3−E1
6、Ψ(x,t)
7、=9ψ1+ψ3−6ψ1ψ3cost,101soax=x
8、Ψ(x,t)
9、2dx0a913E3−E1=x1+x3−costxψ1(x)ψ3(x)dx,101050wherexn=a/2istheexpectationvalueofxinthenthstationarystate.Theremainingintegral
10、isaa2πx3πx12πx4πxxsinsindx=xcos−cosdxa0aaa0aa221a2πxxa2πxa4πx=cos+sin−cosa2πa2πa4πaaxa4πx−sin=0.4πa0Evidently,then,9a1aax=+=.1021022•Page23,Prob.2.37:Becauseoftheiand−iinsertedinEqs.2.52and2.53respectively(seeCorrections#2—June1997),line3onshouldreadasfollows:−i∗x=√ψn(
11、a+−a−)ψndx.ω2mButa+ψn=i√(n+1)ωψn+1[2.52],soa−ψn=−inωψn−1[2.53]1√x=√(n+1)ωψ∗ψdx+nωψ∗ψdx=0nn+1nn−1ω2mdx(bytheorthogonalityof{ψn}).Alsop=m=0.Meanwhiledt2−1−122xˆ=2mω2(a+−a−)(a+−a−)=2mω2(a+−a+a−−a−a++a−),2−1∗22sox=2ψn(a+−a+a−−a−a++a−)ψndx.But2mωa2ψ=a(i(n+1)ωψ)=−(n+1
12、)(n+2)ωψ.
