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12_variation_perturbation_matrixhq

# 2 e v 1 x 2 e v 104 looks just like our analytical

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Unformatted text preview: a quntum number of one, not zero. As a result, this expression adds one to the "i" and "j" indices to shift from the inex to the quantum number 2 H i j 2 hbar ( j 1) 2 m a ( i j) 2 ( i 1) x x sin ( j 1) x dx sin a a a 0.549 0.180 0.000 H 0.014 0.000 0.004 0.180 0.000 0.014 0.000 0.004 0.697 0.195 0.000 0.018 0.000 E eigenvals ( H) 0.195 0.944 0.199 0.000 0.020 0.000 0.020 0.000 0.201 2.277 0.000 0.199 1.290 0.200 0.000 Hamiltonian matrix is not diagonal, it must be diagonalized 0.018 0.000 0.200 1.734 0.201 0.400 0.701 0.976 E 1.314 1.753 2.348 Find Eigenvalues E sort ( E) they don't come sorted by energy Ramp Potential Find Eigenfunctions v c1 eigenvec H E v 1 c0 eigenvec H E c2 eigenvec H E Eigenvectors give the coefficients 0.747 0.614 0.246 c0 0.071 0.020 0.006 v 2 0.622 0.508 0.563 c1 0.191 0.048 0.012 coefficients and basis functions give the wavefunctions 0( x) i 1( x) a 0 Hamiltonian Matrix v 0 0 a 2 2 x 0.1 check normalization 2 ( i 1) x c0i sin a a a 2 0(...
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