Q¡
?li
trHAPTER FOUR
úl
THE POSTULATES AND EENERAL PRINtrIPLES
OF AUANTUI'{ ],IEtrHANICS
!1,'
4-7. We rnurst determine which futnctions can be integrated when their magnitudes are squared (squtare
,
t t t t t
integrable). a) co "d,* =
_.{"-=The normali:ed b)
wave function
3,, ,..rlii
is (?/nlt¡+.-x?
o,
diverges.
Jerxdx
?,.;',i,
¡n1L/2 LZJ
c)
rr
Jrl ,"t o¡ *pi o6 e
Not norrnal izable.
|.
=ld0=!n J r:t
d)
0
,
4-2. a)
dx
d
i verges.
Not normali=able.
,
t , ,
Not normali¡able becatrse
N
jr1*=d* is not finite. r:t
b) c
)
,
t
The angle l:etween v1 and v¡ is given by
coso =
GD
Jsinhrx o
,
t
and
i
, ,
note that = j . j = k . k= 1 x I )qtrosll= 1 = I ){ L xcsr¡r¡2t = (, = i . k = j . k these resurlts to show that = (ayi + arj + a=k) . (bxi * byj + brk) = a¡byi 'i + axbyi. j + ar,b=i.k + ayb;¡j'i + aybyj.j + ayb¡j.k + aEbxk.¡ + azbYk.j + a¡b¡k.k = axb¡l * ayby + azbz Therefore, the dot product of v1 and v2 is V1'V?=2-3+3=s
lrtl = (vr - vr) L/? = JT + 1 +" =,[IA lr"l = (va - va) t/z = \ft=?=f = J6
zff
-[he normalized wave f uncti on i s (2¡r) - 1/ !ei
,'il;; I
4-3- First i . i j i' trle now l¡se a . b
Acceptable.
Not norrnal i zabl e because t Ir: d:{ J(1 - x?)t is not finite.
-1
d)
Acceptable
e) tan-tx is not a single-valned function. For e»i amp 1 e, tan-1(1) = ¡/4, ,t/4 + ?TÍt 1¡/4 + 4n, etc.
Éhti"h
=
É,.
= o'§?7
0 = tros-11).A?7 = 7(l. gg" 4-4, i :* i = j )4 j = k )d ft = (.t becaurse O = r:, in the formulaaHb=absino. i :r ¡ = k becalrse 0 = ¡rltr
lil= l¡l
=
r
and the dírection of k is perpendicLrlar to the plane formed by i and j. El )'i b = (arri + ayj + ark) H (b¡li * byj + b:k) - .r, b;* i r1 .. ayb!.t j ¡":i + ar b), k]{i
- a*brixj + arbrjxj + arbyk¡cj + áNbzi¡:k + ayb¡j:+k + arb¡k:{k = tl - arb,.,iHj - a¡byir.:¡6 ¡ a).rbyi:,4j t) - a=byj3:k + axb¡i¡
0
flt
,ii
= (ayb¡ - arby)i + (a¿b¡,¡ - ar¡b=)j + (axby - arbr,)k This resr-rlt is equrivalent to j k li a
:i
]1
I
b = 1",.l ay
lo,, by
1i
i
i
li
tt vrr*: y2=l!-1
I I
l_.1 11 ?
,
:l, ,t, l,i
I
;tii;
;, I
I
il
l, i
:,' irr
ii lliii
;i: l;i lrlt
,i
i
1l
j
"
u
Iim = at+o
At
utu] * ["tt,"tl.] +
rir[[ Aa ..*' At+rlLL IE da
aT:": 4-6.
b
higher-order termsJ
Start with ai.:a=l)
Then
,- Étrt] *
[,i " ']
=o
4-7. §tart with dlr m¡-r?
=
Operate with r)< to obtain
" $fi
dt -a'b
=
= ,Er[,-
"
#] = r ]{ F
d
eT(rxp)=r:4Ftorque
+ Iim a(t + At).b(t At At) - a(t)'b(t) at+0 ra(t) + Aa(t)l.tb(t) lim
+ Ab(t)l
At
At +r)
lim
4¡+r)
+ [ffi.u rtr
= da aE'b
db * "'E
att).ffi + higher-order terms]
moti on.
4-8.
a
{E}.= J* «,, r H* (x ) d¡¡ a
J*
iI
fJ
d d d
I
EJ
ql
F
r-1.945
,',.
G G G G
d
db +á ,.^aT
If the torque eqr-rals zero, then $¡.«. * pl = 0r and r )": p = angular momentLtfn = constant. For circutlar motion, r ,.: p = rmvLt, where u is a urnit vector perpendicular to the plane of the c i rcurl ar
d
,rl
t- H #.]*(>,)dx
lr
q{
3d 3n 31
q
d
G G G G G G
ill iiil
Eb(t) {- ab (t) l a(t) ::{ b(t)
Then
I
ii
,l
'r{
or
rl
ii
5-t"
.j
+ Aa(t)l
4-5. Start with df (r) = f (t + at) - +(t) dt atifr ---ar
I
:l
ta(t)
,'i
fut. .," .r = [.
,l
.il
SimilarlY
r'.¡
:l
kl .ll =-7i+j+5k
sino = l,vr ¡c, ve,l = f75 = lvrllvel Jm _f¿0 = 7rJ.B9"
i:
'I
bz
The angle between v1 and v3 is given by
ritli
,I
j
l, li
aE
To prove that a I{ ! = -b :{ ár note that we áre interchanging thro rows of the above determinant. To prove that a 1.4 ¿ = rJ r note that two rows áre identical. Lastl y,
,;l
iil
54
55
+
G
lil
= (arbs - a¡br)i + (a2b¡4 - a¡¡b¡)j + (axby - ayby)k This resutlt is equrivalent to
j li a :{ b = la,.< ay lor, by
,il
I
,i'
I
v1
ii
lr
t:t
'
z
kl ElI = -7i + j +
0=
7rf
4-6,
rl
4-5.
:'l
i'i i;
II j.rl rti
;il t,il l'rl;i I
,l:l
bY
dar
rneEE
r;=
-9J..:¿- = .t14 .t¿'
cl
,,. * Ég =,oqT[" " Ei] = r
Iim
At
lim [§.u «tr +
at+(:¡
iI
':l
il
¡E(r!
+ ab(t)l - a(t)'b(t)
aril.ffi + higher-order terms]
db da = aE'o * a.E
If the torqlre equals zero, tfren $¡.tr x pl = {)r and r F: p = angul ar mornentLrm = constant. For circutlar motion, r ){ p = rmvu, where u is a urnit vector perpendicular to the plane of the circlrlar moti on.
4-8.
Á
{E}'
=
t^
J.I{x)Hg(¡r)d¡l ú
a
J*r»,r
rl
t)
,ili
t- * ffin(x)dx
G G
F
t.i
G G G G G G G G G G
i,l'l
i!;ll
ti1,
iitili
F
d
At
Ea(t) + aa(t)l.tb(t)
]{
or
a(t + At).b(t + At) - a(t)'b(t)
C C
G G
F
Operate with rx to obtain
Then
At+Cr
=
.g4S
Start with (t + at) - f (t) df (t) Itfn f dr = at+o At
I
it
it] - [Íi,. a] = r:
Start with
4-7.
C C C
C
Start with aHa=C,
I
I
:;i
hi gher-order- terms]
1l
.89"
9-a'U l im ot' = at+('
* [.rt, " tt]
""5t
+
fut" * "r = [. *
Ek
I I
(t) l b(t)
+ ab :x:
Then
i
i',
tb(t)
At
at+o
da * o ='F
:{
a(t)
§ = lim
+
,tll'
;,
i.' "
+ Aa(t)l
bz
sino=fift#=
:Ii,.
f
ta(t)
Aa rir[[ At x b(t)] = At+OLL
The angle between v1 and vs is given
tl:!i itil¡
SimilarlY
:lrrl
aE
j li :+y3=13-1
l:l
Fl
;j
k
To prove that a !{ § = -b }{ ár note that we áre interchanging two rows of the above deLerminant. To prove that a :4 a = O, note that two rows ere identical. Lastly, I
[il
54
55
r
3
=
-
-
*qt]E' ?ma Jro',"
¡¡) 3(2aÉ
¡:r
- 12a>l + 1!x t)
it is most convenient to
To eval'utate these integrals'
oEE
dx
t4¡-LO. 'v-
{p}
u5e x
[xm(l
J
-x)ndx=6fl!*1¡¡
a
O
b
= jo,,Jor*,-,*n, (x ,v¡ ao ab oi*Jr'Jo, =-l-,
[-it' [ifu
- ,fo]]+n*n, {* ,v)
sin$i"Tg
ñ
tr
Thuts
{E}
=
-
=*?E'[rrc2rBlas
* [i$"'P,i"ftu * :$i"T}"'-u]
- ,r"4# " '=Li+d]
-'*#t- r*-'l
=0 ab 1p,? = Ja*Jav+,.,..,-,, «x,yr ¡f tf
éñ"
= mÁE
qEt:. = J* t:r I Érg
(;< )
dx
=
a
[# É;]'P(x)dx 13é¡l t(4's'2)ax = l=o"' x) ffiL = *# Jx"r"
= i*t»,,
and
oE=={E3}-4E}==;¡;7
9rlh4
4_g.Ifthesystemisinoneofitseigenstates, then
'
= (E3> - {E}'" =
.iEr) = JJ+"§^rff"únnnrdx dY = Etnxny
(x
,y)
sin{}t"Try
#Jo.Jav ÚD
4-f 1. To
show-
that
'|'(x
ry) is
normal ized
abab = Jo,,Jaya"t*,yr tfo6rlt
ffilo* ¡rr(a - xr zJav v,{b - v) " = H*=[H] [%F] ='
= Etnynrtnlny
{E} = JJ+n}nrÉ'l'n*nyd* dY = Enxny
,¡r.rr,r.r,
This result is the two-dimensional extension of Eq.3-6O' op==4pt)-{Pi'a={P"}
and
There{ore
;<
"lqp:-ofi":1sinq*=inryry = ¡rsñr[Sl - B#] = 3:t# - #]
É,+r,,.,.y = Enr,nr'l'n*n,
Ép,tnr,rr,
-ab
[-t,e[# - #=]]
We
have
utsed
the integral
1
J*ntr -
x)mdx =
i. f#+-il, i
dE:' is given
d)
bY
ab
{E} = s3=J0,,{., gcto ^a
J+*ipr+on
-:iis ir,. -'l:-ss *'i*ffi*
=
l--**'E*
-@
:,i_ *[;;:,_,,
I
lm
OJ
zxta-xr]]
b
= ;E63' H:Jo-JaY ', (a-x)Y(b-Y) DLa x t (y (b-y) + x (a-x) l
=
=Htt#l t#l
e)
. ['hd] t++fll
J+**tS* =
1I Eñeft m Lae baJ
-j* [i$]]*+*o*
Not Hermitian
lo
C
I +**+ --i-t,.:s * ¡*]o,n l-co
*-i**3*i-
G G G G
G
Hermi t i an
CE
!
4-L2.
oc
SrSz
= ) ) n=I m=r"
anbm
B lf _t -Ln )"nll
2
'l
)o,.l
Lm=l J =(a r * á2 + as)(br + bt) = a1 br+a1b¡+aqb1 + Babs + asbl * ásbs t ar br + aeba =t
J
4-13. We must deterrnine if the operator satis{ies the condition =_J«Á*,¡*1,¡a*
-i+*Á+a* a)
Go
J+*$;'+a*
=
l--"-
-
| -i*ÉF,.
o)
- J+$;+*o*
Not Hermitian
_(1o
b)
see Page
c)
see ExamPle 4-4
+or, =_J+ x,l l *ax
Hermi t i an
«
_J+**
4-14. The operator is Hermitia.n, and Js*A+an = J+A*e*ax
so
EE Je*«A
- {a})rdx
= Je*Ara*
-
J+«A
+ É)*s*dx
5?
G G
Je*+ox
= J+A*e*oN - {a}*Jfs*dx = J+tÁ - {a})*s*d* because {a} is real «A is Hermitian). To prove that the sum of two Hermitian operators real, consider Je*tA + Br+dx = Je*A+ax + Je*É+ox
=
5É
EE
Then
= JrA*e*lfdx + JrÉ*s*l+oN
125
G G G
II
I5
I I
G
I
e
I I I I
I
ff n = nr then lna sldx = 1 AJ
4-15. Start with c
+ ..*.
1*caJ+*Agdx
rJg*A+aH
D
= c1trr*J+A*g*dx + c1*c=JeA*+*or,
..*."J(f*Ág -
gA*++)dx =
and so we have Proved that
trrt"*JtfA*g* -
If==>l+iyand r. = lí + iy = z* = N - iy then y mutst equal -Y, which can be trute only i{ The integrals on the two y = C,, or i{ ¡ is real. sides of the above equration are comple¡l conjutgates o{ each other. Tht-ts J+*Aqa:t
=
(Eq.4'7?)
JsA*+ux
a
g*A+)dx
If I tr Os------tr nrTl{o5------rl nrn{ x -AJÁá
4-LA. In the r2 ttt Ln'
n=1
the 6n, caLtses al l the terms to be rero, e>lcept the term in which n = rn. Thus
r-
¿l. cnónm =
SimiIarly, ) ) nfn
Let f = É+ and g = + and the probl*m it solved. ,f,n (:r
) = (?/a)
L/ zcas (n¡rx
/a)
.
)
surm
a,.,b*4,..,,
) ) a,.,uro.,, = ) Bmbm = ) "r,o, nm rn ñ
aa J+n*,,.l ) +m
in the
ás we sLrm over n, all terms ere rero except the term in which n = m. Thr-rsn
Then (x
cm
n=
Je*A+ar, = JtA*s*r+ax
Ne have
sutrn
co
GO
4-16. I+ Á is Hermitian, then
4-L7.
onfit
-
tt
dx = 3J."tE-I!l.o=Ep''
_¡lrl
Use the trigonometric cos«cosF. = |o.(«
identitY - P) " |o=(a
+
(n
F)
(n
+ rn) n,{_.. li + =Jco5l--------------flx tl
Unless n = m, these integrals are equtal to zero
J+ 'r(>l)f1(:<)dlr
¡c I
¡d
Now
I = Q = J(a1 + b1x)dx = Ll
{1(x) = Br(1 - Ix) To find a1, require that 11
= 1 = ar"J«r J*rr,*)dx E
a
¡:r
o
N = 1rlr...
requrire that
ot-
becaurse
J.o={rtr, =
ác, = l. 1
r:J
li - m) nx ;Jrl cosir-;:¿-.--Ox
trc
=
1
to get
EE,
Let f¿i(x)
4-19.
or
f 1(x) =
*[3f
1-
I>r)
-
2:<)
ca¡, = $
a1
. br -f_
,tIffi Eútlill l'lr
-¡ill
-i,i ¡rr
5'
c1*caJ+*Ágdx + c"*c1Jg*Afdx
l,
r:l
",1 i l
|,
5,
..
=ll'i
='ri; túr
i $ri',
=ri lr 5, ii
r,,;,
li
r,
;
,Il
}j."'*osufux 4-lB. In the
i' ll
F.
f
a1
sum
cnEnm
the 6n, caurses aIl the terms to be zero, except the term in which n = m. Thr-rs
''
) c,',6,'r. = c¡
Let
n=1
Js*A+or, = J«Á*s*t+ax f = É,¡ and g = + and the problei¡
4-lz.
l^,e
have
,|,n (x
) = (?/a) r/
SimiIarly, in the
zcos (n¡rx
/a)
aa J+n*,r,),¡r(x)dx = 3J."=T+o.{ro* ,fJü Use the trigonornetric identity cos«cosF = l.o.(d - F) * |o=(« + to get aA I ¡ .o= (n - m) ¡n<0, *
.
anb.an,
ás we sLtfn over n, aII terms are rero except the term in which n = m. Thus,
4-f9.
Let {¿¡(x)
=
I
F)
n
Jfo(>r)f1(x)drr ú
ambm
áo = 1. t
= ) "no. nNow
=C,=J«"r*
N = 1r2r...
re that
b1;<)dx = Er * ab.
f 1(x) = ar(l _ Ix) To find a1, reqlrire that 11
= 1 = ar"Jat Jt.",*)d>r rlo
requri
or
e
J.o={fur, =
sum
ar,uror,, = ) _) ) nm ñ
!J.o=ol#aú
AJá t:r
) ) nm
is solved.
or
f 1(x) =
citt
6nm
n=1
becautse ,'1,
=
(o
4-16. f+ Á is Hermitian, then
Unless n = m, these integrals are eqlral to zero
l
¡rl E,
JsA*+ax
Then ,
=ú'llil
=l'
(Eq. 4-79,
a
60
-¡i:.t{iiiir
=li
=
r
Eo h,e have proved that
Ifa=>l+iyand x.=)í*iy=e*=N-iy then y must equal -y, which can be true only if y = 0t or if ¡ ig real. The integrals on the two sides of the above equration are complex conjugates of each other. Thurs
,i
,
|Ja,, =
.r*."J«+*Ág - gA*+*)dx = cr.r*Jt+A*g* - q*A+)ax
J+*Aea:.r
5iu't1:::
=ál
a
= clca*J+Á*g*dx * .r*."JgA*+*dx
l.
= nr then
fn
I
il
=l,i,
,i "
i:
4-15. Start with
.[s-f
1 - Zx)
2x) "d¡r _1?
Now
requrire that I
r
Jf.'(x)f¡(x)dx
=0=""*f*ff
a
bm= ]J+ t*
rf,
)
o
and
t+
f n = .riJ r 1- 2x)(a3 Jfr(x)f3(x)dx rl
+
f (x)
b2x + caxt)dx
L1
x, then a ?r rnrr!{ bm= -lx eJa Sln-tlx =
D
=Q
= {3["
,*F*5e-",-$"-F.]
or
Solving the two equations for aa, br and c3¡
¡
t,
obtai
'11
lr, ,1 ,.¿
,
$n,
,fl,
rI
mrx
A pl
n
Thurs t' !¡.'
d-?
--_g1not of si n (n¡rx /a) i s gi ven i n Fi gutre 3-?. 4-21 . Note that sin(nrr¡la) is even aboutt a/2 when n iE even and is odd abourt c¡lZ when n is odd. The integral leading to Eq. 4-49 is
wE
ba=-CZ=-6ác
".,
ñ ¿t r
l- --f -AL MIf osmr JI
Thuts, o (-1 ) m+1 x =-2as ) ¡lL mÁ ln= 1
I t.
=
?S.q-t)m+1 - ?§(-t¡m = fn'f mIf
c¡ 2. a"*Eb"*f=o
i
G G G G G
sin$x
f 3(x) = Be(1 - 6x + 6xa) The value of a1 can be f olrnd by reqr-riring that
á
Jx«a
1
- x)sin$dx
ü
=1 J*8,*)d;¡ D giving f 3(x) = .l'5t1 - 6x +
EecaLrse ;r (a
- x) is even about a/3, the integral wi 1 I vanish if sin(nn¡¡la) 1s odd. ó¡{a)
4-22.
Ne are given CO
4-2O. Start with
T (x
rt) = )
n,.,exp
(-Iñt) sin$
CO
i'
li
nrx f (>r) = \¿ bn51nn=1
l'lt-tltipJ.y both sides by sin (mnNla) and integrate Otoa d
J+ fJ
t*
»
sin{rtx
t0a =
r
t
n,Ix
fnrrx
2 b"Jsrn;-¡n-{x n=1 rl
6
F .On;'bn¡n aa. ) = p¡ L4 I
f rom
Set t = O, multiply by sin (m¡x/Q) and int=g..t= O to 0 (see Froblem 4-2t-t) a
8,.,
=
?JT
(x rr))
f8/
=3[
sinE{fox
p
&
'' J (a - x)siroto"-] {xsinfr>r 8/c
n=1
t,
6?
6S
+.oln
G G G
q q el q q q .I q ¡l q .=
Ci
d d G E
e
A/
=?[
e,
,f,
Fm(rl) = lCm(Cr) lt = 6mk If A and Íl hav= different eigenvalues, then pm(t) = lcm(t)lr varies with tirne. The qurantity p¡(t) decreages with time from its initial value of r-rnity, and Fr(t) (m I k) increases with time f rorn an initial value o{ zero. If A and É have a mutural set of eigenf ttnctions, then tr¡ (t) = trtlei{p (-iE¡¡t/ñ) and so p¡(t) = lCu(t) l" = l.tl" is independent of time.
n7f
Q/e
Itsin(nnx/{) n
Q/¡
and
Q)
Q- 0cos (n¡rxl0)
+Q Q
0xcos (n¡x/ n,f
(n¡x
-0 "si nn"¡¡"-/ Q)
0¡'rcos(nnx/0)
2¡rt
nrf
f?0" n¡r'l 40 fsin (n¡/2)1 = IE["."¡=rni-J = ;E[--- nE-J
t
Therefore,
r(x,t) = f*["-rttsinp - *"-ttt=inS
,
+
l
4-24.
hle must evalurate
cA,Ér=AÉ-ÉA
I
4-23. l,lurltiply
,
{tir:
i!(:<
ái
by 6r*
(:r
nt) )
)
=
c,..,
ttt {¡ (x )
integrate
and
to obtai
Jii,,,
á
_J***(:r)i[(xrt)dx
ll'
i;;:
Sr-rbsti
t'
; )
n
cn
(t) J+r*,r,
) 0n (x ) dx
-§J
En(t) 6nm = Cm(t)
tute Eq, 4-21
E(xrt¡ = ) c,..,,f,r.,(x)e-iEnt/ñ ,L n
,
into the above integral to obtain
b) AÉ+ = tÉr - .] [5* + x+] = É# + r - ;{E+ = [# - r - xa]+ EÁ+ = [Sn - .] [o* - r,+]
=Í#-r-xz{=tt}-r-xt]+
CO
E,
Cm(t)
,
If 9(>r,tl)
=
) f'l
cnexp
(-i Ent/ñ) JOr*,*
trrn
t úi
(
(1)
=
.llltl
(x ) dx
AÉ-ÉA=e c)
-:
6mlr
I
,f,,..,
-or
Jtr*,*)*(x,r))dx
_J*,n*
ú¡
)
{¡(:<), then
=
ot
,
cli
= =
á,"' l;
n
6f
C§
x
ÁÉ+
= Jar,,$}. = r(x)
- +(())
11
(* ) g¡ (x ) dx
,\
ÉÁ+
=
(;{,) = f (x) kJo.,f rl
tAB - BAI{ (x ) = -f (rl)
li¡
li ll, '|¡; ,i,
l!
ir
ii
d) AB+ = It} - -][3* * ""+] do+
=-+d>¡ 3
:it
i.l i,i I
*"$ff*=*$f + (?-xo)r
¡i .i
* *"] [É#- -*] = [eLcllt ds+
=++d>r s
tA,Él = AÉ
d{ -dtf x'ffi-x;i-(1 - ÉA = +r,$;- + s
rlI ,l
+)ro)+
'full [-'-[=h - .f;]l*
4-zs. [l¡[.r+ = [-tr[vf; -
e'lr e+- xa=J a+] = -ñ3 t'*t - =ovj L"a* aaf aa+ = -ñr ['** - Yzñ 'rY¡E='
,
i
_
i lltr,, it,r
l
:,I
rl .,1
EA+
!,,,
I
I
f'rü,,+
=
ee+ I
.¡-_r ¡ - ^a2+ 0;ráy ""0y0=J
-¡-
[-'"[=3r - ,*;rll [_,"[vf; _ =b]l* a'lr af - a+] -"'[=*-- - *a=l Lv?; =W) a2+ -oz+ - XYTF, -ñ" [y';;*; - r-?r¡t
o+* ,r=ÉñJ a?+ t * *ay
rl
rü,*,ü.r: = -ñEtr*-
-,.;I] = in[-i"[rá- -.87]]
= iñü= The other relations ere proved in a similar rnánner. The pattern involves the cyclic permuttation of xry and z. To prove that [t commutes with Ér,, [y and f.=, Iet's use ü* as án example. trlearly f.¡r commutes with üx. The other terms are given by
fy"ür, - f*üyt = üy"ü* - (ty[.x + inü=lü,
,j
,i
and
Lrtü* - ür,ü=. - f.z.ü* - tü=L* - itürlü= ¡¡e now add these two lines to obtain fy"f* * üz"fx - üy[.]rüy - LEt:xLz + iñ(iñün) = Lrrürün - [.,0ürl * i]"r[]=f* - [.r,tl= t - ñ"f* = üy(-irf,¡l * [z «in[]y) - htt:x = it'tl¡Éy - frü=) - ht[* = iñ(-iñ[r,)
- ht[.¡ = I
4-2É.. Start witn JO1*Añ63ax. Let g* = 01* f = É0" in Eq. 4-?9 ta obtain J+r*Añq"a,.r Now
let
=
*Éq2d:r
JtA+,.r*Éoz¿x
=
"[0"
tÉA¡1r*o;r
G G
Thus we have
J+r*Añ+ua* = J+.rñAotl*ax Notice that AÉ is HerrnÍtian only if
A and É
4-27. tronsider the Taylor series of .A+É = f + «A + Él * |tA + É)a
*
cornrnurte.
f,-«A + É)s + AÉ
...)
+ ÉA + Ér
consider eA=É
= (f + A +
=
S" * -..)(r
+ É * |e" +
...)
r * A + É + ÁÉ + F" " fn, * ...
= I + A * É * |«4" + aAÉ + Ét) + ... The two expansions will be the same only if A and commute.
t
G
eA*É
Let's look at the quadratic term. (Á + fi¡a = tA + E¡«A + Él = Ar + Now
G G G G
G G G G
and
{Ag1l* = g* and use Eq. 4-?q again
J,Aor,
C
eE
G G G G G e¡
G
d
el
i
66
67
q
I
G
4r¡r¡
cil,l rl,lil
4-zB. If rA,É: = Or then the {0¡(x)} and the t+m (N ) ) are the same as in Froblem 4-38. In this case, c,,
J+r*,*
{,'1,
J!l
and
{ )
and
':
) ,¡n (x )
drr
ái l¡util::
5,i' ¡flrr,
E
I
á á tr
ú
I,,l
a'
l'
Jr-r*
i,i
-
_ J+*saxJ+e*ax _
.ir"r*--
!
Cm(t) = Crexp(-iEnt/h)
i,i
:i
-i
trm*cm
Thurs
J+*+ax
r
J+*s¿xJe*+sx
,in*r*
o
ot uPon cancellation t+*+dx
j
a is a conserved
'
ii
4-Zg. First consider r? ^ + + fs*)o*]= = J+e*o>r J U,**n [J+"eax Let J+*edx = a + ib. Then J{*gax * J+g*a* =
iii i
u
l'[**no"
l'
Je*eo*
Dirrerentiate {x, = J**(x,t)ii&(x,t)dN
@
to obtain
%#=JáSi*r.*Jv*iffr*
2tsr
and so
- iEJtÉEr*Í*¿x * LJE*i(É*ro>: = - i"J**ÉiEox * tJ**i «Ésrox = i;-[** lÉ - rii r Eax = iJ** rilÍ - iÉr sox «Ht - tÍtr+ = t- H # - u{x)]x+ =
[J,**n + rs*)a*]" = +lJ+"oo*1" = olJrn*o,,1" Using the first
ineqr-tal ity
given in the problem,
we
r
=!nli:,,i,
á'i
ljr-*r-
,¡
have
á'
tn'n
il
=tl
álll
Iii i,i
..i
= Err,
Fr(t) = lEr
á
tT
I
r
[Jt* t=r'.] [Jtn t"0,-1 lJ+*eo. l" Now consider the intrinsical Iy positive qr-rantity Jt+*rgl*tr+¡s¡6.,
= and let
l.r.l"Js*sax
¿c¡
*
r*J+s*ax + .lJ+*q6¡ +
-,[-Hf#*ur"r+] J+*rd>r
=-=HÍ,.1 ñEi =--:-l-.,=-:
rn
ñ "^
:
*!ur,
Finalty d{x
dt
}
iJ**rri,irpdx *.f**u.va,, =
!*É
-t
i
{*\
lil
trHAPTER FIVE
I+
THE HAR¡'IONItr OSCILLATOR
dF) = J**tx rt)Év(x rt)dx
,i
I
then I
I
d{-.f>
¡#*'*(x,t)dx dt = Jat
i'l
* Jú*{x,uÉffrx
- hJtÉ***¡Éüd»r + |;J**Éff*ax = - ifl**ÉÉ*¿x * LJ**ÉÉ*ax = J**f rÉrF - Éür *ox
=
I
Let É = É,,, fiÉx - P)rill = U(>t)Pr, - Ér,U{x) . -du
S-1. Newton's equtation {or a harmonic oscillator is dtI * Ex = dt n rn" r:¡
= Asint'rt + Bcosr¡tr we find that + Bcosrot) = - Ar¡tsinr.ot - Er¡tcostot + E
Using x(t)
5-2.
Usi d{
F'., }.
dt
=
- I**#üd>, = <#>
a
if we recaLl that r¡3
=
k
/m.
Consider the trajectory x(t) = Csin(r¡t + 0) Becaurse the si ne f utncti on vari es {rom +1 to -l t the valure of x varies from +C to -C. To find the period $re wish to determÍne the smalLest of osciIIationr nonzero val ure o't r that sati sf i es the condi ti on sin (r,¡t + Q) = sin¡u¡1¡ + r) + Ol =sin[{.tt+6+t,t'rJ = Ein(r.rt + 0)cosrnT + cos ( r¡t ¡- { ) si n r.,t'r = sin(r¡t + S) Thuts, the trajectory repeats itsel{ when r satis{ies the two conditions sinr¡r = (l cosr¡'T = 1 or when l¡t = 3¡rn n = 1r3t...
I
l,
tl
,:,i
:,1:,
:ir
ilii,ii
is
ng >l (t ) = Csi n ( r,.tt + E), we find that t, - Cr¡Ési¡(6rl + $) + i:Csin(r¡t+ó)=(:) m
S-5.
t,
harrnonic oscillator
3#-\=tt
dll
Thuts
Newton's equration for
(:l
7t-t
71
The smallest value of r'is
ú
J
5-5. a)
T=;- ln This is the which is the period of the oscillation' time it tahes for the oscillator to undergo one is cyclE. The frequrency of oscillation 1t¡ Ir=-==_ ,TIr 5-4.
T
= +J ú
U(t) = mg(>lr¡ -
mrr¡C3
lr..)f+ 0 |
2"1
3
l0
m,,¡C?
,II!
2t
b)
t:i:
,,rc"
sinz(rrrr+ó)-sin2ol Tt
[,.0r +
(sin?{.rr) (cos?q) + (cos?tor) (sin?ü) _----_4
lr,
2,r
,
[e
_7-= mr..l-U-
I
KCE
42 -=-
,
I'
Similarly
t t
5-6.
D t, ,ilrl
lo
¡vrr-
- rn;a" - m§v¡t =
initiá1 total energy
The forces acting Ltpon particles
1 and Z are
{e = - ñ = ñ = -*. The center-of-mass and relative coordinates are sttch
¡=Orxr*maxe
N = Xl
- Xe
l'lultiply the f irst of these equations by H and the second by m¡ and add to obtain
xr = X + frslr
ú
+
t_au-auaLJ tr = - 7;-i
D
!,
mgxo
K(t) + U(t¡ = "-d. + mgi{(,
kCt fr.,¡.r'l 2r¡¡LE )
ú
I
u(t) = mgli =
{U}=Si=r""(,^rf+S)dt ktra E ? 4 The natural rnotion of a harmonic oscillator is that {l{} = {lJ} over a cycle o{ motion.
energy
dax meEE
K(t) =F"=#+,r,Vot
E
where we have used the fact that r = ?n/u ín evalurating the trigonometric terms.
!.í) = mgx¿
= -mg Integration gives I >l(t) = - Ftz - v,1t + x¿ The instantaneous kinetic energy and potential energy is
+ si n2>l
L=
ft,.r +
- $'
- msv¡t - $1 l{(t) + U(t, = %d + rngxü = initial total
{
úi
is
+ 6)dr
to'*s J cossxax
- -¡;
rng
]d(t) = S1" = $.É + rrev,¡t
t$osa(r.¡r
,r.oE=
=
Integration giYes x (t) = hac + v¡t The instantaneous kinetic energy and potentÍal energy
The vallte of {l'í} is 4t*i).
dtx
maEE
,lllil; ;,ltl
Mr-rltiply the f irst bY l"l and the second by m1 afld slrbtract to obtain
-"-fu-,
itil
lir -^H^
ill
,lil lll rii
ir;i
Fol
Surbstiturte these exPressions for x1 and ¡{r into Newton's equtation=
lowing the maniPutlations in Problem 5-6, we can daY
r'iil ,trj
'lii ilii
iil
i,. ,iI ¡:i :;, t, I
,i'il; .., ri.
.
;tl
rt' I
I
t:¡li
1.,"'
l'" ',:
',,i
,,i,.1 ,,1 iii
iil ,i,i ,, .ii '1"
'ii: 'ri, it
:,i:, tii,
:ll
,',il
AU
firmc d!¡re dlx mrAtn-=meafE-ñ'-
d
AU
3¡l
0x r
and
0»t
fl Lmr
á>r
0y
It is Btráightforward to derive similar equations for the z coordinates, and so we have N ,Éf$ =
,r
dEr
ft¡¡7 = - vu as the three-dimensional generali:ation Problem 5-6.
0>r
of
or d
E¡l
r-raET
=_
5-8.
AU
;;
This is the eqltation o{ motion o{ one body of mass moving urnder a {orce -|lJ/ 0¡t.
mrEEf = - ñ; In terms of the Y-cornponent relative coordinates
-
It is straightforward
zul dr, I
TyE-
the center-of-mass
and
[d
=
(-,
EUI
and that
[dsul
- = lA-á'l -X=D -
.q
q q q C q q d q q d d ¡É
=,r,
LhÉJ.-, = 2DeF'r
become
dEY m.m. d?Y mr.¡fE*T-att=-
,.,
to show that
that
yn=Y-fr',
yr=y*ficy Newton's eqttations
c¡f
=
- [#1"1"- á[#1.1i . h[Í#]*:: * '
Z-7. The equtations in the x-direction are the game as ín problem 5-6. Let's took at the y-direction' aU d Evo d Ev. aU meeTE:
Expand U(x) in a Taylor series aboltt i{ = o
u(¡<) = u(r))
Lt
q q q q
eq
and
l]AU rnel
_au
d2v
tsdt 3
AU
+mr)t$=,r
dE>r dtc
.l
tl
-
_au
which says that the center o{ mass moves uniformly' trle now divide the f irst of Newton's equtations by ml and the second by m¡ and subtract to obtain
riii
T,,
";r dtl -=--=
(m1
.,
ll,;ili
d
ár¡¡ dtt -= We add these two equations to obtain
'i ;:, ',
dex. d3X rllrrn¡ mrAE=:=mIEEE*-M-
=
dtl
G
C
get
iii
,,:Ii
q
0U dtY - mlma day = Ít"Et Í- ATÉ - t
,n óo ousF
aU
0y
G
73
74 J
atr
e
or
Therefore
u(x) = DeFtx= - DeP"x3 ¡ "' (L/?lkxt' we {ind Cornparing this result to U(x) = that 17.3L k 1,1-te,r) (1.82 x l0ro m-1)3 k = 2DeFB ".
484 N'm-1
k = (2ncio6=)
3¡r
12.?78 x 1§ro cm.s-1) (2.60 x lOB cm-1) ]" x t (1 x 7il /8t) amul x tl.66e7 x lQ-27 kg.amur-13 = 393 N'm-l of vibration is given bY period The
=
[27f
vobs
5-1O.
Eq. e-4O
Use
iobs
1.28 :4 t(r-1+= cvobs =
=
L fk1 L/2 ?nc
LLIJ
-t-l
cm's- l ) 1
?4rl¡N - mv f_ (79l?)amur) (1.á6 ¡: ltJ-"7 llg.amur-1) L(
= 3I1 cm-1 rero-point energy i5 given by 1l ilv = fhcvo6= = 3.19 x
Eo = 5-1
1.
The
osci I I ator d
t
ú,1
t
ji
is
|tx
r¡,¡,n
(r'r )
=
tll
From Eqs. 3-4L, 5-43 and Table 5-20 we have ,l,t (x
,
ti
J
Schródinger equation {or a harmonic
* fi$«en -
d>l c
c
t t
3*n
10-31
)
= {t/7oí'''-cttt'/?
and
5-9. Use Eq. §-fi1 -!'obs
1 fhll/É
= E"ALIJ
*a(x) = [¿f,;]"n r=o*E - 2) =-u>t',/? For n = 1r the Schródinger eqLtation becomes ff5<*"-o,,"2=¡
- #[Én" -
fr,,,a]xe-",r2/?
?
r,
]"^
(-E«x + «r¡ro) («"¡rs and this li I
For
ñ=
-
3cr¡rl
- #[fu"
*
doeg eqltal 2r we have
(6cr
-
1I«3>l
r +
'
t'
fr,r, "]r, =
t/"* - *[¡,e 3
¡ero
becaLtse o( =
(
(-)
r-r
Ltll) t/ a¡f..
3/2 !«s»l 4) e-Gu
fr.,,"] <+.,'t
-
t)=-«xc/r -?-
(6« - 12«3>l¡ + Ic{sxa) + (5« - «!;-(r) (4«xl - ?) ? rr 1l¡¿r - ??crt¡rt + 4«3>r4 + (IrJcxt¡ll - 4«s¡a - 1r)« + 3«3xE) = O Thr-ts 'f,3(x) satisfies the SchródingErr eqLtation for harmonic ogcillator.
(_)
5-15. If f (>r) is an even futnction of x, then it can be represented by a power series of the form f (x) = f ,¡ * {:¡tc + faxa + ... The deriv"rtive of f (x ) is {'(x) = ?f¡r: + 4f ¿xo "r ... which is an odd f utnction of ¡l . Similarlyr if {(x) is an odd futnction o{ :r, then it can be represented by á power series of the form f (x) = ft)í * fglfs + {5xs + The derivative of f(x) is {'(x) = {r * Sfgxa + 5.fsx4 + ... which is an even f ttnction of >¡. From Eqs. 5-41 and 5-43, we have
5-74.
r/2 ) = NEe-«x 4,r (¡r) = Nr (3«1/rx)t*«¡lt/e 3/? *s (x ) = N¡ (4s¡r t - 3) *-«x ,|,¿
¡
[ll,ril oin,
T-
f,t¡rtr)
- H[F," -
x'iii; .,..
-
5-12.
tronsider the produtct of two even futnctions F(x) = {(x)g(>r)
Then
F(-x) = f (-x)g(-:r) = f (lt)g(x) = F(x) and so F(¡r) is an even f utnction of lr. If F (¡< ) i s the prodr-tct o{ two odd f utncti ons of then F(¡l) = f(x)g(x)
xt
*e(x) = Ns(E«3/2¡is - L?aL/3")e-c${t/e There are five integrals that we mutst evaluate. Three o{ them are ro o) rf (x),1's(¡l) Jdr,+o(x),f, 1(x) = Jdx,¡'1 -O
and
P(-:<) = {(-»l)9(->t) = -f (x)g(:r) and so F(>l) is an odd furnction of x.
7A
-O co
r
= Jd,,,¡1(»l),Ir(¡l) =
and
F(-x) = f (-:<)g(-x) = (-)l+(x)g(x) = F(lt) (»l) >l . of is an even f utnction and so F Lastly, if P(>l) is a produtct of an even fttnction an odd fr-rnction of xr then F(x) = f(x)g()r)
(x
r:,
-o
becautse in each case the integrand ig an odd futnction
of >r. This leaves two integrals: ol t
co
Jdr,+o(x)'lc(>r ) =
n 2N¡N3Jd>: (4«r lE _ 2)*_«»ll ú
-cü
= =
ZN,rNr
t-"
(:,
7?
th] [*] ""- = [i*] ""]
ll,,1ir
D
ri;
l,i
T Jd*,+,r
-GD
rl ,l rl ¡l
and so forth to obtain
(>r )
= 4«L/"* r*"Tr* (B«a/ rn 4 - L?«L/t* l¡ "-d!(
=
4ar/.N,N" ["o"2 " [6fu] [*]
li,iir
,li
'r'
9.-16. From Eqs. 5-4L and 5-43
=
C6
(a) = Jar, x rne-ax " = "Jr* co
,,
{xr}. =
x
roo,rt
-
?)
"-ax*/?
= [.h]
co
sJdye-aY" ¡" {a) = 4fdxe-ax
Io*(16¿rtx¿
2) 2¡rc"-«xa
-
16«x
a
+
4x"¡s-c{}r'
"Jli"^'t#l
- 16«[u*] - -t?nl] [I]"" 55ñ =il=e.p.
É8 CO
t6ft]t'" ir*(4«:rr -
= [ft]""
cn"-arl r
'f,
For n = t:r, w] write
#rtt
= [¿fi;]"o
ThLtE
even function of 5-15. Because )'3ne-d¡{" ir "n (n is an integer), we have that
I
,lz(>r)
(:,
OD
(',
:F = t-r»rTd* *añ"-ex! = (-1)nI,..,(a)
c
_ rzctL/, [k] [*],r"]
r,.,
D
),¡s
r1
Drl
,i
(>l
CO
= cJ J¿* dy E-a(xe+y=) ¡f L1
D,,I,
I D D D
I I
D
l.at dxdy = rdrd§ and x" * y" = rl and real i¡e that ar¡r integrating oveF the entire first quadrant ((, § r 4 to, 0 < Q S ¡/?l to write ¡/z co
I6t(a) = 4jae Jdr r"-"." rlE
="[k]
or that
'l'e(x) = [#] "oroorra - ?'t=-«»rE/3 Recal
I that É.? ,.
=*
= -n31t= ox
and so
CO
I¡(a) =]dxe-axE= [*]"" Now dl++"r=r,ti'"te Io(a) with respect to
D
,
d2r.-
,il;
From Eqs. 5-41 and 5-4f,'
É¡!'l'3 {x
dI^ = :f
D
5-17.
ol4
=+l¿
? JO* ),!e-ali? =
-Ir(a)
;* 3 = Jldx x¡le-all = Iz(a)
) = teft]
t'o,
,o,, -
2?«r¡r
2+
4«3>r
4) e-cl*? /2
Therefore CT'
a
{p3;" = Ja*+"
= -ñr
(»r )
Éy
l,l,¡
(¡r
t.ft] ""jo*
)
(4«x
c - ?) ( 1O« -
??¿r3>r 3
+ 4«3x 4¡ s-fi)t
t
li = -hz ["fr] = -ñi
"
=
(84«a>r T o*
{or
If we let x = F§r then this equration becomes
4
[.;] "" ["oo" [i"] - "s"" [s$] + 16«at#] - ac,«] [*]"" 3 ?
5-14. This problem is similar lrse the {ormula
Thr-rs,
e6,rsr.
ñ
to Example 5-1t).
For
(trFr
-
.{"F4t"}'l =
=
1.1155 :4 1O-34 J.B
t/2
4n(?.998 x 101o cm.s-1) :c (56O cm-1) ( (J3l?) amu) x (1.66Ct7 2< 1(:,-É7 kg'amu-1) .15 ¡{ 1(-,-1E m = 4.15 pm
the substitution
L/2,
5-2O. Eq. 3-71 is
$3.ti-t"]+=': Sr-rbsti
turte e!§t i nto thi s
eqr-rati
(4ra t ?)ettt [* - *,]*tt' As §t becomes 1Árge (asymptotic solution), this e:rpressi on becomes (4§a - ir)ettt I O
Eq. 5-6? is
obtai ned
]"'
(4Fr - 1) ¡3etF'§r =
0
F = !L/7
5-21. Start with Eq.
=u-71
*#- t*- t=]+='r
:
:,,t
II]]
l
il;
on to get
If we had r-rsed etFlainstead o+ ettE, we would
5#-(\-«">r')+=,:,
iri'ii
rl
converts Eq. 5-69 to Eq. 3-7L.
1.O55 ).: 1r-l-o+ ,.. Arrs =[ 4¡ (2. ?98 ).: 1r) 1D cm . s- 1) ){ (?.35 :{ los cm) (7 amlt) Á (1.661)7 )<:1Q-27 kg.amr-r-1) = 3.?O r< 10-14 m = 5.2O pm
[;;
-
Fz
| = ¿1¡1./t*
,.4N 14N
5-19.
$H
Thr-rs
H3
-1
Hu¡ltiPlY through bY
o
choose F Eo that the coefficient of se is urnity, or crtPa = 1, or F = L/(«L/2)
= [a*o*]""
ñrmg -
(), - ctF=Ea)+ =
We
?tl¡f,TU
1.O55 x 10-s4 J.s 4n(7.998 x lQrD cm.s-1) :< (4.33 x l0s cm-1) ( (1/2)amu) x (1.é,61:t7 j< 1L)-27 kg.amur-1) t = B.B2 x 10-12 ¡ = 8.83 prn For 3strl 3scl Arms
d 2'¡
i*?'+ F-U B-
+ 16«a:<á - ?Q«)e-«)r
_-E- = 5 . ^ = 2(:)flñ2 ¡crñ-
Arms
r-
B3
ES
have
mr!l ñt
Sutbstitute to obtain
t:i* -
,¡(
§) =
=tÍ+
Hn (
ü)e-ia/2 into this
" (t! - ,,]
eqLration
+ t* - *"],
=Q
where we have cancelled a factor of e -1"/a trollecting terms, we have
Éf?-=F5+-
[i-r],=,r
(3-7á'
each power o{ § to zero to obtain l'?ap + (t/« - l)ae = ¡;¡ Il;.¡t'-Tas + (.\./« - 1 - ?)Br = o I.,:l.i iit 5'4aa + (\/«- 1 - ?.2)a3 = r) i,! (\/q - 1- 2.§)as = ¡1 4'Eas ',1,i - + I;i ..0 Eo on. The general term (by inspection) is li
lii I;J
¡tow equrate
f.¡
:,1
(n + 1)(n + 2)a¡.¡2 + (I/« - 1 - Zn)a¡ =
ii;
ii
¡¡
t:1
l:i
iii
t-=o,
From Froblem 5-2§ we have
¡.!
5-22. Eq. 5-76 with (I/q) - t = ?n is
ii
"n*"-(.\/«-1-ln) (n + 1)(n + l) an
i'i
+:H^ dte
- =¡1!-5dt + inHn =
For large values of n, we tran neglect(I./«) -
r-,
If n = C,, Hn(¡) = 1 clearly satisfies Hermite's eqlrat i on
.
For n = 1r we have r) - ?t(2) + 2(2t) = r) Forn=2, a - lt(Bt) + 4(4Ba - ?) = r) Forn=3r (48É) - 3t(?4§3 - 12) + 6(8§3 - t?l) = s 5-25.
Hermite's differential
eqLlation is (Eq. .5-7á,
!|F-=u:P*[[-r],.«tr=e
c ,
t t t I
I
,i
l,r J
tll:
Substiturtion of H(§) = áD -h ar§ + aat! + asts + ...' into this equation gives [2a¡ + ?.3asB + 3.4a¿Ea + ...] - 2§Eai + 2az6 + Sassa +...1 + t),/s - lJta¡ + erl + a¡ta + ..1 = O CoI I ect I i lle powers of I to obtai n [2a3 + (tr/« - l)a¡J + [t.§as + (I/« - 1 - !)ar]t + t3-4aa + (\/« - 1 - Z.I)aellt + t4.Ea5 + (I/« - 1 - ?.3)aslts + .-. =tJ
and get la.,+e
I
=n I""l=ñ7=;
1
2
5-25. From Eq. 5-BO we have (n + 1) (n + 3)an*" + (.\/c - I - 2n)ar.., = n = Crrlr?n...
11
f+
l/a-1=2n for a given valure of n, then Eq. E-Brl becomes án+a = tJ Butt if a¡+r = rJ for a given value of n, then En+4 = án+6 = ñn+B = "' = ad so all the coefficients of the powers of s beyond ¡n are Eero. Thus the f r_rnctions H( §) are polynomials in t. C,
Set (.\./«) - t = ?n in Eqs. E-BZ and E-BI to
5-26. obtai n
aa=--il?na.. 64-
(2n) (?n - 4) a¡ 4l
Using
with n = 4¡ (I/«-1-B)a¿ _------.6_
Eq. E-BO
a¿ =
(2n) (2n
can see that
wE
n = l:
HÉ(i) = 1r Hl(l) = 2lr Hr(i) = 4§a 2,=?t(?E) -413+?=? (4tr-?) -2i(3§) +2(1) 2 = 2(1)
_ (2n - B)a+ 3.é
- 4) (?n -
where we have used the above expression for a4. Therefore, we have for Hn(§)
Hn(i)
- *o[r - St" * 4[t2r.
n = 2:
Hl(l) = Zln Hr(i) = 4§r - 2, Hs(§) = Bis 1?t (e¡s Bi = 2¡«+¡r - ll - 1?§) - BB (Bta-1?l)-zt(4EE-E) + 4(2§) = El§ = 4(?§) = BE
tu+il@r"*...]
,'
I
Starti
i
with Eqs. 5-84 and 5-95, we have
ng
i
ág
l
=
I
,i Itr
i'l 'rl
using
.tl !;¡
it
ill
ili
5!
Eq.
.
67. -
,ll
(?n - 2) - -E=Er-"t (2n - ?) (ln - 6)
6E=
ilr
5-28.
5-BC¡withn=5r (?n -
Thr-rs we
á'7
Hn (
have f or
ot = J+n,*
(p)
'{
G
)Ér,'l'r., (x )dx
c
-co
1t))
CO
Lr5
= -itJ+n,r,, [k]+n (x]dx
_ _ (?n - ?)(?n -t 6)(2n -----a 10)- t 7
'li
=t)
-GD
+n(x) ig an even {urnction of x if n is even and an odd fltnction of x if n is odd, BetraLtse
H ( §)
§) = á1[§ - =tt=f t'u"
. ?3(n -
d t['.,
1) (n 3!
-
3) '
_ 23(n - 1)(n - 3)(n 7l
5-27. lrle shall verify the for n = fJr 1, and ?. n = Oi Hü(l) = 1,
f or
Hr(t) = 2i 0 = ?E(1) - ?i = O 2t - 2§(1) + O = O
murl
as i n
5)
;'*
"']
A;* = an even fltnction of ¡: i{ n is odd = ean odd f utnction of x if n is even (see Problem 5-13). Consequrently the integrand in {p} will be an odd furnction of ¡¡ for any valne of n and so rip| = (t. (t
Prob I em
"ip) = J«"aa f urnction of x ) dx _co Now
consider cr,
dpz).
=
J+r,,r,)Ér,
",1,n
(¡r ) d>l
-co
O=Cr
=
8é
G G G G G G G G G
-*:i*",-, [o}],¡¡ (x ) dx a7
= o
G G G G r¿l
TE
G G G G G ET
G G
a
4fl,tti' ¡eili li¡i
Jt ,rl
lli rlll
ár
li
áll =ti =r
l,l
tl
FÍrst
Replátre >r by obtai n É4,,., t-x ) Thus, we see
let E = aL/zN and write
{pr}. = -orzzr,"i+. , rr [fr]+n Now eval uate ,¡n " « 6i
( r) d r
3t" = r'r" ffutH. ( §) e-t"/"r ] -
fr*"
Nn
[rr,'
, §) e-
t2l2 -
= *n[rn,,(§)e-t'/z
? lHn ( §)
e-
t"r=]
- 2rHn'(§)e-tr/z
* {i= - 1)Hn{tte-§t/e]
Therefore
{pa} = -o1lr¡,r¡¡fi'ja t.-t"Hr, ( i) tHn,, ( i)
-l
=qlili:
á,
:ll,¡,
r[{lii Í!l?,"i,
- ?lHn'(l) + (i"-t)Hn(i)l From Hermite's differential equration, we see that Hn"(t) - SlHn'(l) = -?nHn(l) Surbstitute this resurlt into {pz:r t,6 gs¡
"lpr) = -«lltñ3Nn"iat"-t"un(l)r(i"
- 1 * rn)tHn(t)
ál
= -«ñe t{ FE"'' - (2n + 1) l Burt {ft}.= n + 1/? iran page 1BI , and so {pt.} = -«HaEn + L/? - 2(n + L/2, 1 = (n + 1/2)ñ"« = (n + t/?»ñ(¡rk)l/a
T t
3-29. The Hamiltonian operator of a harmonic oscillator is
-
t I ,. ¡,E H«xr--+h**nx" JFt ,-r
=t
If we replace x by -x, É{x) does not change, or Ilt;rl = lll(-x) Now considerLhe Schródinger eqLtation of a harmonic osci I I ator f¡1;.¡¡,¡,n(x) = E,..,¡.1 (x)
t ú {
I
=D ,li,,
a I
-x and use the fact that É(x) = ii(->l) to
= En'!n (-x ) that +n(-x) is al.so an eígenf urnction of fi(x) belonging to the same eigenvalue En. Since the systern is nondegenerate, *¡(x) "rnd ,f,n(-x) differ by only a multiplicative constant. We can write this condition as ,|,¡ (x ) = c,ln (-x ) Burt '|,¡ (-x ) = tr'ln (x ) , and so ,f,¡(x) = cr+n(-x) which implies that ce = t1 . Thurs ,f,r.,(x) = t+n(-x) trle de{ine a re{lection operator É by fru(n) = ur(->l) The operator R is linear becaurse R(c1r-r1(>:) + crLr!(x)) -..:: 1(-;r) + caur(-¡l) = tr1FLt1(x) + crRure(¡l) Becaurse É(x) = É«-x) Atxlfr,¡,..,{;t) = fftx),+,n(-x) = En,ln(-x) ftÉ tx I ,{,¡ (x ) = frE,.r,¡r., {x ) = En*n (-x ) Becaurse R46 «>l ) - ,f,¡ (-x ) = t+n ()í ) ., we see that the eigenvalttes of fr are t1, and its eigenfutnctions are +h(x). Furthermore, becautse Éi and fr commurte, they have mutual eigenfunctions (page 1SB). 5-5O. Eq. 5-71 is d",l - l¡_ .al drt* [a- E-Jrf'=t"r .
St-rbsti turte
+(t) = BE * ar§ + arsl + aeso +... into the differential equation to obtain [!a¡ + 2..]aet -F 5.4a¿st + 4.5ari3 + . -.l + (\./«) Ea¡ + árt -F ái§3 + agÉs + ...1 - [a,r;3 + arls + "'] = (,
troI I ect I i l;e powert cl'f
E
EIa¡ + (\/3)a,¡l + t?'3a3 + (.\/c{)a1ll + t3.4a4 + (\/a)an - a¡lt! + t4'5a5 + (\/d)as - arlEs + "' = Cr Note tlrat setting the coe'fficients of ¡nr n 2 !t equal to ¡ero Ieads to three-term recutrsion forrnutlas.
Éol l
ect 1 i lle Powerg of >l (2a¡ + li2a¡) + (3'Eas + ktar):r
Set each coefficient general result
k3 aa = - E-as
gi ves the ti me that a trI assi cal harmoni c ostri I I atc¡r Epends between ){ and x + d}í. The fraction of time that the ogcillator spends be'tween ¡.t and x + d¡< is
kt
34=
4;Ea?
51n
-A 2r-|'ÁE-=TE'
5-32.
¿1,:,
* á1)i + a3)13 f áslis
+ E¿x4 + "'
(:)
l':
ó
7Ta r
Thr-rs
[r,
'
k4x4
!,:
4t.
kr>lI f,!
[,:
6¡1
6
1
6!
4x
51
5
J
lr.b;.'.7 7I
1
J
Let a1' = arlll and obtai n y(>l) = anC(h:x) 4 ár's(lt¡l) where
C(t) = [ -
.a ,4 -"6 5- + l- - ..5-ó! 21 4!
"r-
and
S(E) = $ - l"-st-l. r! 5!
ds
di -=
r-l-
^
7l
5(¡) term by term, then we find
.¡+ .ó .+ -L- - -L- + ' ?! 4t é! sl
i,i"
iirlit
7[at
p6
If we differentiate
?o
t,e
á7=
6Ea+
",
y"(x) + [,: ly(¡r] = L) to obtain t2as + ?.Saer: + S'4a¿;<3 + 4'Eas»rs + 5'6a¿¡r4 * ...1 + llcta¡ * ár){ * RexE + asxg + a4)í4 + "'l
,l,i
kt - ilaao
A,l - ;-, cr:
A
i nto
=
EE=
5i-4,
y(x) = -El "o[1 -
Su¡bsti tutte
y(x) =
kt
fr54.
¡*a
[lt
á6 =
and go on. d)Í
we
ág=
4f"o
=
Thr-rs A
= 2r,lr.r.,
n
[,4
=n'l-ÁE-:-lft To show that this elipression is normali=edr we mLtet integrate over one trornplete cycle of vibrationr or twice the path -A to +4.
rJ
nI2
{ind
dli ,+ "'- - r¡tAt - ){E11./§
f (t)dt
G G G G G G
* ... = r) equtal to zero and obtain the
n(n-l)an+klan-"=(r Using this recursion relation for
5-31. The expression
\
+ (3'i';-l=1,"i'i,1",,*"
?1
= tr(t)
G G G G G G G G G G G G G G G G G G
R EDI
and
d
I
differentiate tr(6) term by term gg -f. §t ... = -S ( t) df = -*' * §! - 5! *
if
The power series for C(q) .g
1+iÉ-ft-til*f' r?
á
To prove
we
+ ..4
*ifr-ft-tfu*... .5
.A
\''
Introdurcing i3 = -1, we can rewrite thiE power series et iasa iSts ! ... I + ii * t¡ 2.? -' i§Éo r! - 41 - 5l
which is the power series expansion of ei t. To prove that Cl(§) + st(F) = 1r wE write c3(t)
sÉ(t) to o(§8) tr3(§) = f - §" +
[h.
t - # - +É - TÉ* ot68) ár, - c(zt)r = tr - $t. * fut'+ o(tB) E(?t) =
is
iS(i)
Eomparing this we see that
=, :,
qtrS. ) The normal distribution is and
0(e)dz = Note that
]] u,
/-\ -
- th - fu]86 + o(r.)
sl(i) = ¡¿ - # [h.' c+r=] rá + o(§B) E3(¡) + sa( i)=t+Ot¡8) prove To that S( 2t) = 25(§)tr(i), we {irst multiply S(t) by c(§) to 0(i8) s(§)c(E)
t-
\4
/
á á
=t
S(?t)=2q-q.f'3! and so we see that
¡'
at least to O(18).
¡
D
s(2t) = ?s(t)c(t)
F5 5!
s2
1?B
7l
r7
GD
(!noa) -t/a[z
expt-(z - d=Y»2/?ú"Jdz
(i¡oa) _tt zI g + {r}.) exp(-y"/?u3)dy = {:} -C§
and that .Íz
3l = (?ns3)-tr:j=rexpr-(¡
- "te.!)2/Zü3ldz
ct¡
= (ZnoE)-ttz[(y,
+ ?y{z}. + {=}3)
"-y?/2o"¿"
=sE+o**]r" Introduce the standardized variable
x=
and
c c
"'
-L/? e>tp[- !Qs{]rt
co
=
[h. ]i]*" - [h. i+5.i * 1¡]u'
t - *¡t" * #t' - fft'+
-
(Zncra)
-co
-th*Eh.#-h]r7+...
¡,
to St(i)¡ which was catcutlated above,
s"(l) = (1/21t1 - c(21)l at least to O(§e).
-
ft
that 53(t) =
-
{u }'
to get E(x)dx =
(2¡r)
-L/Le-it'/2d*
derive thÍs resultr we Lrsed the fact that dr = t dx. Note that 4x) =
\
o
/=---_.--="
B H
and that
E u1
,t
GO
áa = (|rt-L/ "J*., . _co
(x
)
f
(x ) dx
§i
ti
= (2n)-rzaitxe - 1)f (x)dx = o
htj
trl
=o"*(zle-12J2-. -!ñ-=
I'
[l
'
I'tultiply f (>l) = e-x"/2t.oH"o(x) + alHer(x) r aaHez(x) + ...1
iil
JHe3
_CO
because {xt} = 1. The equtation for {(x)
becomeE
t,i lr.l ?1
Ii
by He5(x) and integrate to get 60ó
iri Iri
Ct,
f
(x
) =
(2¡r)
-t/ae-rt,"/2 *
[i
(2¡r) t/zil «x)f (x)ax = a5J "-xtlZ¡",r(x)dx = -GD
Ll
ti it
or
ti
1
T..
aj = 6;ft¡-
il
JHe¡{x)f (x)dx
-GO
To evallrate the first first few He5 (x). He¡(x) = I He 1 (n
few a5, we must generate the
If !ji
tl tl
) = (-1 ) elr azz $O=-x 2 /? =
Hes(r.r)
il li
ii lil
¡1
ti
- xa - I
= exz/? #-xz/z
ii [i
Thr-rs
lr!
oo
oD
tsrl = 12rr¡ -rzaJHe6 (x ) f
(¡r )
dx = (2¡l'r/ "J+ t* I ax -O
-S
(2¡l-t/
d
ti i'r
-rleJUer
(x
)
f
l1
il
(D
(x ) dx
l;l
it
-C,O
ri
Ii
co
= (2¡r)-t/ztx f (x)dx = because
!i H
since f (x) is a probability densitY. á1 = (3¡r)
,t
II C¡
ilrl ri
{x} = O.
[1
t'l
il
ri
ti H
fi Íl u
94
li
il
a,.rHen (x )
G
Lr
H
\
L "->:,?/2 n=3
C C C G G G G G G G G G G G G G G C C C C
?5
C É
