Solutions to problem set ); (, ) (

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1 Solutos to proble set.. L = ( yp p ); L = ( p p ); y y L, L = yp p, p p = yp p, + p [, p ] y y y = yp + p = L y Here we use for eaple that yp, p = yp p p yp = yp, p = yp : factors that coute ca be treated as costats. The safe way s to epad all, of course. Here I dd ot eve epad the oetu operators as p =, but used the coutato relato betwee the operators drectly: [ pˆ ] ˆ, =.. ( )( ) [, ] L L L L L L L L L L L L ( ) L + = + = + + = + + = L L + L y y y y y s the prevous ecercse there s o eed to wrte out the operators. It s easer to evaluate the coutators drectly. L, L = [ L, L + L ] = L, L + L, L = L + ( ) L = L + y y y + L, L = [ L, L L ] = L, L L, L = L ( ) L = L y y y 3. Ψ(, t) t t c e Et/ c e Et/ = [ ψ ( ) + ψ ( ) ] = Et/ Et/ ceψ ( e ) + ceψ ( e ) (, ) ( Et/ ( ) Et/ HΨ t = H c e ψ + c e ψ ( ) E / ( / ce H ( )) ce ( H t Et = ψ + ψ ( )) Et = ceψ ( e ) + ceψ ( e ) / Et/

2 ad you see that both evaluate to the sae epresso. Therefore Ψ(, t ) satsfes the t te-depedet Schrödger equato Ψ (, ) = H Ψ (, t). Please ote that t H Ψ(, t) EΨ(, t), ad Ψ(, t ) caot be wrtte as Φ( ) ϕ ( t). Ths s ot a statoary state, uless E = E. Fally Ψ( t, ) does deped o te. Let e ephase statoary states are very specal. There are other ways to satsfy the TDSE. The above proof oly depeds o the fact that t ad H are lear operators (ad H Et/ does ot deped o te). Therefore ay lear cobato of fuctos ψ e satsfes the te-depedet Schrödger equato. It does ot have to be a egestate! 4. d d = Ψ ( t, ) Ψ( t, ) dt dt Ψ( t, ) Ψ( t, ) = [ Ψ(, t) + Ψ( t, ) d dt dt = Ψ t H Ψ t [ (,)( (,)) Ψ(,) t HΨ(,)] t d = [ Ψ(,) t H( Ψ(,)) t Ψ(,) t HΨ(,)] t d = Ψ( t, ) [ H, ] Ψ( td, ) I the secod le we used the cople cojugate of the Schrödger equato, ad the thrd le we used that H s hertea: Χ( )( HΦ( )) d = Φ( ) HΧ( ) d. d d H, = [, ] + V( ), = + 0 = d d P. The rest of the proble s straghtforward ad we fd that the average values quatu echacs satsfy the classcal deftos ad equatos of oto (see als MS 4-7).

3 5. operator O s Hertea f for ay two fuctos f ( ), g( ) that satsfy the B boudary codtos the tegral g ( )[ Of ( )] d = f( )[ Og ( )] d. The tegrato terval s fro 0..a for the partcle the bo, over a perod 0.. π or π.. π for the partcle o the rg, whle for the Haroc oscllator proble t would eted fro... Cosder the oetu operator p = : B df g p f d g d d g f B B f g ( ) ( ) ( ) = ( )( ) = ( ) ( ) + ( ) d B g( ) B = f( )[ ] d = f ( )[ p g ( )] d B (Where we used partal tegrato). The crucal part s the vashg of g f B ( ) ( ). For all of the eaples lsted above, you ay verfy that ths s true. I geeral the boudary codtos QM are always such that the above ter vashes. Ths s where the caveat that the fuctos should satsfy the boudary codtos coes. lso ote the careful use of brackets to dcate what s to be starred (cople cojugato). I these ecersses I a careful to oly star costats or fuctos. I do ot accept the starrg of operators. You ca oly use the foral defto of Hertcty. B 6. Et/ Et/ a. I ecercse 3 we have show geeral that a fucto ce φ ( ) + ce φ ( ) would satsfy the TDSE. Here we have =, =, whle E = ω ad E = 4ω. The wave fucto gve uder a hece satsfes the TDSE ad equals Ψ(, 0 ) at t=0, whch s all that s requred. b. π t = : Ψ(, t) = [ s( π) + s( π)] ω π t = : Ψ(, t) = [ s( π) + s( π)] ω π t = : Ψ(, t) = [s( π) + s( π)] = Ψ(, 0) ω c.... Use pctures of s( π), s( π) ad add the wth proper factors. The agary part of the wave fucto oly shows up for t = π /ω. Please ote that the wavefucto s ero outsde the bo ( < 0; > ) d.

4 Ψ ωt 4ωt ωt 4ωt ( t, ) Ψ( t, ) = [ e s( π) + e s( π)][ e s( π) + e s( π)] 3ωt 3ωt = [s ( π) + s ( π) + ( e + e )s( π)s( π)] = [s ( π) + s ( π) + cos( 3ω t)s( π)s( π))] You see we have a oscllatory behavor of ths probablty dstrbuto wth agular ω frequecy 3ω, (or ν = 3 ), the dfferece oscllato of the two copoets of the π wave fucto. The wave fucto tself s perodc wth frequecy ω / π, as see uder b. (the slowest coo frequecy). e. To sketch the fucto, t s better to square the real ad agary parts of the wave fuctos as draw uder c, ad add the up pctorally. Ths s what t eas to take Ψ ( t, ) Ψ( t., ) 7. a. ˆ S ( ˆ ˆ = S+ + S ) ˆ ˆ ˆ S ( α + β) = ( S+ + S )( α + β) = (0+ α + β + 0) = [ ( α + β)] ˆ ˆ ˆ S ( α β) = ( S+ + S )( α β) = (0 α + β 0) = [ ( α β)] Ths shows that these fuctos are deed egefuctos of S ˆ wth respectve egevalues are + ad BS ˆ wth egevalue. The fucto α s a egefucto of the Haltoa B = B, whle β s a egefucto wth egevalue B. The wavefucto at te t=0 s hece a superposto of statoary states, ad followg the geeral recpe the wave fucto at te t s gve by Bt Bt Ψ ( t) = ( αe + βe ) = ( α[cos( Bt) s( Bt)] + β[cos( Bt) + s( Bt)] = ( α + β )cos( Bt ) ( α β )s( Bt ) I apulated the result for future coveece. You ay otce that the wavefucto s eplctly wrtte ter of the orthooral egefuctos of the S ˆ operator (see ht part b). Both the se ad cose oscllate as Bt. The perod s defed as the te t takes π for ths phase to chage by π, hece T =. B

5 b. The possble values that ca be easured for S ˆ are always oe of the egevalues: + or. The probabltes deped o the wavefucto Ψ () t P( + ) = ( α + β) Ψ ( t) = cos( Bt) P( + ) = ( α β) Ψ ( t) = s( Bt) = s( Bt), ad they are gve by Hece: TIME: P(+/) P(-/) t=0 0 t=t/4 =π /B 0 t= T/ =π /B 0 t=t/8=π /4B / / You see that the probabltes deped o the te of easureet. Ths s true because the wave fucto s ot a statoary state, whle the operator S ˆ does ot coute wth the Haltoa, ad so these operators do ot have a coplete set of coo egefuctos. If I would chage the proble such that you would easure S ˆ, the the possble outcoes would also be + or, but the probabltes would be Bt P( + ) = α Ψ ( t) = e = Bt P( ) = β Ψ ( t) = e = Ths s a geeral result. You ca prove that f the operator you easure, ad the Haltoa coute, the results of your easureet are depedet of te.

6 Proble 8: S&O.: a. b. Oe = e O j j j e Oe = e e O = e e O = δ O = O k k j j k j j kj j k j j j Oa = b Oe a j j j k k k j j k k k k j k k j j j e b e Oe a = e e b = δ b Oa = = b S&O.3: S&O.4 ( B) j = ( B) j = ( jkbk) = jkbk = kj Bk = Bk kj = ( B ) j k k k k a. ( ) ( ) Tr( B) = B = B = B = B = Tr( B) j j j j, j, j j jj b. c. d. e. B ( B) = B ( ) B= B B= B = ( B) B= U U = = = UBU UU UU ( B) ( ) B B = B B B = ( ) = ( ) ( ) = ( ) = ( ) = f. Show by eplct ultplcato that 0 = = 0

7 S&O.6: 6. Iterchagg equal rows does ot chage atr, but does chage sg deterat, hece = = = = = = = = = = U OU = U OU = OU U = OUU = O = O S&O.8: Tr U OU = Tr U OU = Tr OU U = Tr OUU = Tr O = Tr O ( ) ( ( )) (( ) ) ( ) ( ) ( ) S&O.0,.. a+ b 0 = 0 a b S&O.3 f( a+ b) 0 f( ) = 0 f( a b) f( a+ b) f( a+ b) = f( a b) f( a b) f( a+ b) + f( a b) f( a+ b) f( a b) = f( a+ b) f( a b) f( a+ b) f( a b) S&O.4 ε ε ( ) δ ( ) = l ( ) l [ (0) '(0) ''(0)...] ε 0 = ε ε 0 ε ε ε da a d a a a d 3 ε ε 3 ε ε = l[ { a(0)[ ] ε + a'(0)[ ] ε + a''(0)[ 6 ] ε +...} = a(0) l[ +...] = a(0) ε 0 ε ε 0 6ε S&O.5: Bra-ket otato: O ˆ = k ko k j O ˆ = j ko = jko = δ O = O k k jk k j k k k

8 S&O.7 e: ψ( ) Ojψ j( ') = Oj j ' = O j j ' = O ' = O ', j, j, j The other probles are straghtforward substtuto.