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Unformatted text preview: CHEM 120A Problem Set 8 Solutions David Hoffman , Doran Bennett , Tara Yacovitch 4/12/2010 1. We’ll begin this problem by noting that the easiest way to solve this problem is to make use of matrices. Our plan of action will be to form a Hamiltonian matrix using | cov i = | AB i + | BA i and | ion i = | AA i + | BB i 1 as our basis kets (vectors). We can diagonalize H whose matrix elements will be functions of R and find two eigenvalues (which will also be functions of R ) and two eigenkets (eigenvectors) whose coefficients in our original basis will be functions of R . Let’s break this problem into smaller pieces. We know that H can be written as, H = T (1) + T (2) + V (1) + V (2) + Δ V (1 , 2) From symmetry we know that T (1) = T (2) and V (1) = V (2) so we can determine the matrix elements of T (1), V (1) andΔ V and then sum them to find H . Before we start determining matrix elements it is important that we first normalize our kets. h cov | cov i = ( h AB | + h BA | )( | AB i + | BA i ) = h AB | AB i + h AB | BA i + h BA | AB i + h BA | BA i = 2 + 2 e- 2 R/a (1) Similarly we find that, h ion | ion i = 2 + 2 e- 2 R/a (2) This means that we can divide our entire matrix by 2 + 2 e- 2 R/a after determining the matrix elements. Finding the matrix elements is a task that is simple in principle but can get confusing in practice. All four matrix elements for Δ V (1 , 2) will be derived because they are the most complicated. The matrix elements of T and V will be left as an exercise for the reader. 1 I’m using the notation that | AB i is equivalent to | φ A (1) φ B (2) i . 1 Starting with the on diagonal 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 (3) 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 1 + R/a v 2 (4) Now on to the off diagonal terms. Remember that H is hermitian so we only need to calculate one of the elements because the other will be the same. 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 (5) Finally we have, H ( R ) = 2 T ( R ) + 2 V ( R ) + ΔV ( R ) 2 + 2 e- 2 R/a (6) Where the bold indicates that we are talking about matrices and not operators. For the rest of this question you’ll need either a math program like Mathematica, Maple or MatLab or a spreadsheet program like Excel. Examples of how to solve this question with Excel and Mathematica are posted with this solution set....
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This note was uploaded on 05/14/2010 for the course CHEM 120ACHEM taught by Professor Chandler during the Spring '10 term at Berkeley.
- Spring '10