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Unformatted text preview: Homework 2 Solutions TA: Kevin Chiou (original solutions by Matt Block) Problem 1 We wrote down ψ and ψ 1 in the notes. I’ll reproduce them here, where again α ≡ mω ¯ h : ψ ( 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 i and h p 1 i for ψ and ψ 1 respectively: h p n i = Z ∞∞ ψ n ( x )ˆ pψ n ( x ) dx, ˆ p = ¯ h i d dx (3) h p 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 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 )ˆ p 2 ψ n ( x ) dx, ˆ p 2 = ¯ h 2 d 2 dx 2 (7) h p 2 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 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...
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This note was uploaded on 03/01/2009 for the course PHYS 115A taught by Professor Nelson during the Spring '03 term at UCSB.
 Spring '03
 Nelson
 mechanics, Work

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