Problem Set 4 Solution

# thus it is clear that the left and right sides are

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Unformatted text preview: d2 dx € 2 e−α x 2 /2 ( ) = −α + α 2 x 2 e−α x 2 /2 . dx . 11 6 . continued Substituting, the expectation value becomes T =− 1/ 2 2 ȹ α ȹ ȹ ȹ 2 m ȹ π Ⱥ ∫−∞ e−α x =− T 1/ 2 2 ȹ α ȹ ȹ ȹ 2 m ȹ π Ⱥ ∫−∞ (−α + α 2 x 2 ) e−α x =− 1/ 2 2 ȹ α ȹ Ⱥ ȹ ȹ Ⱥ − α 2 m ȹ π Ⱥ Ⱥ ∞ 2 /2 22 ∞ ∞ − x2 / 2 (−α + α x ) e α ∫−∞ e−α x 2 dx 2 dx dx + α2 ∞ ∫−∞ x 2e−α x 2 Ⱥ dx Ⱥ . Ⱥ 2 Ⱥ dx Ⱥ Ⱥ From integral tables, the first integral is € ∞ ∫ ∞ 2 e−α x dx = 2 ∫ 0 −∞ 2 ȹ π ȹ1 / 2 e−α x dx = ȹ ȹ . ȹ α Ⱥ From integral tables, the second integral is € ∞ 2 −α x 2 ∫ xe ∞ dx = 2 ∫ 2 x 2 e−α x dx = 0 −∞ 1/ 2 1 ȹ π ȹ ȹ ȹ 2α ȹ α Ⱥ . Substituting these results into the expression for T we have, € T 1/ 2 2 ȹ α ȹ Ⱥ ȹ ȹ Ⱥ − α 2 m ȹ π Ⱥ€ Ⱥ 1/ 2 Ⱥ 2 ȹ α ȹ Ⱥ =− ȹ ȹ Ⱥ − α 2 m ȹ π Ⱥ Ⱥ Ⱥ =− 2 dx + α2 2 2m =− 2 ȹ α ȹ ȹ − ȹ 2 m ȹ 2 Ⱥ = 2α . 4m This result can be simplified further if desired, using the definition of α, € ȹ m k ȹ1 / 2...
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