4604_Solutions_Set5_fa07

4604_Solutions_Set5_fa07 - PHY4604 Fall 2007 Problem Set 5...

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PHY4604 Fall 2007 Problem Set 5 Solutions Department of Physics Page 1 of 9 PHY 4604 Problem Set #5 Solutions Problem 1 (20 points): Use separation of variables in Cartesian coordinates to solve the infinite cubical well (or “particle in a box”): + = 0 ) , , ( z y x V otherwise L z L y L x < < < < < < 0 , 0 , 0 (a) (10 points) Find the stationary states and the corresponding energies. Answer: ) / sin( ) / sin( ) / sin( 8 ) , , ( 3 L z n L y n L x n L z y x z y x n n n z y x π ψ = and ) ( 2 2 2 2 2 2 2 z y x n n n mL E + + = h Solution: In this case the V(x,y,z) = V(x)+V(y)+V(z) and we look for stationary state solutions of the form ) ( ) ( ) ( ) , , ( z Z y Y x X z y x = and Schrödinger’s equation becomes E Z z V dz Z d m Z Y y V dy Y d m Y X x V dx X d m X = + + + + + ) ( 2 1 ) ( 2 1 ) ( 2 1 2 2 2 2 2 2 2 2 2 h h h and hence x E X x V dx X d m X = + ) ( 2 1 2 2 2 h y E Y y V dy Y d m Y = + ) ( 2 1 2 2 2 h z E Z z V dz Z d m Z = + ) ( 2 1 2 2 2 h where the E i is constants and E x +E y +E z = E. The problem breaks into three one dimensional infinite wells and we know that the solution to the one dimensional problem is ) / sin( 2 ) ( L x n L x X x n x = with 2 2 2 2 2 x n n mL E x h = where n x = 1, 2, 3, … ) / sin( 2 ) ( L y n L y Y y n y = with 2 2 2 2 2 y n n mL E y h = where n y = 1, 2, 3, … ) / sin( 2 ) ( L z n L z Z z n z = with 2 2 2 2 2 z n n mL E z h = where n z = 1, 2, 3, … Thus, ) / sin( ) / sin( ) / sin( 8 ) ( ) ( ) ( ) , , ( 3 L z n L y n L x n L z Z y Y x X z y x z y x n n n n n n z y x z y x = = , and ) ( 2 2 2 2 2 2 2 z y x n n n n n n mL E E E E z y x + + = + + = h . (b) (5 points) Call the energies E 1 , E 2 , E 3 , …, in order of increasing energy. Find E 1 , E 2 , E 3 , E 4 , E 5 , and E 6 . Determine the degeneraries ( i.e. the number of different states with the same energy).
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PHY4604 Fall 2007 Problem Set 5 Solutions Department of Physics Page 2 of 9 Answer: Energy n x n y n z Number of States E 1 =) 2 /( 3 2 2 2 mL π h 1 1 1 1 E 2 2 /( 6 2 2 2 mL h 2 1 1 E 2 1 2 1 E 2 1 1 2 3 E 3 2 /( 9 2 2 2 mL h 2 2 1 E 3 2 1 2 E 3 1 2 2 3 E 4 2 /( 11 2 2 2 mL h 3 1 1 E 4 1 3 1 E 4 1 1 3 3 E 5 2 /( 12 2 2 2 mL h 2 2 2 1 E 6 2 /( 14 2 2 2 mL h 1 2 3 E 6 1 3 2 E 6 2 1 3 E 6 2 3 1 E 6 3 1 2 E 6 3 2 1 6 (c) (5 points) What is the degeneracy of E 14 , and why is this case interesting? Answer: The degeneracy of E 14 is 4. Solution: The combinations after E 6 are E 7 (322), E 8 (411), E 9 (331), E 10 (421), E 11 (332), E 12 (422), E 13 (431), and then E 14 (333 and 511). Simple combinatorics account for the degeneracies of 1 (n x = n y = n z ), 3 (two the same, one different), and 6 (all three different). But in the case of E 14 there is a numerical “accident”: 3 2 + 3 2 +3 2 = 27 and 5 2 +1 2 +1 2 = 27, so the degeneracy is greater than combinatorial reasoning alone would suggest. Problem 2 (20 points): A particle of mass m and E < 0 is placed in a finite spherical well ⎧− = 0 ) ( 0 V r V R r R r > (a) (10 points) Find the ground state by solving the radial equation with l = 0.
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This note was uploaded on 05/29/2011 for the course PHY 4064 taught by Professor Fields during the Spring '07 term at University of Florida.

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4604_Solutions_Set5_fa07 - PHY4604 Fall 2007 Problem Set 5...

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