Exam_2_Crib

Exam_2_Crib - CHEM 370 EXAM 2 Thursday March 6 2008 in Nome...

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Unformatted text preview: CHEM 370 EXAM 2 Thursday, March 6, 2008 in . - Nome This exam consists of four (4) questions on fourteen (.14) pages. A Formula Sheet and the Figure needed to do Problem 1 are included at the end of the exam. You must Show all work and reasoning to obtain any credit. Use answer boxes where provided. Finally there are three extra credit questions in Problem 4 worth 15 points. 1. (15 points) Using the figure at the back of the exam, explain why the actual behavior of the ground state (V = 0) of the one-dimension harmonic oscillator (1D H0) is radically different from classical behavior. You should especially describe phenomena like'zero point motion and its relation to the Heisenberg Uncertainty Principle, classically forbidden behavior, and so 011. 0.; C\ was \i C 03 ABC Zero {Ashe 01‘”) in grams 2. (25 points) This problem is concerned with the probability distributions for energy P(E, t) given in Figs. 221-20 for the 1D HO. 3. (8 pts) First assume that the oscillator is in a nonstationary state with wavefunetion Il’(x,t). The probability distribution P(Ev,t) for its energy E at time t = t‘ is plotted on Fig. 2a ' P(E‘..t‘) H2 3/2 5/2 7/2 99/2 E m Fig. 2a Using the plot determine (Ext where (E)! E:EVP(EV,t). (2.1) ,.. 3.. . . - 75" it . “l «(t-:29. : (zfior‘tt “35% 9‘! Jr EXG‘L T EMJ) l1” :; MW b. (10 pts) Next assume that oscillator is in the stationary state LP(x,t) = exp[—%Ev:nt)wv_0 (x) (2.2) where wwo (x) is the ground state V = 0 eigenfunction of the 1D HO Hamiltonian (see formula sheet). (i) ( pts) Show that at t = 0 , the probability distribution for energy P (E, t) has the form plotted in Fig. 2b. P(E,i=0) L0 E Pm) 1/2 3/2 5/2 7/2 99f2 Fig. 2b mm. au'_ A i ) I k {e < r [ LH’N) 43:0) H tho}, {7:- o) a“ ‘R km” in i K A I '_ ,.. :1 X {BXNGl (fats) H Wmo) (fiwm AVA” ‘3‘" if 91% Ec- o’VK ., (ii) .(7 pts) Show for all t 2 0 , the probability distribution foi‘ energy P (EJ) has the form plotted in Fig. 20. P(E.l) Ifl E hm 1/2 3/2 5/2 7!? 99/2 ebmtaw rig,— 3. (25 points) This problem involves applications of quantum mechanics to two simple systems. a. (10 pts) Prove the following two results for all 1D H0 stationary state wavefunctions l|JV (y), v = 0,1,2,. ... (Hint: The results follow from just the even or odd natures of the integrands.) (i) (5 ptS) x 0 pm. w: (M: r _ u_ ‘ 21K- ! 6/2,: %{r)v~jl{;,éy)ot\j z. Wm / Pi -—“i’3 P (We (7‘)]; 115 was gamma , is: {-3 0M WWW . >6 ‘_ '- _ v . <>x>v : o b. (15 pts) Using spherical polar coordinates, Show that the 23 and 2p wavefunctious of the H atom are orthogonal. That is, Show jw:s(r)w2p(r)dr=0. (3.1) The information you need, including the forms of the wavefunctions, are on the formula sheet. . x - ' ’t I a» ' r V a H35“ VG“) A Mr , 7 fig: w V; { fiat-acmgta :2. L 5mg (lame) a; m (5} a Z 0 lth WW : 4. (35 points) Consider the nonstationary state of the 11) H0 LP(y,t) =—j—§[w0(y)exp[—%Eot]+wl (y)exp[—%Elt]] (4.1) I where 0 denotes the v = 0 ground state and where 1 denotes the v :1 first excited state. a. (17 pts) The Born rule for the position probability distribution P for a particle is P = shy . (i) (10 pts) Show when the Born rule is applied to ‘P(X,t) of Eq. (4.1) it gives a: 1 ‘1 7 P(yat)=‘P (ystfl’bnt)=§[wa(y)+wr(y)+2wo(y)wt(floowtl- (4-2) (Hint: You might be helped by going through steps similar to those leading to Eqs. [4.6] and [4,7].) \ Hl‘f—Iraao o Eli-1in ._ #1; :EEmoyl 1 Lif- (3\Jr “8:83 4 QOBW‘LMEKYQ(& ‘ Bi 0 wit-aw ' ‘ ~ mom EV;3#E\/c-O;.%:W.. Uswédhxfadwb . .k. QKYC‘ED‘V “R933: We 36’- efiw Do I . 2 Sim adorn/e (ii) (7 pts) Notice that P(y,t) is time-dependenr reflecting the nonstationary nature of the state in question. However at all times t, normalization requires the integral of P(y,t) be time—independent and equal to unity. That is, we must have [:P(y,t)dy:1. Prove Eq. (4.3) from Eq. (4.2). (4.3) farms; ‘ 5).. Giffméy’if‘fibfif #0,; lb 2‘ rob ll W Cg, \UIEU (3):} 3 d O b. (18 pts) For the wavefunction of Eq. (4.1), the average momentum at times t is - , d‘I’ , (15):: [:11 (y’t Ody. In part b, you will prove that titgiflpmgw (1W0 (Y) dy dW: (Y) dy dy]. ' (i) (5 pts) To start, using Eq. 4.1) and the 1D HO energy level formula Ev = (V+1/2)h(n, show Vfigtflmyi) zihdflexppmjwi(y)exp[+3imflx 1 dy 2 2 2 dy + exp(-imt) Eu! 0 iidWS§Y)eXp(-“°%l+dwaiy)e><p[‘3é‘“‘]i' _ Pwa 12V; 1 Angie EVA) , 2 \Y Gum 3C \5 ex 2— 2/» KW”: QELWJQ Xvi”; 4~ A 1? 10 (4.4) (4.5) (iii) (8 pts) Using arguments from Eq. (4-7). -—.-r'- y exp(icot) + \po (y) 11 dull 6X 1 based on evenness and oddness only derive _ 109v + “8° K a (3' New; 33" A Y W a? 123: was 393* X Ar \ngtbeflifii/Z ea“ °é¢io e 0* MG? 0 gfiischB + KW '23“? 3 ".00 AA. “23 as 3%(9 c. (15 pts) Part c is extra credit. Attenlpt it only if you have time. (i) (5 pts) Use the fact that (py >t is a real number to Show from Eq. (4.5) that r £W:(Y)dwo(y)dy=—[:\Vo(Y)dw'(y)dy. dy dy (Eq. [4.8] also follows from an integration by parts.) ' “ 7.7 PM" _- . 2:. 2i} [(eswéfigmmltm .4};de Hmm owl/@330 yr, . ) _ ‘ . l”); y A "1:. lacojmelaljol~4£yclf+w$w“ML v'l/ol'l Real, m a ‘ .Lnl (Lfifl’LW 7} aft;— S/lm 5’3 fliE In’ttlfi’flm“j “yaw-b '65 O 29' , ~ ‘” My) .\ z \ {Chm %g)fllj +i‘t0) 5U a Jana ‘_ K \\ icky) div-)oly : ~— [7%(7’) T), 0” 12 (4.8) “F0544: (ii) (3 pts) Using Eq. (4.8), show that Eq. (4.5) simplifies to (p3,)t = hsinmt £11}, (y) dwo (y) dy. firm C (3;) We get-’- A h 3.— ~-— - i"(‘3 ‘ ’ <lly>fi b T: Iju’ngi/‘Uté: \“""""“'1‘- - wags-{,fchw ') at)" 13 (4-9) 14 (iii) (7 pts) Using the facts that dip0 (y)/dy is proportional to w,(y) and that w,(y) is normalized, evaluate the integral in Eq. (4.9) to show (use forms for mu (y) and w] (y) on the formula sheet) (p > ={mgafifsinmt (410) A “‘2 1 W0 ‘2‘ E “ 2,210 6K flittlbg (An 3 l 0‘3 l #939 ZmVOUl: +3 W A C An v0 QZPOVQ L “m f C, : «If/E: ’lfi \ Formulas 1. Heisenberg Uncertainty Principle [Ax]t[npx]1 2; 2. One Dimensional Harmonic Oscillator (1D HO) a. Energy Levels Ev = [v +%)hm,v = 0,1,2,... b. Hamiltonian 0. Lowest energy eigenfunctions of 1:1 Wv=o (Y) = [Egyd Cxp[—1/2[%9]y2] = er en“ yexptwemn 1/4 1/2 mu) 1 mo) 2 me) = — — 2 m ~1 e #12 ————~ WHO) [Tm] [2] “E ]y ]XP[ /[h d. Orthonorrnality ' Kw (Y)wv(y)dy 35...] 6. Even and odd behavior _of wv (y) \M-y) =(—1)"wv(y) 3. Even fe (—X) = fe (x) and odd f0 (—x) : —f0(x) functions a. fe (x)fe(x) : even function f0 (x)f0 (x) 2 even function f0 (x)fc (x) 2 odd function dfe (X) dX df0 b. 3 odd function : even function dx Efo(x)dx :0. it] 4. Classical Observables and Quantum Operators CM QM x fizx- ~ __1 px px—idx 2 2 3 p n F1 (51 H: " U H:—— U . 2m+ (X) 2mdxz+ (X) '5. Calculation of Average Values (A)t = Iw*(r,t)rz\ly(r,t)dr and similarly for (142)l . Also [AA]! : [(Az)‘ —(A)f]v2. 6. Spherical Polar Coordinates a. x = rsinScosrl) y=rsinesin¢ z=rcos9 b. If f(x,y,z) = g(r,6,¢), then defldxfldz f(x,y,z)= Erldrfsinedfifindd) g(r,e,¢). 7. Wavefunctions of the 2s and 2pz states of the hydrogen atom. )— 1 [II/20— ) —/2a) 11125“ —4\/fi ; r/a0 exp( r 0 and 3/2 map (1') = 4J2? [ijexpfi-r/Zaflcose _ where a0 = 5.29x10’”m is the Bohr radius. 8. Complex Exponentials a. exp(iimt) = cos mt i i sin mt b. coscot = é-[exp (imt) + exp(—imt)] . l . . 0. sm oat : 2—i[exp (1mt)- exp(—rmt):| Arm Ear? 30518303 varoiow om .Q <no 73305.8 89:29 ...
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Exam_2_Crib - CHEM 370 EXAM 2 Thursday March 6 2008 in Nome...

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