hws05 - PHYS 115A HW 5 Solutions 1 First we need...

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Unformatted text preview: PHYS 115A HW 5 Solutions 1. First we need expressions for x and p in terms of creation and annihilation ops. A 1 h + x: (a +a) me0 fizi‘ hwoghf—a) ,and r2 5 x: (a+a++a+a+aa++aa) 2mm0 Thus for the state |2> we get: <56>2=<2|5cl2>=1l 2 h <2|(a++a)l2>:c0nst*<2|3>+const*<2|1>=O =><5c>3=O Similarly: hwom 2 <2l(a+—a)l2>=const*(2l3>—c0nst*<2ll>=0 <13>2=i =><iy>3=0 For the x2 term we can automatically eliminate the (a+)2 and a2 terms. Thus, £2>22C<2l(a+a+aa+)l2>,where C: h 2mw0 £2>2=¢3C<2la+l1>+6C<2lal3>=fixfic<2lz>+J§J§C<2I2>=c(2+3)=5cz2 5” mwo 7h 2mm0 32>3zc<3l(a+a+aa+)l3>:\/§C<3|a+l2>+JEC<3|a|4>=3C+4C27C= 2) a) At time t = 0 the state is: Iw>=A(211>+3i|2>) <wl=(i<|l—3i<2|) <wlw>=1=lAl (4<:I,|1>+9<212>)=13|A| 3A:— £3 Where we have used the orthonormality of the enero c,y eigenstates. b) By expanding in energy eigenstates we can operate with the time evolution operator: _,- , 1 —i l —: 1 IY’(x,t)>=e ”m: —(2 “+"Il> 3,-e E’”l2>) JE 3fiw0 5th where E1: 2— 2, 2 c) Now we can use the same procedure we did in question 1. We must remember that once we've acted H on an energy eigenstate then we get the corresponding eigenvalue which is just a number. Thus a+ and a do not operate on them. <x(t)>:<‘I’(x,t)lx|‘I’(x, t)> 1 h l I I I —l l —l (26 E’”<1|—3ie“’<2|)(a +a)(2e E’”|1>+ 3ie Wm) _13 2mm <x(t)>=1—\/§(2e’E"’”< 1I—3ieiE’l/fi<2|)(2Jae—iEl'/h|2>+3ifie_iEZ'/h| 1 >) For brevity I‘ve kept only the terms I know will not go to zero. C is defined as in problem 1. l? <x(t)>=—(6ix/Ee_i(Ez_E')I/h<1|1>—6i6ei(EZ_E‘)l/h<2l2>) 13 12V2C . 12 h <x(t)>: 13 sm(AEt/fi)=E mw 0 sin(AEt/h) For <x2> again the (a+)2 and a2 terms won't contribute as they increase the eigenstate by two and thus give orthogonal states to our <2| and <].|. (x2(t)>=%(2e5'<1I—3ieE2<2l)(a+a+aa+)(2e_E‘|1)+3ie_E2|2>) Where I use E 1 and E 2 to stand for the entire exponent with those respective energies <x2(t)>:%(2e5'<1|—3ie52<2|)><(2e‘5'|1>+121e‘EZ|2>+8e‘E'|1>+9ie‘52|2>) (x2(t)>=£(4+18+8+27)=57—C= 57h 13 13 26mw0 3) I think this problem is best done using the functions in the book. Either way you have to do an integral and so bra—ket notation isn't so useful here as a calculational tool but it helps set up the process without having to write out all the integrals. If we label the eigenstates of the original potential as In> and the eigenstates of the new potential as |n'>. Then the probability of going from the |l> state to the |m'> state is |<m'|1>|2. The primed state with energy hw'/2 = hw0/4 is the |0'> state and the state with energy 3hw'/2 = 3hw0/4 is the |1'> state, simply because w' = w0/2. Then we just have to do the integrals and square them. I'll go over some ways to do Gaussian integrals that you may not have seen before. I1>=w,(z)=A1H,(c)e‘Cz’2 l0'>:‘l/ '0(CI):AOIHO(C')€_§'2/2:A0'Ho(C/\/E)e_zzl4 since C’z—f: 2 l1'>=w'1(§/l3)=Alwig/6);?” 2 (I) AIAO’ i LH1(§)H0(C/\/E)e_3gll4dz; 711(00— Prbllo.=l<0'll>lz= 2 h °° _2 AIAO' —f2:e “Md: =0 H1600— Since the integrand is an odd function over symmetric bounds. 2 Prblfll,:l<1’|1>|2=AA’ h —TH(?;) H(lj/\/—) e‘3z’4dc mwo 8T2 °° _ AHA' _J‘Cze 3:22/4dg mwo °° Once way to do this integral is to note: That integral is easy to remember: 00 —0(x2 e—ux7 1 7T :[oe dx— \/:=>:|: x2 dng 5 Thus for our problem we can quickly integrate to get the answer: 81Th 2 X16 \/_ 8 5 3mm AlAl’ " ><i =AfAl'2—O—zl6 2~0.84 3mm0 3 9 27 4) OK, this problem is easily done using bra—ket notation (Dirac notation for those of you who come to discussion session). 3) Although you don't have to expand in energy eigenstates to do this part and the next part it's a good idea because then you can use the orthonormality of the eigenstates to simplify our calculations. It turns out we'll need to do this for part (c) anyway so we might as well do it here. You can expand the given function in energy eigenfunctions in many different ways. You can do the general process of using Fourier‘s trick to get the constant coefficients or you can use a trig table to expand and see that it‘s expansion just happens to be in terms of energy eigenfunctions or you can derive the trig identity yourself which is what I'll do here. 3 iX+ —iX 1 ix — ix ix —ix 1 , cos3(x)= e 26 :§(e3 +e 3 +3(e +e ))=§(2cos(3x)+6cos(x)) cos3(x)=i(cos(3x)+3cos(x)) =>|w>=Acos3 fl =Ay/i(l3>+3|1>) a 32 2 5|AI a 4 _1_ <w|w>=1=lAlzi<1+9>= =A= =Iw>= 32 16 [5—61 (|3>+3|1>) J17) b) Since we've already broken down the state into it's energy eigenstate components this part is easy: Prb.l.=l<2lw>lz=o c) Having the state as a superposition of energy eigenstates makes this trivial: 2 2 2 2 —iF//h —iE//h 977' 71 'IT h ' ) where E3: and E1: 2 Zma 2ma2 ...
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