Chapter 3 Problem Solutions

# Chapter 3 Problem Solutions - Intermediate QMI Ch 3 Problem...

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Intermediate QMI Ch 3 Problem Solutions Problem 3.5 Given the rotation operator R ` H q j L = ‰ S ` y q ê which rotates spin states by an angle q counterclockwise about the y axis. ü Part a In the ± z \ basis, we have S ` y = ° 2 0 Â 0 S ` y 2 = ° I 2 M 2 1 0 0 1 = I 2 M 2 1 ` We see from the second equality that even powers of S ` y are proportional to the identity, and thus odd powers of S ` y are propor- tional to S ` y itself. Taylor expanding R ` H q j L : R ` H q j L = / n = 0 1 n ! J S ` y q ê N n ¬ Separate sum into even/odd powers of S ` y = / m = 0 1 H 2 m L ! H -Â q ê L 2 m S ` y 2 m + / m = 0 1 H 2 m + 1 L ! H -Â q ê L 2 m + 1 S ` y 2 m + 1 ¬ S ` y 2 m = I 2 M 2 m 1 ` S ` y 2 m + 1 = I 2 M 2 m S ` y ; H L 2 m = H - 1 L m H L 2 m + 1 = -Â H - 1 L m = / m = 0 H - 1 L m H 2 m L ! I q M 2 m I 2 M 2 m 1 ` + / m = 0 H - 1 L m H 2 m + 1 L ! I q M 2 m + 1 I 2 M 2 m = 2 J q 2 N 2 m + 1 S ` y = B/ m = 0 H - 1 L m H 2 m L ! I q 2 M 2 m F cos H q ê 2 L 1 ` - 2 Â B/ m = 0 H - 1 L m H 2 m + 1 L ! I q 2 M 2 m + 1 F sin H q ê 2 L S ` y therefore R ` H q j L = cos I q 2 M - 2 Â sin I q 2 M S ` y ü Part b We have from (a) R ` H q j L = ° cos I q 2 M 1 0 0 1 - 2 Â sin I q 2 M ÿ 2 0 Â 0 = cos H q ê 2 L - sin H q ê 2 L sin H q ê 2 L cos H q ê 2 L Thus Printed by Mathematica for Students

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R ` H q j L + z ^ = ° cos H q ê 2 L - sin H q ê 2 L sin H q ê 2 L cos H q ê 2 L 1 0 = cos H q ê 2 L sin H q ê 2 L = cos I q 2 M + z \ + sin I q 2 M - z \ therefore R ` H q j L + z ^ = cos I q 2 M + z \ + ‰ Â H 0 L sin I q 2 M - z \ which is the state + n \ with f = 0. Similarly R ` H q j L - z ^ = ° cos H q ê 2 L - sin H q ê 2 L sin H q ê 2 L cos H q ê 2 L 0 1 = - sin H q ê 2 L cos H q ê 2 L = - sin I q 2 M + z \ + cos I q 2 M - z \ therefore R ` H q j L - z ^ = - B sin I q 2 M + z \ - ‰ Â H 0 L cos I q 2 M - z \F which is the state - - n \ with f = 0.
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Chapter 3 Problem Solutions - Intermediate QMI Ch 3 Problem...

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