Problem Set 4 Solution

Recall that the wavefunctions a re 0 x 2 1 4

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Unformatted text preview: . 4 a2 = or a 0 € 4 5 2 . continued Substituting a 2 = −48, we have 1 a2 4 1 = − ⋅ (−48) 4 = 12 . a0 = − € a0 Since H 4 (Q) = a 4 Q 4 + a 2 Q 2 + a 0 = 16Q 4 − 48 Q 2 + 12 , we see that the coefficients of the v=4 Hermite polynomial given are correct. € € 3 . Show b y e xplicit i ntegration t hat t he g round s tate h armonic o scillator w avefunction, ψ 0 ( x ) , i s o rthogonal t o t he f irst e xcited s tate w avefunction, ψ1 ( x ) . Recall that the wavefunctions a re € ψ 0 ( x) 2 ȹ α ȹ1 / 4 = ȹ ȹ e −α x / 2 ȹ π Ⱥ For orthogonality, we must show that € ∞ ∫−∞ ȹ 4α 3 ȹ1 / 4 € 2 ψ1( x ) = ȹ ȹ x e −α x / 2 . π Ⱥ ȹ and ∞ * ∫−∞ ψ0 ( x) ψ1( x) dx = 0 . Substituting, 1/ 4 * ψ0 ȹ α ȹ1 / 4 ȹ 4α 3 ȹ x (€) ψ1( x) dx = ȹ ȹ ȹ ȹ ȹ π Ⱥ ȹ π ȹ ȹ Ⱥ ȹ 2α 2 ȹ1 / 2 = ȹ ȹ π ȹ ȹ ȹ Ⱥ ∞ ∞ ∫−∞ e−α x 2 /2 x e−α x 2 /2 dx 2 x e−α x dx . ∫−∞ Since the integrand is an odd function, the integral must be zero, so € ∞ * ∫−∞ ψ0 ( x) ψ1( x) dx = 0. Alternately...
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