Introduction to Quantum Mechanics, 3rd Edition Solution Manual

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Instructors’ Solution ManualIntroduction to Quantum Mechanics, 3rd ed.David Griffiths, Darrell SchroeterReed CollegeAugust 3, 2018SOLUTIONSGUIDE

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2Contents1The Wave Function42The Time-Independent Schr¨odinger Equation163Formalism784Quantum Mechanics in Three Dimensions1095Identical Particles1686Symmetries and Conservation Laws1977Time-Independent Perturbation Theory2358The Variational Principle3019The WKB Approximation33310 Scattering35411 Quantum Dynamics37212 Afterword420ALinear Algebra427

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4CHAPTER 1. THE WAVE FUNCTIONChapter 1The Wave FunctionProblem 1.1(a)〈j〉2= 212=441.〈j2〉=1N∑j2N(j) =114[(142) + (152) + 3(162) + 2(222) + 2(242) + 5(252)]=114 (196 + 225 + 768 + 968 + 1152 + 3125) = 643414=459.571.(b)j∆j=j− 〈j〉1414−21 =−71515−21 =−61616−21 =−52222−21 = 12424−21 = 32525−21 = 4σ2=1N∑(∆j)2N(j) =114[(−7)2+ (−6)2+ (−5)2·3 + (1)2·2 + (3)2·2 + (4)2·5]=114 (49 + 36 + 75 + 2 + 18 + 80) = 26014=18.571.σ=√18.571 =4.309.(c)〈j2〉 − 〈j〉2= 459.571−441 = 18.571.[Agrees with (b).]

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CHAPTER 1. THE WAVE FUNCTION5Problem 1.2(a)〈x2〉=∫h0x212√hx dx=12√h(25x5/2)∣∣∣∣h0=h25.σ2=〈x2〉 − 〈x〉2=h25−(h3)2=445h2⇒σ=2h3√5 = 0.2981h.(b)P= 1−∫x+x−12√hx dx= 1−12√h(2√x)∣∣∣∣x+x−= 1−1√h(√x+− √x−).x+≡ 〈x〉+σ= 0.3333h+ 0.2981h= 0.6315h;x−≡ 〈x〉 −σ= 0.3333h−0.2981h= 0.0352h.P= 1− √0.6315 +√0.0352 =0.393.Problem 1.3(a)1 =∫∞−∞Ae−λ(x−a)2dx.Letu≡x−a,du=dx,u:−∞ → ∞.1 =A∫∞−∞e−λu2du=A√πλ⇒A=√λπ.(b)〈x〉=A∫∞−∞xe−λ(x−a)2dx=A∫∞−∞(u+a)e−λu2du=A[∫∞−∞ue−λu2du+a∫∞−∞e−λu2du]=A(0 +a√πλ)=a.〈x2〉=A∫∞−∞x2e−λ(x−a)2dx=A{∫∞−∞u2e−λu2du+ 2a∫∞−∞ue−λu2du+a2∫∞−∞e−λu2du}=A[12λ√πλ+ 0 +a2√πλ]=a2+12λ.σ2=〈x2〉 − 〈x〉2=a2+12λ−a2=12λ;σ=1√2λ.

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6CHAPTER 1. THE WAVE FUNCTION(c)Axal(x)Problem 1.4(a)1 =|A|2a2∫a0x2dx+|A|2(b−a)2∫ba(b−x)2dx=|A|2{1a2(x33)∣∣∣∣a0+1(b−a)2(−(b−x)33)∣∣∣∣ba}=|A|2[a3 +b−a3]=|A|2b3⇒A=√3b.(b)xaAb^(c)Atx=a.(d)P=∫a0|Ψ|2dx=|A|2a2∫a0x2dx=|A|2a3 =ab.{P= 1ifb=a,XP= 1/2 ifb= 2a.X(e)〈x〉=∫x|Ψ|2dx=|A|2{1a2∫a0x3dx+1(b−a)2∫bax(b−x)2dx}= 3b{1a2(x44)∣∣∣∣a0+1(b−a)2(b2x22−2b x33+x44)∣∣∣∣ba}=34b(b−a)2[a2(b−a)2+ 2b4−8b4/3 +b4−2a2b2+ 8a3b/3−a4]=34b(b−a)2(b43−a2b2+ 23a3b)=14(b−a)2(b3−3a2b+ 2a3) =2a+b4.

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CHAPTER 1. THE WAVE FUNCTION7Problem 1.5(a)1 =∫|Ψ|2dx= 2|A|2∫∞0e−2λxdx= 2|A|2(e−2λx−2λ)∣∣∣∣∞0=|A|2λ;A=√λ.(b)〈x〉=∫x|Ψ|2dx=|A|2∫∞−∞xe−2λ|x|dx=0.[Odd integrand.]〈x2〉= 2|A|2∫∞0x2e−2λxdx= 2λ[2(2λ)3]=12λ2.(c)σ2=〈x2〉 − 〈x〉2=12λ2;σ=1√2λ.|Ψ(±σ)|2=|A|2e−2λσ=λe−2λ/√2λ=λe−√2= 0.2431λ.|^|2hm<m+x.24hProbability outside:2∫∞σ|Ψ|2dx= 2|A|2∫∞σe−2λxdx= 2λ(e−2λx−2λ)∣∣∣∣∞σ=e−2λσ=e−√2= 0.2431.Problem 1.6For integration by parts, the differentiation has to be with respect to theintegrationvariable – in this case thedifferentiation is with respect tot, but the integration variable isx. It’s true that∂∂t(x|Ψ|2) =∂x∂t|Ψ|2+x ∂∂t|Ψ|2=x ∂∂t|Ψ|2,but this doesnotallow us to perform the integration:∫bax ∂∂t|Ψ|2dx=∫ba∂∂t(x|Ψ|2)dx6= (x|Ψ|2)∣∣ba.

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8CHAPTER 1. THE WAVE FUNCTIONProblem 1.7From Eq. 1.33,d〈p〉dt=−iℏ∫∂∂t(Ψ∗∂Ψ∂x)dx. But, noting that∂2Ψ∂x∂t=∂2Ψ∂t∂xand using Eqs. 1.23-1.24:∂∂t(Ψ∗∂Ψ∂x)=∂Ψ∗∂t∂Ψ∂x+ Ψ∗∂∂x(∂Ψ∂t)=[−iℏ2m∂2Ψ∗∂x2+iℏVΨ∗]∂Ψ∂x+ Ψ∗∂∂x[iℏ2m∂2Ψ∂x2−iℏVΨ]=iℏ2m[Ψ∗∂3Ψ∂x3−∂2Ψ∗∂x2∂Ψ∂x]+iℏ[VΨ∗∂Ψ∂x−Ψ∗∂∂x(VΨ)]The first term integrates to zero, using integration by parts twice, and the second term can be simplified toVΨ∗∂Ψ∂x−Ψ∗V∂Ψ∂x−Ψ∗∂V∂xΨ =−|Ψ|2∂V∂x.Sod〈p〉dt=−iℏ(iℏ) ∫−|Ψ|2∂V∂x dx=〈−∂V∂x〉.QEDProblem 1.8Suppose Ψ satisfies the Schr¨odinger equationwithoutV0:iℏ∂Ψ∂t=−ℏ22m∂2Ψ∂x2+VΨ. We want to find the solutionΨ0withV0:iℏ∂Ψ0∂t=−ℏ22m∂2Ψ0∂x2+ (V+V0)Ψ0.Claim: Ψ0= Ψe−iV0t/ℏ.Proof:iℏ∂Ψ0∂t=iℏ∂Ψ∂te−iV0t/ℏ+iℏΨ(−iV0ℏ)e−iV0t/ℏ=[−ℏ22m∂2Ψ∂x2+VΨ]e−iV0t/ℏ+V0Ψe−iV0t/ℏ=−ℏ22m∂2Ψ0∂x2+ (V+V0)Ψ0.QEDThis hasnoeffect on the expectation value of a dynamical variable, since the extra phase factor, being inde-pendent ofx, cancels out in Eq. 1.36.Problem 1.9(a)1 = 2|A|2∫∞0e−2amx2/ℏdx= 2|A|212√π(2am/ℏ) =|A|2√πℏ2am;A=(2amπℏ)1/4.(b)∂Ψ∂t=−iaΨ;∂Ψ∂x=−2amxℏΨ;∂2Ψ∂x2=−2amℏ(Ψ +x ∂Ψ∂x)=−2amℏ(1−2amx2ℏ)Ψ.Plug these into the Schr¨odinger equation,iℏ∂Ψ∂t=−ℏ22m∂2Ψ∂x2+VΨ:VΨ =iℏ(−ia)Ψ +ℏ22m(−2amℏ) (1−2amx2ℏ)Ψ=[ℏa−ℏa(1−2amx2ℏ)]Ψ = 2a2mx2Ψ,soV(x) = 2ma2x2.

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CHAPTER 1. THE WAVE FUNCTION9(c)〈x〉=∫∞−∞x|Ψ|2dx=0.[Odd integrand.]〈x2〉= 2|A|2∫∞0x2e−2amx2/ℏdx= 2|A|2122(2am/ℏ)√πℏ2am=ℏ4am .〈p〉=m d〈x〉dt=0.〈p2〉=∫Ψ∗(ℏi∂∂x)2Ψdx=−ℏ2∫Ψ∗∂2Ψ∂x2dx=−ℏ2∫Ψ∗[−2amℏ(1−2amx2ℏ)Ψ]dx= 2amℏ{∫|Ψ|2dx−2amℏ∫x2|Ψ|2dx}= 2amℏ(1−2amℏ〈x2〉)= 2amℏ(1−2amℏℏ4am)= 2amℏ(12)=amℏ.(d)σ2x=〈x2〉 − 〈x〉2=ℏ4am=⇒σx=√ℏ4am;σ2p=〈p2〉 − 〈p〉2=amℏ=⇒σp=√amℏ.σxσp=√ℏ4am√amℏ=ℏ2. Thisis(just barely) consistent with the uncertainty principle.Problem 1.10From Math Tables:π= 3.141592653589793238462643· · ·(a)P(0) = 0P(1) = 2/25P(2) = 3/25P(3) = 5/25P(4) = 3/25P(5) = 3/25P(6) = 3/25P(7) = 1/25P(8) = 2/25P(9) = 3/25In general,P(j) =N(j)N.(b)Most probable:3.Median: 13 are≤4, 12 are≥5, so median is4.Average:〈j〉=125[0·0 + 1·2 + 2·3 + 3·5 + 4·3 + 5·3 + 6·3 + 7·1 + 8·2 + 9·3]=125[0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] =11825=4.72.(c)〈j2〉=125[0 + 12·2 + 22·3 + 32·5 + 42·3 + 52·3 + 62·3 + 72·1 + 82·2 + 92·3]=125[0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] =71025=28.4.σ2=〈j2〉 − 〈j〉2= 28.4−4.722= 28.4−22.2784 = 6.1216;σ=√6.1216 =2.474.

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10CHAPTER 1. THE WAVE FUNCTIONProblem 1.11(a)12mv2+V=E→v(x) =√2m(E−V(x)).(b)T=∫ba1√2m(E−12kx2)dx=√mk∫ba1√(2E/k)−x2dx.Turning points:v= 0⇒E=V=12kb2⇒b=√2E/k;a=−b.T= 2√mk∫b01√b2−x2dx= 2√mksin−1(xb) ∣∣∣b0= 2√mksin−1(1)= 2√mk(π2)=π√mk .ρ(x) =1π√mk√2m(E−12kx21)=1π√b2−x2.∫baρ(x)dx= 2π∫b01√b2−x2dx= 2π(π2)= 1.Xρ(x)xb-b(c)〈x〉= 0.〈x2〉= 1π∫b−bx2√b2−x2dx= 2π∫b0x2√b2−x2dx= 2π[−x2√b2−x2+b22 sin−1(xb)] ∣∣∣∣b0=b2πsin−1(1) =b2ππ2 =b22 =Ek .σx=√〈x2〉 − 〈x〉2=√〈x2〉=b√2 =√Ek .Problem 1.12(a)ρ(p)dp=dtT=|dt/dp|dpTwheredtis now the time it spends with momentum in the rangedp(dtis intrinsically positive, butdp/dt=F=−kxruns negative—hence the absolute value). Nowp22m+ 12kx2=E⇒x=±√2k(E−p22m),

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CHAPTER 1. THE WAVE FUNCTION11soρ(p) =1π√mkk√2k(E−p22m)=1π√2mE−p2=1π√c2−p2,wherec≡ √2mE.This is the same asρ(x) (Problem 1.11(b)), withcin place ofb(and, of course,pinplace ofx).(b)From Problem 1.11(c),〈p〉= 0,〈p2〉=c22, σp=c√2 =√mE .(c)σxσp=√Ek√mE=√mk E=Eω .IfE≥12ℏω, thenσxσp≥12ℏ, which is precisely the Heisenberguncertainty principle!Problem 1.13x@t_D:=Cos@tDsnapshots=Table@x@pRandomReal@jDD,8j, 10 000<D8-0.977661, 0.371147,-0.0389023,-0.0257086,-0.968682,-0.681951,-0.874704,Ü9987á, 0.584236,-0.974667, 0.10798, 0.996311,-0.767106, 0.926913<Histogram@snapshots, 100, "PDF", PlotRangeÆ80, 2<D-0.50.00.51.00.51.01.52.0r@x_D:=1p1-x2Plot@r@xD,8x,-1, 1<, PlotRange->80, 2<Dx@t_D:=Cos@tDsnapshots=Table@x@pRandomReal@jDD,8j, 10 000<D8-0.977661, 0.371147,-0.0389023,-0.0257086,-0.968682,-0.681951,-0.874704,Ü9987á, 0.584236,-0.974667, 0.10798, 0.996311,-0.767106, 0.926913<Histogram@snapshots, 100, "PDF", PlotRangeÆ80, 2<D-0.50.00.51.00.51.01.52.0r@x_D:=1p1-x2Plot@r@xD,8x,-1, 1<, PlotRange->80, 2<D1.52.0

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12CHAPTER 1. THE WAVE FUNCTION-0.50.00.51.00.51.01.5r@x_D:=1p1-x2Plot@r@xD,8x,-1, 1<, PlotRange->80, 2<D-1.0-0.50.51.00.51.01.52.0Show@Histogram@snapshots, 100, "PDF", PlotRangeÆ80, 2<D,Plot@r@xD,8x,-1, 1<, PlotRange->80, 2<DD-0.50.00.51.00.51.01.52.021p13.nbProblem 1.14(a)Pab(t) =∫ba|Ψ(x, t)|2dx,sodPabdt=∫ba∂∂t|Ψ|2dx.But (Eq. 1.25):∂|Ψ|2∂t=∂∂x[iℏ2m(Ψ∗∂Ψ∂x−∂Ψ∗∂xΨ)]=−∂∂x J(x, t).∴dPabdt=−∫ba∂∂x J(x, t)dx=−[J(x, t)]|ba=J(a, t)−J(b, t).QEDProbability is dimensionless, soJhas the dimensions 1/time, and unitsseconds−1.(b)Here Ψ(x, t) =f(x)e−iat, wheref(x)≡Ae−amx2/ℏ, so Ψ∂Ψ∗∂x=f e−iat dfdxeiat=fdfdx,and Ψ∗∂Ψ∂x=fdfdxtoo, soJ(x, t) = 0.

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CHAPTER 1. THE WAVE FUNCTION13Problem 1.15Use Eqs. [1.23] and [1.24], and integration by parts:ddt∫∞−∞Ψ∗1Ψ2dx=∫∞−∞∂∂t(Ψ∗1Ψ2)dx=∫∞−∞(∂Ψ∗1∂tΨ2+ Ψ∗1∂Ψ2∂t)dx=∫∞−∞[(−iℏ2m∂2Ψ∗1∂x2+iℏVΨ∗1)Ψ2+ Ψ∗1(iℏ2m∂2Ψ2∂x2−iℏVΨ2)]dx=−iℏ2m∫∞−∞(∂2Ψ∗1∂x2Ψ2−Ψ∗1∂2Ψ2∂x2)dx=−iℏ2m[∂Ψ∗1∂xΨ2∣∣∣∣∞−∞−∫∞−∞∂Ψ∗1∂x∂Ψ2∂xdx−Ψ∗1∂Ψ2∂x∣∣∣∣∞−∞+∫∞−∞∂Ψ∗1∂x∂Ψ2∂xdx]= 0.QEDProblem 1.16(a)1 =|A|2∫a−a(a2−x2)2dx= 2|A|2∫a0(a4−2a2x2+x4)dx= 2|A|2[a4x−2a2x33+x55]∣∣∣∣a0= 2|A|2a5(1−23 + 15)= 1615a5|A|2,soA=√1516a5.(b)〈x〉=∫a−ax|Ψ|2dx=0.(Odd integrand.)(c)〈p〉=ℏi A2∫a−a(a2−x2)ddx(a2−x2)︸︷︷︸−2xdx=0.(Odd integrand.)Since we only know〈x〉att= 0 we cannot calculated〈x〉/dtdirectly.(d)〈x2〉=A2∫a−ax2(a2−x2)2dx= 2A2∫a0(a4x2−2a2x4+x6)dx= 21516a5[a4x33−2a2x55+x77]∣∣∣∣a0=158a5(a7)(13−25 + 17)=15a28(35−42 + 153·5·7)=a28·87 =a27.

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14CHAPTER 1. THE WAVE FUNCTION(e)〈p2〉=−A2ℏ2∫a−a(a2−x2)d2dx2(a2−x2)︸︷︷︸−2dx= 2A2ℏ22∫a0(a2−x2)dx= 4·1516a5ℏ2(a2x−x33)∣∣∣∣a0= 15ℏ24a5(a3−a33)= 15ℏ24a2·23 =52ℏ2a2.(f )σx=√〈x2〉 − 〈x〉2=√17a2=a√7.(g)σp=√〈p2〉 − 〈p〉2=√52ℏ2a2=√52ℏa .(h)σxσp=a√7·√52ℏa=√514ℏ=√107ℏ2>ℏ2.XProblem 1.17(a)Eq. 1.24 now reads∂Ψ∗∂t=−iℏ2m∂2Ψ∗∂x2+iℏV∗Ψ∗, and Eq. 1.25 picks up an extra term:∂∂t|Ψ|2=· · ·+iℏ|Ψ|2(V∗−V) =· · ·+iℏ|Ψ|2(V0+iΓ−V0+iΓ) =· · · −2Γℏ|Ψ|2,and Eq. 1.27 becomesdPdt=−2Γℏ∫∞−∞|Ψ|2dx=−2ΓℏP.QED(b)dPP=−2Γℏdt=⇒lnP=−2Γℏt+ constant =⇒P(t) =P(0)e−2Γt/ℏ,soτ=ℏ2Γ.Problem 1.18h√3mkBT > d⇒T <h23mkBd2.(a)Electrons (m= 9.1×10−31kg):T <(6.6×10−34)23(9.1×10−31)(1.4×10−23)(3×10−10)2=1.3×105K.Silicon nuclei (m= 28mp= 28(1.7×10−27) = 4.8×10−26kg):T <(6.6×10−34)23(4.8×10−26)(1.4×10−23)(3×10−10)2=2.4 K.

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CHAPTER 1. THE WAVE FUNCTION15(b)P V=N kBT; volume occupied by one molecule (N= 1, V=d3)⇒d= (kBT /P)1/3.T <h23mkB(PkBT)2/3⇒T5/3< h23mP2/3k5/3B⇒T <1kB(h23m)3/5P2/5.For helium (m= 4mp= 6.8×10−27kg) at 1 atm = 1.0×105N/m2:T <1(1.4×10−23)((6.6×10−34)23(6.8×10−27))3/5(1.0×105)2/5=2.8 K.For atomic hydrogen (m=mp= 1.7×10−27kg) withd= 0.01 m:T <(6.6×10−34)23(1.7×10−27)(1.4×10−23)(10−2)2=6.2×10−14K.At 3 K it is definitely in the classical regime.

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16CHAPTER 2. THE TIME-INDEPENDENT SCHR ¨ODINGER EQUATIONChapter 2The Time-Independent Schr¨odingerEquationProblem 2.1(a)Ψ(x, t) =ψ(x)e−i(E0+iΓ)t/ℏ=ψ(x)eΓt/ℏe−iE0t/ℏ=⇒ |Ψ|2=|ψ|2e2Γt/ℏ.∫∞−∞|Ψ(x, t)|2dx=e2Γt/ℏ∫∞−∞|ψ|2dx.The second term is independent oft, so if the product is to be 1 for all time, the first term (e2Γt/ℏ) mustalso be constant, and hence Γ = 0.QED(b)Ifψsatisfies Eq. 2.5,−ℏ22md2ψdx2+V ψ=Eψ, then (taking the complex conjugate and noting thatVandEare real):−ℏ22md2ψ∗dx2+V ψ∗=Eψ∗, soψ∗alsosatisfies Eq. 2.5. Now, ifψ1andψ2satisfy Eq. 2.5, sotoo does any linear combination of them (ψ3≡c1ψ1+c2ψ2):−ℏ22md2ψ3dx2+V ψ3=−ℏ22m(c1d2ψ1dx2+c2d2ψ2dx2)+V(c1ψ1+c2ψ2)=c1[−ℏ22md2ψ1dx2+V ψ1]+c2[−ℏ22md2ψ2dx2+V ψ2]=c1(Eψ1) +c2(Eψ2) =E(c1ψ1+c2ψ2) =Eψ3.Thus, (ψ+ψ∗) andi(ψ−ψ∗) – both of which arereal– satisfy Eq. 2.5.Conclusion:From any complexsolution, we can always construct tworealsolutions (of course, ifψis already real, the second one will bezero). In particular, sinceψ=12[(ψ+ψ∗)−i(i(ψ−ψ∗))], ψcan be expressed as a linear combination oftwo real solutions.QED(c)Ifψ(x) satisfies Eq. 2.5, then, changing variablesx→ −xand noting thatd2/d(−x)2=d2/dx2,−ℏ22md2ψ(−x)dx2+V(−x)ψ(−x) =Eψ(−x);so ifV(−x) =V(x) thenψ(−x)alsosatisfies Eq. 2.5. It follows thatψ+(x)≡ψ(x) +ψ(−x) (which iseven:ψ+(−x) =ψ+(x)) andψ−(x)≡ψ(x)−ψ(−x) (which isodd:ψ−(−x) =−ψ−(x)) both satisfy Eq.2.5. Butψ(x) =12(ψ+(x) +ψ−(x)), so any solution can be expressed as a linear combination of even andodd solutions.QED

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