{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PSet4Solutions

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

This preview shows pages 1–18. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...“ ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern