This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY251 Homework Set 9 Reading: Chapter 10, with exception of Sections 105 and 106 Homework: Chapter 10, Question 4, Problems 1,3,5,6,7,11 Hints and Solutions Question X.4 (10 points) According to the exclusion principle, in any system involving more than one electron, the wavefunction of the system must be antisymmetric under the interchange of the coordinates (and spin labels) of any pair of electrons. At the same time, it does not appear reasonable that in describing the 2 p state of the electron in New York, it should be necessary to take into account the wave function of an electron in the ground state of hydrogen located on the Moon. Give an argument as to why we can forego antisymmetry in this instance. When does antisymmetry need to be taken into account? Hints: See Problem 4. Solution: The Pauli principle only operates on indistinguishable fermions, thus their wave functions have to significantly overlap otherwise the fermions stay distinct. In cases where overlap exists, like in the hydrogen atom where the electron "clouds" are as large as the classical Bohr orbits, electrons overlap significantly, and the Pauli principle operates. Problem X.1 (15 points) Show that for the case of two particles in an infinite well, N A = N S = 2 in Eqs. (1022) and (1023). In order to do this you need to show that ( 2 / L ) ∫ L sin( n π x/L ) sin( m π x/L ) dx = 0 for m ≠ n, and = 1 for m = n . Hints: See example 96. Solution: The normalization condition for the wave function Eq. 1022/23 is: 1 = ∫ L ∫ L  u ± ( x 1 , x 2 )² dx 1 dx 2 = ( 1 / N ) ∫ L ∫ L  u m ( x 1 ) u n ( x 2 ) ± u m ( x 2 ) u n ( x 1 )² dx 1 dx 2 = phy251_hw09_solutions file:///C/DOCUME~1/PHY_CO~1/PHY251/PHY251_HW09_solutions.html (1 of 5) [4/30/2001 12:13:41 AM] = ( 1 / N ){ ∫ L ∫ L  u m ( x 1 )² u n ( x 2 )² dx 1 dx 2 + ∫ L ∫ L  u m ( x 2 )² u n ( x 1 )² dx 1 dx 2 ± 2 ∫ L ∫ L u m ( x 1 ) u n ( x 2 ) u m ( x 2 ) u n ( x 1 ) dx 1 dx 2 } = = ( 1 / N ){1×1 + 1×1 ± 2( ∫ L u m ( x 1 ) u n ( x 1 ) dx 1 )( ∫ L u m ( x 2 ) u n ( x 2 ) dx 2 )} = = ( 1 / N ){2 ± 2 K mn ²} = 1, and thus: N = 2 ± 2 K mn ² Note that for fermions , in case m = n , the wave function vanishes automatically,...
View
Full
Document
 Spring '01
 Rijssenbeek
 Physics, Atom, Work, Sin, Pauli exclusion principle, wave function

Click to edit the document details