12_variation_perturbation_matrixhq

444 e 0790 1234 1777 find eigenvalues e sort

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Unformatted text preview: H E c2 eigenvec H E v Eigenvectors give the coefficients v 1 1.000 0.000 0.000 c0 0.000 0.000 0.000 v 2 0.000 1.000 0.000 c1 0.000 0.000 0.000 coefficients and basis functions give the wavefunctions 0( x) c0i 2 a 1( x) ( i 1) x a a 2 0( x) dx 1.000 0 c1i 2 a ( i 1) x a 0 a each eigenfunction has only one coefficient, i.e. that basis function is the eigenfunction a 2 1( x) dx 1.000 sin i 2( x) check normalization sin i 0.000 0.000 1.000 c2 0.000 0.000 0.000 0 c2i 2 a ( i 1) x a a 2 2( x) dx 1.000 sin i 0 0.6 x 0( x) .2 E v 1( x) .2 E v 10.4 Looks just like our analytical solution 2( x) .2 E v 2 Ev E v 1 0.2 E v 2 0 0 2 4 6 8 10 x 13 10/7/2013 Ramp Potential nmax 6 pick the max number of basis functions i 0 nmax 1 set up indices over the basis functions j 0 nmax 1 m 1 a 10 set constants for problem, atomic units hbar 1 set ramp to nonzero value Hamiltonian Matrix Elements - Note that the eigenvector and eigenfunction routines need the origin of matrices to be zero. Our first wavefunction has...
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