4604_Exam3_solutions_fa07

# 4604_Exam3_solutions_fa07 - PHY4604 Fall 2007 R D Field PHY...

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PHY4604 Fall 2007 R. D. Field Department of Physics Page 1 of 16 Exam 2 Solutions PHY 4604 Final Exam Take Home Exam (Total Points = 100) Problem 1 (20 points): (4 points each). (1a) An electron is in a one-dimensional infinite square well with zero potential energy in the interior and infinite potential energy at the walls and size L has a ground state energy E 0 . If the size of the square well is decreased to L/2 what is the new ground state energy? ( circle the correct answer ) (a) 0 (b) E 0 /2 (c) E 0 /4 (d) 2E 0 (e) 4E 0 Solution: The groud state energy is 0 2 2 2 2 / 2 2 2 0 4 ) 2 / ( 2 2 E L m mL E L L = = π π h h . (1b) Which of the following angular momentum states has the smallest lower bound on the product ( Δ L x )×( Δ L y )? ( circle the correct answer ) (a) l = 2 m = 2 (b) l = 2 m = 1 (c) l = 2 m = 0 (d) l = 2 m = -1 (e) l = 2 m = -2 Solution: We know that from the uncertainty principle that > < Δ Δ ] , [ B A i B A and hence > < = > < Δ Δ z y x y x L L L i L L h ] , [ . The <L z > is zero in the state l = 2 m = 0 which is the smallest. (1c) A “glue ball” is a particle constructed entirely from gluons? Gluons carry intrinsic spin 1 ( i.e. they are vector particles). Which of the following correspond to possible “glue ball” spins? ( circle all the correct answers ) (a) 0 (b) 1 (c) 2 (d) 3/2 (e) 1/2 Solution: 3 × 3 = 5 + 3 + 1 but cannot form 3/2 or ½ from any number of spin 1 particles even if you add in orbital angular momentum. (1d) Consider the angle, θ , between the angular momentum vector and the z-axis for the following states. For which state is the angle θ the smallest? ( circle all the correct answers ) (a) l = 2 m = 2 (b) l = 2 m = 1 (c) s = 3/2 m = 3/2 (d) s = 1/2 m = 1/2 (e) l = 3 m = 2 Solution: We know that ) 1 ( cos + = l l m θ or ) 1 ( cos + = s s m θ and hence State cos θ l =2 m = 2 0.816 l = 2 m = 1 0.408 s =3/2 m = 3/2 0.775 s = ½ m = 1/2 0.577 l = 3 m = 2 0.577 (1e) What is 4 × 3 × 2 × 1 in SU(2)? Solution: 4 × 3 × 2 × 1 = 4 × (4 + 2) = 4 × 4 + 4 × 2 = 7 + 5 + 3 + 1 +5 + 3 = 7 + 5 + 5 +3 + 3 + 1

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PHY4604 Fall 2007 R. D. Field Department of Physics Page 2 of 16 Exam 2 Solutions Problem 2 (40 points): Consider the case of two non-interacting particles both with mass m in a one-dimensional infinite square well given by V(x) = 0 for 0 < x < L, and V(x) = . For one particle we know that the stationary states of Schrödinger’s equation are given by h / ) ( ) , ( t iE n n n e x t x = Ψ ψ and 0 2 E n E n = , and n is a positive integer and 2 2 2 0 2 mL E h π = and where ) / sin( 2 ) ( L x n L x n π ψ = . For two ( non-interacting ) particles we look for a solution of the form h h / 2 1 / 2 1 2 1 ) ( ) ( ) , ( ) , , ( iEt iEt e x x e x x t x x = = Ψ ψ ψ ψ with E m p m p x x = + 2 ) ( 2 ) ( 2 2 2 1 . (a) (4 points): Show that ) ( ) ( ) , ( 2 1 2 1 x x x x β α αβ ψ ψ ψ = , is a solution to the two particle non- interaction Schrödinger equations, where α and β are positive integers and ψ α (x) and ψ β (x) are the one particle stationary state solutions. Show that the allowed energy levels are given by 0 2 2 ) ( E E β α αβ + = . This solution corresponds to the case where the two particles are distinguishable. Define the following three probabilities. Let LL P αβ be the probability of finding both particles in the left 1/2 of the box as follows: ∫ ∫ = 2 / 0 2 / 0 2 1 2 2 1 | ) , ( | L L LL dx dx x x P
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