HW1Solutions - 1.1 a Begin with y 1 = 3 4 To normalize introduce an overall complex multiplicative factor and solve for this factor by imposing the

HW1Solutions - 1.1 a Begin with y 1 = 3 4 To normalize...

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1.1. a) Begin with 1 3 4 y = + + - To normalize, introduce an overall complex multiplicative factor and solve for this factor by imposing the normalization condition: ( ) ( ) { } ( ) { } ( ) ( ) 1 * 1 1 * * 2 3 4 1 3 4 3 4 9 12 12 16 25 1 25 C C C C C C C C y y y = + + - = = + + - + + - = + + + + - + - + + - - = = Because an overall phase is physically meaningless, we choose C to be real and positive: 1 5 C = . Hence the normalized input state is 3 4 1 5 5 y = + + - . Likewise: ( ) ( ) { } ( ) { } ( ) ( ) 2 2 * * 2 1 2 5 5 2 1 2 2 4 5 i C i C i C i C C C y y = + + - = + - - + + - = + + + - - = = + + - and ( ) ( ) { } ( ) { } ( ) ( ) 3 3 2 * 3 3 * 3 3 1 3 10 10 3 1 3 3 9 1 10 i i i i C e C e C e C C C e p p p p y y - = + - - = + - - + - - = + + + - - = = + - - b) The probabilities for state 1 are ( ) 2 2 2 2 3 3 3 9 4 4 1, 1 5 5 5 5 5 25 y + = + = + + + - = + + + + - = = P ( ) 2 2 2 2 3 3 16 4 4 4 1, 1 5 5 5 5 5 25 y - = - = - + + - = - + + - - = = P For the other axes, we get ( ) ( ) ( ) ( ) 2 2 2 3 3 49 1 1 4 1 1 4 1, 1 5 5 5 5 50 2 2 2 2 2 2 2 3 3 1 1 4 1 1 4 1 1, 1 5 5 5 5 50 2 2 2 2 x x x x y y + - = + = + + - + + - = + = = - = + - - + + - = - = P P ( ) ( ) ( ) ( ) 2 2 2 3 3 1 4 1 4 1 1, 1 5 5 5 5 2 2 2 2 2 2 2 2 3 3 1 4 1 4 1 1, 1 5 5 5 5 2 2 2 2 2 i i y y i i y y y y + - = + = + - - + + - = - = = - = + + - + + - = - = P P
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The probabilities for state 2 are ( ) 2 2 2 2 1 1 1 2, 2 5 5 5 5 i y + = + = + + + - = = P ( ) 2 2 2 2 2 1 4 2, 2 5 5 5 5 i i y - = - = - + + - = = P ( ) ( ) ( ) ( ) 2 2 2 2 2 1 1 1 1 1 2, 2 2 2 2 5 5 10 10 2 2 2 2 2 1 1 1 1 1 2, 2 2 2 2 5 5 10 10 i i x x i i x x y y + - = + = + + - + + - = + = = - = + - - + + - = - = P P ( ) ( ) ( ) ( ) 2 2 2 2 9 1 1 1 2 2, 2 10 2 2 5 5 10 10 2 2 2 2 1 1 1 2 1 2, 2 10 2 2 5 5 10 10 i i y y i i y y y y + - = + = + - - + + - = + = = - = + + - + + - = - = P P The probabilities for state 3 are ( ) 2 2 2 3 3 3 9 1 3, 3 10 10 10 10 i e p y + = + = + + - - = = P ( ) 2 2 2 3 3 3 1 1 1 3, 3 10 10 10 10 i i e e p p y - = - = - + - - = - = P ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 1 1 1 3, 3 2 2 10 10 2 3 3 9 3 7 1 1 20 20 20 3 20 20 20 2 2 3 3 1 1 1 3, 3 2 2 10 10 2 3 3 9 3 13 1 1 20 20 20 3 20 20 20 2cos 2cos i x x i i x x i e e e e p p p p p p y y + - = + = + + - + - - = - = + - = = - = + - - + - - = + = + + = P P ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 1 1 3, 3 2 2 10 10 2 3 3 9 3 1 1 20 20 20 3 20 20 20 2 2 3 3 1 1 3, 3 2 2 10 10 2 3 3 9 3 1 1 20 20 20 3 20 20 20 2sin 10 3 3 0.24 2sin 10 3 3 0.76 i i y
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