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# HW2-sol - PHYSICS 4455 QUANTUM MECHANICS Problem Set 2...

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PHYSICS 4455 — QUANTUM MECHANICS Problem Set 2 — Solutions. First of all, here is a general comment. Liboff uses Gaussian units throughout the book - so, for example, Coulomb’s law for the force between two charges, q 1 and q 2 separated by a distance r , would be written as F ( r ) = q 1 q 2 r 2 r r Note the absence of the 4 π o factor in the denominator which is familiar from SI units. If you want to convert expressions in Liboff from Gaussian to SI units, look for factors of e 2 (i.e., the square of the electron charge - not Euler’s constant!!). Replace each e 2 by e 2 /( 4 π o ) , and you get the expression in SI units. I’ll write the solutions in SI units (at least until further notice). 1. Recapping complex variables. Liboff Problem 1.21 , without parts (b), (d), (f), (j). We start with the representation of a complex number (of magnitude 1): e i θ = cos θ + i sin θ which also implies cos θ = 1 2 ( e i θ + e i θ ) = Re e i θ sin θ = 1 2 i ( e i θ e i θ ) Im e i θ The first few parts just check trigonometric identities. (a) cos 1 + θ 2 ) = 1 2 ( e i 1 2 ) + e i 1 2 ) ) = 1 2 ( e i θ 1 e i θ 2 + e i θ 1 e i θ 2 ) = 1 2 ( cos θ 1 + i sin θ 1 )( cos θ 2 + i sin θ 2 ) + 1 2 ( cos θ 1 i sin θ 1 )( cos θ 2 i sin θ 2 ) = 1 2 ( cos θ 1 cos θ 2 + i cos θ 1 sin θ 2 + i sin θ 1 cos θ 2 sin θ 1 sin θ 2 ) + 1 2 ( cos θ 1 cos θ 2 i cos θ 1 sin θ 2 i sin θ 1 cos θ 2 sin θ 1 sin θ 2 ) = cos θ 1 cos θ 2 sin θ 1 sin θ 2 (c) 2sin θ 1 cos θ 2 = 1 2 i ( e i θ 1 e i θ 1 )( e i θ 2 + e i θ 2 ) = 1 2 i ( e i 1 2 ) + e i 1 −θ 2 ) e i 1 −θ 2 ) e i 1 2 ) ) = 1 2 i ( e i 1 2 ) e i 1 2 ) ) + 1 2 i ( e i 1 −θ 2 ) e i 1 −θ 2 ) ) = sin 1 + θ 2 ) + sin 1 − θ 2 )

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(e) 2sin θ 1 sin θ 2 = − 1 2 ( e i θ 1 e i θ 1 )( e i θ 2 e i θ 2 ) = − 1 2 ( e i 1 2 ) e i 1 −θ 2 ) e i 1 −θ 2 ) + e i 1 2 ) ) = cos 1 − θ 2 ) − cos 1 + θ 2 ) (g) Use part (e): 2sin 2 θ = cos (θ − θ) − cos (θ + θ) = 1 cos2 θ (h) e i θ 1 = e i θ/ 2 ( e i θ/ 2 e i θ/ 2 ) = 2 ie i θ/ 2 sin (θ/ 2 ) (i) 1 2 | e i θ 1 + e i θ 2 | 2 = 1 2 ( e i θ 1 + e i θ 2 )( e i θ 1 + e i θ 2 ) = 1 2 ( 1 + e i 1 −θ 2 ) + e i 1 −θ 2 ) + 1 ) = 1 + cos 1 − θ 2 ) (k) Let’s write z = | z | e i θ : 2 i Im z = 2 i Im | z | e i θ = 2 i | z | sin θ = | z |( e i θ e i θ ) = z z (l) ( e z ) = [ exp (| z | e i θ )] = [ exp (| z | cos θ + i | z | sin θ)] = [ e | z | cos θ e i | z | sin θ ] = e | z | cos θ e i | z | sin θ = e | z | cos θ− i | z | sin θ = e z (m) | e z | 2 = | exp (| z | e i θ )| 2 = [ e | z | cos θ e i | z | sin θ ][ e | z | cos θ e i | z | sin θ ] = [ e | z | cos θ e i | z | sin θ ][ e | z | cos θ e i | z | sin θ ] = e 2 |
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