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HW6-sol - PHYSICS 4455 QUANTUM MECHANICS Problem Set 6 due...

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PHYSICS 4455 — QUANTUM MECHANICS Problem Set 6 — due 10 / 13 / 2005, in class. 1. More practice with one-dimensional problems. Liboff Problem 4.33 (a) A particle in a one-dimensional box on the interval (0, a ) is in the superposition state ϕ( x )= b 1 e i λ 1 ϕ 1 ( x )+ b 3 e i λ 3 ϕ 3 ( x ) To find the probability density in terms of λ 1 , λ 3 we compute ϕ ( x )ϕ( x )= [ b 1 e i λ 1 ϕ 1 ( x )+ b 3 e i λ 3 ϕ 3 ( x )][ b 1 e i λ 1 ϕ 1 ( x )+ b 3 e i λ 3 ϕ 3 ( x )] = b 1 b 1 ϕ 1 ( x 1 ( x )+ b 1 b 3 e i λ 1 e i λ 3 ϕ 1 ( x 3 ( x )+ + b 1 b 3 e i λ 1 e i λ 3 ϕ 3 ( x 1 ( x )+ b 3 b 3 ϕ 3 ( x 3 ( x ) Using the fact that b 1 , b 3 are real and sum according to 1 = b 1 2 + b 3 2 , we continue to = b 1 2 ϕ 1 ( x 1 ( x )+ 2Re [ b 1 b 3 e i 1 −λ 3 ) ϕ 3 ( x 1 ( x )]+( 1 b 1 2 3 ( x 3 ( x ) = b 1 2 1 ( x )| 2 + 2 b 1 b 3 cos 1 −λ 3 1 ( x 3 ( x )+( 1 b 1 2 )|ϕ 3 ( x )| 2 Inserting the explicit form of the eigenfunctions, we obtain ϕ ( x )ϕ( x )= 2 a b 1 2 sin 2 ( k 1 x )+ 4 a b 1 b 3 cos 1 −λ 3 ) sin ( k 1 x ) sin ( k 3 x )+ 2 a ( 1 b 1 2 ) sin 2 ( k 3 x ) Answer to (a): the probability density is a function of cos 1 −λ 3 ) . That is actually a more detailed statement than the one in Liboff. You may want to check that the condition 1 = b 1 2 + b 3 2 ensures that this state is normalized. To avoid lots of integrals, exploit the orthonormality of the ϕ n ( x ) .Thatgives you the desired relation in one line: 1 = 0 a ϕ ( x )ϕ( x ) dx = b 1 2 0 a 2 a sin 2 ( k 1 x )+ 2 b 1 b 3 cos 1 −λ 3 ) 0 a 2 a sin ( k 1 x ) sin ( k 3 x )+ b 3 2 0 a 2 a sin 2 ( k 3 x ) = b 1 2 + b 3 2 (b) Now, we do the same for a more complicated state, namely, ϕ( x )= b 1 e i λ 1 ϕ 1 ( x )+ b 3 e i λ 3 ϕ 3 ( x )+ b 7 e i λ 7 ϕ 7 ( x ) Again, computing the probability density gives us ϕ ( x )ϕ( x )= [ b 1 e i λ 1 ϕ 1 ( x )+ b 3 e i λ 3 ϕ 3 ( x )+ b 7 e i λ 7 ϕ 7 ( x )][ b 1 e i λ 1 ϕ 1 ( x )+ b 3 e i λ 3 ϕ 3 ( x )+ b 7 e i λ 7 ϕ 7 ( x )] = b 1 b 1 ϕ 1 ( x 1 ( x )+ b 3 b 3 ϕ 3 ( x 3 ( x )+ b 7 b 7 ϕ 7 ( x 7 ( x ) + 2Re [ b 1 b 3 e i 1 −λ 3 ) ϕ 3 ( x 1 ( x )] + 2Re [ b 1 b 7 e i 1 −λ 7 ) ϕ 7 ( x 1 ( x )] + 2Re [ b 3 b 7 e i 3 −λ 7 ) ϕ 7 ( x 3 ( x )] = b 1 2 ϕ 1 2 ( x )+ b 3 2 ϕ 3 2 ( x )+ b 7 2 ϕ 7 2 ( x ) + 2 b 1 b 3 cos 1 −λ 3 3 ( x 1 ( x )+ 2 b 1 b 7 cos 1 −λ 7 7 ( x 1 ( x ) + 2 b 3 b 7 cos 3 −λ 7 7 ( x 3 ( x ) where we have used the fact that the b i and the ϕ n are real. Again, the probability density only depends on the cos of the phase differences.

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Liboff Problem 4.35 An electron in a box with walls at x = 0and x = a is in the state
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HW6-sol - PHYSICS 4455 QUANTUM MECHANICS Problem Set 6 due...

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