Physics 505 Homework No. 5 Solutions S Angular momentum uncertainty relations. A system is in the lm eigenstate of L 2, L z.


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1 Physics 55 Homewok No. 5 s S5. Angula momentum uncetainty elations. A system is in the lm eigenstate of L 2, L z. a Show that the expectation values of L ± = L x ± il y, L x, and L y all vanish. ψ lm L ± ψ lm = h ll + mm ± ψ lm ψ l,m± =, since ψ lm ψ l,m± ae othogonal. Note in the special cases m = ±l fo L ±, the squae oot kills the expectation value! L x = L + + L /2 and L y = il + L /2, so these expectations must be zeo as well. b Detemine L i = L 2 i L i 2. Veify the genealized uncetainty elation holds fo all pais of angula momentum components. Comment on L x L y fo the cases m = and m = l. L z = m h, Lz 2 = m2 h 2, L z =. The state ψ lm is symmetic in x and y, so Lx 2 = L 2 y = L 2 L 2 z /2 = h 2 ll + m 2 /2. L x = L y = h ll + m 2 /2. L x L z = [L x, L z ] /2 = i hl y =. This is not vey exciting! Similaly, L y L z =, also not exciting. Howeve, L x L y = h 2 ll + m 2 /2 [L x, L y ] /2 = i hl z = h 2 m/2. In the case m =, thee is no angula momentum about the zaxis. The paticle is found as close to the zaxis as possible. In this case the minimum uncetainty poduct fo the x and y components is and this is well satisfied since all the angula momentum is in the x and y components but with vanishing expectation value. The only case whee the equality sign would be used in the uncetainty elation is the case l = in which case all components and thei squaes ae zeo. The case m = l coesponds to the maximum angula momentum component along the zaxis. One might visualize the paticle in the xyplane otating about the zaxis. Of couse, it can t be exactly in the xyplane and its out of plane motion poduces some components of L x and L y which aveage to, but have some spead aound the aveage. The uncetainty elation becomes L x L y = h 2 ll + l 2 /2 = h 2 l/2 = [L x, L y ] /2 = i hl z = h 2 l/2. In this case, the angula momentum components in the x and y diections have the minimum possible uncetainty poduct.
2 Physics 55 Homewok No. 5 s S Fun with angula momentum commutatos. a Suppose the vecto opeatos A and B commute with each othe and L. Show that [A L, B L] = i ha B L. We use the summation convention and the completely antisymmetic tenso and just gind it out! [A L, B L] = A i L i B j L j B j L j A i L i = A i B j L i L j L j L i = A i B j [L i, L j ] = A i B j i hǫ ijk L k = A i B j i hǫ ikj L k = +A i B j i hǫ kij L k = i ha B k L k = i ha B L. b Suppose V is a vecto opeato which might be a function of x and p, so it doesn t necessaily commute with L. Show that [L 2, V ] = 2i hv L i hv. You might need the elation deived in lectue fo a vecto opeato, [L i, V j ] = i hǫ ijk V k. Again, ginding it out woks: [L 2, V ] = L i jl j V i V i L j L j = L j V i L j + L j [L j, V i ] L j V i L j + [L j, V i ]L j = i h L j ǫ jik V k + ǫ jik V k L j = i h ǫ jik L j V k + ǫ jik V k L j = i h ǫ jik V k L j + ǫ jik [L j, V k ] + ǫ jik V k L j = i h 2ǫ jik V k L j + i hǫ jik ǫ jkl V l = i h 2ǫ ijk V k L j i hǫ ijk ǫ jkl V l = i h +2ǫ ikj V k L j + i hǫ ijk ǫ jlk V l = i h +2ǫ ikj V k L j i hǫ ijk ǫ ljk V l = i h +2ǫ ikj V k L j 2i hδ il V l = i h +2ǫ ikj V k L j 2i hv i = 2i h V L i hv i.
3 Physics 55 Homewok No. 5 s S5 Note that we used the fact that ǫ ijk ǫ ljk = 2δ il. This is easy to see. Pick an i, say. Then the only nonzeo values of ǫ ijk ae those with j, k = 2, o j, k =, 2. In the second tenso, the only nonzeo values will occu fo l =, the sign will be the same as the fist, and thee ae two contibutions.. Classically, a paticle moving in a spheically symmetic potential has the Hamiltonian H = p2 2m + L2 2m 2 + V, whee p = p/. Fo quantum mechanics, we must define with the Hemitian opeato p = 2 p + p p = h i +,. 2 Show the opeato defined in equation is the same as that in equation 2. Show that p in equation 2 is Hemitian conside ψ, θ, φ and ϕ, θ, φ and that when used in the Hamilton, p of equation 2 gives the coect Schoedinge equation. Also show that the opeato h/i/ is not Hemitian! = e. The momentum opeato in spheical coodinates is h i = h i e + e θ θ + e φ sin θ φ The fist tem in equation becomes h/2i/. The second tem is a bit moe wok. It s most easily evaluated in a mix of Catesian and spheical coodinates. Note that the momentum opeato in the second tem opeates on, /, and whateve might be to the ight. When it opeates on whateve is to the ight, we get a tem that s the same as the fist tem. So let s just evaluate the second tem when the p opeates on and /. 2 p = h 2i x x + y y + x z = h 2i x2 + y2 + z2 = h 2i x2 + y 2 + z 2 = h i,.
4 Physics 55 Homewok No. 5 s S54 and putting this togethe with the two othe pieces, we see that indeed, the opeato in equation is the same as that in equation 2. The adjoint of p is We conside p ϕ ψ = 2 d π sin θ dθ p = h i 2π + dφ h i. + ϕ, θ, φ ψ, θ, φ. The angula integations can be done independently of the adial integation, so we just assume they have been done o will be done and dop the angula vaiables fom the discussion. We ll use an integation by pats to move the deivative fom the ϕ to ψ. The deivative also acts on 2. p ϕ ψ = h i + ϕ ψ 2 d = h ϕ i ψ 2 d + ϕ ψ 2 d = h ϕ ψ 2 ϕ ψ i 2 d ϕ ψ 2 d + ϕ ψ 2 d = h + ϕ ψ i 2 d + ϕ 2 ψ 2 d ϕ ψ 2 d h = ϕ i + ψ 2 d = ϕ p ψ, and p is Hemitian. Note that the evaluation of the limits gives since the integand must convege at both ends! If we do the same calculation with p bogus = h/i/ we find p bogus ϕ ψ ϕ p bogus ψ = h i which is not zeo in geneal, so p bogus is not Hemitian. Finally, the Laplacian, in spheical coodinates, is 2 = sin θ ϕ 2 ψ θ sinθ θ + 2 sin 2 θ, 2 φ 2.
5 Physics 55 Homewok No. 5 s S55 If we multiply this by h 2, we get p 2 fo the Schoedinge equation expessed in spheical coodinates. We ve aleady seen in lectue that the angula pat is L 2 / 2. So it emains to show that the fist tem times h 2 is the same as p 2. The fist tem is h = h Squaing p, p 2 = h2 + + = h = h What if we like x instead of z? a Find the eigenfunction, ψ, of L 2 and L x with eigenvalues 2 h 2 and h, espectively. Since the eigenvalue of L 2 is 2 h 2, the eigenfunction has l =. The eigenfunctions of L z with l = ae Y,+ = sin θ e+iφ = Y, = + 4π cos θ = + 4π Y, = + sin θ e iφ = + x + iy z x iy. If we otate the coodinate system about the yaxis so the new xaxis is along the old zaxis then thee will be one unit of angula momentum along the new xaxis. This means eplacing z by x, y by y, and x by z. With these eplacements, the eigenfunction Y,+ will have one unit of angula momentum along the xaxis and of couse, it will still be an eigenfunction of L 2 with eigenvalue 2 h 2. The desied eigenfunction is then ψ = z + iy = cos θ + i sin θ sin φ.
6 Physics 55 Homewok No. 5 s S56 We can check this by opeating with L x in pola coodinates. L x ψ = h i sin φ cos φ cotθ θ φ cos θ + i sin θ sin φ = h sin φ sin θ i sin 2 φ cos θ i cos θ cos 2 φ i = h cos θ + i sin θ sin φ = hψ. b Expess the ψ just found as a linea combination of eigenfunctions of L 2 and L z. Of couse, if we want to wite ψθ, φ = +l l= m= l c lm Y lm θ, φ, we find the c lm by pojecting ψ onto the Y lm, c lm = π sin θ dθ 2π dφ Ylm θ, φ ψθ, φ, but in this case, it s easie just to look at ψ and the Y,+, Y,, and Y, and decide what we need: ψ = 2 Y,+ + 2 Y, + 2 Y,. The sum of the squaes of the coefficients is, as it should be. 5. Algeba all the way. We used algebaic techniques in lectue to deduce that L 2 and L z could be simultaneously diagonalized that is, eigenfunctions could be eigenfunctions of L 2 and L z at the same time and that the eigenvalues ae ll+ h 2 and m h, espectively, with l and m integes o half integes and with l and m = l, l +,...l, l. We then abandoned the algebaic technique and solved a diffeential equation to find the obital angula momentum eigenfunctions. Hee we will outline the use of algebaic techniques to deduce the obital angula momentum eigenfunctions. We stat by intoducing x ± = x±iy.
7 Physics 55 Homewok No. 5 s S57 a Show that the following commutation elations hold you may use elations aleady deived in lectue: [L z, x ± ] = ± hx ± [L ±, x ± ] = [L ±, x ] = ±2 hz [L 2, x + ] = 2 hx + L z + 2 h 2 x + 2 hzl +. [L z, x] = i hy and [L z, iy] = ii hx = hx, so [L z, x ± iy] = h±x + iy = ± hx ±. [L ±, x ± ] = [L x ± il y, x ± iy] = ±i[l x, y] ± i[l y, x] = ± hz + ± hz =. [L ±, x ] = [L x ± il y, x iy] = i[l x, y] ± i[l y, x] = +± hz + ± hz = ±2 hz. Fo the final commutato, we use the following, [AB, C] = ABC BAC = ACB + A[B, C] [C, A]B ACB = A[B, C] + [A, C]B. Also, fom lectue L 2 = L L + + hl z + L 2 z. So [L L +, x + ] = L [L +, x + ] + [L, x + ]L + = 2 hzl +, h[l z, x + ] = h 2 x +, [L 2 z, x +] = L z [L z, x + ] + [L z, x + ]L z = hl z x + + hx + L z = hx + L z + h 2 x + + hx + L z, and adding it all up, we have [L 2, x + ] = 2 hx + L z + 2 h 2 x + 2 hzl +. b Show that L z x + l, l = x + L z l, l + hx + l, l = hl + x + l, l, and L 2 x + l, l = h 2 ll + x + l, l + 2 h 2 l + x + l, l = h 2 l + l + 2x + l, l. This means that x + is the ladde o aising opeato fo states in which m = l.
8 Physics 55 Homewok No. 5 s S58 This is a piece of cake afte pat a. L z x + l, l = x + L z l, l + [L z, x + ] l, l = x + L z l, l + hx + l, l = x + l + l, l. Similaly, L 2 x + l, l = x + L 2 l, l + [L 2, x + ] l, l = x + L 2 l, l + 2 hx + L z + 2 h 2 x + 2 hzl + l, l = x + h 2 ll + + 2l l, l = h 2 l + l + 2x + l, l. So, any state l, m can be found by applying the opeato x + to the state, l times and then applying L l m times. l, m = CL l m xl +,, whee C is a nomalization constant. We can show that, is independent of angle. L, =, so otating the state with U δϕ intoduced in lectue just gives back,. This means,, must be a constant. c Detemine l, l equivalently, Y ll θ, φ up to a phase using the x ±. Hint: commutes with L, L 2, and x, so it is just a constant as fa as all these opeatos ae concened. l, l = Cx l + = Cx + iyl = C sin θ cos φ + i sin θ sin φ l = C l sin l θe ilφ, whee C is a nomalization constant to be detemined and we have eplaced the constant, by which will be omitted in the subsequent discussion. 2π l, l l, l = = C 2 2l dφ π + sin 2l θ sin θ dθ = 2π C 2 2l 2 l dx.
9 Physics 55 Homewok No. 5 s S59 One can look up the integal o integate by pats l times. + 2 l dx = 2 l x + = 2l + + 2l 2 l x 2 dx 2 + l x + 22 ll + 2 l 2 x 5 = 22 ll 5 = We deduce that, up to a phase, and + 2 ll l l l! 5 2l 2 l l! = 2 5 2l + 2 l l! = 2 2l +!!. + + x 2l dx C = l 2l +!! 4π 2 l l! 2l +!! l, l = sin l θ e ilφ. 4π 2 l l! + 2 l x 6 dx, 2 l 2 x 4 dx Note that this is almost Y ll θ, φ. It s missing l, but this phase is detemined by convention! Fom hee one could go on to use L to detemine up to a phase all the angula momentum eigenfunctions fo intege l. The nomalization constants wee given in lectue. Howeve, it s unlikely that this will lead to new insights, so this poblem ends hee!
10 Physics 55 Homewok No. 5 s S5 Appendix. Since we wee somewhat ushed with the coveage of P lm s and Y lm s, I include a few items hee. Much moe can be found in any efeence on mathematical functions such as Abamowitz and Stegun o any quantum text. Associated Legende equation. Afte sepaation of vaiables θ and φ so the solutions fo φ ae exp±imφ the equation fo θ becomes sin θ + ll + m2 sin θ θ θ sin 2 fθ =. θ This is the associated Legende equation and it s customay to change the vaiable to µ = cos θ. µ 2 d2 dµ 2 2µ d m2 + ll + dµ µ 2 P lm µ =, whee the nonsingula at µ = ± solution has been witten as P lm µ which is known as an associated Legende function. When m =, the equation is known as the Legende equation with egula solutions P l µ = P l µ which ae Legende polynomials. P l µ = l 2 l l! l d µ 2 l. dµ P l µ is an l th ode polynomial in µ and is eithe even o odd depending on whethe l is even o odd. The nomalization be caeful when consulting efeences, not eveyone uses the same is P l =. The associated Legende functions ae given by m, m l+m d P lm µ = µ 2 m/2 P l µ = l d µ 2 m/2 µ 2 l. dµ 2 l l! dµ Fo m = l, the associated Legende function is paticulaly simple. The fathest ight facto is a polynomial in l in which the highest powe is µ 2 l Since the polynomial is diffeentiated 2l times, only this tem suvives. The l cancels the l in font. The l deivatives poduce a facto 2l! which combined with the othe factos in font poduces 2l 2l 2l 5 which is often abbeviated 2l!! whee!! is ead double factoial. So P ll µ = 2l!! µ 2 l/2 = 2l!! sin l θ. The associated Legende polynomials with diffeent l, but the same m, ae othogonal on the inteval to +, + dµ P lm µp l mµ = 2 2l + l + m! l m! δ ll. The nomalized, othogonal, complete eigenfunctions of L 2 and L z fo obital angula momentum ae the spheical hamonics which ae defined in tems of the associated Legende functions fo the pola angle and the azimuthal wave fo the azimuthal angle. Fo m, these ae Y lm θ, φ = m 2l + 4π l m! l + m! P lmcos θe imφ,
11 Physics 55 Homewok No. 5 s S5 and fo negative m, we use The othonomality elation is Y l, m θ, φ = m Ylm θ, φ. π sin θ dθ 2π dφ Y lm θ, φy l m θ, φ = δ ll δ mm. The completeness elation is +l l= m= l Some of the low ode spheical hamonics ae, Y lm θ, φy lmθ, φ = sin θ δθ θ δφ φ. Y = + 4π Y = + 4π cos θ Y = sin θeiφ 5 Y 2 = + cos 2 θ 6π 5 Y 2 = sin θ cos θeiφ 5 Y 22 = + 2π sin2 θe 2iφ 7 Y = + 5 cos θ cosθ 6π 2 Y = 64π sinθ 5 cos 2 θ e iφ 5 Y 2 = + 2π sin2 θ cos θe 2iφ 5 Y = 64π sin θe iφ
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