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The first can be evaluated as follows 2 1 6 24 6 6 6 2

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Unformatted text preview: t should be 2 × 2: − − 0= − − Knowing that the HO solutions are an orthogonal set allows us to simplify the determinant, because = → = = 1 and = =0 − ∴0= − However, the integrals must still be evaluated. The first, , can be evaluated as follows: = = = − ℏ 2 + − + = + 1+ 6 + 24 6 + 6 + 6 24 24 + 24 In this expression, and are the expected values of and for the product × , respectively. Using Mathcad to evaluate the two integrals gives = 0 (this can also be seen by inspection, since the integrand is odd and the integral is taken over all space) 3 = 4 1 3 ℏ ∴ = + + =ℏ 0+ + 0 + = + 6 24 2 6 24 4 2 32 The remaining integrals can be evaluated in similar fashion: = 6 + 2 24 = = + + 2 + ℏ 2 = − ℏ 2 + 2 + 6 + 24 − = ℏ 2 + + + = + 0+ =0 = ∴ = 3 + 6 6 = 6 0+ = = − ℏ 2 − + + = + 0+ =0 = = 6 3 + 6 3 ℏ 2 = 2 + 8 2 + 2 + 6 + 24 6 + 6 + 6 = 24 = 24 24 24 2 + 24 24 = = 24 2 + = + 6 24 24 24 = ∴ + 6 2 + 6 24 = = + 2 6 0+ − 24 ℏ 2 3 = 2 + 2 8 +...
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