4604_Solutions_Set7

4604_Solutions_Set7 - PHY4604 Problem Set 7 Solutions PHY...

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PHY4604 Problem Set 7 Solutions Department of Physics Page 1 of 14 PHY 4604 Problem Set #7 Solutions (Total Points = 120) Problem 1 (30 points): The state of a two particle system is described by the wave function ) , , ( 2 1 t r r r r Ψ . The time evolution is given by Schrödinger’s equation t i H Ψ = Ψ h with ) , , ( 2 2 2 1 2 2 2 2 2 1 1 2 t r r V m m H r r h h + = where 2 2 2 2 2 2 2 k k k k z y x + + = for k = 1,2. For time-independent potentials, we obtain a complete set of solutions of the form h r r r r / 2 1 2 1 ) , ( ) , , ( iEt e r r t r r = Ψ ψ , where ) , ( 2 1 r r r r satisfies the time- independent Schrödinger equation E r r V m m = + ) , ( 2 2 2 1 2 2 2 2 2 1 1 2 r r h h . Typically the interaction potential depends only on the vector 2 1 r r r r r r = ( i.e. the separation between the two particles). Suppose ) ( ) , ( 2 1 r V r r V r r r = and we change variables from 1 r r and 2 r r to 2 1 r r r r r r = and ) /( ) ( 2 1 2 2 1 1 m m r m r m R + + = r r r ( i.e. the center-of-mass vector). (a) (5 points) Show that r m R r r r r ) / ( 1 1 µ + = and r m R r r r r ) / ( 2 2 = , where ) /( 2 1 2 1 m m m m + = is the “reduced mass”. Solution: We see that r m r m m r r m r m r m r m R m m r r r r r r r r 2 1 2 1 1 2 1 1 2 2 1 1 2 1 ) ( ) ( ) ( + = + = + = + and hence r m R r m m m R r r r r r r 1 2 1 2 1 ) ( + = + + = . Also, r m r m m r m r r m r m r m R m m r r r r r r r r 1 2 2 1 2 2 2 1 2 2 1 1 2 1 ) ( ) ( ) ( + + = + + = + = + and hence r m R r m m m R r r r r r r 2 2 1 1 2 ) ( = + = . (b) (5 points) Show that r R m + = r r r ) / ( 2 1 and r R m = r r r ) / ( 1 2 , where ) /( 2 1 2 1 m m m m + = is the “reduced mass”. Solution: Let ) , , ( Z Y X R = r and ) , , ( z y x r = r then ) /( ) ( 2 1 2 2 1 1 m m x m x m X + + = and 2 1 x x x = and we see that x r x R x m m m x X m m m x x x X x X x ) ( ) ( ) ( 2 1 1 2 1 1 1 1 1 1 + + = + + = + = = . Also, ) /( ) ( 2 1 2 2 1 1 m m y m y m Y + + = and 2 1 y y y =
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PHY4604 Problem Set 7 Solutions Department of Physics Page 2 of 14 and we see that y r y R y m m m y Y m m m y y y Y y Y y ) ( ) ( ) ( 2 1 1 2 1 1 1 1 1 1 + + = + + = + = = Also, ) /( ) ( 2 1 2 2 1 1 m m z m z m Z + + = and 2 1 z z z = and we see that z r z R z m m m z Z m m m z z z Z z Z z ) ( ) ( ) ( 2 1 1 2 1 1 1 1 1 1 + + = + + = + = = . Hence, r R r R m m m m + = + + = r r r r r 2 2 1 1 1 µ . Similarly, x r x R x m m m x X m m m x x x X x X x ) ( ) ( ) ( 2 1 2 2 1 2 2 2 2 2 + = + = + = = . y r y R y m m m y Y m m m y y y Y y Y y ) ( ) ( ) ( 2 1 2 2 1 2 2 2 2 2 + = + = + = = z r z R z m m m z Z m m m z z z Z z Z z ) ( ) ( ) ( 2 1 2 2 1 2 2 2 2 2
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This note was uploaded on 04/19/2008 for the course PHY 4604 taught by Professor Field during the Spring '07 term at University of Florida.

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4604_Solutions_Set7 - PHY4604 Problem Set 7 Solutions PHY...

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