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Unformatted text preview: CHEM 120A Problem Set 8 Solutions David Hoffman , Doran Bennett , Tara Yacovitch 4/12/2010 1. Method 1: Matrices This method is attractive when there are a lot of basis functions. Our plan of action will be to express our Hamiltonian matrix using some orthonormal basis. We might be tempted to use the following basis:  cov i =  AB i +  BA i and  ion i =  AA i +  BB i 1 . But we can’t use this basis because the kets are not orthogonal. 2 After forming our Hamiltonian in our new basis – the creation of which will be described later – we can diagonalize it to find two eigenvalues (which will be functions of R ) and two eigenkets (eigenvectors) whose components will also be functions of R . First let’s create an orthonormal basis from the { cov i ,  ion i} basis. To do this we’ll need to employ the GramSchmidt method . The first vector will just be a normalized version of  cov i ,  ξ 1 i =  cov i p h cov  cov i =  cov i √ 2 + 2 e 2 R/a (1a)  ξ 2 i =  cov i + c  ion i p ( c 2 1)(2 + 2 e 2 R/a ) (1b) Where c is, c = h cov  cov i h cov  ion i = cosh R a (1c) The easiest way to attack the rest of this problem is to calculate the following integrals, h cov  H  cov i h cov  H  ion i h ion  H  cov i h ion  H  ion i (1d) Notice that the second and third integrals are identical so there are really only three to calculate. You can further break down these integrals into smaller pieces. We know that H can be written as, H = T (1) + T (2) + V (1) + V (2) + Δ V (1 , 2) (1e) 1 I’m using the notation that  AB i is equivalent to  φ A (1) φ B (2) i . 2 If you remember when we talked about forming matrices from operators we required that our basis functions be orthonormal (orthogonal would have sufficed but we imposed normality as well to make the math easier 1 From symmetry we know that T (1) = T (2) and V (1) = V (2) so we can determine the integrals of T (1), V (1) andΔ V and then sum them to find H . Finding the integrals is a task that is simple in principle but can get confusing in practice. All four integrals for Δ V (1 , 2) will be derived because they are the most complicated. The integrals of T and V will be left as an exercise for the reader. Starting with the on symmetric terms, h cov  Δ V  cov i = ( h AB  + h BA  )Δ V (  AB i +  BA i ) = h AB  Δ V  AB i + h AB  Δ V  BA i + h BA  Δ V  AB i + h BA  Δ V  BA i Here it’s important to note that h AB  Δ V  BA i = h AA  Δ V  BB i = 2 1 + 1 1 + R/a v 2 e 2 R/a (1f) h ion  Δ V  ion i = ( h AA  + h BB  )Δ V (  AA i +  BB i ) = h AA  Δ V  AA i + h AA  Δ V  BB i + h BB  Δ V  AA i + h BB  Δ V  BB i = 2 ( 2 + e 2 R/a ) v 2 (1g) Now on to the cross terms, h cov  Δ V  ion i = ( h AB  + h BA  )Δ V (  AA i +  BB i ) = h AB  Δ V  AA i + h AB  Δ V  BB i + h BA  Δ V  AA i + h BA  Δ V  BB i = 4 v 2 e R/a (1h) We can then use all of these integrals in conjunction with our orthonormal basis to express our Hamiltonian matrix,...
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 Spring '10
 CHandler
 Energy, wave function, covH cov

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