4604_Exam3_solutions_fa07

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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern