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Unformatted text preview: HOMEWORK ASSIGNMENT 8 PHYS852 Quantum Mechanics II, Spring 2009 New topics covered: Green’s functions, Tmatrix . 1. The full Green’s function : A system with hamiltonian H has a Green’s function defined by G H ( E ) = ( E − H + iǫ ) − 1 For case H = H + V , there is also a ‘background Green’s function’: G H ( E ) = ( E − H + iǫ ) − 1 What is the relationship between G H ( E ) and G H ( E )? To answer this, start from the Schr¨odinger equation ( E − H − V )  ψ s ) = V  ψ ) and operate on both sides with the full Green’s function G H ( E ). Then compare your result with the Tmatrix definition  ψ s ) = G H T H ,V ( E )  ψ ) to show that G H ( E ) V = G H ( E ) T H ,V ( E ) . Insert the general solution T H ,V ( E ) = (1 − V G H ( E )) − 1 V and derive an expression for G H ( E ) in terms of G H ( E ) and T H ,V ( E ). Now use the definition of the operator inverse to derive the expression (1 − V G H ( E )) − 1 = 1 + T H ,V ( E ) G H ( E ) Plug this into your previous expression for G H ( E ) to show finally, that G H ( E ) = G H ( E ) + G H ( E ) T H ,V ( E ) G H ( E ) 1 Answer: The full Schr¨odinger equation is: ( E − H − V )(  ψ ) +  ψ 2 ) ) = 0 . Since ( E − H )  ψ ) = 0 this is equivalent to: ( E − H − V )  ψ s ) = V  ψ ) . Operating on both sides with G H ( E ) then gives:  ψ s ) = G H ( E ) V  ψ ) . Comparing with  ψ s ) = G H ( E ) T H ,V ( E )  ψ ) , we see that since  ψ s ) equals itself, we get G H ( E ) V  ψ ) = G H ( E ) T H ,V ( E )  ψ ) . Since this needs to be true for any incident state, it follows that: G H ( E ) V = G H ( E ) T H ,V ( E ) . Replacing T H ,V with its solution then gives: G H ( E ) V = G H ( E )(1 − V G H ( E )) − 1 V. To be true for any V requires: G H ( E ) = G H ( E )(1 − V G H ( E )) − 1 . The definition of the inverse is: (1 − V G H ( E )) − 1 (1 − V G H ( E )) = 1 . Distributing the inverted quantity gives: (1 − V G H ( E )) − 1 − (1 − V G H ( E )) − 1 V G H ( E ) = 1 . Recognizing the expression for the Tmatrix in the second term we find: (1 − V G H ( E )) − 1 − T H ,V ( E ) G H ( E ) = 1 , or (1 − V G H ( E )) − 1 = 1 + T H ,V ( E ) G H ( E ) . Putting this into our equation for G H ( E ) then gives: G H ( E ) = G H ( E )(1 + T H ,V ( E ) G H ( E )) , or G H ( E ) = G H ( E ) + G H ( E ) T H ,V ( E ) G H ( E ) . 2 2. Finding new boundstates : Consider a system described by H which has no bound states, but has a continuum of states for E > 0. This means that G H ( E ) = integraldisplay ∞ dE ′  E ′ )( E ′  E − E ′ + iǫ , where we are assuming no degeneracy for now. Now we are going to add to this system a perturbation V , so that H = H + V . It is often the case that H may have boundstates, even though H does not....
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 Fall '08
 MichaelMoore
 mechanics, Work, U0, bound state, GH0

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