hw6_solution - Homework #6 Solutions 1.(a) Define...

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Unformatted text preview: Homework #6 Solutions 1.(a) Define Unnonnalized Wave Function m[5];= W[x_, y__, z__] := N *Exp[-a * (x"2 + y"2 + z"2)] 1: 2 a: (x +y) Integrate[llr[x, y, 2] *IHx, y, 2], {x, ~Infinity, Infinity}, (y, —Infinity, Infinity}, {2, —Infinity, Infinity}, Asshmptions -» {a > 0}] N2 7T3/2 Ouif6]: 5—— ‘ 16 V2 a"2 N2 "3/2 We]: Solve == 1, N] 16 ’\f2_a7/2 4 X 21/4 a7/4 4 X 21/4 017/4 {{N» -—~——-————}, {Na ~—-—-——}} 7,3/4 “3/4 Define Normalized Wave Function and Check Normalization 1n[14];= ¢2[x_, y_, z_] := (4 x2A (1 / 4) (IA (7 / 4)) /7r" (3 / 4) *Exp[—a* (x"2 +y"2 + 2"2)] * 2* (x+y) Integrate[qll2[x, y, 2] *llr2 [x, y, z], (x, -Infinity, Infinity}, {Yr —Infinity, Infinity}, {2, -Infinity, Infinity}, Assumptions -) {a > 0}] Out[15]= l (b) Expectation Values of L and L"2 ' First Let's calculate Lx = y*p_z - z*p_y. Let's begin by calculating the Derivatives |n[17]:= DWIXI y, 2], Y] 0mm]: e'(x2*y2*zz) ‘1 N'z — 2 e‘(x2+y2+zz) “‘N y (x + y) z 01 In{18]:= DUNK, y, z] I 21 Out[18}= e'(x2+yz+zz) 0‘ N (x + y) — 2 e‘("2*yz+zz) 0‘ N (x + y) z2 a Now Let’s do the Integrals V Integratehflx, y, 2] *y* (-I) *hbar .(E‘ (- (x"2 +y"2 + 2‘2) * a) *N_ * (x+y) -2E" (- (x"2+y"2 + 2‘2) * a) *N* (x+y) 2‘2 * a), (x, —Infinity, Infinity}, (y, -Infinity, Infinity}, {2, -Infinity, Infinity}, Assumptions -y {a > 0}] Integrate [W[x, y, 2] w 2 * I *hbar (EA (- (x"2+y"2+2"2) a) *N* 2—2 *E" (- (x"2+y"2+z"2) a) * N *y-k (x+y) * 2* a), {x, —Infinity, Infinity}, (y, -Infinity', Infinity}, {2, -Infinity, Infinity}, Assumptions -> {a > 0}] Ou1[25]= 0 Oul[26]= 0 So the expectation value of L_x is 0. Now let's do Ly=z*p__x-x*p_z 2 I hw6_1.nb In[21]:= DUNK, Y! z]! x] DUI/[XI Y: z]: z] ‘ Quip”: e—(x2+y2+zz) a N Z _ 2 e—(x2+y2+zz) a N x (x + y) z a (Burma: e—(xz+y2+zz) a N (X + y) _ 2 e—(x2+y2+zz) a N (x + y) 22 a v; In[32]:= Integratehflx, y, z] * z * (—I) *hbar (N*z*Exp[-a* (x"2+y"2+z"2)] -2*N*a*x*z* (x+y) *Exp[-a* (x"2+y"2+z"2)]), (x, -In:Einity, Infinity}, {y, -Infinity, Infinity}, (z, —Infinity, Infinity}, Assumptions -> {a > 0}] Integrate[tlr[x, y, z] *x-A- (—I) *hbar \ (Exp[—a* (x"2+y"2+z"2) ] * N-k (x+y) —21:‘.xp[— (x"2+y"2+z"2) a] *N * (x+y) * z"2 *a), {x, -Infinity, Infinity}, (y, -Infinity, Infinity}, {2, —Infinity, Infinity}, Assumptions —) {a > 0}] OM32]: O Out[33]= 0 So the expectation value of L_y is 0' Now for L_z=x*p_y-y*p_x ln[34]:= D[¢[xl Y! z]! Dlll’lx, Y: z]: 3‘] Out[34]= e'("2*yz"22) ‘1 N z — 2 e'lxz”y2”2) °‘ N y (x + y) z (x was]: e’(“2*yz*zzl,°‘ N z — 2 e'(XZ+Y2+22) 0‘ N x (x + y) z a |n[39];= Integrate[—Ilr[x, y, z] *x*I*hbar*D[nk[x, y, z], y], {x, ~Infinity, Infinity}, {y, -Infinity, Infinity}, (z, -Infinity, Infinity}, Assumptions —» {a > 0}] Integrate[-tll[x, y,»z] *y*I *hbar*D[dt[x, y, z], x], (x, —Infinity, Infinity}, (y, -Infinity, Infinity}, {2, ~Infinity, Infinity}, Assumptions -> {a > 0}] 0mm]: 0 Out[40]= 0: SO the Expectation Value of L is (0,0,0) Now for L"2 - First Convert the Wave Function to Spherical Coordinates In(45];= PS[r_, e_, ¢_] := N*Exp[-a*r"2] r*Cos[6] * (r*Sin[e] Cos [4)] + r*Sin[6] Sin[¢]) Integrate[PS[r, t, f] *PS[r, t, f] *r"2*Sin[t], (r, 0, Infinity},. {t, 0, Pi}, (f, O, 2*Pi}, Assumptions 4 {a > 0}] N2 "3/2 16 «5127/2 I have the same normalization Factor, so I have a good function that I am integrating correctly. Let's Compute The derivatives. Out[47]= l)1[48]1= D[D[Ps[rl tr f], t]! t] bums]: —2ce‘r2°‘Nr (rCos[f] Cos[t] +rCos[t] Sin[f]) Sin[t] + e-rzaurcOsm (—rCos[f] Sin[t] —rSin[f] Sin[t]) — e"2°‘NrCos[t] (rCos[f] Sin[t] +rSin[f] Sin[t]) In[49};= Cot[t] *D[PS[r, t, :1, t] 0mm}: Cot[t] (e'rzaNrCos[t] (rCos[f] Cos[t] + rCos[t] Sin[f]) — e‘”2°‘NrSin[t] (rCos[f] Sin[t] +rSin[f] Sin[t])) hw6_1.nb |3 mm];= 1/ (Sin[t])“2*D[D[PS[r, t, f], f], f] ou‘qso]: e'rz‘erCotH] Qsc[t] (—rCos[f] Sin[t] «rSin[f] Sin[t]) 'x' Now We Can Integrate ln[51]:= Integrate[—hbar"2*PS[r, t, f] * g (D[D[PS[r, t, f], t], t] +Cot[t] *D[PS[r, t, f], t] +1/ (Sin[t])x2*D[D[PS[r, t, f], f], f]) * r"2*Sin[t], (r, O, Infinity}, (t, 0, Pi}, (f, 0, 2*Pi}, Assumptions —) (a > 0}] 8 .\/2_a7/2 ‘ Replacing with Correct N value 1n[52];= (3hbar"2* ((4x2"(1/4) a" (7-/ 4)) /7r" (3/4))‘2i- 7r"(3/2)) / (85qrt[2] a" (7/2)) Out[52}= 6 hbar2 (c) Variances in these Quantities Variance of L is given by AL=(<L"2>-<L>"2)’\( 1/2) So AL = Sqrt(6)*hbar Now Calculating the Variance in L"2 Means we need the Expectation Value of L".4 4 I hw6_1.nb |n[56}:= D[(D[D[PS[r, t, f], t], t] +Cot[t] *D[PS[r-, t, f], t] +1/’§(Sin[t])"2*D[D[PS[r, t, f], If], f]), (t, 2}] +Cot[t] *D[(D[D[PS[r, t, f], t], t] + Cot[t] *qusu, t, f], t] +1/ (Sin[t])‘2*D[D[PS[r, t, f], f], f]), t] + 1/ (Sin[t])‘2*D[(D[D[PS[r, t, f], t], t] +Cot[t] *DEPS[r, t, f], t] + 1/ (Sin[t])‘2*D[D[PS[r, t, f]: 5], f1), {fl 2}]? Out[56]= —2 e'rzaN (r (—rCos[f] Cos[t] —rCos[t] Sin[f]) Sin[t] — r (r Cos[f] Cos[t] +rCos[t] Sin[f]) Sin[t] +2 rCos[t] (—rCos[~f] Sin[t] —rSin[f] Sin[t])) + e'rzaNr (—2 (—rCos[f] Cos[t] — rCos[t] Sin[f]) Sin[t] — Cos[t] (—rCos[f] Sin[t] —rsin[f] Sin[t]) +Cos[t] (rCos[f] Sin[t] +rSin[f] Sin[t])) — e'r2“N (—2 r (rCos[f] Cos[t] +rCos[t] Sin[f]) Sin[t] + rCos[t] (—rCos[f] Sin[t] —rSin[f] Sin[t]) —rCos[t] (rCos[f] Sin[t] +rSin[f] Sin[t])) — 2Csc[t]2 (—2«e'r2°‘Nr-(rCos[f] Cos[t] +rCos[t] Sin[f]) Sin[t] +e'r2aNrCos[t] (—rCos[f] Sin[t] —rSin[f] Sin[t]) —e‘r2°‘NrCos[t] (rCos[f] Sin[t] +rSin[f] Sin[t])) + e“2°‘Nr (2 (—Cot[t]2Csc[t] ;Csc[t]3) (—rCos[f] Cos[t] —rCos[t] Sin[f]) + (4 Cot[t] Csc[t]3+Cot[t] (Cot[t12Csc[t] +Csc[t]3)) (—rCos[f] Sin[t] —rSin[f] Sin[t]) + Cot[t] Csc[t] (rCos[f] Sin[t] +rSin[f] Sin[t])) + 2Cot[t] Csc[t]2 (e'rz‘erCos[t] (rCos[f] Cos[t] +rCos[t] Sin[f]) —' e"2“NrSin[t] (r Cos[f] Sin[t] +rSin[f] Sin[t]))'+ Csc[t]2 {—Ze‘r2“Nr (—rCos[f] Cos[t] —rCos[t] Sin[f]) Sin[t] —e‘r2°‘NrCos[t] (—rCos[f] Sin[t] —rSin[f] Sin[t]) +e‘rzaNrCos[t] (rCos[f] Sin[t] +rSin[f] Sin[t]) + e'IZO‘NrCotPc] Csc[t] (rCos[f] Sin[t] +rSin[f] Sin[t]) + Cot[t] (e‘rzaNrCos[t] (—rCos[f] Cos[t] —rCos[t] Sin[f]) — )+ Cot[t] (6“2“Nr (Cos[t] (~rCos[f] Cos[t] —r-Cos[t] Sin[ )- Cos[t] (rCos[f] Cos[t] +rCos[t] Sin[f]) —28in[t] (—rCos[f] Sin[t] —rSin[f] Sin[t])) — ' e‘rz'J‘N (2 rCos[t] (r Cos[f] Cos[t] +rCos[t] Sin[f]) +rSin[t] (-rCos[f] Sin[t] ~rsin[f] Sin[t]) —rSin[t] (rCos[f] Sin[t] +rSin[f] Sin[t]))) + e‘rzaNrSinfi] (—rCos[f] Sin[t] —rSin[f] Sin[t])) f] Cot[t] (e‘rzaNrCos[t] (—rCos[f] Cos[t] —rCos[t] Sin[f]) +e‘r2“NrCot[t] Csc[t] (—rCos[f] Cos[t] —rCos[t] Sin[f]) —3e‘rZ“NrCos[t] (rCos[f] Cos[t] +rCos[t] Sin[f]) — e‘rzaNrCot[t]2Csc[t] (—rCos[f] Sin[t] ~rSin[f] Sin[t]) — «3"2"‘NrCsc[t]3 (—rCos[f] Sin[t] —rSin[f] Sin[t]) —3e'rz°‘NrSin[t] (—rCos[f] Sin[t] ~rSin[f] Sin[t]) +e'rzaNrSin[t] (rCos[f] Sin[t] +rSin[f] Sin[t]) + Cot[t] (—2 e'r2“Nr (r Cos[f] Cos[t] +rCos[t] Sin[f]) Sin[t] +e'r2aNrCos[t] (—rCos[f] Sin[t] —rSin[f] Sin[t]) —e‘r2°‘NrCos[t] (rCos[f] Sin[t] +rSin[f] Sin[t])) — Csc[t]2 (e'IZQNrCos[t] (r Cos[f] Cos[t] +rCos[t] Sin[f]) — e"2°‘NrSin[t] (rCos[f] Sin[t] +rSin[f] Sin[t]))) [M62]; Integrate[ i hbar‘4*PS[r, t, f] * (D[(D[D[PS[r, t, f], t], t] +Cot[t] *D[PS[r, t, f], t] +1/ (Sin[t]) A2* DEDEPSIr, t, f], £1, £1), (t, 2}] +C°t[t] *D[(D[D[PS[r, t, f], t], t] + Cot[t] *D[PS[r, t, £1, t] +1/ (Sin[t])*2*D[D[PS[r, t, f], £1, £1), 1:] + 1/ (Sin[t])‘2*D[(D[D[PS[r, t, f], t], t] +Cot[t] *D[PS[r, t, f], t] + 1/ (Sin[t])‘2*D[D[PS[r, t, f], £1, £1), {f, 2}]) *r‘2*Sin[t], (r, 0, Infinity}, (t, 0, Pi}, (f, 0, 2*Pi), Assumptions —> {a >'0}] 9 hbar4 N2 n3” 443—017” Inserting the correct value for N Out[62]= ln[63]:= (9hbar‘4 ((4x2"(1/4) a"(7/4)) /7r" (3/4))"27r" (3/2)) / (4Sqrt[2] a"(7/2)) owes]: 36 hbar4 In[64];=~ variance = Sqrt [36 * hbar‘4 - (6 * hbar"2) A2] Out[64]= 0 hw6_1.nb |5 we;ng ‘ a B’Mf/Ie, OOMVPOWQ‘N‘K'; ‘7'; _ ‘ " M i 433 g; from +(f5e/ rash/k”? wag LW Hm} XX [‘7 Z CaWwO‘L Sde-«e, a 00MM6- §e+ €\‘38§/\7C‘Ww<.‘1"\ov\§ 733 31 Cc) M W‘GV‘ {axse‘wfimmgw for ML We. awe-- a; 92+ L f/ ~— + mm (mm .WV’C , /J ‘f , M» w' "mam; e:gwmxwggwg *3“ 5% b1 :71" ~ -- .- I Dwr<§© 37 72/. ‘/ .man 1‘— 9 - a? :2 9w “av ‘fi: ??‘VSF * Q ' AZ 5-”? 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This note was uploaded on 01/18/2012 for the course CHEM 550/475 taught by Professor Davidj.masiello during the Fall '11 term at University of Washington.

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hw6_solution - Homework #6 Solutions 1.(a) Define...

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