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Chapter2_7

Chapter2_7 - sin 2 2 n L x n L L L x n L L x x n n − =...

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PHY4604 R. D. Field Department of Physics Chapter2_7.doc University of Florida A symmetry implies a conservation law! The Infinite Square Well (5) Case II: Another set of solutions comes from taking A' = 0 and sin(kL/2) = 0 , which implies that kL/2 = n - π with n - = 1, 2, 3, ... . Thus, ) / 2 sin( ' 2 ) ( L x n B x n π ψ = and = = = 0 2 2 2 2 2 2 2 ) ( ) ( 2 2 E n mL n m k E n π h h where 2 2 2 0 2 mL E π h = . These wavefunctions are odd under x -x . Namely, ) ( ) ( x x n n = ψ ψ . Ground State (Lowest Energy State): The “+” wavefunctions ( i.e. parity even) give the lowest energy given by 0 2 2 2 0 1 2 E mL E E = = = + + π h . 1 st Excited State (Next Lowest Energy State): The “-” wavefunctions ( i.e. parity odd) give the next lowest energy given by 0 2 2 2 0 1 4 2 E mL E E = = = π h . We, of course, get the same energy levels as before. In fact, we could have taken our result for 0 < x < L , and made the shift x x – L/2 and arrived at the same result as above:
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Unformatted text preview: / ( sin( 2 ) 2 / ( ) ( n L x n L L L x n L L x x n n − = − = − = ± Parity Operator: The parity operator is the operator that takes x →-x as follows: ) ( ) ( x x P op − = . In the above problem the wavefunctions ψ ± are eigenfunctions of the parity operator with eigenvalues ±1 as follows: ) ( ) ( x x P n n op ± ± ± = . If the Hamiltonian H is invariant under x →-x then the eigenvalues of the parity operator are constants of the motion. In this case we say that “parity is a good quantum number” ( i.e. it is conserved ) and as a function of time “+” states remain “+” states and “-” states remain “-” states. E 1 = E n + = 1 Energy Levels n-= 1 E 2 = 4E E 3 = 9E n + = 2 + - +...
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