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PSet4Solutions - Physics 131 Problem Set 4— Due February...

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Unformatted text preview: Physics 131 Problem Set 4— Due February 8, 2008 Problem 1 (10 points) Consider a harmonic oscillator with a Hamiltonian of 2 2 2 p mw a: H = — 1 2m 2 ( ) using a trial wavefunction N we) — $2 + b2 (2) a.) Using dimensional analysis, determine how N and b depend on the dimensionful param— eters of the Hamiltonian (and h). b.) Determine N by demanding that 1/2(a:) is normalized. c.) Determine b for the ground state. (1.) How does the ground state energy compare with the exact answer. Problem 2 (10 points) Following Section 7.2 in Griffiths, ShOW that the ansatz used for He leads to the fact that H “ is not stable and would decay to H + 6‘. Problem 3 (10 points) a.) Following the discussion in class, show that the (90?) WKB approximation satisfies 1 1 060’; + 5012 + 50,1, 3 0 (3) b.) Using that F = pp’/m (where F = —%¥), show that 1mF 1 m2F2 =——— — d 4 02 4p3+8/ p61” () Notice that the condition that we found in class for the consistency of the WKB approxi— mation hmF/p3 << 1. c.) Show that this leads to a modified WKB wave function of the form N 2' hmF 2' #77121172 pdcc 2' = — 1 — — — — — — d % g t n E ti E :ezé‘kzdfim’fifltzia «amaayswagw aawgm mazeemawms new;ea.e.at-,»mw.¢myawmtmmaww, stage-enter»): Problem 4 (20 points) Consider a quantum mechanical particle bouncing on the floor. The Hamiltonian is given by H = 1’1 + ve) mm) = {2” :3 (6) a.) Solve the Schroedinger equation exactly using Airy functions. There is no need to normalize the wavefunctions. b.) For a neutron, what are the first four energy eigenvalues? (This experiment has been done at a reactor in Grenoble). c.) Using the WKB approximation and implenting the connection formula discussed in class, solve for the energy eigenvalues for the first four levels. d.) Compare the errors between the WKB results with the exact answer to hmF/p3. Problem 5 (10 points) Classically, show that the probability to find a particle in a harmonic oscillator potential is proportional to P(:c) oc 1/v(x) by explicitly constructing P(33)clx. wavazflalfififlmar xK/re'ama«Wmew—7<zrv&7m7rrxflumV®;;,W-V,>r.m;7:<5 .16 M75 m M mmmag“wwmmeumx 4W0 a9 00 1 w I: S ““1375 = S ('$M;‘:)Lz>z.d% 2 ‘Mh, goo iii??? .. Z w 7’ — M g “Ax m m mey bag INA; Agfi ~00 b (l‘4rxt/bz') b“ ~00 (1+ 3L 7‘ b -03 (H’flz’) L86 3‘ {ta/mu, ‘37 «by seaz‘mdumwn, Hf: l+£am1pc=sacfw 11/; s =. w? 55%“ _ [M11 fit a“ _ ML 1% z. b -E (3661“ 2: b3 41% 5869“, " ‘03, 811% £106 MAM = M“: '7': 1.. M J E” “WM“ ‘ 'le ‘5’ <T>= ”“1 1 ’L 0%... m S \ L. 1 3‘3" m .00 X14: ‘0" dxz X1+ 5" ‘ 5%: W?“ {M ‘Exz ._ 2 Age am 3am (thy (xz+bz)§ E 3 8 .¢ 6}; mo (XI—t £061 <V>= i z 00 3mm [MRS 442x : imwflmWE): W, <H>= <77+4xj>= i7. + muoibz ”like, W W baa; Wit.» Jgfiw.<H>maa :05 a—W “’€ 7/921 (9%. g g 3%} ’- l . . 0 “mm W \mk’Ws m. \c pus/e. WW \n, 00 $7. 5 . 6% =3 g £2 a 7. - (W355 ‘ “:7. a? w 3:. {gaW‘bz} = % 41.....42 C CE+255LC%-i’b>2 : ’L \ 7. g 2 /C%+Z‘O) &% :: L (2rd»); ._ 4172 < :7; > = 11:... H b ab \l/PKW dam‘tWS “aorta 00 VFW “W Lg 3r: 1 / ,5 \ z 2: , “ ~ a, {351. v: - 31/35 7 ‘ \ ‘24: ‘fi 5: 2 “fl“ ”fW x “as; m 3“ 1:5 ”d 9 ?r0\o\€/m 2, H“? = [ J7} 2. Z a .»_v‘-§_.L]/q3+—L 1-311 Z 3 0 v2. ‘______ ~—- .6...— ” .- W W 5 3M») ; El ltPtua(r‘D/LP!°D ((3‘) E“! “P300031 )quJFZ) ng :- (‘Qe\‘\. “3&3 [$0 32%, <H>= sic-3t + (we) {13% ANCEMQQKVWO «:mxa khm3Mé W A?“ m M flaw Much!) 0km gm.» Mumiflww WE: .9. W M W FL MW; ‘ <Vee>= L J... M89, Zr‘/&<‘ __ (\+_CL>€-Z¢l/a’> r‘[email protected]\/fic\,®.é<€; a, Q1565 6W3; r1 : $3. 3.3: arm + Jon/0.] ,1 (if 41567, W33 [1"( i (4)3 ' “ 9’1 8’; 5‘ ' 5:) ‘5‘” — * JAE i 5 + ~ ~—-~ ' * W‘ =~——— : )‘7r\/ Lhréc of 3/2<3 9% %& L‘Xd [46‘ e, : " 3&5; $1 ”WW “WNW... M‘w..-‘Mwmm.. Vw‘wfl zufi 0”" ‘ (0")2 ~+ $1 = O (5“) 3‘ Si #2 "a E" MWS :fiw, 0W (533-pr I‘M QW <0_/>Z. = P1. » (g-S) x W” ,8 W \S‘3 W XZWWWQ ‘3me ifimhaifim’mW O 7') “in” , Oz’a'.’ ‘ “W 00) W Wmmm (3:3) ~. 073/0”! 3' §(U.‘)z+ i0“.”33€9 (Ase; F: 12%! P01, ”2: HEY“ §<'§§L;<‘F§'>:O =7 pm: 2 P 1 3+w%—:::@%O P 33- =::$‘;:‘3x33§<; 3:;:>3 3:33 (:313339 :3 g g; 6 )5 g a morn/mi? M‘HM é wmsm‘!‘ Kr- - ‘l ‘ _ s [10.5% mo 2“ kobxolsflfl M4501 (9.615% 10‘3“ 36?] [\jmgl-le? \948 (c.2’4’ZKtO‘? 2;! 5m wmvw‘i‘ VHH’S = - gotwwom' W '3 l3 p 'x) 52 (c vm) am[g_m3o¢3 = mézggm MW 9C1," E/m%- (m Hi4}! X7, «2A W Elm/W977 PLWJ“ FCXYD:O> @W‘m W. W”‘ NEW-m- WWW 649a 32(70 W (14.0% M :2); 6‘ 3 4.?52)‘ :O-iZ av 53 : \‘\30x 10"” CV 3 El: (5340 X 10-115” Eq = MSW)»: 20“”aV; {ELM WWW/MW WWW W Wkfi W gy— ,{jig Mwefi' WywaWzM AE‘ ::. £6}!th .. Elwyfivfl :1 0,036 ”lov‘z av :15? - = 6-02ch A CZ El _E:"Mfi : 0.0\Q_ X ‘O-‘IZ 6V 7% - 3 2. cf‘ W303 __ A55 éém E3 ‘ O-OO} wo‘neu :) vane: aeq =g;““““‘—€n,t =30 . ‘ . :7 {Equal L [0'4 X [0‘12 ev W w 44‘ ) MM MMMAI ,(r—e; m 46 MW Wiéfi WWMW JA W. #W‘A W 2C3» W m «Wgain-nymmyxazwmugzm zmwymvwxmgmn.“ , W. acan-Mn:mm;mac/{Myarm:,.‘.A.»:.«.-r.»,»x.l.rmmrm.»“my“,rr.,.,«..,...«mm.,,/«..;>.=,.r,;~;.l.“my, 4 AN4...,I“.,may”.M»,«m(Mu,..,WA..W.»..>..».,..u,.mwp A “WM...“ ...
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