townsend, quantum physics, chap_5, ppios of qm

8/4/2019 Townsend, Quantum Physics, CHAP_5, Ppios of QM

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CHAPTER

Principles of Quantum MechanicsOperators playa fundamenta l role in quantum mechanics. So fa r we have introduced thepos ition , momentum , and energy ope rators. After adding the pari ty opera tor to our qui verof operators , we go on to examine so me of the fundamenta l properties of these so-calledHermitian ope rators. We will also exa mine the fund amental role played by commuta tionre lations of operators in qu antum mec hanics.

5.1 The Parity OperatorWe have noted th at the time-independent SchrMinger equ ation

fz' a'l/!---- + V(x)", = E",2111 ax'

isalso the energy eigenvalue equation

wherefi2 a2H = ---+ V(x )2111 ax 2

(5. 1

(5.2)

(5.3)is th e energy operator and we have added a subscript E on the wave functi on in (5.2)to emph as ize that thi s particular eige nfunction co rresponds to the eigenva lue E. As wehave seen in Chap ter 4, the energy eigenfuncti ons and eige nva lues ca n be de termin edonce we know the potential ene rgy Vex). We have also noted in the prev ious chap terthat when the potential ene rgy V(x) is an even function ofx [ V(x) = V(-x)]. thenthe energy eigenfunct ions are either eve n or odd fu nct ions. Thi s behavior of the wavefunctions under x -> - x is closely connected with the pari ty operato r.

The parity operator n is de fin ed byn ",(x) = " '( -x) (5.4)

153


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154 Chapter 5: Princ iples of Quantum Mechanicsthat is, when the parity operator acts on a wa ve function it inverts th e coordinates. Theeigenvalue equation for th e parity operator

(5.5)where Ais a constant, can be eas ily solved by noting th at inverting the coordinates twicemust get LI S right back where we started, that is, n 2 is the identity operator. I Since

(5.6)and n2= I, we see that

), ' = I or A = I (5.7)The eigenfunction corresponding to the eigenvalue A = +1 satisfies

(5.8)but given (5.4), this means

(5.9)Thus th is eigenfunction is an even function. Sim ilarly, = - 1

(5.10)and hence

l / t ; . I = " ~ (5. 11 )Thus the eigenfunct ion corresponding to ), = - I is an odd function. The eigenfunctionsof the parity operator are simply even and odd func tions.

It is easy to show that any fun ct ion can be wri tten as a superposition of even and oddfunct ions. Note that1 I"' (x ) = 2" ["'(x) + ", (- x )]+ 2" ["'(x) - ",(-x)] (5.12)

and th e first term in the brackets is an even fun ction whil e th e second term is an odd oneunder the tra nsformation x --+ -x . Thus the eigenfunctions of th e pa rity operator forma complete set.

EXAMPLE 5.1 i\ rgue that the eigenfunctions of the parity operator corresponding todistinct eigenval ues satisfy the orthogonality condition

I A couple of words about our notation arc pro bably in order. We usc the symbol n in stead of Pfo r the parity operator to avo id confusion with lllomCIlIUlll . Since the e igenvalucs are dimensionless .there is no o bvious symbol 10 ass ign to them. In linea r algebra. the symbol ), (not 10 be confused withwavelength) is o ftell used in the general eigenvalue problem. and we have followcd sllit here.


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Section 5.2 Observables and Hermit ian Operators 155

SOLUTION Since '/!i.=1x) is an even funct io nofx and >/1,.=- 1x) is an odd function ofx, the product > / I ' = _ I (x) >/1>=1x) is an odd fu nct ion. The integral in the orthogonalitycondition is the area under the curve of >/1:=-1 (X)>/I,.=I(X) from -00 to 00. Sin cethe integrand is an odd function and the integral is over all x, there is exactly asmu ch negati ve area as pos itive area und er the cur ve .

Alterna tive ly, make the change of variables x' = -x :

In go ing from the first line to the second line, we have switched the limits ofintegrat ion (wh ich absorbs one minus sign) and taken advantage of the fact that>/1 ,.=1 (- x') = >/1,.=1 (x ' ) and '/!!.=_I -x') = -'/!J.= - I x ' ), which yie ld s a secondminus sign. Thu s, the ex pression on the right is eq ual to the negative ofitself(youcan now set dumm y va riable x' to x on the right -hand side to make thi s morcevident) and consequentl y mu st be zero.

5.2 Observables and Hermitian OperatorsTh e pari ty operator has real eige nva lues . The eigenfun ctions co rresponding to dist incteige nva lues are orthogona l. And the eigenfunctions fo rm a complete set. These propertiesare true fo r a general cla ss of operators ca lled Hermitian operators. We postu late thatthe opcrators in quant um mechanics that co rrespo nd to ob serva ble s, to the things we canmcasure (suc h as energy), are Hermitian operators. A linear operato r ! l op co rrespondingto the observab le A is Her mitian provided it sa ti sfies

(5.13)whe re

/I and that we arctaking the complex conjugate of the function Aop . We could rewrite the left-hand sid eas ~ c o "Aop >/I dx , without the parentheses shown on the lefthand side of (5. 13). Theparent heses are n't strictly necessary since the on ly funct ion to the right of A op on whi chthe operator ca n act is l/J .

The defining relationship (5. 13) for a Hermiti an operator looks pretty abstract. But ifwe take the special case of = >/I , then (5.13) reduces to

(5.14)Recall that the left-hand side of(5 .14) is the expectation va lue of A, while the righthandside is just the comp lex conj ugate of the left-hand side since the co mplex conj ugate of>/I " is >/I and the complex conj ugate of the fun ction Aop>/l is (Aop /Ir . herefore (5.14)state s that

(A ) = (A )" (5. 15 )


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156 Chapter 5: Principles of Quantum MechanicsThus a Hermiti an operator is one that yield s real expectation va lues, which is ce rtainlya necessary condi tion fo r an obscrvab\c. 2

Nex t consider the eigenvalue equation(5. 16)

where Aop is a Hermit ian operator and a is an eigenfunction of Aop with eigenvalue a . Ifwe calculate the ex pectation value of A assuming the wave function is an eigenfunction,we obta in

(5. 17)presumi ng th at the e igenfunction is norma lize d. Thus the requirement that (A) = (A ) 'means th at the eigenvalues arc real: a = a*,

It is also straightforward to show that the eigenfunctions of a Herm itian operatorcorrespond ing to distinct eigenva lues are o rthogonal. If we assume that l/Il and 1/12 afCeigenfunctions correspondi ng to distinct eigenva lues 01 and a2 . respective ly

and (5. 18)th en

(5.19)since Aop is presumed to be Hermitian. But tak ing advantage of (5. 18), this equationbecomes

(5.20)

or(5.2 1)

Th erefore, since the eigenvalues are distinct, we must have(5.22)

This is the same defi nition oforthogonal ity th at we used in our di scussionof he propert iesof the energy eigenfunc tions for the particle in a box in Chapter 3 [see (3 .45) ]. But nowwe see that it appli es to the eigenfun ctions of any Herm itian operator corresponding toan observable.Th e eigenfu nctions [1/1" l form an orthonormal basis, at least if th e eigenva lues aredi screte. Orthonorm ality is a mix of orthogonali ty and normalization and can be conve ni ently ex pressed as

(5.23)

2Whil c (5. 14) looks less genera l than (5. 13), we can start from the requirement that the expectationvalue be real fo r an arbitrary square-integrable wave function and derivc (5 .1 3). See Problcm 5.6.


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Section 5.2 Observabtes and Hermitian Operators 157where again

8"." = { III = 1111/ # /I (5.24)is the Kronecker delta . The fact that the eigenfunctions forms a basis means that anywave function I.JJ can be exp ressed as a superposition orlhe eigenfunctions

IjJ = c,1/I, + c21/12 + C31/13 + ... = 2:: c" 1/1" (5.25)"th at is, the e igenfunctions are complete. In Section 5. 1 we demonstrated that the eigen-functionsof the pari ty operator are complete.

111 general , proving completeness is not as easy as proving or thogonali ty, but com-pl eteness is an essential par t of quantum mechanics. For if IjJ is nor malized, then

COO 1jJ ' '\I dx = 10 (2:: c ~ , 1/1,;, ) (2:: c" v,,,) dx.1-00 00 III "

(5.26)m.n " "

Thus it is natural to identify Ic" 12 as th e probab ility of obtaining the eigenvalue a" if a mea-surement of the observable A is ca rried out. While we made a sim il ar statement earlier.inSec tion 3.3 forth e particle in a box, here we arc ex tending this ansatz to the eige nfun c-tions of any ope rator corresponding to an observable. If the eigenfunction s were not acomplete sct, i f we could not presume thatwe could wri te any wave fUllction as a superpo -sition of the eigenfunctions as shown in (5.25), then we would have no way of ca lculat ingthese probabi li ties fo r an arbitrary wave functio n. Becau se of th e orthonorm a li ty of theeigenfunctions, it is straightforward to de termine c" given a pa rticular wave fUllc tion \lJ:1/I; '\Idx =1/1,; (2; Cm 1/Im ) dx

= L Cm 1::0 I / I I/Im dx = L Cm8 1111 = Cnm - 00 m

Of co urse, since Ic,1 2 is the probability of obtaining the eigenvalu e a,,expecta tion va lue of the observa ble A is given by

(A ) = 2:: Ic"1a,,"

Howe ver, 10 ljJ' AopjJdx = 1" ( 2 : : c ~ , 1 / I , : , Aop (2:: c,1/I,, ) dx00 00 IIJ I I

= 10 ( I (2:: c"a" /1,, ) dx00 m 1/

!/J . /J " "

(5.27)

then th e

(5.2 8)

(5.29)


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158 Chapter 5: Principles of Quantum Mechanicsand the refore we see quite generally thatj 'OO(A) = \jJ ' Aop \jJ dx-0 0 (5,30)Similarly

(5,3 1)and therefore

(5.32)

EXAM PLE 5.2 Show that the linear momentum ope ratorIi aP - - -XOf> - i or

is a Hermitan operator.SOLUTION Our goal is to show that10 rP ' Px", t/! dx = 10 (Px", rP )'t/! dx-0 0 - 00Start with the express ion on the left and integrate by parts (see footnote 12 inSec tion 2,9)10 , Ii at/! fll '" (a rP ) ' fI , 100rP -:- - dx = - -:- - t/! dx + -:- rP t/! - 00-0 0 faX [ - 00 ax I

= 10 ( ~ a r P ) t/! dx-0 0 I axwhere we have taken advantage of the ract that rP and t/! vani sh as Ix I ---+ 00 toelim inate the "surface" term. Notice tha t the fac t tha t the linear momentum operatoris H ermi tian depends on the natllre of the ope rato r itse l f (consid er wheth er th isoperator wou ld be Hermitian without the "i", for ex ampl e), but it also depe nds onthe behavior of the wave func tions on which this operator ac ts, nam ely that thesewave functions vanish at infini ty.

EXAMP LE 5.3 Ass ume the operator Aop co rrespo nd ing to an observab leofa particlehas j ust two eigenfunctions t/!, (x) and t/!, (x) with dist inct eigenvalues a, and a"respectively. Thusa wave fun ction co rrespondin g to an arbitrary stateof he particlecan be wr itten as

An operator Bop is defined accord ing to

What are th e eigenva lue s and norma li zed eigenfunctionsof Bop?


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Section 5.2 Observables and Hermi tian Operators 159

SOLUTION Start with th e eigenvalue equation Bop 1/1 = bl/l , that is

Since 1/1, and '/Iz are orthogona l basis func tions, the coefficients multiplying 1/1,on both sides of th is equation must be equal (a lternatively, mult iply bo th sides byI / I ~ and integrate ove r a ll x, taking advantage of the orthogonali ty of 1/1, and 1/1,).Similarly, the coefficients multiplying 1/1, must be equal. Thus we rea lly have twoequations

C2 = be] andIf we substitute th e second of these eq uations into the first, for example, we obtainc, = b'c,. Th us b' = I, or b = 1. Fo r b = I, c, = CJ, while for b = - I,C2 = -c ] . Thu s the two normalized eigenfunctionsare

and

There is an alternative strategy for solvin g this problem, a strategy that introduces some of the key ideas of matrix mechanics. Although thi s example mayseem to be an abstract one, it will turn out to be ve ry useful to LI S when we tllrnour attention to intrinsic spin angular momentum in the nex t chapter. To set thestage, note that we have determined the eigenfunctions and eigenvalu es of Bopwit hout actua lly know ing th e exp licit form of the e igenfunctions VI , and VI,. In th esame way that we can write an ordinary two-dimensiona l vector V = + asV = (V(, ~ I ' ) we lise a more compact' notation that expresses 1/1 = c]l/I] + C21/12as a two -dime nsiona l column vector

1/1 = (c,)C2

Sim ilar ly, we can write the operator Bop as a 2 x 2 matrix , name ly

sinceB 1/1 _ (0 I) (c, ) (c, )op - I 0 C2 Cl

con sistent with Bo"I/I(x) = c,I/I, (x) + c,I/I, (x)In this matri x mechanics, the eigenvalue problem Bop 1/1 = bl/l takes the fam i iar

form

which can be rewritten as

(OI)( C' ) (bO)( C' )10 c, = c, or ( -b I) (c,) =0I - b c,


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160 Chapter 5: Princip les of Quantum MechanicsFor a nontrivial solution the determinant of the coefficients of this las t 2 x 2 matrixmust van ish. Hence

I-b II 0I -bThus b' = I. As before c, = CI when b = I and c, = -C I when b = - I. Thu sthe norma li ze d eigenvectors arc given by the column vec tors

andcon sistent with o ur ea rlier results.

5.3 Commuting OperatorsWhen one multipli es nUlJlbers suchasa andb together, the order in whi ch we carry out themultiplication does not matter, namely ab = ba. We say thatthe num bers commute undermu ltiplication. But in quantum mechanics observables are associated with Hermitianoperators and the ordering of these operators mutters a great dea l. As we will see inthis section and the next, whether the operators corresponding to obscrvables commuteor not is of utmost importance. In fact, one can make the case that th e commutationrelations satisfied by these opera tors are at the very hea rt of quantum mechanics. Thecommutator of two operators Aop and Bor is defined by

(5.33)Let's start with an example in whi ch the comlllutator is zero, that is AopBop = BopAop.Consider the commutator of the parity operator and the Hamiltonian for the harm onicosci llator. In order to eva luate th e commutator, we apply the commutator, which is afterall an operator, to an arbitrary wave function ",(x):

[HSIIO, n ] "'(x)= + ~ I I I W x 2 ) n V/(x) _ n ( _ Ii ' a' + ~ I I I W X ' ) ",(x)2111 ax ' 2 2111 ax' 2= ( _ I i ' + ~ I I I W X ' ) V' (-x) _ (_ I i ' _a_'_ + ~ I I I W ( - X ) ",(-x)2111 ax ' 2 2111 a(_x)2 2

(1i2 a I ' )--- + - IIIW.C " , ( -x ) -2m ax 2 2 (

h' a' I ' )---, + -IIIWX- ",( - x)2m ax- 2=0 (5.34)

where the key step is noting that the transform at ion x ----* - x leaves the Hamiltonianun chan ged. Thu s the commutator va nishes. Thi s will be true for any Hamilton ian forwhich Vex) = V(-x).

What are the implications of two Hermitian operators Aop and Bop cOlll llluting? If weapply the operator Bor to the eigenvalue equation

(5.35)


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Section 5.3 Commuting Operators 161we obtain

(5.36)But s ince Aop and Bop commute (i.e. , AopBop = BopA op) , we can flip the order ofoperators on th e left-hand side to ob tain

(5.37)To make th e meaning of this equation clearer, let 's add some parentheses to guide theeye :

(5.38)Thi s equation tells li S that Bop 1/l(l is also an eigenfunction of the operator A op witheigenva lue a.

Now there arc two cases to cons ider. First, lel 's a SSLIme there ex ists a single eigen-function 1/Iu with eigenva lue o. In this case we say the eigenfunction is nondegenerate .Then since Bop1{la is an eigenfunction of Aop with eigenva lue a as well , Bop 1/J(1 differsfrom 1fra by at mo st a multiplica tive constant. Ifwe ca ll thi s constant b. then

(5.39)But (5.39) simply states that 1/1" is an eigenfunction of Bop with eigenvalue b. Hence 1/1"is simultan eously an eigenfunctionof A op with eigenva luea and of Bop with eigenva lueb. We can therefore relabel theeige nfu nction as Vla.h, indicating the eigenvalues for eac hof the operators. The simple harmoni c oscillator provi des a nice illustration. Since theHamiltonian for the harm onic osc i llator commutes with the pa rity operator and the re ex-istsa single energy eigenfun ction for each energy eigenvalue , each energy eigenfun ctio nmu st also be an eige nfunction of he parity opera tor. That is, eac h eigenfunction musl beeit her an eve n or an odd function, as we saw in Sec tion 4.3.If there is more than one eigenfunction with the eigenva lue ( ll we say there isdegen-eracy. What then can we conclu de? Let 's take an example involving the Hamiltonian fora free particle to illustrate. As Example 5.4 shows, this Ham il tonian commutes with thelinear mome ntum operato r. But there arc two eigenfuncti ons of the Hamiltonian wi th aparticular eigenva lue E, namely,

1/1 (x ) = A sin kx and 1/I(X) = Beoskx (5.40)where

(5.4 1We say, in this case, there is tlVo-fo ld degeneracy. You might at first be tempted 10thin k that these ene rgy eigenfun ctions shou ld also be eige nfunctions of the momentulllope rato r, but thi s is clearly not th e case si nce

. IiP. "" A Sil l kx = -:-k II cos kxI andIiP Acoskx = -- kl ls in kxx"" i (5.42)

A single derivative does not retu rn the same fun ction whe n app li ed to a sine or a co-sine. Nonetheless. there are linear combinationsof these energy eige nfu nctions that are


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162 Chapter 5: Principles of Quantum Mechan icss imul tan eously eigenfunctions of the mome ntum ope rato r, name ly the compl ex ex po-nentials

1/I (X) = Ae ik x and 1/I(X) = Be - ik x (5.43)that we used in our ana lysis of scattering. Th ese two func tions are not only eigenfunc-tions of the Ham iltonian with the eigenvalue 1i 2k2 j 2111 but they are also eigenfunctionsof the mom entum operator with eigenvalues hk and -f ik , respective ly. This illustratesthe genera l result: when the re exists degeneracy of the eigenfunction s of Aop , it is alwayspo ss ible to choose a linear combination of the degenerate eigenfunctions that are si mul-taneou s eigenfullction s of Bop when Aop and Bop com mute. We wi ll not give a ge neralproof. The interested reader is referred to a textbook on lin ear alge bra.

In summary, when two Hermitian operators commute, they have a compl ete set ofeigenfunctions in common. We can then know both of these dynamica l va riables simul-tan eously, wi th out uncerta inty.

EXAMPLE 5.4 Show that the Ham il tonian for a free particle and the linear momen-tum operator COlll mute.SOLUTION For a free particle, we can sc i the polentia l energy V to zero. Then

H = (fl ,,,,,),2m

To eva luate the commutator of thi s Ham iltonian and the linear mo mentumoperato r, let the commutator act on an arbitrary wave funct ion t/I(x):

[H, flx",,] 1/I(x) = [ 1i 2 a2 fi a]- --, - - 1/1 (x)2111 ax' i ax1i3 ( a a a a )=- - - ---- 1/1 (x)2111i ax 2 ax ax ax 2

= (a1/1(x) _ a1/1(x)) = 02111i ax 3 ax 3

5.4 Noncommuting Operators and Uncertainty RelationsWhat if the commutator of two operators is not zero? What are the implications? Aclass ic exa mple ofa nonva llishin g com mutator is the conunutato r of the po sition and themome ntum ope rators:

(5 .44)To eva luate thi s com mu tator: we again let the conunutator act on an arbitrary function1/1 (x ).

(5.45)


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Section 5.4 Noncommu ting Operators and Uncertainty Relations 163Therefore

[ X , ~ l 1 / l = ( x ~ ~ - 1/1i ax i ax i ax"a fla= x 7 - 1/I - 7- (X1/l)I ax I ax" a /1 " " a1/lx- - - -1/1 - x- -i ax i i ax= i" 1/1 (5.46)

Si nce 1/1 is arbitrary, we have shown that(5.4 7)

Notice that the key element in this derivation wa s that the momentum operator differe n-ti ates everyth ing to its right. In the first term in the commutator, this corresponds to thewave function IjJ whi le for the second term of the commu tator the function to the rightof the mom en tum ope rator is x Ijr.

The product of two operators, say Ao"B op , is itse lf an operator. Conseq uently, thecommutator oftwo opera tors isalso an operator. We will now show that iftwo Hermitianoperators do not commute and their commutator is exp ress ed as

[Aop, Bop] = iCopwhere Cop is a Hermitian opera tor, then an uncertainty relation of the form

!o.A!o.B > I(C)I- 2

(5.48)

(5.49)must hold 3 Comparing (5.4 7) with (5.48), we sec for the position- momentum commutator that Cop = " and the uncertainty relation (5.49) becomes

"o.x !o.p .< ::: '2the famous Heisenberg uncerta in ty relat ion.

(5.50)

We now turn to the proofof(5.49). Thi s gcneral uncertainty relation is a very importan tonc. Thc proof is presented so that you wi ll have confidence in the result , but in the end itis more worthwhile to foc us on the result itself rather than on the details o f the derivation.Reca ll from (5.32) that(!o.A)' = A - (A))2 ) = (A') - (A )2 and (!o.B )' = B - (B))' ) = (B' ) - (B )'

(5.5 I)We start by defining two operators

Uop == A op - (A ) and Vo p == Bop - (B ) (5.52)Then1/I ' U;;p1/l dx = ( !o.A)' and (5.53)

l it is not hard to show that Cop = - i [Aop . Bp] is Hermitian prov ided /fop imd Bop are JIcnnilian .See Problem 5.11.


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164 Chapter 5: Principles of Quantum Mechanicsand since

we also have

Now we know fo r any complex funct ion tj> that1j> ' tj>dx "': 0Let's choose as our tj>

where the constant A is presum ed to be rca l, and define

Since Aop and Bop arc Herm itian , so afC Uor and Vor' ThusI(A) = 1Uop 1/! + OYop1/!) ' (Uop 1/! + iA Vp 1/!)dx

=1/! ' (uJp+ A'V;p + iA [Uop , Vp ]) if(lx= (t.A)' + A' (t .B)' - A(C) "': 0

The minimum occurs whendl- = 2i.. (t . B)' - (C) = 0dA

A _ (C )- 2 (t.B)'

Substituting this va lue for A into the la st line of(5.59), we obtain(C)'

2 (t . B) ' ",:0or

(5.54)

(5.55)

(5.56)

(5.57)

(5.58)

(5.59)

(5.60)

(5.6 1)

(5.62)Fin ally, taki ng the square root of this equation, we obta in the general uncertaintyrela tion

t .At .B> I(C)I- 2 (5.63)As we have o t e ( ~ not on ly docs this result lead to the Heisenberg un certa inty pr incipl e,

but it al so can be llsed to derive many other important uncertainty relations. In thenext section , we examine another such relation: the Heisenberg energy- tim e uncertaintyrelation.


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Section 5.5 Time Development 165

EXAMPLE 5.5 Determ ine the matrix that r e p r e the operator Aop in Example 5.3. Eva luate the co mmu ta tor of A op and Bop in matrix me chanic s. What doyou conclude about the eigenstates of Aop and Bop?SOLUTION Since VI and 1/'2 are eige nstates of A op w ith eigenvalues {II and {I 2,respec ti ve ly, the ma tr ix representat ion of Aop is given by

Aop = (ar0 )oaas can be verified by apply ing the matrix to the column vectors

andFrom Examp le 5.3

Bop=Therefore the commutator of Aop and Bop is g iven by

A opBop - BopA op = ~ , ) ) -=Uar) U , a ~ ) = (G' ar ar G, ) 0

Since the opera torsdo not commute (preslImin g (I I =I (I2), th ey do not have eigcn -states in C0 I11111 0 11 , as we have seen in working out Example 5.3.

5.5 Time DevelopmentThe time-dependent Schrodinger equation

r, ' a' \jJ (x , I) . a\jJ(x, I)- 2111 ax ' + V(x)\jJ (x , I) = In al (5.64)is the equation of Illotion for one-d imensiona l wave mechanic s. It can also be writtensimply as

a\jJ(x, I)H\jJ(x , I ) = in al (5.65)where H is th e Hami lton ian . It is n structi ve to ask how expectation va lues vary with time.

Let's start withd (A) d 10 -- = - \jJ Aop\jJ dxdl dl - oo (5.66)

Since the limits of int eg rat ion are ind epe ndent oft im e, we can mo ve the time deri vati veunder th e integ ral sign, being careful to repl ace the ordinary time derivative with a


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166 Chapter 5: Pr inc iples of Quantum Mechanicspartia l time der iva tive since the wave function depends on both x and 1 and we arc onlydifferentiating with respect to I. Thu s, using the chain rule,

d eAl = r oc a\jJ*Aop \jJ dx +10 \jJ * aA op \jJ dx +10 \jJ* Aop aa\jJ dxdt .too at - 0 0 at - 0 0 t (5.67)From (5.65)

a\jJ I- = - H\jJat in (5.68)and therefore

a \jJ * I- = - - ( IN) *at in (5.69)Substituting (5.68) and (5.69) in to (5.67), we obtaind (A) i [ 00 10 aA i 10- - = - ( /-/\jJ )* Aop \jJ dx + \ j J dx - - \jJ ' Aop /-\jJ dxdt n. 0 0 - 0 0 at n - 0 0 (5.70)Since the operator H is Hermitian, we can rewrite thi s resu lt asd eAl i 10 i 10 1" aAo- = - \jJ ' I-IA op\jJdx - - \jJ ' AopH\jJdx + \jJ, - -P\jJdxdt n - 0 0 n -c o - 0 0 at (5.71)

Finally, note th at the first two terms ca n be rewritten in terms of the co mmutator[1-1, Aop] = I-IAop - Aop H, leadi ng to the final re suitd eAl i ;'00 [ '" aAo >- - = - \jJ ' [/-I, Aop ] \jJ dx + \jJ'-- ' \jJ dxdt n -00 .-00 at (5.72)

Inmost cases, the ope rato r corresponding to an observab le, such as position , mom entlllll,or parity, does not depend on time explicitly and the term invo lvi ng aAop / at vanishes,in which case

d eAl i [ "" - = - \jJ [/-I, Aop] \jJ dxdl n. -00 (5.73)Expression (5.73) show s us that ir the operator corresponding to an observab le A

commutes with the Hamiltonian, then that va riabl e is a constant of the motion- itsexpectat ion val ue does not va ry with tim e. In Sec tion 5.3 we saw th at the Hamiltonianfor the simp le harmonic osci ll ator and the parity operator commute. Thus pa rity is aco nsta nt of the mot ion; we say parity is co nserved for this Hamilton ian.

On the othe r hand , the Hamiltonian and th e momentum operator do not ge nerall yco mmute. As before, the co mmutator ca n be evaluated by lett in g it act on an arbitrarywave fun ct ion rjJ:

[H, Px 1 " = - --, + V(x), - , - - ' "[ n' a n a]"" 2111 ax- f ax= [vex) , ! : . ~ l '"ax

naif' na= V(x) -,- -. - -,- - V(x)",1 ax f a X

= _!: . (av)" ,1 ax (5.74)


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168 Chapter 5: Principles of Quantum Mechanics

,,,,,,,- '

,I,,,,,,,,,,,1'1,,,, ,,,,,,,,,,,,,,,,,,,, "

x

Figure 5.1 A wave packet with an uncertainty in the position of the pilrt icle can beclearly said to have changed ils location when the packet hassh ifted from its originalposition by an amount equal 10 6.x .

Let's take an exampl e to illustrate the meaning of :). (. Con sider a wav e packet witha position uncerta in ty I:;x = 10 - 3 m, as depicted in Fig. 5. 1. Suppose that d (x )/dl =I mi s, that is th e wave packet is movin g along the x direct ion with a s peed of I mls. Thetime 1:;( = I:;x / ld (x )/dl l = 10 - 3 s. T his is the time it takes the wave packet to shiftits position by an amount equal to the unce rtainty in the position. We can say this is thetime it is necessary to wa it before we can be confident that th e position of the particlehas changed. We can ca ll th e time 1:;1 an evo lutionary time, the time required for theparticle's position to change. Clea rly if the wave fun ct ion of the particle had been anenergy eigenfunction instead of a superposition of energy eigenfunctions, then 1:; = o.But such a wave func tion co rresponds to a stationary state since the lime we would needto wa it fo r the sys tem to evol ve is infini te.

Example 3 .3 provides a good illustratio n of a wave funct ion th at is a superpositionof two energy eigenfunctions. In thi s exa mple , we examined the time depe ndence of thewave function fo r a particle in a box that initia lly is in the state

(5.83)with a 50% probab il ity that a meas urement of the energy yields , and a 50% chancetha t it yields 2. The wave function al time I is given by

W(x , I) = (5.84)which can be rew ritten as

(5.85)In Examp le 3 .3 . we saw that 1:; = (2 - ,) / 2. Thus we can a lso write the wavefunction as

(5.86)


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Section 5.6 EPR, Schr6dinger's Cat, and All That 169We can say the time we mu st wait before the wave fu nction ha s clearly changed from itsinit ial fo rm (5.83) to a different wave function is the tim e t.1 necessary for the re lativephase to become of ordel' unity, that is, 2t.Et.I / fI '" I , or t.Et.1 '" fI / 2, cons istentwith our ear lier discussion.

Thi s exa mp le ill ustrates how a non zero 6.E leads to a fini te evolutionary tim e f:j"f.We ca n, in fac t, turn this argulllent around and, at the same tim e, mak e our discllss ion ofexcited states of the systems we have treated so far more rea li st ic. Cons ider, fo rexample,an electron in an exci ted state of an atom. Since the electron wi ll eventua lly make arad iative transition to a lower ene rgy state with a lifetime r, there is a finite evolutiona rytim e fo rthe system in thi s exc ited state. Forthe first exc ited state ofhydrogen, forexamplc,T = 1.6 ns. Co nsequent ly, according to the energy- time uncerta in ty relation , the excitedstate mu st have a non zero uncertainty in its energy, as indi cated in Fi g. 5.2. Therefore,the photon that is emitted in the transition between this exc ited sta te and the ground statedoes not have a definite frequen cy. Th is spread in freque ncies, or wavelengths, is referredto as the natural Iincwidth for the state.

EXAMPLE 5.6 Calcul ate the naturallinewidth for the first excited state of hydrogen.Co mpare thi s spread in wave lengt hs wi th the prim ary wave length of the transition.Note: E, - E, = 10.2 eV.SOLUTION

E, - E, he=/ iv= -ieThe principal wavelength for the transition is 12 1.5 nm. But since the excited statehas a lifetime r = 1.6 x 10- 9 s, the uncerta inty in energy for thi s excited state is

fI6.rTh e uncertainty in the energy cor responds to a spread in wave length

het.E ' " - t.ieie'Given the uncertainty in 6., we see that 6.A = 5 x 10- 15 m, or roughly 4 pmt sin lOs of the wavelength of the transition. Thu s the natural linewidth is tough toobserve in th is case. Of course, if the lifet ime were shorter, the effect wo uld beeasier to observe. We will see a striking example in Chap ter 10.

5.6 EPR, Schrodinger's Cat, and All ThatOne of he arresting feature sofquantummcchanics is that particles or systemsof part iclesdo not in general have definite attributes. \Vhcn we exp ress the wave functio n as

(5.87)where 1/1[ and 1/12are eigenfunction s of an operator A op cor respondin g to an ob servab leA , we are say ing that before we make a measurement of A the particle does not have

t:.EtE, ____ _

Figure 5.2 An energy -l eve ldiagram fo r the ground staleand the first excited staterequires modi fica tion whenthe effect of the finite lifetimeof the excited state is takenint o accoun t. The s pread incncrgy for the excited state isnot drawn to sca le.


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170 Chapter 5: Principles of Quantum Mechanicsa de fi nite value of th at attr ibute. The probability that a measurement yields (I I is, ofco urse, IC112. Aft er a meas urement yieldin g ai , the wave fun ction co llapses to 1/11 , sincea measurement o f A immediately the reafter aga in yie lds the va lue (I I . How this co llapsehappens is a mystery. It is referred to as the measurement problem. We will return to thi sisslIe at the end of thi s seclion.4

a t everyone has been happy with th e idea that particles do not necessari ly havedefini te attributes before a me as urem ent is mad e. The mo st famous malcont ent wasalmost certa in ly Albert Einstein. In his v iew, writing a wave function in the form (5.87)was really expressing our lack of knowledge of the state of the particle. Some particleshave attribute ti l an d ot hers have attribute 0 2 , but we arc not able to distingui sh oneparticle from the other. It is as if th e attribut e, or variable, that wou ld allow us to makethi sdistinction is hidde n fram lls, hence a hidd en var iable theory of quantum mechan ics.Such a view , whi le no doubt appeali ng to some, seem s to raise troub ling issues as we ll.A fter a ll , if the helium atom in the double-slit experiment desc ribed in Section 2. 1 rea llyhad a defin ite pos it ion berore it s position was measured in the detection plane , then onecould log ica lly presume it also had a defini te posi tion ju st moments before this, andso forth. Thus we woul d be led to presum e th at the atom followed a definite trajectorybetween the source and the detecto r. passing through one slit or the o ther, which makesit hard 10 und erstand how interfe rence ca n occur. onethc lcss, Einstei n persisted in hi svicw that the descri ption of nature prov ided by quantum me chanics was incomplete , thatthere was more to nature than quantum mechanics pres umed.

To sharpen hi s critici sm of quantum mechanics, Einstein, together with BorisPodol sky and Nathan Ro sen, proposed a thought exp eri ment that , in his view, showe dhow crazy quantulll Illcchanics rca ll y is.5 Th e key aspect ofquantulll Illcchanic s that EP Rfo cused on wa s something that we now call an entangled state. Consider the two -parti clewave function

I\j I = Ji [1/1", (I )1/1", 2 ) - 1/1",(I 1/1", (2)1 (5.88)

as an example. In this wave fun ction 1/1" ,( 1) means that one of the particles, particle I,is in sta te wi th va lue a, of the observable A . The wave fun ction (5.88) shows there isa 50% proba bility that a measurement of A on particle I will yie ld this va lue. But thenwe will su bseq uent ly know that pan ic le 2 is in a stale with va lue Q2 fo r the observa bleA . What was really di stress ing, at least to Ein ste in , was th at not only do particles I and2 not ha ve de fin ite attribute s, but the va lue of the observable that particle 2 takes on isdetermined by a mea surement on particle I. After all , i f the mea surement on particle Iyielded the value 0 2 instead ofa t, thcn we wo uld know that particle 2 is in the state withvaluc a t . And parti cles I and 2 do not have to be situa ted together in spac e for this tooccur.

A partic ularly intere sting examp le is the two-p hoton "wave funct ion"(5.89)

..\ If the operato r Aop corresponding to the obse rvable A com mut es with the Ham iltonian, then A isaconstant of the motion. Thu s once the system collap ses to a part icu lar eigenfu llct ion. it wi ll rema in inthis sta te.

SA. Einstei n, B. Podo lsky. and N. Rosen, Phys. Rev. 47 , 777 ( 1935).


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Section 5.6 EPR, Schrbdinger's Cat, and All That 171where the subscripts Rand L refer to right and left circula rly polarized photons. Such atwo-photon sta te is generated in the cascade decay of the excited state of he calcium atomthat we discussed in Sect ion 1.5. Thus if a measurement on one of the photons finds it tobe right-circularly polarized, then the other photon must be right-circularl y polarized aswell. But if we express the circul arly polarized states in terms of the approp ri ate linearlypolarized states:

(5.90)and

(5.91 )assum ing that photon I is travel ing along the z axis, we find

(5.92)as can be readily verified by substituti ng the expressions (5 .90) and (5.91) for the ci rcularly polarized wave functi ons for photon I and the cor respondi ng expressions forphoton 2 into (5.89). See Problem 5. 12. In thi s form the wave function indicates that ifa measurement of the linear po larizat ion shows photon I to be x po lari zed (and there isa 50% chance of obta ining thi s result), then photon 2 wi ll necessarily be x polarized ifa measuremen t of its lin ear po larizat ion is carr ied out. Simil arly, if a measurement oflinea r polarization on photon I fin ds it to be y polarized, then photon 2 is necessarilyy polar ized. The state (5.92) exhib its the same sort of en tang lement that we saw in ourdiscussion of c ircular polar ization for the state (5.89). But what is strik ing here is thatphoton I cannot be both right circular ly polarized and at the same time linearly polarizedalong the x-axis. Thu s it is impossibl e to think of the photons as having a definite sta teof pola ri zation before a measure ment is ca rried out. In the ori gi nal EPR argume nt, thetwo incompatible var iables were position and momentum, not ci rcula r and linear po larizatio n. As we have seen, the operators corresponding to position and momentum donot comm ute with each other. Thus it is impossible to imag ine a part icle as hav ing botha definite posit ion and a de fini te momentum, just like we cannot think of a photo n ashaving a defi nite circu lar and linear polari za ti on . Nonetheless, states of the fo rm (5.89)or (5.92) seem to be saying that not only do part icles not have definite attri butes but theseattributes can be determined by measurements on a different particle al together, one thatmay be spat ially qui te separated from the other. To Einste in , thi s "spooky action at adi stance" was unacceptable, so met hing that "no reasona ble definition of rea li ty" sho ul dpermit.

In the 1960s John S. Be ll rea lized that two-partic le states such as (5.88) could provide an experi men tal test as to whether a particle rea ll y has defini te attrib utes before ameasu rem ent is ca rried out. It was ju st necessary to measu re the correlatio ns betweenmeasurements carried out on the two partic les. These were challengi ng experiments, atleast initially. But as the technology has improved, the resu lts have become more andmore striking, to the point that the most rece nt res ults are in clear agreement with thepredictions ofquantum mechanics and disagree, at the level of 250 standard deviations,with local hidden variable theory.

A classic example that seems to ill ustrate the problem with taking this line of reasoningthat particles do not necessari ly have defin ite attributes too far is Sch rodinger's cat. As


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172 Chapter 5: Principles of Quantum Mechanics

Figure 5.3 Schrodinger's cat.

a thoug ht experiment, Schrodinger suggested placi ng a cat in a box with a rad ioacti veisotope that had a half-life of one hou r. As indicated in Fig. 5.3, the decay of a singlcnucleus, with the emission of, say, an alpha part icle, wo uld trigger a hamme r to breaka fl ask of prussic acid, a hi ghly volatile and toxic substance . Thus, in the languageof qu an tum mechanics, one mig ht wri te the wave fu nct ion for the cat" at the one-hourmark as

I I\jl (i hour) = .J2 l / t ~ l i : , + .J2 l / t d ~ ; d (5.93)Qu antu m mecha nics seemed to say that only when the box was ope ned at the one-hourmark and the ca t was observed to be ali ve or dea d was the wave funct ion co lla psed toone orthesc two stales. Thi s indeed seemed si lly, s ince there was little doubt that the ca twas truly alive or dead indepen de nt of th e measurement. Thu s think ing ofa mac roscopicobject such as a cat as ge nu inely not hav ing a defin ite attri bute, such as being alive ordead, seems not to be correct. Bu t on the other hand, we might thin k of the cat as amacroscopic measuring dev ice that has become en tangled with the corresponding stateof the nucleus, say in the form

\jl( 11 .) 1 ",nuclcus",cat 1 ."nucleus ." cat10U I = .J2 'f' nodecay 'f'alive + .J2 'f' decayed 'f'dc3d (5.94)Such a wave funct ion wo uld st ill suggest that the ca t was ge nuin ely in a su per position oftwo states . But how would we know if this were indeed the case?

Let's go back to the doub le-slit experiment with helium ato ms (Sec tion 2.1). S incethere arc two paths each ato m can take between the source and the detector, we mi ghtwr ite the wave funct ion as

(5.95)where


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conduct the appropri ate expe ri ment. The largest objec ts for whi ch we have seen inter-ference fr inges in a double-slit exper im ent are Coo molecules, bl'ekyballs. Phys icists arefrying to push the envelope here , to mo ve towa rd mac roscopi c objects that would trul yexhibit interference effects. Such states , if they ex ist, are ca lled Schriid inger cat states.But so far, no one has been abl e to observe in terference effec ts with even Schrodinge rkittens. Why not? Our best guess is that interactions with the envi ron men t, whi ch areinev itable for a macroscop ic obj ec t such as a cat , cau se the wave fun ction to lose thi srelative ph ase in fo rmation. We say th e state decoheres. And perhaps thi s decoherencein some as ye t unexplained way leads to the co llapse of the system to one or the otherof the states mak ing up the superpos ition. But thi s seems inconsistent with the funda-mental pr in ciples of qua ntullI mechanics , since tile equat ion of time developm ent, theSchr6d inge r equati on, isa linear di fferential equat iOIl. I f the in itial state isa superpo sitionof states , so too should be the state at a later t ime. This is the crux of the measurementproblem.

5.7 SummaryLe t us summa ri ze the prin cip lesof qua ntum mechani cs tha t we ha ve discussed in thi schapter:

I . In one-dim ensional wave mechan ics, the sta te of a particle is gi ven by a wavefunction ll1 (x , f) that conta ins all that can be kn own abollt the parti cle .6

If th e question is "wh ere is the parti cle," then w' wdx is th e probabili ty offindin g the parti cle between x and x + dx if a measurem ent of the po sitio n ofthe particle is carri ed out at timc I , prov ided the wave function is normalize d,namely

1" w' wdx = I-0 0 (5.97)Note that the particle does not have a defi nite location before a measurement iscarried out. Bu t wha t i f we want to know something else abo ut the pa rti cle suchas its encrgy? We start with a second principle:

2. Eac h dynam ica l va riable, or observab le, A is assoc iated with a lin ear,Hermitian operator Aop , an operator for which

The eigenva lue equ atio n isof the formAop1{l(l = a 1{l(/

(5.98)

(5 .99)where the cons tant a is the eigenvalue and 1{Ia is the eigenfunction. The onlypossible resu lt of a measurement fo r the observable A is one of th e eigenva lue s.

6 [n Cha pter 6 we extend our di sc lission of qUlI ntum mcchanics to inclu dc three-d im ensional sys tems.at which poin t we di scliss intrinsi c spin. a 'deg ree of freedom" tha t isnot spec ified by a wave funct ion.

Section 5.7 Summary 173


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174 Chapter 5: Principles of Quantum MechanicsSo far, we have cons idered observab les such as the momentum and the

energy. For the energy the corresponding operator is referred to as theHamiltonian H and the energy eigenva lue equation is the tim e-independentSchr6dillger equation. In the next cha pter, where we will ven ture into threedimensio ns, we will introduce additional operators such as the orbital angul armomentUIll operators.

3. The eigenfunctions 1/10 form an orthonorma l basis. Therefore any wave funct ion\lJ ca n be written as

(5. 100 )The probab ili ty of obtain ing eigenvalue a is given by

(5.101)where 1/1" is the eigenfunction of A op with eigenva lu e a7 The average va lu e, orexpectat ion va lue, of the observable in this state is then given by

(A ) = L Ic,,12a =1" \jJ ' Ao,,\jJ dx(/ -0 0 (5. 102)The results th at follow from the commutat ion relations oFtwo operators can bederived from the general properties of Hermitian operators. Nonetheless , theseresu lts arc of stich importance tha t it is wo rth singling them out.

4. The commutator o f two operators Ao" and Bop is defined by(5. 103)

Irthe operators commute. that is the comm utator va nishes, then it is possible tolabel th e basis states as 1/I".b, namely as simultaneous eigenfunctions of the twooperators. which we are assuming to be Hermitian. If the operato rs do notcommute , but

thentl A t lB> 1(c) 1- 2

The mo st famous example is the Heisenberg unce rtainty prin cip le

ntlx tlp, :: 2wh ich follows from the cO lllmuta tor

(5. 104)

(5. 105 )

(5. 106 )

(5 .107)

71fthe ci genvaluc spectrulll is continuous ratherthan discrete, then the probab ility of obtai ning a resultbetween (I and a + dll is given by 1c,, 1(/a provided the eigenfunctions sat isfyJ-oo VI; 1/1/1' dx = &(lI - a')where o(a - (I' ) is a Dirac delta fun ction.


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5. Time dependence is determined by the Schriid inger equ at io nf-/'lJ(x , I) = itJ _a_lJ7,...:.,-'.I)al (5.108)

whe re the Hamil to nian H is the energy ope rato r. In one-dimensional wave mechan icstJ2 a'

/-I = - 2111 ax ' + V(x) (5. 109)We can usc the Sc hriidinger equat ion to show that expec tation va lues vary with timeaccording to

cI (A ) i loo 10 aA- -= - 41*[ /-/ , Aop ] l/Jdx+ 4 l . f J dcli tJ . -0 0 - 00 al (5 .110)Thus if the Hamiltoni an commutes w ith the operator corresponding to th e observabl e IIand aAop / al = 0, then (A ) is independent of time and A is re ferred to as a con stant ofth e motion. One consequence o f (5.1 10) th at is obta ined by c hoosi ng th e o pera to r Bop in(5. 104) to be the Ham il toni an is

which can be wri ne n as

t.A tJt .-- >-~ 2d,

tJt . t .1>- 2

(5 . 11 1)

(5. 112)th e Heise nberg energy- tim e unce rta inty re lat ion. In thi s relat ion, t.1 is called an evolutionary time fo r the sys tem , the time that is necessary fo r the system to change in asignificant way.

In ge nera l, the numb er of eigenfunctions is infinite and the space spanned by thesebas is functions isan infinite dim ensional vec tor space ca lled a Hilbert space,s

Problems

Proble ms 175

5 .1 . (a) Prove that the parity operator is Hermitian.(b) Show that the e igenfunc tions of th e parity operatorcorresponding to di fferent eigenva lues are orthogonal.

5.4. A part icle of mass 11/ moves in the potential ene rgyVex) = ~ 1 1 l w 2 x 2 . The gro und-state wave fu nction is

5.2 . Show that the operato r

isno! Hermitian.

aax

5.3. We have argued that a Hermitian operatorcorresponds to each obse rvabl e. Phys ica lly, why is itessential th at the eigenvalues be rea l?

(a)' /4 ,1/10(.') = ;: e- ax /2and the first exci ted-state wave function is

where a = mwjli. What is the ave rage va lue of tileparity for the stat e

J3 1 - i'lJ (x) = - 1/Io(x) + r-; 1/I ,(x )2 2-;2gNote this is not ord inary space. It is sometimes said that in Hilben space no one can hearyoll sc ream.


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176 Chap1er 5: Principles of Quantum Mechanics5.5. For a pa rticle in a harmonic osci llator potential, it isknown that there is a one-third chance of obta ining theground-state energy Eo, a one-thi rd chan ce of obtai ningthe first-excited-state energy E" and a o ne-third chanceof ob taining the second-excited-state energy 2 if ameasurement of the ene rgy is carried out. If ameasurement of the parity is carried out and the va llie- I is ob tained, what va lue wi ll a subseq uentmeasurement of the energy yield? If a measurement ofthe parity yields the va lue + , what values can asubseque nt mea suremen t or lhe energy yield? What arethe probabili ties of obtain ing those ene rgies?5.6. By writing the wave fun ction \{I in

as t/J + At/> where A is a n arbitrary compl ex number,show that

Thus the requirement that an operator corresponding toan observab le has rea l expec tati on va lues is eq ui va lent tothe deAnitio n ofa Hermiti an operator g iven in (5.98).Suggestion: Take advantage of the fact that and A arelinearly independent.5.7. Show that the two wave funcr ions

and

from Exa mple 5.3 ca n be expressed in th e formt / = cosO t/J, + s in O t/J,

andt / b = sin O t/J, - cos O t/J,

if the appropriate cho ice for th e angle 0 is made. What isthe va lu e of0 for these wave fu nctions?5.8. Let th e operato r Ao" co rrespond to an observable ofa particle. It is assum ed to have j ust two eigenfunctionst/J ,(x) and t/J, (x) with dist inct eigenvalues. Th e functioncorresponding to an arb itrary state of the particle can bewritten as

An operator Bop is defined acco rding to

Prove tha t Bop is Hermiti an.5.9. Show that if the operator Aop correspond ing 10 theobservab le A is Hermitian then

(A' ) ;0: 05.10. I f !lop and Bop are Hermi tian operato rs, prove thatA op Bop is Hermitian on ly if !l op and Bop comm ute.5.11 . Supp ose that Aop and Bop are Hermitian opera torsthat do not commute:

Prove tha t Cop is Hermitian.5.12. Use the defi nitions (5.90) and (5.9 1) of the rightand left circular polarized states to show that thetwo-photon state (5 .89) becomes (5.92) when expressedin term s of the linearly pol arized state s. Calilioll : Sincethe photons arc trave ling back to back and photon I istraveling in the posit ive z direction, photon 2 is trave lingin the negative =direction. Consequently, for photon 2

It/JH (2) = Ji [t/J . (2) - it/J, (2)j

andI

t/JL(2) = Ji [1ft. 2) + it/J.(2) j

townsend, quantum physics, chap_5, ppios of qm

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