Question

Answer all of the following questions:

5. A molecule in a pa orbital is placed in region of uniform magnetic field Bo that points in the +z-axis. The t 0 state is given by b(0)) = |Px) = (1,-1) 1,1)), where the basis states are the orbital angular momentum eigenstates |l, m). The Hamiltonian in this situation is given by H = L B = woL2, where е Во 2m wo 2m (a) (10 points) Find |(t)) (b) (10 points) Find (0,øb(t)) = b(0,ø, t). Simplify the expression as much as possible (c) (5 points) Calculate (L2) and (L2) as a function of time. Explain how these calculations are related to (E) (d) (5 points) Explain how this result can be interpreted as the the pr orbital precessing around the (1, 1)1,-1)) z-axis. Note that the p, orbital state is given by py) =

Answer 1

121 lm7 = mhlom) ů = wolz eignvalue can Elem) = Ĥ lemy - wohnlem) Elom) = womt 1 lm) so Em = wohm) £²1 km) = Alanily! State Em 111) hwo 242 € 116) two at2 t @ 190) = [11-1)-1,3) - 0121 147 = 1/2 (11-1) E DE +/h - 1,17 € 14) = 111, -1) est _ 11,00 woto 6 41 = + 1 % 10,93 e ilgo - Y, etwol] U-1) 410,8, 4) = 4 x 50 m sinoeto e inol e toot7 + sino e' i woted] ---- 410, 0, 1) = 3 1410,8d) = 12 sino I i (Wood) sino con lwoord) 7, +1 = 1 0 m sine et 10 ② Probability of getting at? P= l<+601462) K-1 e il c) way (11) etwor_11, 128 kuota P [<1,-111,-) + <!!,>= [1 +13, (12y = Pxp = 02x1 = 3? Probability of getting hert in P OP- respectively. Pe = k1, 1/4(712

Pe=th K., 11 -17 Wol - 11,) etwol/2 = +1 <11/6 te wol - Lust & Woh/2 = + 1 - e-of/2 = t . 16/0/=) & (l'm') em>= des d'un P = K 1/4 (7) ² 2 when dort s it uw lo if non (22) = txt + (-2XP- - txt - t x 2 = og LED Probability of getting Ex = two P = K 1 11 + (37) 2 - 2 for E = -two P = K 1,-1/+(2)/² £ 2E) = Ex XP, + E_XP2 = ² ( two otwa) = og e E = (22)=0 as lod- &