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Exercise

# Exercise - =-= h h h h h h h This is the final result Note...

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The solution of the exercise problem in Chapter 1.3.5: For free particles, we have of course ( 29 ( 29 , n n ik q ik q n n q e q e φ φ - = = , where n k is related with the momentum p by the relation / n n k p = h . In fact, for free particles, k and p are continuum, and the summation should be replaced by integral. Thus we have (keep in mind ( 29 ( 29 ( 29 2 2 / 2 / 2 n n n E p m k m = = h ): ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 , ; , exp exp exp exp exp exp 2 exp exp 2 1 exp exp 2 2 n n n n n n n n n n n n n n n n n E t t K q t q t q q i E t t ik q ik q i t t k ik q q i m t t ik q q i k m t t dk ik q q i k m φ φ π -∞ - ′ ′ = - - = - - - = - - - = - - - = - - h h h h h h You could take the above integral and find the result: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 1 , ; , exp exp 2 2 1 2 exp 1 2 2 2 exp 1 2 2 2 n n n t t K q t q t dk ik q q i k m im q q m i when csgn t t i t t t t m im q q m i when csgn t t i t t t t m π π π π -∞ - ′ ′ = - - - = - =
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Unformatted text preview: - =-= -- h h h h h h h This is the final result. Note that in above the csgn is a special function with this property: ( 29 1 Re , Im 1 when x or if x is an imaginary number x csgn x otherwise = - Of course, in our case, because t t , ( 29 1 2 i csgn t t m -= h holds. You could substitute this result into the equation (13) of Dr. Los lecture notes (Chapter 1.3.5) and you will find for a free-particle-solution in the RHS, the integral will give you a free-particle- solution in the LHS....
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