# hws02_r - Homework 2 Solutions TA Kevin Chiou(original...

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Homework 2 Solutions TA: Kevin Chiou (original solutions by Matt Block) Problem 1 We wrote down ψ 0 and ψ 1 in the notes. I’ll reproduce them here, where again α 0 ¯ h : ψ 0 ( x ) = α π 1 / 4 e - α 2 x 2 (1) ψ 1 ( x ) = 4 α 3 π ! 1 / 4 xe - α 2 x 2 (2) First, we will calculate h p 0 i and h p 1 i for ψ 0 and ψ 1 respectively: h p n i = Z -∞ ψ n ( x n ( x ) dx, ˆ p = ¯ h i d dx (3) h p 0 i = Z -∞ ( α π ) 1 2 ¯ h i ( - αx ) e αx 2 (4) h p 1 i = Z -∞ ( 4 α 3 π ) 1 2 ¯ h i e αx 2 ( x - αx 3 ) (5) Note that the integrals are all odd over an even interval, and therefore are zero. Therefore: h p 0 i = h p 1 i = 0 (6) To calculate h p 2 i we have the following expressions: h p 2 n i = Z -∞ ψ n ( x p 2 ψ n ( x ) dx, ˆ p 2 = - ¯ h 2 d 2 dx 2 (7) h p 2 0 i = Z -∞ ( α π ) 1 2 ( - ¯ h 2 ) e αx 2 ( - α + α 2 x 2 ) (8) h p 2 1 i = Z -∞ ( 4 α 3 π ) 1 2 ( - ¯ h 2 ) e αx 2 x 2 ( - 3 α + α 2 x 2 ) (9) Now, to help us do the integrals, notice the following facts: Z -∞ e - αx 2 = r π α (10) Z -∞ x 2 e - αx 2 = - ∂α Z -∞ e - αx 2 (11) - ∂α Z -∞ e - αx 2 = - ∂α r π α (12) Z -∞ x 2 e - αx 2 = 1 2 r π α 3 (13) 1

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In calculating R -∞ x 4 e - αx 2 one can simply do the previous trick twice. This is left as an exercise to the student. We will simply quote the result here: Z -∞ x 4 e - αx 2 = 3 4 r π α 5 (14) Putting all parts of the integrals together, we get the following results: h p 2 0 i = 1 2 ¯ hmω (15) h p 2 1 i = 3 2 ¯ hmω (16) Problem 2
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