

![Hence, we finally write ea. (A) as Ru (d) = cosa I - isine (coso O₂ + sino (cos & Ox+ singsy)] no cosø Ox tsino sino Oy acoso](//img.homeworklib.com/questions/c88c5420-447a-11eb-8403-bdc6f0418f5f.png?x-oss-process=image/resize,w_560)


Answer: The rotation by an angle & about an arbitrary axis nis giver by Ru (2) = R₂ (8) Ry (0) R₂ (2) Ry (-0) R₂ (-4) .. (1) = R₂ (8) Ry (0) R₂ (2) R (0) R₂ (8) . (2) from righet The rotation R₂ (-0) first rotates d into the x-z plane, then Ry(-0) rotates it into the z-axis, Rq (2) performs the desired rotation about ñ, and finally Ry (0) and Rz (8) rotate in back to its original orientation. In the second line, we've used Ry (-0) = Ryt (6) and Rz (-0) = R (0) which is evident from their matix representation Ry (0) = ( cos of – sin % . - 1012 o 1 cos ) : R₂ (8) = (042 10/2 Isin - (3) as can be written Now the third term R₂ (2) 1 (cost-isire/ le o Rz (2) = ial IIo costisind To eo a cost/ I - i sin %02 sep
identity matrix and Oz is the where I is the Pauli z-martix. So from ear. (2), we write B: () = R2 CA) Ry(0) {cose I - ising or] R760) RICK) @ (5) we can also prove that Ry (0) Sz Ry 10) = coso Oq+sin Ox (6) eq. (3). cusing the massix multiplication by using also we're - which can be also checked using (3) fence aan. (5) becomes. Rü (d) = R2 (8) (cosa I - isina (coses & sin 05) R (8) we also use the following identities R₂(0) Iz R (0) = Oz Ra (C) One Rico) - cessed, cos O One tsino oy which can be again checked by simple matrix multiplication eq-3). using
Hence, we finally write ea. (A) as Ru (d) = cosa I - isine (coso O₂ + sino (cos & Ox+ singsy)] no cosø Ox tsino sino Oy acoso oz) cosa I - isina ( On tony Oy at Mz8z) (as given n in the problem) - cosa I - isina n.o .. (9) To write it in terms of operator exponential, notice that. (5.5)² = (5.5) (5.0) = (n.) I tillö using (2.5) (5.5) = (a I) I + i (5x5). reference gu Griffithe wikipedia ( i j is unit vector, ras). so. S takiy A= n o in the identity iOA - coso I tising A - (10)
we can write -(LV Ru(d) = cos & I - isina n.o. (12) exp (id 5.5) SO . hơ (Proved). ( K3 ()).. The equivalence between ev. (11) & ea. (12) can. also be seen by answering the 2nd part of the question. we have. we have. (5.5) - 16. (5.577 K20 2x+1 { cal", a : 34807 ko (seperating out the and imaginary real part ) - * (.) co* k20 (2*+1)! 2x) (14)
where were used the identity 66" +(68.333" 3. (: (6-3)*: 1) ano (m. 3j*** (.6) (.6) 2* = (ń.) The first term in ear (14) is the cosine term, where as the second term is the sine term. Hence, eia (n.) = cosa I isina (ño) (Proved)