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Unformatted text preview: PHYS 507 Answers to Homework V (Fall ’05) 1. (a) h ϕ n  [ H,A ]  ϕ n i = h ϕ n  HA  ϕ n i  h ϕ n  AH  ϕ n i = E n h ϕ n  A  ϕ n i  h ϕ n  A  ϕ n i E n = ( E n E n ) A nn = 0 . (b) [ T,A ] = 1 2 m [ p 2 ,xp + px ] = 1 2 m ( [ p 2 ,x ] p + p [ p 2 ,x ] ) = 1 2 m ¯ h i ((2 p ) p + p (2 p )) = 4 ¯ h i p 2 2 m = 4 ¯ h i T . [ V,A ] = [ V ( x ) ,xp + px ] = x [ V ( x ) ,p ] + [ V ( x ) ,p ] x = i ¯ h ( xV ( x ) + V ( x ) x ) = 2 i ¯ hxV ( x ) . Part (a) gives h [ T,A ] i n + h [ V,A ] i n = 0. Simplifying this we get 2 h T i n h xV ( x ) i n = 0 . (c) We have V ( x ) = c  x  α (note that we should have cα > 0 if the potential is bounding). In this case, xV ( x ) = xc ( α  x  α 1 sgn( x )) = αc  x  α = αV . The virial theorem then gives 2 h T i n = α h V i n . Since E n = h T i n + h V i n we then get, h T i n = α α + 2 E n , h V i n = 2 α + 2 E n ....
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