量子力學課件25答案

O?dhpi/dt. lhpi= (Ψ??Ψ)dx??

(Ψ

?,5

?2Ψ=

?|

?t

+ i~?2Ψ

iV

+Ψ?

[i~

2m ?x2m?x ?2Ψ??Ψ]+i(V

?Ψ?

2m[Ψ?x3?

???1??????? ???0 ? V

—Ψ?V?Ψ—Ψ??VΨ

?

1?? ?f?uXd A a ?~?(a)O?8?z?fA

(b)Ψ?v? ?? V(x)(c)O?Xe?" x,x2,p, (d)O?σx y?????v?(?' Z

1r 1 0? A=(2am

e?2amx2/~dx=

=

?2am(Ψ+x?Ψ)=?2am(1? 2r ?f? Schrodinger??

?~2

+VΨ V

i~(?ia)Ψ+2m(??? V(x)=

~)(1 )Ψ=[~a?~a(1 )]Ψ=2amxZhxi x|Ψ|2dx=

?????? ?∞

2

2

π~ ~ hx = x0

dx=

2am

hpi=m =

~

)2Ψdx=

?? ??

?d?

=?

(1

h

p

2amh2i)

2am~ 2am~ (2)3

=hx2i?

=

?σx

r ~ x hpi?hpi

4am?

= σ σ r

σxσp?? /????( n

√am~=24

Pab(t)L?3??t3??(a<x<b)? ? V?(a)y?

J(a,t)?J(b, J(x,

i~(Ψ

–Ψ??Ψ J(x, ????o (b)??Ψ(x,t)=f(x)e?iωt?? f(x) λe?λ|x| V?6 Pab(t)?

Z a

Zb

a?2

(Ψ??x ?xΨ)]=??xJ(x, ?xZ?x

b ?xJ(x,t)dx=?(J(x,t))|a=J(a,t)?J(b, √ (b)Ψ(x,t)=f(x)e?iωt,? f(x) λe?λ|x|,?

=fe?iωtdf

=fdf,? Ψ??Ψfdf,? J(x,t)=

5b????????????f???3??/?? ?mτSg??) 3??e T? V?????~? ????P ZP(t) |Ψ(x,t)|2dx= ?y??( ??o ?{Xe"3Schrodinger?????b?V(?U)? .,?? ?? ?V??? XJ???? ??J?V=V0?d V0?? ? Γ?? ??(a)y

(b)?)P(t),????f?? 'X (a)Schrodinger??C?

?2ΓP=~=

?Ψ?

i~?2Ψ?

iV 6?????????

?2m ?

i|Ψ|2(V??V)=

i|Ψ|2(V0+iΓ?V0+iΓ)=

C

2ΓZ

?

|Ψ|dx=6

~ ?2Γdt? τ=~

~ln ?2Γt+constant~P(t) P 7y ?uz???8? ?m? Schrodinger?? )? ??? V(x) ?.????;?A ? d2ψ=2m[V(x)?E]ψ,X E< ,@oψ? ψo?k? ?? ??Xψo l?I? x→±∞?o??u0(?K??8?z) XJ??m??0(?K???8?z) ?7L O??(3d?? ? ? ~??(3d??? )—?????e????0.u??=?0?????k??¨?ψ (????)8?,?? N??vk?n?? O?? ???g? ),X ???%??? ?.~X,b???UCe? ψ1 ???? A[ψ1(x)+ ? φ?,?~? ? Ψ(x,t),|Ψ(x,t)|2 hxi,? ( φ=π/2 φ=1 √

?iωt[sin(πx)+sin(2πx)e?3iωt a|Ψ(x,t)|2 1[sin2(πx)+sin2(2πx)+2sin(πx)sin(2πx)cos(3ωt? l hxi 2[1?9π2cos(3ωt?? ul? ?m:?m?O?? (???? t=0:)

2X φ π(?22Ψ(x,0)=A(ψ1(x)+iψ2(x))),cos(3ωt?φ)=sin(3ωt);hxi? X φ=π(2Ψ(x,0)=A(ψ1(x)?ψ2(x))),cos(3ωt?φ)=?cos(3ωt);hxi?

a(1

3297.5 ??????5?K?#N??O!????? b?V0?? ?,??)?.(a)?O?Xe??x? ?? ψ1?1?- ??? (i)b=0,(ii)b≈(b)? b ∞??5???AU Cz(E E2)? 3?????x ? ???(c)V????? f?f?> ?~ ?? (???? f??). X f???gd? ,????k?$U ?. ??(b)?(? >f?u???.3????????m ,?? ??f?????L@?,???? (a)(i)b=0?~5k ?? ??P~ ?S (cos? ψ1,sin?ψ2).ψ1?!: ψ2k??!: (ii) ???? ??P~ ?S ,?m???? ??cosh. 1?- ????–?^?? u?? ψ1?!: ψ2k??: (iii)?u a, (ii) ?

?3h

^??? ?? u??? 2 ?? 2

EV0=

????

?5|?

d h?? ? ? E?+0? a,z?? ??

2m(2a)2=

E+V≈E+

π2~2=

0

(2 $ud?) ???/.(^ ?3z?????? ?,U p ???/?g .)

=?2m(V0+E)ψ,???3?.(i)?U???$ —???3 b→0.??>f?u??? f?.3??, J, ?PU.?

V(x) msech da??~?”sech”?Lda??~?”sech”?L?(ax?d?

ψ0(x)

????U? 8? ψ0,?x?????(c)y??

ik?ψk(x) √

ik+

(? k?1)

2mE/~)?? U?E? Schrodinger? .(5 z→?∞ tanhzψk(x)?u??L??x?ψk(x)5g?? ?vk????o ??R?T?(???vexp(?ikx)?). 3

ψ0

~2

2—msech22mm m2

m ~2a2A[?sech(ax)tanh2(ax)+sech2(ax)?2sech3(ax)]m 2m?

sech(ax)[tanh2(ax)+sech

sinh2 sinh2θ+

θ+

θ)

cosh2θ+cosh2θ cosh2

= ? ?2m ZE —Z∞

r

1

sech2(ax)dx=

tanh(ax)

|A|2?A

A [(ik?atanh(ax))ik?a2sech2(ax)]e ik+a

{ik[(ik?atanh(ax))ik?asech(ax)]?aiksecha+a32sech2(ax) ~2d2ψk+Vψ

2— ax?a

{?

sech )]+2maisech 2m

ik+

—msech(ax)tanh(ax) msech(ax)(ik?atanh(ax))}e ik+a·

3—ak2tanh(ax)+ah)iam?2a3sech2(ax)tanh(ax)?2iak2sech2x)+2as3ch2x) ik+a·2mk(ik

=

d x→+∞? tanh(ax)→+1,? ψk(x)→A(ik?a)eiAx,?L??? ?.R=T=|ik?a|2=(?ik?a)(ik?a)=

(c)?u5g? =0),ψ(x)??? :S22=0,S21=

Aeikx+B(0) xik? 0

,?

=0,S21

..S (ik+ 1

?u k→iκ.? S=( 0

,? κ=a??,, E=?~2κ2=?~2a2 1

??: (a)??? d??w?k?? ?? n?w????3???,V(x)=0, Aeikx+ ? k3?m??V(x)2g?0?

√2mE~

Feikx+

3?m? ??? ?N/??c? ?? O(&E ?Schrodinger????? ?5???????)?k/ψ(x) Cf(x)+ ??f(x)?g(x)????5??A)"?ko?>^?????^?C?D? ??K??^5??B,F?A,'XB S11A+ F=S21A+ ?o??6u X?Sij| ??2×2?S??? S?w??? (B F)? (A G)?m 'X 3; 5g? ?/?? =0 ? X? |B|2

|A|2G=0

11| Tl

|A|2|G=0= ?u5gm ?A=0?

|S

T=|B|2

|S

|G|2A=0

|G|2A=0 ?,8?~???^?,8?~???^??transfermatrix,M,????m ?(?G)?? ?(A B)'X

(a)? o? , L? ??_C? ? Rl,Tl,Rr L /?(b)b???k????d????|? L? ??O? ??M=M=(c)?l??delta????:V(x) | (b)??{ ??Vdelta??? V(x) ?α[δ(x+a)+δ(x? ???? D?X? B=SA+

G?G=

(B

A)= A+

B? =?S11,M =

F=SA+S SA+S22(B A)=

—S12S21)A

S22B=

A+ B

=?detS

1 ?detS

??

G M21A+M22B?B

M

(G?M21A)=S11A+S12G

?M21

F M11A+M12B=

A+M12(G (M11M22?M12M21)A+M12G=

A+SG?

detM;

M

S 1

Schrodinger? ?m???C5?V?????M22=M?,M21=M?, detM=1 1R=|S|2

M21|2;T=|S|2

detM|2;R=

|2

M12|2;T=

=

|M

?

2; 2;

=M

1BG M2M 1BG

=M

Aeikx+ (x<

Feikx+

(x> d ?Y5:Aeika+Be?ika=Feika+dψ

??Y

2~2

2=?2mα(Aeika+Be2~Fe2ika+G=Ae2ika+ Fe2ika?G=Ae2ika

+i~2k

Ae2ika+

???\

2Fe2ika=2Ae2ika+i2mα(Ae2ika+B) F=[1+ ]A+ e?2ikaB=M11A+M ? M

=(1+iβ);M

=iβe?2ika;β

???~

2G=2B? A?2iβBG=(1?iβ)B? A=M21A+ ? M21= M22=(1?(1+

(1?(1+

M2,UC ??=?

?iβe2ika(1? =(1+ = ?iβe?2ika(1?[1+2iβ+β2(e4ika? 2iβ[cos(2ka)+β

M2M1

?2iβ[cos(2ka)+β

[1?2iβ+β ?

T

=Tr

?T?1=[1+2iβ+β2(e4ika?1)][1?2iβ+β2(e?4ika?T

2 3 2 3=1?2iβ+β ?β+2iβ+4β+2iβ ?2iβ+β ?β?2iβ+2iβ3+β4(1?e4ika?e?4ika+ 1+2β2+β2(e4ika+e?4ika)?2iβ3(e4ika?e?4ika)+2β4?β4(e4ika+=1+2β2+2β2cos(4ka)?2iβ32isin(4ka)+2β4?2β4==1+2β2(1+cos(4ka))+4β3sin(4ka)+2β4(1?cos(4ka))1+4β2cos2(2ka)+8β3sin(2ka)cos(2ka)+4β4sin2(2ka)=1+4β2(cos(2ka)+β?/? X??? (a)?U????o(b)dX ??ag? ??????/.ophr??0,??L En?1(3???r ?;U? ?{?y n r0/ag.?? ?/ ?f nb?/ ?$U (n?1).??? U? 1f???f ? (a)V(r)=?GMm.? →GMm=?@^ f(J(b)a=(4π0)

4π

=2.34×

(6.6726×10?11m3)(1.9892×1030kg)(5.98×1024kg)

=?[m(GMm)2] MmEc

mv2? GMm=mv2

1mv2=GMm.? E=?

=?[m(GMm)2]

?n2

GMm2rr2p

0r0?n r

=F-/??? =1.496×1011m?n

1.496×1011=2.34×

6.39×

=2.53×4

G2M ( ][n+1)2?n2(

n(n+1)2=n2(1+1)2 n2(1?n

? [(n+1)2?n2 n2(1?n?1)= 4E

G2M 4

(6.67×10?11)2(1.99×1030)2(5.98×(1.055×10?34)2(2.53×

=2.09× ν=E λ

c=

λ (3×108)(6.63×10?34)/(2.09×10?41)=9.52×

(a)y???u?u?V(r) ?f? ??L?? UC u? ??tdhLi td? K?quD??dA? ^=?/"(b)y?????????kdhLi/dt=0"?d=?f? ? ? ??L?")?(a)

xx

+[V,Lx] ??

][V,pz] [V,py]

[H,Lx]

—

=i~[r×(?V

?

?y,z??ka L????d

hr×(??)i= y.(b)?u???

?kV(r)=V(r),K?V

?Vr?qr×r= y(a)L+Yll ?? ?(?I?O?!| (a)?(J ?L ?NL??L =~eiφ(?+icotθ?) ? LzYl=~lYl,OlYl(θ,φ), 8?z~?l?L ??O?8?z~?

L+Yll=

LzY

?~

Yl=~lY

i?φ ?ll l l

lYl f l[5? f(θ)3d??~ —?? φ...? ????6 LYl 0?~eiφ(?+icotθ?)[f(θ)eilφ]= ??

dfθeilφ+ifcotθileilφ ??

df df Zdf Zcos = sinθdθ?lnf=lln(sinθ)+

ln ln(sinlθ)+K?ln( )=Ksinll∴Yl(θ,φ) A(eiφsinl

sinl

=constant?f(θ)=Asinlθ Z

1 sin2lθsinθdθdφ=2

sin(2l+1)θdθ=101

(2l+=4πA2

=4πA2 (2l+∴A

(2l+

????? ? ?3??? f5\?? ?f\?7??%gd=??% ??y?Tf5= U? =~2n(n+1) ? n=0,1,2, ?TX 8? ????O?1n?U {??)?(a)X?M???H=2(1mv2)= 2f5 ?? ?dM????L

|L|=22mv=

HH ??? ??~2l(l+1)??d ~2n(n+ n n

(n=0,1,2, (b)?L ??? ??L1n?U {???2n+1

ψnm(θ,φ)=Ym(θ, 14.?g^???f???g^?f?u???X?g^?n?^z????~"XJ?g^ ?z? g^? ????k? ?fT?Vψ?? 10??>fg^?"\U??>??????g^??? ??k T V?? )?(a)d2×1CGX?Lr r r |31i 1|22i|1(?1)i 8|21i|10i+ 6|20i|11i ?d? ??9?V??2~(1),~(8),0(6)2(b)d1×2

CG

r r1 2i?\l(l+ 15~2(2),3~2(1)

23 i3

11|3

a15.b???g^ ?f?u ??? S(1) 1???fg ab? ??? ?????a??????? a , S(2)? ??fg ?? b?? ?? y ? θ ?

–4–

????? ?? z??? b xz??S @o

=S(1) S(2)cosθS(2)+sinθS(2).h00|S(1)S(2)|00i????O

S(1)S(2)|0 1(S(1)(cosθS(2)+sinθS(2)))(↑↓? √ 1 √2{(Sz↑)(cosθSz↓+sinθSx↓)?(Sz↓)(cosθSz↑+sinθSx↑)} √2{2↑(cosθ(?2↓)+sinθ(2↑))?(?2↓)(cosθ(2↑)+sinθ(2↓))}

(?↑↓+↓↑)+sinθ(↑↑+↓↓)} 4{?cosθ|00i+sinθ(|11i+|1(?1)i)}

?hS(1)S(2)i 4h00|[?cosθ|00i+sinθ(|11i+|1(?1)i)]=?4cosθh00|00i+ d ?? hS(1)S(2)i=?~2cos 16. f >16. f >f? 1/3Y0χ+ Y1χ?(a)X????(L2),??o? ?V??O??o(b)???z? Lz?9V?()??g??? (S2)??V?()??g??z? ??V?.J≡L+S ??(e)?J2,?o? V?O? (f)(g)? ??f??V??? ?V?????

(h)XJ? ?g^z??? ? (5???????? ),@? g^??? :?l? V???? ) (a)?u?????k l=1,? ~2(1)(2)=2~2,P=0,P=1/3,? ~,P=43~2,P=4~/2,P=1/3,? ?~/2,P= 1

1Clebsch-GordanL? r

r r r 1 21? 23 1 3 21 |22i|10i 2i|1

√ i?√ i] 32 2

|22i 3

32

2)|31i+(1)|11i ? S=3? ??? 15~2,P=8,? 3~2

2

221~,P=22√2 |Y0|2(χ?χ+)+ (Y0?Y1(χ?χ)+Y1?Y0(χ?χ+))

|Y1|2(χ?χ—)}

{z1

1

0

1

0

1 |2(|Y0|2+2|Y 1·

3·

·

[4πcosθ+2·8πsin r2e?r/a·

(cos2θ+sin23·24·

1 1 |Y0|2sin2θdθdφ |R21|2

r2e?r/a

? ??'X#N ?? ? 3 ??u ?? ???y ?d7 3, ?? L=r×p?? ? a?,? ?? ~ (~X???? f | Bohr??),???? q1≡√[x+(a p1≡√[px?(~/a q2≡ [x?(a p2≡ [px+(~/a (a)?'X[q1,q2]=[p1,p2]=0;[q1,p1]=[q2,p2]=i~.L?q’sp’svaq???K??'X(b)y

z 2a2(q1?q2)+2~(p1? (c)u Lz=H1?H2,?? H????m =~/a2???ω=1 M????i=1,2?LgC?q? ?I? ? fM?? ? (n+1/2)~ω,? n=0,1,2,...|^??:? ? (? q,q

a2

,x

a2

]=[x,x]=

,p]=[ 2]

+(~)

(~

=

[p,p]

~y,

~ = ∵[y,px]=[y,y]=[px,px]=

( +

[q1,p1] 2[q2,p2] 2

x+(~)py,px?(a2 =2{[x,px]?[py,y]}=2(i~?(?i~))= x?(~)py,px+(a2 =2{[x,px]?[py,y]}=

1{x

2

a221?q2

+~(xpy+pyx)+(~)py?x+~(xpy+pyx)?(~) ~ p2?p2 1{p2 ~(pxy+ypx)+(~)2y2?p2 ~(pxy+ypx)?(~

? ? 2a2(q1?q2)+2~(p1?p2)=xpy?ypx=

2 a2 @o?

H 2mp

2mωx

2~p

2a2x=H(x,

~q2≡HH(q1,

2~ 2a2

H(q,p) a2p2 ~q2≡ 2~

2a2

∴Lz

H1?

(d)

? (n1

1)~,

? (n2

~,? Lz ? (n1

~?(n221)~=(n?n)~=

m

,? n

n ? ? ?? ?f? ??gd?(~X,?3??L??w?f)y? ? ?Xe/ √L (?L/2<x< ? n=0,±1,±2,...,# U? En 5 (n=0) ? ?{ b????\?H0

?V

0? L.(T? x=0????? /€?0. |^{??6n?? 0 .J? ?O??? ??| L^????? ±L/2* ±∞;.? H0?a<x< ???0?ud?K?ψn ?5|???o y?|^ ??l?{??6? (a)? a→n,b→?n,Wij=hψ0|H0|ψ0i, VZ

VZ

V0

=Wbb=?LZ

e?x

dx ?L

e?x

dx=?L

VZ

?L

e?x

dx ?L

/a+4πnix/L) ?

e?(2πn/L) La3d?/

Wab? ,???????

?E

=Waa±|Wab|,?±E ?√±E

(E1?W

±√πV0ae?(2πn/L)?? ?6? ?β=

=

}=?α.@o ?5|??

—πV0ae?(2πn/L)LrL

?i2πnx/L]

ψ+=αψn?αψ?n=√2√L[e?

— r

|^ ?6U??? E1=hψ|H1|ψi

2(?V)

0 E1=hψ|H1|ψi=2(?V

L/2e?x2/a2cos2(2πnx

2| sin2θ=1(1?cos 2cos2θ=1(1+cosE± E±

Z0 ?∞

e?x2/a2(1?Z

L L

dx

)dx}

L ?√

(b)???

L????kn??5? ?fX? b? / M????(1)00H01 ? V0?~? ?, (a)e??6M?? ? =(b)?) ? ? ?? ?g? (c)|^ ?{??6n???C ? (a)|?

?

'O.???{ ? ????((J? (a)χ1 ? V0;χ2 ? V0;χ3 ? )? (b)????:det(H?λ) [V0? = [2V0?2[V0(1 )?λ]{(V0?λ)(2V0?λ)?(V0)}=0 λ1=V0(1 (V?

–λ)?(

λ2?3Vλ+(2V2

2V2)=0

= 9V2?4(2V2 2V √ 'λ '

[3

√1+4

[3±(1+ λ=

(3

√1+ )2

V(1 );λ2

2(3V0V2

1+4'

V0(2 E300=001;010E300=001;010EE0000=000010=0011=0101033E1=0(vk )3X3E23

E0?E

?10

1000010=1001=010100 0101=0E0?E0=2V0?V0= 3? E23

=(=

2V0. ??? =E0+E1+E2=2V0+0 2V0=V0(2

(?c?(b) λ3??)001=1000010=1000=—010000000=0100011=0100=0010010000=1000011=1000=001001?\?????

q ?V+0

2V2+ (?V±V)

— Cq E1=V0?V0,E2=V0,?c (JgU f?3??1???? ?? Balmer? 5 n= n= [ k,??Bohrn????????? ?^?du?[ ?A ?;?K?:k ^,?m?? ?J? ?k? n= U? ? ^fU? ???fs^ ?[ U?E1, eV??? n=3????? x?U???fs???? ln= n= .? U?(?1f/? (E3?E2)+4E,????u? ??? 4E(du?[ )???? ??z? 4(?? eV). ,???1f?? ??? ??? m?(?? Hz)—5????6(J??6(J? ? (???? ? ??? m??? ?(JA?Xe/? ”Balmer? (???)^? U???, ^S ??5 j=(???) j=(???),(2)j=(???) j=(???),....3 (1)? (2)? ??m?(???)Hz,3 (2)? (3)? ??m? (???) E0?E

=hν=

=E1(1?1)=?5

?λ=?36 E1= ?

~c=1.97×10?11MeV λ36(2π)(1.97×10?11×106eVλ

=

×10?5cm=

22 3.00×ν λ=6.55×

=4.58×

EE= EE=

(3

j+

? n=2:l=0 l=1,? j=1 3.? n=2 ???U?

5 5

j E1= (3

)=? 2=? )2 =

=?5.66×

2 2 32(0.511×1 j E1= (3

)=? = (3.62×10?4eV)=1.13×

2

? n=3:l=0,1, 2,? j=1,3,? 5.? n=3 ?n^U?2 1 j E1= (3 )= = = (3.62×10?4eV)=?2.01×

3E2

92 1j E1= (3

)=?

= (3.62×10?4eV)=?0.67×2j=2

3 E13

E2E

(3

2 1)=? 2

=?

(3.62×10?4eV)=?0.22×

k8 [;? U??(E0+E1)?(E0+E1)=(E0?E0)+4E,? 4E≡E1?E 1 β≡E2/mc2=3.62×1@o

4E=[(?

)?(—

)]β=?

β=?8.80×10

4E=[(?1) 1)]β=

β=4.61× — —

4E=[?1

1]β=

β=9.08×

1→

4E=

—1]β

β=36.45×

3→

4E=

+

]β=119β=49.86×

5→

4E=[?1

5]β

β=54.33× (? k8^?? ? (1→3)?k$u???6? ?? ???^?? u???6? ??.

,?? #2:3

3;#3:5

3;#4:

→1;#5:3→1;22#6:522

1.??m??

?^

ν2

4E2?4E

=3.23× ν3

4E3?4E2

=1.08×

ν4?ν3=ν5?ν4=ν6?ν5=

4E4?4E32π~3.23×1091.08×10

=6.60× ?l n= ,|2lmlmsi.??z? 3r|Zeeman U? ?z?JL??n? BohrU?,?[ ' α2),? ' μBBext).X? ?[ ? k ?? U?,z?{??? n=2,l=0(j=1) l=1(j=1or3).l |1i=|2011i21|2i=|20 2|3i=|2111i

(1/2)(3/2)+ gJ=[1 ]=1+3/2=(1/2)(3/2)?(1)(2)+

21|4i=|21 2

gJ=[1

]=1 = ?13.6eV[1

α2(2?3)]=?3.4eV(1+5

|5i=|2133i2|6i=|213123|7i=|2

gJ=[1

(3/2)(5/2)?(1)(2)+(3/4)

]=1+5/2= |8i=|213

α2(2?3)]=?3.4eV(1

1 U??

E1 ?3.4eV(1+E2 ?3.4eV(1+

α2)+μBB α2)?μBB ?3.4eV(1+5α2)+1μBBext

?3.4eV(1+5α2)?1μ

E5 ?3.4eV(1+

α2)+2μBB ?3.4eV(1+1α2)+2μBBext

?3.4eV(1+1α2)?2μ

E8 ?3.4eV(1+

α2)?2μBB y Kramers’'Xs+1

hri?(2s+ i

4[(2l+

—s]a i ??f??>n??g(s,s?1,s?2)r?"??X?5J??????? [l(l+

—ar

|^?

(ur¨)drL? hrihr i, i. |^R??? ? y(uru)dr=?(s/2)hrs?1i,? (urs)dr=?[2/(s+ (¨rs+1 Schrodinger?? ??? l(l+1)?

— 1¨=

~2(4π0)r

=1,? ?2mEn

2mm(

)21=14π ?

~22~24π ¨=[l(l+.

—ar

¨)dr

l(l+1)?2+

]udr=l(l+1)hrs?2i

i

—d—

? Z

Z

0 (ursu0)dr Z

(urs)udr=

(uru)dr? ∴

ss2? (uru)dr=2

hrs?1? d (¨rs+1u0)dr Z

u0r(rs+1u0)dr=?(s+dZd

u0rs+1¨dr∴ (¨rs+1u0)dr ?(s+ 0s 2 R ? (uru)d=

(¨ u? 3:3? 2?| *?,?|^? (J Z l(l+ 2 1 (u0rsu0)dr

(s+

Z

–ar

? {l(l+1)(s+1)

a

ursu0 1

? {l(l+1)(?(s?

hrs?2i)

srs?1i)+

(s+1)–a(? 2–(s+ n s l(l+

)hrs?2i

2( )hrs?1i+ s+

as+

?? 1?? 3^u l(l+

2rs?1i+ hrsi=?l(l+1)(s?

)hrs?2i

2( i 2 n

s+

as+—

hrsi+s(s?2

hrsi

2[1+ ]hrs?1i+{l(l+1)[1

s?

]?s(s?

}hrs?2i=

a| z

| z+ 2s2(s+1)hrsi

2s+

s?1i+2s[l2+l?(s2?

]hrs?2i= ?

(s+1)hrsi?a(2s+

s?1i

4l2+4l+1?s2)hrs?2i= 4(| {z 3?N? l >|K?? U? ????o . ??? n?:>?????X??? K??g^.b d2,and d3,yH0=V0+3(β1x2+β2y2+β3z2)?(β1+β2+ ?

≡?eqi,? i4π0i

(b)??? U ? (c)O?1?- (n=2)U Cq 3 ??eTo?{?U??O ?^,(i)β1=β2=β3;(ii)β1=β2=β3;(iii)??? (n????) (a)>f3(x,y,z)?U??=5 q’s x ±d? V ?eq{√ +4π

}.?m(??

2

2

=1(1

2x

r2 (x±d)2+y2+

±2dx+d+

+z

=

±2dx+r

3 d(1?d?2d2+8d2)=d[1?d+ ?r

∴V

eq[1

x+

(3x2?r)2+1

x+

(3x2?4π ?

— (3x2?4π 4π 2βd2+3βx2? ? β≡? 4π0?d? ?8?>?k H0=2(β1d2+β2d2+β3d2)+3(β1x2+β2y2+β3z2)

h100|H0|100i

e?2r/aH0r2sinZ

+

e?2r/a(β1x2+β2y2+β3z2)r2sinZβ1+β2+

r2e?2r/ar2sin

|

{z 4π∞r4e?2r/adr=4π4!(a)5=3π Z e?2r/a(β1x2+β2y2+β3z2)r2sinZ r4e?2r/a(β1sin2θcos2φ+β2sin2θsin2φ+β3cos2θ)drsin R2πcos2φdφ=R2πsin2φdφ=π,R2πdφ=2π.? Z Z [π(β1+β2)sin2θ+2πβ3cos2θ]sin 4!(a)5[4π(β1+β2)+4πβ3]=πa5(β1+β2+

??

Rπsin3θdθ=4,Rπcos2θsinθdθ= (β1+β2+∴10|H0|100i V πa5(β+β+β) 3πa5=

o

|200i

0R20Y0

11|211i|210i

R21Y11R21Y11

|21(?1)i=R21Y

(a)|^C ny? ?{??6n?o?L O U?(b) (a) *: ?N???" o? .(@??:):(a)|^??6 (ψ0)??&???.C n? H0+H ?

(??6 U?),? ?? U ,? E0+ ≥

(E2) E2 m ??f?w ,?1Ko? (??m m=m

Eg?E u?km E0<E0),? E2 b????fX?H0=k? ,ψa(U??Ea), ψb(U??Eb).??8 ,???{?(b? ).y3?m??? H0,?kXe ? h?,~?(a)?)TX?? ? | ?6n U?(c)|^C ? U?

?&???

ψ

? φ??N?? 5? X ?5|? ψ8? ????{ ??(J ??C?{Xd?( H

det(H?λ)=

a?

2=0 λ2?λ(Ea+Eb)+EaEb?h2=q λ {Ea+Eb E2+2EaEb+E2?4Eab+4h} p

2{Ea+Eb (Ea?E)b2+ (b) :E0=Ea,E0= :

=hψa|H0|ψai=0,E

=hψb|H0|ψbi=

:E2a

|hψ|H0|ψ =

;E

=

? E

E+ Eb

hcosφψa+sinφψb|(H+H)|cosφψa+sin Eacos2φ+Ebsin2φ+2hsinφcos

?Ea2cosφsinφ+Eb2sinφcosφ+2h(cos2φ?sin (Eb?Ea)sin2φ+2hcos2φ=0tan2φ=

=

∴ 1?sin2

Eb? Eb?Ea ?sin22φ 2(1?sin2sin22φ(1

sin2φ=

1+

cos22φ 1?sin22φ=1

1+ 2

1+ 2cos2φ √1+

(??dtan2φ=sin

= ?? cos2φ=sin2φ

1(1+cos2φ)=1(1? 1(1?cos2φ)=1(1±

1 2111 211

? 2Ea(1?√1 2)+2Eb(1±√1 2)±h√1 1{Ea+Eb

(Eb

Ea+2h) ?(Eb?Ea+2h

(Eb?Ea)+

√ 1+ (Eb?Ea)2+ p = 1+ ?

1+(Eb?Ea)21+1

= (Eb?Ea)2+p

(Eb?Ea)2+hHimin=2{Ea+Eb (Eb?Ea)2+?u ???F K? ?????)p 2{Ea+Eb (Eb?Ea)2+ (d)X h (a) ?((J???m?s 2E± 1{(Ea+Eb)±2

– 1

}(Eb?Ea) 1{E+E±(E?E)[1

(Eb?Ea) 1{E+E±

–E)

(Eb?? Eb

(Eb

E—'=Ea?(Eb

( (b)??6n (J C ? U?(E?)—???&????? (5???Uk??????? ????K?u? ?{ ??w? ~f, ?!r^|B =Bzk?????? f, ?M????: y3???m???6 x?? ??!r|H0 m(a)? ,?(@???k?(206 "? h? (b)|^?K(b)?(J ?6Y??? U?(c)|^?K(c)?(J O?C n??

u

??,

.±= ??KgU Eb>Ea,?χb=χ+

,χa=χ—

,Eb=

=eBz~,Ea=E?=?

hχa|H0|χai

0m

0 1

=

0 0

=

0 1 1

= 0hχa|H0|χbi

0 0

0

1 hχb|H0|χai∴h???K^?

1 01

=

l?K(b) ,E E

=?eBz~—(eBx~/2m)2=

e~(B+B2 (E?E

(eB

l?

(c)?

,Eg

1{Ea+Eb (Eb?E)2+4h2}(?S????( ).2r2Eg ?

(

)2+4(

)2=

p B2+B

?, Schrodinger?? ?{?(?) 3#N?( ”a? ??{ ~f??”rubber-band?,”????? ???O?H=?~2(?2+?)+

2(r2+r2)?

2r?

L?lC r1, C1

4mω| 1u≡√2(r1+O??M???C?????n f

v≡√2(r1? H=[?~2?2+

22]+[?~2

2mω

v+2(1(b)??X ?( U?? (c)XJ?? ?() ???U??u ?/ M???(226)$^~ fC?{ T(J??()2 (a)r12

uv);

r2

uv).2∴r2+r2 1u2+uv+v2+u2?~v+v2)=u2+ 2=2(2(?

)f

,

) (?2

+?2

+?2

+?2

+?2

+?2f

?f?x1

?f?ux?ux?x

?f?vx?x

=2

(?f?ux

?f

?f ?f?ux+?f

?f ?f ?u ?v

=√2(?u —?v ?211

?f ?f2(+) 2(+)

[?2f?uxx?u2?xx

?2 ?vx?ux?vx?x

?2 ?ux?vx?ux?x

?2f?vxx?v2?x1xx 1(?2f+x

?2

+?2

2 x?2x22

?f2( 2(

—?f

[?2f?uxx?u2?xx

?2 ?ux?vx?x

?2 ?vx?ux

—?2f?vxx?v2?x2xxx 1(?2f ?2 +?2xx

2 ? (?2f+?2f)=(?2f+?2f)—a y z

∴?2+ ?2+

? H=?~2(?2+?2)+1mω2(u2+v2)?λmω22v2,? H=

~2?2+

2]+[?

22?

2

2mω

2mω

2λmω√ √

3~ω(? u?? 3 1?λ( v??):Eg=3~ω(1 1 ψ

mω1/4

n f

ψ

mω3/4(c)? ?

0()=(π~

?

)=(π~ mω3/2?mω(r2+rψ(r1,r2)

π~

∴hHi

3~ω+3~ω+hVi=3~ω+hV

2

?mω(r2+r

?mω(

π~

r1?r dr1d r?rr r1r

????0,du??5 r2/??

12??? ?12 2 ? )22 e—

(r1+r2)r2 2mω

Z

Zr0 ?mω(r0

π~)

r2dr2

r1dr1 8m4ω5 ][

~

π~]=

[4

8 ∴

44? hHi=3~ω(14

λ).C yd??u? U

4)

3~ω(1+2

–λ)??λ2?

>1+

–λ??λ?1?2 1?λ?? 1?λ+4>1? ?S??XJ?? ?g?m?() Eg

3~ω(1+1?1λ)=3~ω(1?λ), ?C(J

28.f? n=3,l=0,m=>f?L??>??[??.(a)??J????k ?Xe???|300i→|nlmi→|n0l0m0i→···→|10 (b)XJk???u f z? V??O? (c) ??? ?J? ??1? [ ???2? |30 3z? =k1???? O?k' ?L??P ? [???\ |300i

|21|21 →|10|21(?1)i

(|300i→|200i |300i→|100i? 4l=±15K).???J?K?20|~|300i=20|z|300i2(±1)|~|300i=2(±1)|x|300i?i+2(±1)|y|30±h21(±1)|x|300i=ih21(±1)|y|30? |h310|~|300i|2=|h210|z|300i|2? |h21(±1)|r|300i|2=2|h21(±1)|x|30?IO???

2 2 √ √ a3/2(1?3a+27a2)e?r/3a ψ210

r4π

27

r

3sin

r

√ √

h21(±1)|x|30 Z

27 2 sinθe±iφre?r/2a(1?3a+27a2)e?r/3arsinθcosφsinθr Z∞ 2 2 Z

r4(1 r

2r2)e sin3

(cosφ±isinφ)cos0 √

{4!(6a5

25!(6a

6!(6a7} π)

5

5 +

5 (3)( 2

4

(25?3a·5·5·6a+27a2·6·5·6a4a ?3√357(25?100+80)=

h210|z|300i

r ZZ

272 2 cosθre?r/2a(1 r r2)e?r/3arcosθr2sin

2 2

r(1?3ar+27a2r

cosθsin (1/a4)

)(2π)

∴

2

0|~|3

1|~|3

512

=

?wn [?? ,?dz ?1/3.?z??P ? A=ω32|hri|2;d ω=E3?E2

1(E1—E1)=?5E1 oPC??3π

36R

5E1

(21537a2)=6(2)9(

E1)2(c)2

8929 6()

)

3×

)/s=6.32×10 0.511× 0.529× 1R

=1.58× 29.???f(3? t=0?)?u? ? N .y3? U?6?,p S ???x?'??k/? V0(t)???V0(0)=V0(T)=(a)|^???6??)? cm(t),?L????UC?? ?v [u)???UCφ(T) 'X?(b)3 ?6Y???????K (J ?(J.n?P?{?3??\???~?(' x?~? ???6 t)??~ (a)˙m=?~

ei(Em?En)t/~.d H

=hψm|V0(t)|ψni=∴˙m

?i~

~~

=?iV0(t)dt?lncm=?

-

—iRt

1Z≡

t)dt0 ? |cm(t)|r=|cm(0)|2,v [

1Z~ ~

0

cN(t) 1?Z

Z~V0(t)dt0=1+~t0t

N(t)=1+

0i(Em?EN

cm(t)= (m=m(t)= δmNV0(t

dt= (m=N?((J cN(t)=eiΦ(t);cm(t)=0,??? eiΦ 1+iΦ,@o??3 Y