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 z-axis. The paticle is found as close to the z-axis 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 z-axis. One might visualize the paticle in the xy-plane otating about the z-axis. Of couse, it can t be exactly in the xy-plane 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 anti-symmetic 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 non-zeo values of ǫ ijk ae those with j, k = 2, o j, k =, 2. In the second tenso, the only non-zeo 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 S5-4 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 S5-5 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 y-axis so the new x-axis is along the old z-axis then thee will be one unit of angula momentum along the new x-axis. 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 x-axis 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 S5-6 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 S5-7 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 S5-8 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 S5-9 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 non-singula 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 otho-nomality 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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Poblems on oce Exeted by a Magnetic ields fom Ch 6 TM Poblem 6.7 A cuent-caying wie is bent into a semicicula loop of adius that lies in the xy plane. Thee is a unifom magnetic field B Bk pependicula to
Unit Vectos What is pobabl the most common mistake involving unit vectos is simpl leaving thei hats off. While leaving the hat off a unit vecto is a nast communication eo in its own ight, it also leads
Psychology 282 Lectue #2 Outline Review of Peason coelation coefficient: z z ( n 1) Measue of linea elationship. Magnitude Stength Sign Diection Bounded by +1.0 and -1.0. Independent of scales of measuement.
Model Question Pape Mathematics Class XII Time Allowed : 3 hous Maks: 100 Ma: Geneal Instuctions (i) The question pape consists of thee pats A, B and C. Each question of each pat is compulsoy. (ii) Pat
Univesal Cycles 2011 Yu She Wial Gamma School fo Gils Depatment of Mathematical Sciences Univesity of Livepool Supeviso: Pofesso P. J. Giblin Contents 1 Intoduction 2 2 De Buijn sequences and Euleian Gaphs
Magnetic Field and Magnetic Foces Young and Feedman Chapte 27 Intoduction Reiew - electic fields 1) A chage (o collection of chages) poduces an electic field in the space aound it. 2) The electic field
Algeba and Tig. I 4.1 Angles and Radian Measues A Point A A B Line AB AB A point is a location o position that has no size o dimension. A line extends indefinitely in both diections and contains an infinite
Infinite-dimensional äcklund tansfomations between isotopic and anisotopic plasma equilibia. Infinite symmeties of anisotopic plasma equilibia. Alexei F. Cheviakov Queen s Univesity at Kingston, 00. Reseach
Solution Deivations fo Capa #8 1) A ass spectoete applies a voltage of 2.00 kv to acceleate a singly chaged ion (+e). A 0.400 T field then bends the ion into a cicula path of adius 0.305. What is the ass
Appendi A Review of Vectos This appendi is a summa of the mathematical aspects of vectos used in electicit and magnetism. Fo a moe detailed intoduction to vectos, see Chapte 1. A.1 DESCRIBING THE 3D WORLD:
Lectue 16: Colo and Intensity and he made him a coat of many colous. Genesis 37:3 1. Intoduction To display a pictue using Compute Gaphics, we need to compute the colo and intensity of the light at each
LECTURE 3 Maxwell Boltzmann, Femi, and Bose Statistics Suppose we have a gas of N identical point paticles in a box of volume V. When we say gas, we mean that the paticles ae not inteacting with one anothe.
Lesson 7 Gauss s Law and Electic Fields Lawence B. Rees 7. You may make a single copy of this document fo pesonal use without witten pemission. 7. Intoduction While it is impotant to gain a solid conceptual
M13914 Questions & Answes Chapte 10 Softwae Reliability Pediction, Allocation and Demonstation Testing 1. Homewok: How to deive the fomula of failue ate estimate. λ = χ α,+ t When the failue times follow
lectic Fields in Matte page. Capstones in Physics: lectomagnetism. LCTRIC FILDS IN MATTR.. Multipole xpansion A. Multipole expansion of potential B. Dipole moment C. lectic field of dipole D. Dipole in
Candidates should be able to : Descibe how a mass ceates a gavitational field in the space aound it. Define gavitational field stength as foce pe unit mass. Define and use the peiod of an object descibing
Chapte Two Some text, some maths and going loopy In this Chapte you ae going to: Lean how to do some moe with text. Get Python to do some maths fo you. Lean about how loops wok. Lean lots of useful opeatos.
Physics 202, Lectue 4 Today s Topics Review: Gauss s Law Electic Potential (Ch. 25-Pat I) Electic Potential Enegy and Electic Potential Electic Potential and Electic Field Next Tuesday: Electic Potential
The Binomial Distibution A. It would be vey tedious if, evey time we had a slightly diffeent poblem, we had to detemine the pobability distibutions fom scatch. Luckily, thee ae enough similaities between
Dynamical Systems: Pat 2 2 Bifucation Theoy In pactical applications that involve diffeential equations it vey often happens that the diffeential equation contains paametes and the value of these paametes
MULTIPLE SOLUTIONS OF THE PRESCRIBED MEAN CURVATURE EQUATION K.C. CHANG AND TAN ZHANG In memoy of Pofesso S.S. Chen Abstact. We combine heat flow method with Mose theoy, supe- and subsolution method with
CHAPTER DIPOLE AND QUADRUPOLE MOMENTS Intoduction + τ p + + E FIGURE III Conside a body which is on the whole electically neutal, but in which thee is a sepaation of chage such that thee is moe positive
36-632 Intenational Monetay Economics Note Let me biefly ecap on the dynamics of cuent accounts in small open economies. Conside the poblem of a epesentative consume in a county that is pefectly integated
Physics 107 HOMEWORK ASSIGNMENT #14 Cutnell & Johnson, 7 th edition Chapte 17: Poblem 44, 60 Chapte 18: Poblems 14, 18, 8 **44 A tube, open at only one end, is cut into two shote (nonequal) lengths. The
Gauss Law in dielectics We fist deive the diffeential fom of Gauss s law in the pesence of a dielectic. Recall, the diffeential fom of Gauss Law is This law is always tue. E In the pesence of dielectics,
Moment and couple In 3-D, because the detemination of the distance can be tedious, a vecto appoach becomes advantageous. o k j i M k j i M o ) ( ) ( ) ( + + M o M + + + + M M + O A Moment about an abita
Continuous Compounding and Annualization Philip A. Viton Januay 11, 2006 Contents 1 Intoduction 1 2 Continuous Compounding 2 3 Pesent Value with Continuous Compounding 4 4 Annualization 5 5 A Special Poblem
Poiseuille Flow Jean Louis Maie Poiseuille, a Fench physicist and physiologist, was inteested in human blood flow and aound 1840 he expeimentally deived a law fo flow though cylindical pipes. It s extemely
Oveview Fluid kinematics deals with the motion of fluids without consideing the foces and moments which ceate the motion. Items discussed in this Chapte. Mateial deivative and its elationship to Lagangian