Question

Exercise 2:Commutators Given (AB, C) - ABC - ACB + ACB-CAB - ABC] + [A, CJB. 1- Show that the commutator[L.L]is equal to zero. 2-Computethecommutator(Ly, Lx]. 3-Compute the commutator 3.Lx]. 4-Compute the commutator[LLY.Lx]. 5- Add the results from (2)-(4) to compute [LLyLx]. Lxercise 3: Matrix elements The angular momentum operators acting on the angular momentum eigenstates. Il determined by L-11.m) = (1 + 1) - m(m + 1)2.m +1) L_11.m) = /(1+1) - mm - 1)/1.m-1) L211,m) = hm|1,m) 1- For 1=1 compute the three matrices, (1, m'|Lall, m)for a = x,y,z. These are 3 with top left position being m' = m = 1 and the bottom right position being m' = m / (1,1121,1) (1,1|L0|1,0) (1,1|L0|1,-1) La = (1,0|40|1.1) (1,040|1,0) (1.0 La 11.-1) \1.-114|1,1) (1.-1|2a|1.0) (1.-14.11.-1)/ 2- Check that these matrices obey the angular momentum commutation relations: [Lx Ly] =inly [L. 42] = ihLyand (LxLx) = ihly, 3-Consider a quantum particle defined by the modulus of angular momentum L2 = 21 Determine the possible values of the orbital quantum number / relative to t momentum. Deduce the basis Ilm), Where m is the quantum magnetic number 4- Find the matrices of Ly, Ly and Lg. 5- Find the eigenvalues of Ly. 6- Find the normalized eigenvector (eigenfunction) corresponding to the maximum ei Ly. We denote this vectorlu). 7- Assume the system is initially in the state ) = lu). 7-1- If we measure Ly. What are the possibilities and probabilities of a measurement 7-2- What is the expectation value of Ly

Answer 1

Given that CAB, CJ = ACB, CJ + [AC]B we have to show that [ 1², L₂J :o (L2, L₂] = [lukt lý + 22², L2] & Chh, L2] + [lŷ , [2] + [ LZ , L2] ده لعابداع + [علیجاگیا۔ العليوا خوا [Ly, L2] Ly & L2 [ 12 L 2] + CL 2 , L2] 22 As Ln, LZ) = -ctly [Ly, L2J ihla CL2,12) =0 [2², L₂] = La citly) + (1²hLy] La it Ly Lietln) + Litlu) Ly + L2 (0) + (0) mit Ly Ly - it Lyle tih Lyle tehny So [ L², L₂J 20 we have to compute the commutator [urly, Ln] [LuLy , Lz] - [h[Ly, Lu] & [h, Lally- we have med CAB, C = A[B, CJ +(A, 26

Page No. Le Cly, 2nd + ln [ln, Lallyt de [ n, l. n lnly Since [ly, Lu] = alth Lz (2n, ru ] zo [ & Ly, Lx] = 2 m (- 8h22) + ln (0) Ly + = - it in iz Ir taily, bz) : -et Lube to) Zri Ly 3. We have to - Commutator compute the [ 13, 2x] [L}, Ln] = [ Ly Ly, and - lý (Ly, Lm] + ll}, La Ily. = Lylly, lntlly Ly, Ln] Ly Ly [Ly, Lon] + Ly [Ly, In] Ly t - [ly, hn] Ly Ly Sincez tý Since [ly , Lon] - it lz [Ly, Lx] = Lý (-11 h L2) + Ly ( 19 h Lz) Ly + (rit Lz) L - it by L2 + - it Ly Lzly - it L z LZ

[ly, un] = - ith Ly L2 - it Lylzly - 10th 22 lg We have to compute the Commututor (hely, Lu] Ly, Ln] = 12 [Ly , Ln] + [ 1² LeJ Ly - = LZ [Ly, Ly] + L2 [ L2, Ln] Ly + [ L2, Luis I AS (Ly ; Ln] = -8 h Lz L [22, LM] it Ly [ 12 Ly , Lond: L2 (-it Lz) + L2 Lit Ly Ly tih Ly 22 Ly - eh 17 Lz tit L2 Ly + 1h Ly La Ly 11 22 Ly, Lx] = -lit L² + 1h Lalytit Ly laze 5. we have to compute the commutator [2 Ly , La] - [han thy the Ly, in [hn Ly, Ln] + [Lý Ly Ln] + [L² ly, in)

= [ln Ly, Lx] + (LJ, Ln] + [h Ly, Lu] putting value of Clin Ly, Lu] from 12 FCW Lu] from (21 and [ 12 Ly , and from (4) [ 2² Ly, Load = -1h Ln Lz - ch LÝ L2 - et Ly Loty - vħ Laly - ħ L² t con Laky + 10h Lyraly = -1°h Ln L2 - let ý lz - ceh L2 = -18h (Lu Lz thy Lz + 22 22) = -10th (in th} +22) Lz - - lt L² Lz { L² Ly, Ln] = -1°t