(b)(ii-37+2)x2=(x+2)2-3x+1)2+2x2=-2x+1 3.22eDfx)=1+D+212+D331+)fx)=fx)+fx)+f"(x)/21+f"(x)/31+. Tif(x)=f(x+1).The Taylor series(4.85)in Prob.4.1 with x changed to gives f(=)=f(a)+f'(a)(=-a)/1!+f"(a)(=-a)2/21+.Letting h==-a,the Taylor series becomes f(a+h)=f(a)+f'(a)h/!+f"(a)h2/2!+.Changing a to x and letting h=1.we get f(x+1)=f(x)+f(x)/1!+f"(x)/2!+.which shows that ef=ff. 3.23 (a)(d2/dx2)e*=e*and the eigenvalue is 1. (b)(d2/dx2)x2=2 and x2 is not an eigenfunction of d2/dx2 (c)(d2/dx2)sinx=(d/dx)cosx=-sinx and the eigenvalue is-1 (d)(d2/dx2)3cosx=-3cosx and the eigenvalue is-1. (e)(d2/dx2)(sinx+cosx)=-(sinx+cosx)so the eigenvalue is-1. 3.24 (a)(2/x2+210y2(ee)=4e2ey+9e2e=13e2ey.The eigenvalue is 13. (b)(2/x2+02/0y2)(xy3)=6xy+6xy.Not an eigenfunction. (c) (2/x2+2/2)(sin 2xcos4y)=-4sin2xcos4y-16sin 2xcos4y=-20sin 2xcos4y The eigenvalue is -20. (d)(2/x2+2/2)(sin 2x+cos3y)=-4sin 2x-9cos3y.Not an eigenfunction 3.25 -(h2/2m)(d2/dx2)g(x)=kg(x)and g"(x)+(2m/h2)kg(x)=0.This is a linear homogenous differential equation with constant coefficients.The auxiliary equation is s2+(2m/h2)k=0 and s=ti(2mk)2/h.The general solution is gce If the eigenvalue k were a negative number.then would be a pure imaginary number,that is,=ib,where bis real and positive.This would make ika real negative number and the first exponential ing would go toas xo and the second exponential would go to oo as x>oo.Likewise,if k were an imaginary number (k=a+bi=re,where a and b are real and b is nonzero).then would have the form c+id,and ik would have the form-d+ic,where c and d are real.This would make the exponentials go to infinity as x goes to plus or minus infinity. Hence to keepg finite asx the eigenvaluekmust be real and nonnegative,and the allowed eigenvalues are all nonnegative numbers. 3-4 Copyright014 Pearson Education,Inc
Copyright © 2014 Pearson Education, Inc. 3-4 (b) 2 2 22 11 1 ˆˆ ˆ ( 3 2) ( 2) 3( 1) 2 2 1. TT T x x x x x − + = + − + + =− + 3.22 ˆ 2 3 ˆ ˆ ( ) (1 /2! /3! ) ( ) ( ) ( ) ( )/2! ( )/3! . De fx D D D fx fx f x f x f x =++ + + = + + + + " " ′ ′′ ′′′ 1 ˆTf x f x ( ) ( 1). = + The Taylor series (4.85) in Prob. 4.1 with x changed to z gives 2 fz fa f a z a f a z a ( ) ( ) ( )( ) / 1! ( )( ) /2! . =+ − + − + ′ ′′ " Letting , h za ≡ − the Taylor series becomes 2 f a h f a f ah f ah ( ) ( ) ( ) / 1! ( ) /2! . += + + + ′ ′′ " Changing a to x and letting h = 1, we get ( 1) ( ) ( ) / 1! ( )/2! , fx fx f x f x += + + + ′ ′′ " which shows that ˆ 1 ˆ . De f Tf = 3.23 (a) 2 2 (/ ) x x d dx e e = and the eigenvalue is 1. (b) 2 22 (/ ) 2 d dx x = and 2 x is not an eigenfunction of 2 2 d dx / . (c) 2 2 ( / )sin ( / )cos sin d dx x d dx x x = =− and the eigenvalue is –1. (d) 2 2 ( / )3cos 3cos d dx x x = − and the eigenvalue is –1. (e) 2 2 ( / )(sin cos ) (sin cos ) d dx x x x x + =− + so the eigenvalue is –1. 3.24 (a) 2 2 2 2 23 23 23 23 ( / / )( ) 4 9 13 . x y xy xy xy ∂ ∂ +∂ ∂ = + = x y ee ee ee ee The eigenvalue is 13. (b) 2 2 2 2 33 3 3 ( / / )( ) 6 6 . ∂ ∂ +∂ ∂ = + x y x y xy x y Not an eigenfunction. (c) 22 22 ( / / )(sin 2 cos 4 ) 4sin 2 cos 4 16sin 2 cos 4 20sin 2 cos 4 . ∂ ∂ +∂ ∂ =− − =− x y xy xy xy xy The eigenvalue is −20. (d) 22 22 ( / / )(sin 2 cos3 ) 4sin 2 9cos3 . ∂ ∂ +∂ ∂ + =− − x yxy x y Not an eigenfunction, 3.25 2 22 − = ( /2 )( / ) ( ) ( ) = m d dx g x kg x and 2 g x m kg x ′′( ) (2 / ) ( ) 0. + = = This is a linear homogenous differential equation with constant coefficients. The auxiliary equation is 2 2 s mk + = (2 / ) 0 = and 1/2 s i mk = ± (2 ) / . = The general solution is 1/2 1/2 (2 ) / (2 ) / 1 2 . i mk x i mk x g ce c e− = += = If the eigenvalue k were a negative number, then 1/2 k would be a pure imaginary number; that is, 1/2 k ib = , where b is real and positive. This would make 1/2 ik a real negative number and the first exponential in g would go to ∞ as x → −∞ and the second exponential would go to ∞ as x → ∞. Likewise, if k were an imaginary number ( , i k a bi re θ =+ = where a and b are real and b is nonzero), then 1/2 k would have the form c id + , and 1/2 ik would have the form −d ic + , where c and d are real. This would make the exponentials go to infinity as x goes to plus or minus infinity. Hence to keep g finite as , x → ±∞ the eigenvalue k must be real and nonnegative, and the allowed eigenvalues are all nonnegative numbers
3.26 (dx)f=fdx=kf.Differentiation of both sides of this equation gives (dldx)jf dx=f=kf'.So dfldx=kf and (f)df=kdx.Integration gives Inf=kx+c and f=e=ek,where A is a constant and k is the eigenvalue.To prevent the eigenfunctions from becoming infinite asxkmust be a pure imaginary number.(Strictly speaking,Ae is an eigenfunction of x only if we omit the arbitrary constant of integration.) 3.27 d2f/dx2+2df/dx=kf and f+2f-kf =0.The auxiliary equation is s2+2s-k=0 and s=-1(1+k)2.So f=Ael-(+Bel-1-(where A and B are arbitrary constants.To prevent the eigenfunctions from becoming infinite asx the factors multiplying x must be pure imaginary numbers:-1(1+k)=ci,where c is an arbitrary real number.So(+k)2=1+ci and 1+k=(1+ci)=1+2ic-c and k=2ic-2 3.28(ap=(h3(a/@y)3=iia31ay23: (b)炉，-p.=x(h/i)8/a-yh/i)8/ax (c[x(h/i)8/afx,y)=-(xa/@xaf1y)=-2(x2a2f1@2). Hence ()2=-h2(x202/0y2). 3.29 (h/i)(dg/dx)=kg and dg/g =(ik/h)dx.Integration gives Ing=(ik/h)x+C and g=ec=Ae,where C and A are constants.If k were imaginary (k=a+bi. where a and b are real and b is nonzero),then ik=ia-b,and the e factor ing makes g go to infinity asx goes to minus infinity if b is positive or asx goes to infinity if b is negative.Hence b must be zero and k =a,where a is a real number. 3.30 (a)[]f=(hli)[xalax-(8lax)x]f (h/i)[xof/ax-(alax)(xf )] (h/i儿xf1ax-f-xf1a]=-(h/f,so[住，户]=-(h/) (b)[i,p2lf=(h/i)2[xa2/ax2-(021ax2)xlf=-h2[x82flax2-(821ax2)f )] -[x02f/ax2-x02flax2-20f/0x]=2h2of/ex.Hence [.]=2h20/0x. (c)[民，p,lf=(h/i)xa/ay-(a/a)xf=(h/i儿xaf1ay-x(af1@y】=0,so[E,产，】=0 (@)[R,(x,y,f=(xΨ-x)f=0 (e)Let A=-h2/2m.Then [ {4a21ax2+a21ay2+a21z2)+门-[4a21a2+a21a2+a21az2)+V]xf= Copyright2014 Pearson Education,Inc
Copyright © 2014 Pearson Education, Inc. 3-5 3.26 () . ∫ ∫ dx f f dx kf = = Differentiation of both sides of this equation gives (/ ) . d dx f dx f kf = = ′ ∫ So 1 df dx k f / − = and 1 (1/ ) . f df k dx − = Integration gives 1 ln f kxc − = + and / / , c xk xk f = = e e Ae where A is a constant and k is the eigenvalue. To prevent the eigenfunctions from becoming infinite as , x → ±∞ k must be a pure imaginary number. (Strictly speaking, x/k Ae is an eigenfunction of ∫ dx only if we omit the arbitrary constant of integration.) 3.27 2 2 d f dx df dx kf / 2/ + = and f f kf ′′ ′ + 2 0. − = The auxiliary equation is 2 s sk + −= 2 0 and 1/2 s k =− ± + 1 (1 ) . So 1/2 1/2 [ 1 (1 ) ] [ 1 (1 ) ] , kx kx f Ae Be −+ + −− + = + where A and B are arbitrary constants. To prevent the eigenfunctions from becoming infinite as , x → ±∞ the factors multiplying x must be pure imaginary numbers: 1/2 −1 (1 ) , ±+ = k ci where c is an arbitrary real number. So 1/2 ± + =+ (1 ) 1 k ci and 2 1 (1 ) + k ci =+ = 2 1 2 + ic c − and 2 k ic c = − 2 . 3.28 (a) 3 3 3 33 3 ˆ ( /)( / ) / y p = ∂∂ = ∂ ∂ = = i yi y ; (b) ˆˆ ˆˆ ( /) / ( /) / ; y x xp yp x i y y i x − = ∂∂ − ∂ ∂ = = (c) 2 2 [ ( / ) / ] ( , ) ( / )( / ) x = = i y fx y x y xf y ∂ ∂ =− ∂ ∂ ∂ ∂ = 2 22 2 − ∂∂ = ( / ). x f y Hence 2 2 22 2 ( ) ( / ). ˆˆ y xp xy =− ∂ ∂ = 3.29 ( / )( / ) = i dg dx kg = and / ( / ) . dg g ik dx = = Integration gives ln ( / ) g ik x C = = + and / / , ikx C ikx g e e Ae = = = = where C and A are constants. If k were imaginary ( , k a bi = + where a and b are real and b is nonzero), then ik ia b = − , and the bx/ e− = factor in g makes g go to infinity as x goes to minus infinity if b is positive or as x goes to infinity if b is negative. Hence b must be zero and , k a = where a is a real number. 3.30 (a) [ , ] ( / )[ / ( / ) ] ( / )[ / ( / )( )] ˆ ˆ x x p f i x x x x f i x f x x xf = ∂∂ −∂∂ = ∂ ∂ −∂∂ = = = ( / )[ / / ] ( / ) , = = i xf x f xf x if ∂ ∂ − − ∂ ∂ =− so [ , ] ( / ). ˆ ˆ x x p i = − = (b) 2 2 22 22 2 2 2 22 [ , ] ( / ) [ / ( / ) ] [ / ( / )( )] ˆ ˆ x x p f i x x x x f x f x x xf = ∂ ∂ − ∂ ∂ =− ∂ ∂ − ∂ ∂ = = = 22 2 2 2 2 − ∂ ∂ −∂ ∂ −∂∂ = ∂∂ = = [ / / 2 /]2 /. x fx x fx fx fx Hence 2 2 [, ] 2 / . ˆ ˆ x x p x = = ∂ ∂ (c) [ , ] ( / )[ / ( / ) ] ( / )[ / ( / )] 0 ˆ ˆ y x p f i x y yxf i x f y x f y = ∂∂ − ∂∂ = ∂ ∂ − ∂ ∂ = = = , so [, ] 0 ˆ ˆ y x p = . (d) ˆ [ , ( , , )] ( ) 0. x V x y z f xV Vx f ˆ =− = (e) Let 2 A ≡ −= /2 . m Then ˆ [, ] xˆ H f = { } 22 22 22 22 22 22 x[( / / / ) ] [( / / / ) ] A x y z V A x y z Vx f ∂ ∂ +∂ ∂ +∂ ∂ + − ∂ ∂ +∂ ∂ +∂ ∂ + =
Alx82flax2+x82floy2+x2f102-x2flax2-20flax-x82flay2-x82f1ae2+ xAVf-AVxf =-2Aoflax=(h2/m)of lox,so=(h2/m)olax. 田[2，f= -h[xy8flax2-(82lax2xy]=-h[xy2flax2-x8flax2-2yoflax]= 2h2zf1,s0[2,p]=22za/a ˉ2max++厂2ma+a好+】 3.32A-(2/2m)72+c(x2+y2+22),where72=a21ax2+a21@y2+a21z2 3.33(a)61Ψ(x,)2dk b)巴Ψ(x,y,)Pd止； 付巴Ψ(x,h,h,2,fd血d2 3.34 (a)dx is a probability and probabilities have no units.Since dx has SI units of m. the SI units of are m (b)To makeddyd=dimensionless,the SI units of are m (c)To makeldx dy ddx dy d dimensionless.the SI units of w are m 3.35 Let the x,y,and=directions correspond to the order used in the problem to state the edge lengths.The ground state has nquantum numbers of 111.The first excited state has one quantum number equal to 2.The quantum-mechanical energy decreases as the length of a side of the box increases.Hence in the first excited state,the quantum-number value 2 is for the direction of the longest edge,the=direction.Then 8ma己b+c28maF3 36.626×1034Js) 8mce89.109x10-kg600×10-i0mP=758×104s v 3.36 (a)Use of Eqs.(.74)and (A.2)gives4mdxdyd== 4mm(2/a)sin2(zx/a)dx (2/b)sin2(zy/b)dy (2/c)sin2(/c)d= 3-6 Copyright014 Pearson Education,Ine
Copyright © 2014 Pearson Education, Inc. 3-6 22 2222 22 22 22 Ax f x x f y x f z x f x f x x f y x f z [ / / / / 2/ / / ] ∂ ∂ + ∂ ∂ +∂ ∂ − ∂ ∂ −∂ ∂− ∂ ∂ − ∂ ∂ + 2 xAVf AVxf A f x m f x − =− ∂ ∂ = ∂ ∂ 2 / ( /) / , = so 2 ˆ [, ] ( / )/ . xˆ H mx = = ∂ ∂ (f) 2 [, ] ˆˆ ˆ ˆ x xyz p f = 2 2 2 22 2 2 2 2 2 − ∂ ∂ − ∂ ∂ =− ∂ ∂ − ∂ ∂ − ∂ ∂ = = = [ / ( / )( )] [ / / 2 / ] xyz f x x xyzf xyz f x xyz f x yz f x 2 2 /, = yz f x ∂ ∂ so 2 2 [ , ]2 /. ˆˆ ˆ ˆ x xyz p yz x = ∂∂ = 3.31 22 2 2 22 2 2 222 222 1 2 111 222 ˆ 2 2 T m m x yz xyz ⎛ ⎞⎛ ⎞ ∂∂∂ ∂∂∂ =− + + − + + ⎜ ⎟⎜ ⎟ ∂∂∂ ∂∂∂ ⎝ ⎠⎝ ⎠ = = 3.32 2 2 2 22 ˆH m cx y z =− ∇ + + + ( /2 ) ( ), = where 2 22 22 22 ∇ =∂ ∂ +∂ ∂ +∂ ∂ / / /. x y z 3.33 (a) 2 2 0∫ Ψ| ( , )| x t dx ; (b) 2 2 0 | ( , , , )| x y z t dx dy dz ∞ ∞ −∞ −∞ ∫ ∫ ∫Ψ ; (c) 2 2 0 1 11 2 2 2 1 1 1 2 2 2 | ( , , , , , , )| x y z x y z t dx dy dz dx dy dz ∞∞∞∞∞ −∞ −∞ −∞ −∞ −∞ ∫ ∫ ∫ ∫ ∫ ∫Ψ . 3.34 (a) 2 | | ψ dx is a probability and probabilities have no units. Since dx has SI units of m, the SI units of ψ are m–1/2. (b) To make 2 | | ψ dx dy dz dimensionless, the SI units of ψ are m–3/2. (c) To make 2 111 | | nnn ψ dx dy dz dx dy dz " dimensionless, the SI units of ψ are m–3n/2. 3.35 Let the x, y, and z directions correspond to the order used in the problem to state the edge lengths. The ground state has x y z nnn quantum numbers of 111. The first excited state has one quantum number equal to 2. The quantum-mechanical energy decreases as the length of a side of the box increases. Hence in the first excited state, the quantum-number value 2 is for the direction of the longest edge, the z direction. Then 22 2 2 22 2 2 222 222 112 111 8 8 h h h m m abc abc ν ⎛ ⎞⎛ ⎞ = ++ − ++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 34 14 1 2 31 10 2 3 3(6.626 10 J s) 7.58 10 s 8 8(9.109 10 kg)(6.00 10 m) h mc ν − − − − × = = =× × × 3.36 (a) Use of Eqs. (3.74) and (A.2) gives 3.00 nm 2.00 nm 0.40 nm 2 2.00 nm 1.50 nm 0 ∫∫∫ | | ψ dx dy dz = 0.40 nm 2 0∫ (2/ )sin ( / ) a x a dx π 2.00 nm 2 ∫1.50 nm (2/ )sin ( / ) b y b dy π 3.00 nm 2 ∫2.00 nm (2/ )sin ( / ) c z c dz π =
sin(2x)40ysin()sin()om a 2 6 Jh.so nm Lc 2π J200nm 「0.40_sin(2m:0.40/1.0072.00-1.50_sin(2m-2.00/2.0)-sin2m-1.50/2.00 1.00 2π 2.00 2π 3.00-2.00_sin(2π3.00/5.00)-sin(2π2.00/5.00) L5.00 2π (0.30650.090850.3871)=0.0108 (b)They and=ranges of the region include the full range ofyand=,and they and factors in y are normalized.Hence the y and integrals each equal 1.The x integral is the same as in part(a),so the probability is 0.3065. (c)The same as(b),namely,0.3065. 3.37 p.=-iha/ex.(a)(sin kx)/x=k cos kx,so y is not an eigenfunction of p (b)((-nla)where is given by Eq.(3.73).The eigenvalue is hn2/4a2,which is the value observed if p2 is measured (c)(((n.and the observed value is h2n214c2 (d))=x(const.3),so w is not an eigenfunction of 3.38 Since n=2,the plane y=b/2 is a nodal plane within the box;this plane is parallel to thex=plane and bisects the box.With n.=3.the function sin(3/c)is zero on the nodal planes ==c/3 and ==2c/3;these planes are parallel to the xy plane. 3.39 (a)I is a maximum where lwl is a maximum.We have=f(x)g(y)h(=).For n=1 f(x)=(2/a)sin(x/a)is a maximum at x=a/2.Also.g(y)is a maximum at y=b/2 and is a maximum at ==c/2.Therefore is a maximum at the point (a/2,b/2,c/2),which is the center of the box. (b)f(x)=(2/a)sin(2zx/a)is a maximum at x=a/4 and at x=3a/4.g(y)is a maximum at y=b/2 and (=is a maximum at ==c/2.Therefore is a maximum at the points (a/4,b/2,c/2)and (3a/4,b/2,c/2). 3.40 When integrating over one variable,we treat the other two variables as constant:hence ffIF(x)Gy)H(e)dkd止=j[F(x)Gy)He)]yd止=Gy)H(e)IF(x)d]yd店 3-7 Copyright2014 Pearson Education,Inc
Copyright © 2014 Pearson Education, Inc. 3-7 0.40 nm 2.00 nm 3.00 nm 0 1.50 nm 2.00 nm sin(2 / ) sin(2 / ) sin(2 / ) 2 22 x xa y yb z zc abc πππ π ππ ⎡ ⎤⎡ ⎤⎡ ⎤ −−− ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦⎣ ⎦ = 0.40 sin(2 0.40/1.00) 2.00 1.50 sin(2 2.00/2.00) sin(2 1.50/2.00) 1.00 2 2.00 2 π ππ π π ⎡ ⎤⎡ ⎤ ⋅ − ⋅ −⋅ −− × ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦ 3.00 2.00 sin(2 3.00/5.00) sin(2 2.00/5.00) 5.00 2 π π π ⎡ ⎤ − ⋅ −⋅ − ⎢ ⎥ ⎣ ⎦ = (0.3065)(0.09085)(0.3871) = 0.0108. (b) The y and z ranges of the region include the full range of y and z, and the y and z factors in ψ are normalized. Hence the y and z integrals each equal 1. The x integral is the same as in part (a), so the probability is 0.3065. (c) The same as (b), namely, 0.3065. 3.37 ˆ / . x p =− ∂ ∂ i x = (a) ∂ ∂= (sin )/ cos , kx x k kx so ψ is not an eigenfunction of ˆ . x p (b) 2 22 2 2 2 (3.73) (3.73) (3.73) ˆ ( / ) ( 1)( / ) x x p x na ψ ψ πψ =− ∂ ∂ =− − = = , where ψ (3.73) is given by Eq. (3.73). The eigenvalue is 22 2 /4 , x hn a which is the value observed if 2 x p is measured. (c) 2 22 2 2 2 (3.73) (3.73) (3.73) ˆ ( / ) ( 1)( / ) z z p z nc ψ ψ πψ =− ∂ ∂ =− − = = and the observed value is 22 2 /4 . z hn c (d) (3.73) (3.73) (3.73) x x ˆψ = ≠ ψ ψ (const.) , so ψ is not an eigenfunction of xˆ. 3.38 Since 2, y n = the plane /2 y = b is a nodal plane within the box; this plane is parallel to the xz plane and bisects the box. With 3, z n = the function sin(3 / ) π z c is zero on the nodal planes z c = /3 and 2 /3; z c = these planes are parallel to the xy plane. 3.39 (a) 2 | | ψ is a maximum where | | ψ is a maximum. We have ψ = f ( ) ( ) ( ). x gy hz For 1, x n = 1/2 f ( ) (2/ ) sin( / ) x a xa = π is a maximum at x = a/2. Also, g y( ) is a maximum at /2 y = b and h z( ) is a maximum at z c = /2. Therefore ψ is a maximum at the point ( /2, /2, /2), abc which is the center of the box. (b) 1/2 f ( ) (2/ ) sin(2 / ) x a xa = π is a maximum at x = a/4 and at x = 3 /4. a g y( ) is a maximum at /2 y = b and h z( ) is a maximum at z c = /2. Therefore ψ is a maximum at the points ( /4, /2, /2) abc and (3 /4, /2, /2), abc 3.40 When integrating over one variable, we treat the other two variables as constant; hence ∫∫∫ ∫∫ ∫ ∫∫ ∫ F x G y H z dx dy dz F x G y H z dx dy dz G y H z F x dx dy dz () () () () () () () () () = = ⎡ ⎤ ⎡⎤ ⎣ ⎦ ⎣⎦
=JF(x)dx G()H(=)dy d=JF(x)dxH(=)JG(y)dy d= ∫F(x)dxJG(y)∫H(e)d 3.41 If the ratio of two edge lengths is exactly an integer,we have degeneracy.For example,if b=ka,where k is an integer,then nla+n=(n+)/a2.The (n:ny:n) states (1,2k,n.)and (2,k,n.)have the same energy (x,y,=)=F(x)G(y)H(=).Substitution into the Schrodinger equation followed by division by FOM givesd 2m F dy H =E and 2(1d2F） Ia=E+2(Eq:1).Let Ex =-2mF2 2m G dy2 H d22 Then,since Fis a function ofx only,E is independent ofyand=.But Eq.1 shows E is equal to the right side of Eq.1,which is independent of x,so E is independent of x. Hence E is a constant and-(h2/2m)(d2F/dx2)=E,F.This is the same as the one- dimensional free-particle Schrodinger equation(2.29),so F(x)and E are given by(2.30) and(2.31).By symmetry,Gand H are given by(2.30)withx replaced byyand by= respectively. 3.43 For a linear combination of eigenfunctions of to be an eigenfunction of H,the eigenfunctions must have the same eigenvalue.In this case,they must have the same value of n+n+n.The functions (a)and (c)are eigenfunctions of H and (b)is not. 3.44 In addition to the 11 states shown in the table after Eq.(3.75).the following 6 states have E(8ma21h2)<15 n”:123132213231312321 E(8ma21h2)141414141414 These 6 states and the 11 listed in the textbook give a total of 17 states.These 17 states have 6 different values of E(8ma/h),and there are 6 energy levels. 3.45 (a)From the table after Eq.(3.75),there is only one state with this value,so the degree of degeneracy is 1,meaning this level is nondegenerate. (b)From the table in the Prob.3.44 solution,the degree of degeneracy is6 3-8 earson Education,Inc
Copyright © 2014 Pearson Education, Inc. 3-8 = == ⎡ ⎤⎡ ⎤ ⎡ ⎤ ∫ ∫∫ ∫ ∫ ∫ F x dx G y H z dy dz F x dx H z G y dy dz () () () () () () ⎣ ⎦⎣ ⎦ ⎣ ⎦ ∫∫∫ F x dx G y dy H z dz () () () . 3.41 If the ratio of two edge lengths is exactly an integer, we have degeneracy. For example, if b = ka, where k is an integer, then 22 22 2 22 2 / / ( / )/ x y xy na nb n nk a + =+ . The (, , ) x y z nnn states (1, 2 , )z k n and (2, , )z k n have the same energy. 3.42 With V = 0, we have 22 2 2 222 2 E m xyz ψψψ ψ ⎛ ⎞ ∂∂∂ − ++ = ⎜ ⎟ ∂∂∂ ⎝ ⎠ = . Assume ψ ( , , ) ( ) ( ) ( ). x y z F xG yH z = Substitution into the Schrödinger equation followed by division by FGH, gives 22 2 2 22 2 111 2 dF dG dH E mF G H dx dy dz ⎛ ⎞ − ++ = ⎜ ⎟ ⎝ ⎠ = and 22 22 2 2 22 1 11 2 2 dF dG dH E mF mG H dx dy dz ⎛⎞ ⎛ ⎞ − =+ + ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ = = (Eq. 1). Let 2 2 2 1 . 2 x d F E m F dx ⎛ ⎞ ≡ − ⎜ ⎟ ⎝ ⎠ = Then, since F is a function of x only, Ex is independent of y and z. But Eq. 1 shows Ex is equal to the right side of Eq. 1, which is independent of x, so Ex is independent of x. Hence Ex is a constant and 2 22 ( /2 )( / ) . − = m d F dx E Fx = This is the same as the onedimensional free-particle Schrödinger equation (2.29), so F(x) and Ex are given by (2.30) and (2.31). By symmetry, G and H are given by (2.30) with x replaced by y and by z, respectively. 3.43 For a linear combination of eigenfunctions of Hˆ to be an eigenfunction of Hˆ , the eigenfunctions must have the same eigenvalue. In this case, they must have the same value of 222. x y z nnn + + The functions (a) and (c) are eigenfunctions of Hˆ and (b) is not. 3.44 In addition to the 11 states shown in the table after Eq. (3.75), the following 6 states have 2 2 E ma h (8 / ) 15 : < x y z nnn 123 132 213 231 312 321 2 2 E ma h (8 / ) 14 14 14 14 14 14 These 6 states and the 11 listed in the textbook give a total of 17 states. These 17 states have 6 different values of 2 2 E(8 / ) ma h , and there are 6 energy levels. 3.45 (a) From the table after Eq. (3.75), there is only one state with this value, so the degree of degeneracy is 1, meaning this level is nondegenerate. (b) From the table in the Prob. 3.44 solution, the degree of degeneracy is 6