Daily_009.SOLN - Q5.10 and.11 on page 91 Equation Q5.11 in...

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Phys 274 – c Eric B. Szarmes 7 February 2011 Daily Homework 9 SOLUTIONS Q5B.5 We are asked to consider the following series of Stern-Gerlach devices: SG(z) SG( ! ) + + z y ? z y z y z y ! z y ! P(+ ! ) = cos ! 2 2 P( ! ) = sin ! 2 2 I have also indicated schematically the various spin states for the quantons at the input and output channels. We require that the probability for the quantons to appear in the + θ channel of the 2nd device be 1/3; therefore, the angle θ of this 2nd device must satisfy cos 2 ( θ/ 2) = 1 / 3 , so that θ = 109 . 47 . (1) (Similarly, we require sin 2 ( θ/ 2) = 2 / 3, and again we have θ = 109 . 47 .) Evidently, to get the requested probabilities, the final angles appear as follows: z y ! z y ! P(+ ! ) = 1 3 P( ! ) = 2 3 SG( ! ) + Q5B.9
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Unformatted text preview: Q5.10 and .11 on page 91. Equation Q5.11 in particular gives us the prescription for calculating the inner product. Thus, we have the following (!!! don’t forget to conjugate | u i ): a) h u | w i = ± 1-i ² * · ± 2 i 3 ² = (1) * (2 i ) + (-i ) * (3) = (1)(2 i ) + (+ i )(3) = 5i b) h u | w i = ± 1-2 ² * · ± i-5 ² = (1) * ( i ) + (-2) * (-5) = (1)( i ) + (-2)(-5) = 10+i c) h u | w i = ± 1 + i-2 + i ² * · ± i 2-i ² = (1 + i ) * ( i ) + (-2 + i ) * (2-i ) = (1-i )( i ) + (-2-i )(2-i ) = i + 1-4 + 2 i-2 i-1 =-4+i 1...
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