(c)The following nnn values have E(8ma2/h2)=27:115,151,511,333.The degree of degeneracy is4. 3.46 (a)These are linearly independent since none of them can be written as a linea combination of the others. (b)Since 3x2-1=3(x2)(8).these are not linearly independent. (c)Linearly independent. (d)Linearly independent. (e)Since e=cosx+isinx,these are linearly dependent. (f)Since 1=sinx+cosx,these are linearly dependent. (g)Linearly independent. 3.47 See the beginning of Sec.3.6 for the proof. 3.48 (a)(x)=xf(x)Plg(y))Pdxdyd== xf(x)xg(y)dy(Pd,where f.g.and h are given preceding Eq. (3.72).Since g and h are normalized,(x)=xf(x)Pdx=(2/a)o xsin2(n zx/a)dx 2 x2 ax af4m2naa-aro2naxl回-受whee4a3ssad a (b)By symmetry,(y)=b/2 and ()=c/2. (c)The derivation of Eq (3.92)for the ground state applies to any state,and (p=0. (d)Since g and h are normalized, (x2)=x2f(x)Pdx=(2/a)x2sin2(nxla)dx= a6气4n,R875im2,rxo)- 2x3ar2a3） n cos(2x/a) 32mx where Eq.(A.4)was used.We have (x)2=a2/4(x2).Also. (xy)=f(x)Pg(y))P dxdyd== oxf(x)Pdxoylg(y)PdyhP=(xXy). 3.49(A+B)=∫平*(A+B)Ψdr=∫平*(A平+Ψ)dr=∫平*Ψdr+∫平*平dr= (A)+(B).Also(cB)=∫平*(cB)Ψdr=cf*Bydr=c(B) 3-9 Copyright2014 Pearson Education,Inc
Copyright © 2014 Pearson Education, Inc. 3-9 (c) The following x y z nnn values have 2 2 E ma h (8 / ) = 27; 115, 151, 511, 333. The degree of degeneracy is 4. 3.46 (a) These are linearly independent since none of them can be written as a linear combination of the others. (b) Since 2 2 1 8 3 1 3( ) (8), x x −= − these are not linearly independent. (c) Linearly independent. (d) Linearly independent. (e) Since cos sin , ix e xi x = + these are linearly dependent. (f) Since 2 2 1 sin cos , = +x x these are linearly dependent. (g) Linearly independent. 3.47 See the beginning of Sec. 3.6 for the proof. 3.48 (a) 2 22 000 | ( )| | ( )| | ( )| c b a 〈 〉=∫ ∫ ∫ = x x f x g y h z dx dy dz 2 22 0 00 | ( )| | ( )| | ( )| , a b c ∫ ∫∫ x f x dx g y dy h z dz where f, g, and h are given preceding Eq. (3.72). Since g and h are normalized, 2 2 0 0 | ( ) | (2/ ) sin ( / ) a a x 〈 〉=∫ = ∫ x x f x dx a x n x a dx π = 2 2 2 2 0 2 sin(2 / ) cos(2 / ) 44 2 8 a x x x x x ax a a n xa n xa a n n π π π π ⎡ ⎤ ⎢ ⎥ −− = ⎣ ⎦ , where Eq. (A.3) was used. (b) By symmetry, 〈 〉= y b/2 and 〈z c 〉 = /2. (c) The derivation of Eq. (3.92) for the ground state applies to any state, and 0. x 〈 〉= p (d) Since g and h are normalized, 2 2 2 22 0 0 | ( ) | (2/ ) sin ( / ) a a x 〈 〉=∫ = ∫ = x x f x dx a x n x a dx π 3 23 2 33 22 0 2 sin(2 / ) cos(2 / ) 6 4 8 4 a x x x x x x ax a a x n xa n xa a n n n π π π π π ⎡ ⎤ ⎛ ⎞ ⎢ ⎥ −− − = ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ 2 2 2 2 3 2 x a a n π − , where Eq. (A.4) was used. We have 22 2 〈x〉 = ≠〈 〉 a x /4 . Also, 2 22 000 | ( )| | ( )| | ( )| c b a 〈 〉=∫ ∫ ∫ xy xy f x g y h z dx dy dz = 2 22 000 | ( )| | ( )| | ( )| a b c ∫∫∫ x f x dx y g y dy h z dz = 〈x〉〈 〉 y . 3.49 ˆˆ ˆ ˆˆ ˆ 〈 + 〉= Ψ + Ψ = Ψ Ψ+ Ψ = Ψ Ψ + Ψ Ψ = AB AB d A B d Ad Bd *( ) *( ) * * τ ττ τ ∫ ∫ ∫∫ 〈 〉+〈 〉 A B . Also ˆ ˆ 〈 〉= Ψ Ψ = Ψ Ψ = 〈 〉 cB cB d c B d c B *( ) * . τ τ ∫ ∫
350(a)Not it is not quadratically integrable.This is obvious froma graphor frome dx=-(1/2a)e2 (b)This is acceptable.since it is single-valued,continuous,and quadratically integrable when multiplied by a normalization constant.See Eqs.(4.49)and(A.9). (c)This is acceptable,since it is single-valued,continuous,and quadratically integrable when multiplied by a normalization constant.See Egs.(4.49)and (A.10)with n=1. (d)Acceptable for the same reasons as in(b). (e)Not acceptable since it is not continuous atx=0. 3.51 Given:ihaΨ，/at=Ψ1 and ihaΨ，lat-Ψ，.Prove that iha(c出1+c2Ψ2)/at=H(cΨ1+c2Ψ2).We have iho(c出1+c2Ψ2)/a= ia(GΨ1)/at+a(c2Ψ2)/a]=ciha,/a1+c2iha2/at=GΨ+c2HΨ2= H(cΨ1+c2Ψz),since H is linear. 3.52(a)An inefficient C++program is #include <iostream> using namespace std; int main({ intm,i,j,k,nx,y,nz,L[4001.N[400],R[4001,S[400]: i=0: for (nx=1:nx<8:nx=nx+1){ for (ny=1;ny<8;ny=ny+1){ for (nz=1;nz<8;nz =nz+1){1 m-n> x+ny*ny+nz*nz if (m> continue; i=i+1 L0=m; N[i]=nx; R[i]=ny; Slil=nz: for(k=3;k<61;k=k+1)1 0时*)1 for (j= j<- cout<<N[]<<""<<R[]<<""<<S[]<<""<<L[j]<<endl: return 0; 3-10 Copyright2014 Pearson Education,In
Copyright © 2014 Pearson Education, Inc. 3-10 3.50 (a) Not acceptable, since it is not quadratically integrable. This is obvious from a graph or from 2 2 (1/2 ) | . ax ax e dx a e ∞− − ∞ −∞ −∞ ∫ =− =∞ (b) This is acceptable, since it is single-valued, continuous, and quadratically integrable when multiplied by a normalization constant. See Eqs. (4.49) and (A.9). (c) This is acceptable, since it is single-valued, continuous, and quadratically integrable when multiplied by a normalization constant. See Eqs. (4.49) and (A.10) with n = 1. (d) Acceptable for the same reasons as in (b). (e) Not acceptable since it is not continuous at x = 0. 3.51 Given: 1 1 ˆ i tH = ∂Ψ ∂ = Ψ / and 2 2 ˆ i tH = ∂Ψ ∂= Ψ / . Prove that 11 2 2 11 2 2 ˆ i c c t Hc c = ∂ Ψ+ Ψ ∂= Ψ+ Ψ ( )/ ( ). We have 11 2 2 ic c t =∂( )/ Ψ+ Ψ ∂= 11 2 2 ic t c t =[ ( )/ ( )/ ] ∂ Ψ ∂ +∂ Ψ ∂ = 11 2 2 ci t c i t = = ∂Ψ ∂ + ∂Ψ ∂ = / / 1 12 2 ˆ ˆ cH c H Ψ + Ψ= 11 2 2 ˆHc c ( ) Ψ+ Ψ , since Hˆ is linear. 3.52 (a) An inefficient C++ program is #include <iostream> using namespace std; int main() { int m, i, j, k, nx, ny, nz, L[400], N[400], R[400], S[400]; i=0; for (nx=1; nx<8; nx=nx+1) { for (ny=1; ny<8; ny=ny+1) { for (nz=1; nz<8; nz=nz+1) { m=nx*nx+ny*ny+nz*nz; if (m>60) continue; i=i+1; L[i]=m; N[i]=nx; R[i]=ny; S[i]=nz; } } } for (k=3; k<61; k=k+1) { for (j=1; j<=i; j=j+1) { if (L[j]==k) cout<<N[j]<< " "<<R[j]<< " "<<S[j]<< " "<<L[j]<<endl; } } return 0;
egrated development environn ent(IDE)to debug and run C++progr available at www.codebloc k .org.Fo mingw-set p.ex as par de the for C+ ompiler .Free user guides and manuals for Code:Blocks can be found by searching th Internet. Alternatively,you can run the program at ideone.com. (b)One finds 12 states 3.53 (a)T.(b)F.See the paragraph preceding the example at the end of Sec.3.3 (c)F.This is only true iffi andf have the same eigenvalue. (d)F.(e)F.This is only true if the two solutions have the same energy eigenvalue. (f)F.This is only true for stationary states. (g)F.(h)F.x(5x)(const.)(5x). ①T.平=i(eahw）=eBiw=EeBw=EΨ )T.kT.①F. (m)T.f=(4)=A(af)=a4f=a2f.provided A is linear.Note that the definition of eigenfunction and eigenvalue in Sec.3.2 specified that A is linear ()F.(o)F. 3-11 Copyright2014 Pearson Education,Inc
Copyright © 2014 Pearson Education, Inc. 3-11 } A free integrated development environment (IDE) to debug and run C++ programs is Code::Blocks, available at www.codeblocks.org. For a Windows computer, downloading the file with mingw-setup.exe as part of the name will include the MinGW (GCC) compiler for C++. Free user guides and manuals for Code::Blocks can be found by searching the Internet. Alternatively, you can run the program at ideone.com. (b) One finds 12 states. 3.53 (a) T. (b) F. See the paragraph preceding the example at the end of Sec. 3.3. (c) F. This is only true if f1 and f2 have the same eigenvalue. (d) F. (e) F. This is only true if the two solutions have the same energy eigenvalue. (f) F. This is only true for stationary states. (g) F. (h) F. x(5 ) (const.)(5 ). x x ≠ (i) T. // / ˆˆ ˆ () . iEt iEt iEt H H e e H Ee E ψ ψψ −− − Ψ= = = = Ψ == = (j) T. (k) T. (l) F. (m) T. ˆ ˆˆ ˆ ˆ 2 2 A f A Af A af aAf a f = = == ( ) () , provided Aˆ is linear. Note that the definition of eigenfunction and eigenvalue in Sec. 3.2 specified that Aˆ is linear. (n) F. (o) F
Chapter 4 The Harmonic Oscillator 4.1 Taking (d/dx)"of(4.84)gives(x)=Cm(n-1Xn-2).(m-m+lXx-a)"- The factors n,(n-1),.make the terms with n=0,n=1,.,n=m-I vanish,so ()(-2)(m)(xa)(q.1).(If this is too abstract for you,write the expansion as f(x)=co+cx+czx++c+.and do the differentiation.)With x=a in Eq.1,the (x-a)"-m factor makes all terms equal to zero except the term with n=m,which is a constant.Equation(1)withx=a gives f(m(a)=cmm(m-1)(m-2).(m-m+1)=cmm!and cm=fm (a)/m!. 4.2 (a)f(x)=sinx.f'(x)=cosx,f"(x)=-sinx.f"(x)=-cosx,f(i(x)=sinx. a=0andf(0)=sin0=0,f'(0)=cos0=l,f"(0)=0,f"(0)=-l,fm(0)=0,.The Taylor series is sinx=11)! (b)c0sx=1/1l-3x2131+5x/51-=1-x2/21+x4/41-.=∑(-1)x2*1(2k)1 4.3 (a)We use (4.85)with a=0.We have f(x)=e*and f(m(x)=e*.f(m(0)=e=1. S0e=1+x/1+x2121+x2/3+=∑0x1ml (b)e°=1+(i0)/1+(i0)2/21+(i0)3131+(i0)4/41+(i0)3/51+.= 1-82/21+84/41-.+i(0/1l-83/31+85/51-)=cos0+isin0. 4.4 From (4.22)and (4.28).dx/dt 2vAcos(2xvt+b)and T=2mn2v242 cos?(2xvt+b). From (4.22)and (4.27),V=22v2mA2 sin2(2xvt+b).Then T+V=2x2v2mA2,since sin20+cos20=1. 4.5(a)Let y=∑ecnx”.Theny=∑ncnx"and y'=∑nn-l)cnx"-2.Since the first two terms in thesum are zero,we havey")c Let j=n-2.Then y'=∑，U+2U+l)e+2x/=∑e(n+2n+1)c+2x.Substitution in the differential equation gives 4-1 Education.Inc
4-1 Copyright © 2014 Pearson Education, Inc. Chapter 4 The Harmonic Oscillator 4.1 Taking (/ )m d dx of (4.84) gives ( ) 0 ( ) ( 1)( 2) ( 1)( ) m nm n n f x cnn n n m x a ∞ − = = − − −+ − ∑ " . The factors n, ( 1),. n − make the terms with 0, n = 1, n = ., n m= −1 vanish, so ( )( ) ( 1)( 2) ( 1)( ) m nm n n m f x cnn n n m x a ∞ − = = − − −+ − ∑ " (Eq. 1). (If this is too abstract for you, write the expansion as 2 01 2 ( ) k k f x c cx c x c x = + + ++ + " " and do the differentiation.) With x = a in Eq. 1, the ( )n m x a − − factor makes all terms equal to zero except the term with n m= , which is a constant. Equation (1) with x = a gives ( )( ) ( 1)( 2) ( 1) ! m m m f a c mm m m m c m = − − −+= " and ( )( )/ ! m mc f am = . 4.2 (a) (iv) f ( ) sin , ( ) cos , ( ) sin , ( ) cos , ( ) sin , x xfx xf x xf x xf x x = = =− =− = ′ ′′ ′′′ .; a = 0 and (iv) f f ff f (0) sin 0 0, (0) cos0 1, (0) 0, (0) 1, (0) 0, = = = = = =− = ′ ′′ ′′′ . The Taylor series is 3 5 sin 0 / 1! 0 / 3! 0 / 5! xx x x =+ +− ++ + = " 2 1 0 ( 1) / (2 1)! k k k x k ∞ + = ∑ − + . (b) 2 4 24 2 0 cos 1/ 1! 3 / 3! 5 / 5! 1 / 2! / 4! ( 1) / (2 )! k k k x x x x x xk ∞ = = − + −= − + − = − " ∑ . 4.3 (a) We use (4.85) with a = 0. We have ( ) x f x e = and ( )() . n x f x e = () 0 (0) 1. n f e = = So 2 3 0 1 / 1! / 2! / 3! / ! x n n e x x x xn ∞ = =+ + + + = " ∑ . (b) 2345 1 ( ) / 1! ( ) / 2! ( ) / 3! ( ) / 4! ( ) / 5! i ei i i i i θ =+ + + + + + = θθ θ θ θ " 24 35 1 / 2! / 4! ( / 1! / 3! / 5! ) − + −+ − + − θ θ θθ θ " " i = cos sin . θ + i θ 4.4 From (4.22) and (4.28), dx dt A t b / 2 cos(2 ) = πν πν + and 22 2 2 T m A tb = 2 cos (2 ). π ν πν + From (4.22) and (4.27), 22 2 2 V mA t b = + 2 sin (2 ). π ν πν Then 22 2 T V mA + = 2 , π ν since 2 2 sin cos 1. θ θ + = 4.5 (a) Let 0 n n n y c x ∞ = = ∑ . Then 1 0 n n n y nc x ∞ − = ′ = ∑ and 2 0 ( 1) n n n y nn cx ∞ − = ′′ = − ∑ . Since the first two terms in the y′′ sum are zero, we have 2 2 ( 1) . n n n y nn cx ∞ − = ′′ = − ∑ Let j n ≡ − 2. Then 2 2 0 0 ( 2)( 1) ( 2)( 1) j n j n j n y j j cx n n cx ∞ ∞ = = + + ′′ = ++ = ++ ∑ ∑ . Substitution in the differential equation gives
∑，n+2n+1c2-∑n0n-lc,-2∑c+3∑cn=0. We have∑ln+2n+lcn+2+(6-n-产cnr”=0.Setting the coffcnt of equal to zero,we have c2 =(n2+n-3)c/[(n+2)(n+1)]. (b)The recursion relation of (a)with n=0 gives c2=-3co/2 and with n=2 gives ca=3c2/12=c2/4=(-3co/2)/4=-3co/8.With n=1 and n=3 in the recursion relation,we get c3=-c/6 and cs=9c3/20=9(-c/6)/20=-3c/40. 4.6 (a)Odd;(b)even;(c)odd;(d)neither,(e)even;(f)odd;(g)neither,(h)even 4.7 Given:f(-x)=f(x),g(-x)=g(x),h(-x)=-h(x),k(-x)=-k(x). Let p(x)=f(x)g(x).We have p(-x)=f(-x)g(-x)=f(x)g(x)=p(x).so the product oftwo even functions is an even function.Let (x)=h(x)k(x).Then g(-x)=h(-x)k(-x)=-h(x)[-k(x)]=h(x)k(x)=g(x),so the product of two odd functions is an even function.Let r(x)=f(x)(x).Then r(-x)=f(-x)h(-x)=f(x)[-h(x)]=-f(x)h(x)=-r(x). 4.8 (a)Given:f(x)=f(-x).Differentiation of this equation gives f'(x)=df(-x)/dx=f'(-x)d(-x)/dx]=-f(-x),so f'is an odd function. (b)Differentiation of f(x)=-f(-x)gives f'(x)=-(-1)f(-x)=f'(-x). (c)Differentiation of f(x)=f(-x)gives f'(x)=-f(-x).as in (a).Putting x=0 in this equation,we get f(0)=-f(0),so 2f(0)=0 and f(0)=0. 4.9 T)=fv*ivdr=-(12m)(alr)2eP(dld )e-un ds= -(h2I2mXalz)eaP(ax-a)e-wiRdx= -(h2/2m)al)2 (a2x2-a)e-ax dx= -(h21ma/π)2[a20/4(π21a32)-a1/2πla)2]=h2a/4m= h2(2vm/h)/4m=hv/4.where (A.9)and (A.10)were used. V)=SwVydr=(alrewRvmxe-win dx= (al)25 (2n'v2mx2)e-a dx=4rav2m(U/4n2 la)='v2mla π2v2m/2πvmh)=hv/4=(T). 4.10 From (4.54).1=Pea dx=21aeads=21c P2/a2.where (4.49)and (A.10)with n=1 were used.We get Ic=4.From(4.56). 4-2 Copyright2014 Pearson Education.Inc
4-2 Copyright © 2014 Pearson Education, Inc. 2 0 0 00 ( 2)( 1) ( 1) 2 3 0 n n nn n n nn n n nn n n c x n n c x nc x c x ∞ ∞ ∞∞ ∑ ∑ ∑∑ = = == ++ − − − + = + . We have 2 2 0 [( 2)( 1) (3 ) ] 0. n n n n n n c n ncx ∞ ∑ = + + + −− = + Setting the coefficient of n x equal to zero, we have 2 2 ( 3) / [( 2)( 1)]. n n c nn c n n + = +− + + (b) The recursion relation of (a) with n = 0 gives 2 0 c c = −3 /2 and with n = 2 gives 42 2 0 0 cc c c c = = = − =− 3 / 12 /4 ( 3 / 2)/4 3 /8. With n = 1 and n = 3 in the recursion relation, we get 3 1 c c = − /6 and 53 1 1 cc c c = 9 /20 9( /6)/20 3 /40. = − =− 4.6 (a) Odd; (b) even; (c) odd; (d) neither; (e) even; (f) odd; (g) neither; (h) even. 4.7 Given: f ( ) ( ), ( ) ( ), ( ) ( ), ( ) ( ). − = − = − =− − =− x f x g x gx h x hx k x kx Let p( ) ( ) ( ). x f xgx ≡ We have p( ) ( ) ( ) ( ) ( ) ( ), −x f xg x f xgx px = − −= = so the product of two even functions is an even function. Let qx hxkx ( ) ( ) ( ). ≡ Then q x h xk x hx kx hxkx qx ( ) ( ) ( ) ( )[ ( )] ( ) ( ) ( ), − = − − =− − = = so the product of two odd functions is an even function. Let rx f xhx ( ) ( ) ( ). = Then r x f xh x f x hx f xhx rx ( ) ( ) ( ) ( )[ ( )] ( ) ( ) ( ) − ≡ − − = − =− =− . 4.8 (a) Given: ( ) ( ). f x fx = − Differentiation of this equation gives f ′′ ′ ( ) ( )/ ( )[ ( )/ ] ( ), x df x dx f x d x dx f x = − = − − =− − so f ′ is an odd function. (b) Differentiation of ( ) ( ) f x fx =− − gives ( ) ( 1) ( ) ( ). f ′ x fx fx = −− − = − ′ ′ (c) Differentiation of ( ) ( ) f x fx = − gives ( ) ( ), f ′ x fx = − −′ as in (a). Putting x = 0 in this equation, we get (0) (0), f f ′ ′ = − so 2 (0) 0 f ′ = and (0) 0 f ′ = . 4.9 2 2 2 1/2 /2 2 2 /2 ˆ * ( /2 ) ( / ) ( / ) x x T T d m e d dx e dx α α ψ ψ τ απ ∞ − − −∞ 〈 〉= =− ∫ ∫ = = 2 2 2 1/2 /2 2 2 /2 ( /2 )( / ) ( ) x x m e x e dx α α απ α α ∞ − − − ∫− −∞ = = 2 2 1/2 2 2 0 ( /2 )( / ) 2 ( ) x m x e dx α απ α α ∞ − − ∫− = = 2 1/2 2 1/2 3/2 1/2 − −= ( / )( / ) [ (1/4)( / ) (1/2)( / ) ] = m απ α π α α πα 2 = α/4m = 2 = = (2 / )/4 /4, πν ν m mh = where (A.9) and (A.10) were used. 2 2 1/2 /2 2 2 2 /2 ˆ * ( / ) (2 ) x ax V V d e mx e dx α ψ ψ τ απ πν ∞ − − −∞ 〈 〉= = = ∫ ∫2 1/2 2 2 2 0 ( / ) 2 (2 ) ax απ πν mx e dx ∞ − ∫ = 3/2 1/2 2 1/2 3/2 2 2 4 (1/4)( / ) / π α ν π α πν α m m = = 22 1 π ν πν ν mm h T /(2 ) /4 . − = = =〈 〉 4.10 From (4.54), 2 2 2 2 2 2 2 1/2 3/2 1 1 10 1 4 1 | | 2| | 2| | / x x c x e dx c x e dx c α α π α ∞ ∞ − − == = ∫ ∫ −∞ , where (4.49) and (A.10) with n = 1 were used. We get 1/2 3/4 1/4 1 | |2 c α π− = . From (4.56)