Question

1. (20 points) Using the eigenstates of S, as the basis, (a) determine the eigenvalues and eigenstates of Sy; (b) determine the eigenvalues and eigenstates of S.ñ, where S is the spin-1/2 angular momentum, ñ is an unit vector. 2.(30 points) Consider a system with j = 1. (a) Explicitly write down <j = 1, m'J j = 1, m > in 3 x 3 matrix form. (b) Determine the eigenstate and eigenvectors of Jr. (c) Consider the eigenstate of J, with eigenvalue +1h, lj, m >= [1, +1 >= (1,0,0) ("T" is for transpose). If the state |1, +1 > is rotated about y-axis by angle #/2, what is the final state?

Answer 1

Ansla) eigenstates of Sq corresponds to i =% and m=1/2 1-1/2 So there are two possible eigenstates isimy. let us denote them by li>= 1/2, ½> . : )2) = 1/2,-1/2) I now Szli>= ½ ħ li) Sz 127 /2 to 127. We know that St = Sxt i Sy S.=. Sge-iso 4 = St-San 21.

nów St.[jm) = vci-m) (J+m+1) Tum+1) S-lj,m> = Vj+n) (j-m+) 15, m-> I The . . $+11) - Sulyo, Ye>=0 St12) = $1190-12>= 1 1212>=1"> S... 11): S11169) illy 1-122712 S... 12) S-112,-1,3-0 The states 1 and 127 are Orthonomai iei <ili> 1. <212> -) 49.2112> - <2102.0 : NO sul represents e, jith element in In the mateix sepseientation of sy. (Syber </S, S. (<1154/i> $-<1/5-13] (0 - <1123) = 0 Sy he > <11512= [<[S+12) - <1]2>] T - 4 fairyol = 2 - 2 1

(54.), y = <2]54113 -1. [<2};)> - <2}s.)] fp = <21237- Lil : : 27 (54).. = <2]sy 12574521sxl26215:12) - [<2113 - 01:0 1. 1. Sy i 10 minuti Sy = 2 lio found eigenvalues of Sy can by als below : Thy - /I] = 0 ; -2 / 1/2 - WIN # 1 - / - o t . 1 2 - IT eigen vector corresponding to eigenvahire = /. is found as shown below 2 e o 22 1 22

- 1x = % x da = 1 ) = la = x, ) dx = x₂ the two equations above are same yra let X - i ...7 X₂ nel Our fisst Eigensture is it aftu normalizing So we get i . Vart Similarly ripenstate Colles pondere to the eigenvalue =-% is. 2 Li oltr 21 22 - 1x = -14X; =) dx =28, tx - - - - 2 =) ix- X again the above two equations are sammen Let dial3 = ti Second eigensfalle isti after normalizing we get til

Hence the eigenstates of Sy are te lat) and a la cassespording to ejjenvcileres +\, and -1/2 Amb l 11' is a unit vector = (x ; My inz) = ('since cosa, sinosing, 2050 O and ¢ are polas and azimuthal angles. - s = 2 2 6 matrices. Son= k In. Sxt nysyth, sal - [sinocosa foi o ju singsingle-t S + Coscollo 10-11 = h Toso Sin OCSP - isine sind sino caso tisince since -Cosco 1 - t T Cosco sinoptat I sincere - Cosco

eigenvalues of S. ñ are the seots of the equation 75. ñ- dll zo Tül2 Caso-d we sino e 4 t ty singer at coso-xl do tal ( Cosio + sino) = *-* -O hann to 2 eigen vector coliesponding to do ti/2 is found as shown below to foso sino els hat 12 Isinoeta a coscola 2a aith MOV x cos 0 + 1, sin c ñ ® - - x, sinoel & Cosco = x₂ w anotto oo -Xi (1- Cosco) + 7, sincetaso a singeli – X2 614 Caso ) = 0

from 2 3 equation ③ we have. #1, sino eil . X a +32.2 Sinon, cos0%, éty . 2 sino/ = + H2 caso, ela Sinon tallation we have from x = x (coscoti sino eta .

- 122 cosol e la 2 sinco caso x - X COS2 elt. - Singh hence equation and 2 are sarie.

let xi= cos/2) = singhe since, eta hence the first eigensture is I Casco/2 its citeady l sinon, esp , nomalized similarly the other eigenstate corresponding to a= -t can be I shown to be equal to - 7514042 .I - COS0j, etot Hence the eigenstater of sñ are | Cos/ land esinor I singelp and f-cosmeie) corresponding to eigenvalues ti/ and -t/2

Ans.2

We are given Liz E I hence the possible values of lm are alol - let the state of the system with } = 1 be denoted by lo,m>= 11, my These states are ortho normal lie <jmiljm) = 5mm we know that t he Tx = 1 / (J + + J.) and It lum) =/(J-m)(1+m+1) fj, m+) TO) . 2 | J- lj,m) =/(+m) 6-m+I) J, m-12 Ansa) We want to write <= 1, m' l Ju 1 j = 1, m> in matrix form: The possible states of the system I are ljal, mal) = 16 17 جوش ولعل زهرا Ilmanl> 311-12

hence Jo = ť to. Vzo 0 0 0 Ve Toool Similarly using equation and Toithonormality of states we can Tobtain makix for J DJ) = <1,11 J 1,1)=12 21,111,0)

Augbl Jo = À VOLON for eigenvalues of Ja we must have. To 1o Tail , To! Ta TEA I - To to dal where (a is an eigen rector corresponding to eggenvalued. equation can be written as 1 1 0 1 dl bz - 10 io land / T for non trivial solutions we must have 1 lot so o In (a²-17-11-2) = 0 + atd .o -1(x-2)=0 =) = 0, 212

eigenvecta corresponding to an different eigenvalues can be found using et eigenvector corresponding to dzo is found as below: To io al 10" 1 0 22 lo long 2 =0 xitHz = 0 - M2 70 - -Uz choosing x, we have luiten . .. ako affer normalizing it wie obtain our first eifennecter 5 to fort eigenvalue Q Isimilarly other eigenrectors can be found out the पप्प and 4 = 1/2 for eigenvalue (2

1.Pa = /2 this for eigenvalue = 12 Hence eigenrectors of Jp are 14 Y and l with eigenvalues i o fexto a ħ - 2 x t = -t.

Ausch 1jim>= 11,+1) = (1,0,0) - la Jo 11,+1) = ħ 11,17 rector lil can be shown as below. S a finalstate Afinal - - hence final state can be written as 1o. . which can be written in Ket form as 11,-1).