By David J. Griffiths

Options handbook for creation to Quantum Mechanics

**Read Online or Download Solutions Manual for Introduction to Quantum Mechanics (2nd edition) PDF**

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**Extra resources for Solutions Manual for Introduction to Quantum Mechanics (2nd edition)**

2. sixty seven, 1 1 ψ2 = √ (a+ )2 ψ0 = √ 2 2 (b) 1/4 mω π 2mω x2 − 1 e− 2 ψ0 mω π =− mω 2π =− mω 2π x2 . ψ1 (c) considering the fact that ψ0 and ψ2 are even, while ψ1 is peculiar, we have to payment is ψ2∗ ψ0 dx: 1 ψ2∗ ψ0 dx = √ 2 mω ∞ 2mω −∞ ∞ e− mω ψ2 ψ0∗ ψ1 dx and x2 − 1 e− x2 −∞ dx − mω dx ∞ 2mω π 2mω − mω 2mω x2 ψ2∗ ψ1 dx vanish instantly. the one one x2 e− mω x2 dx −∞ π mω = zero. challenge 2. eleven (a) observe that ψ0 is even, and ψ1 is abnormal. In both case |ψ|2 is even, so x = x|ψ|2 dx = zero. as a result p = md x /dt = zero. (These effects carry for any desk bound nation of the harmonic oscillator. ) √ 2 2 From Eqs. 2. fifty nine and a couple of. sixty two, ψ0 = αe−ξ /2 , ψ1 = 2αξe−ξ /2 . So n = zero: 2 x =α 2 ∞ 2 −ξ 2 /2 x e −∞ 3/2 dx = α 2 mω ∞ 2 −ξ 2 ξ e −∞ 1 dξ = √ π √ mω π = . 2 2mω c 2005 Pearson schooling, Inc. , top Saddle River, NJ. All rights reserved. This fabric is secure lower than all copyright legislation as they at the moment exist. No part of this fabric can be reproduced, in any shape or whatsoever, with no permission in writing from the writer. ¨ bankruptcy 2. THE TIME-INDEPENDENT SCHRODINGER EQUATION p2 = 2 d i dx ψ0 ∞ m ω =− √ π −ξ 2 /2 ξ −1 e 2 −∞ ∞ mω ψ0 dx = − 2 α2 −∞ e−ξ 2 d2 −ξ2 /2 dξ e dξ 2 /2 √ π √ − π 2 m ω dξ = − √ π 21 = m ω . 2 n = 1: x2 = 2α2 ∞ 2 mω −∞ ∞ mω p2 = − 2 2α2 ξe−ξ 2 ∞ ξ four e−ξ dξ = √ 2 −∞ 2 d2 ξe−ξ /2 dξ 2 /2 −∞ 2mω =− √ π ∞ 3/2 x2 ξ 2 e−ξ dx = 2α2 dξ √ π three√ π−3 four 2 2 2mω ξ four − threeξ 2 e−ξ dξ = − √ π −∞ √ 2 three three π = . 2mω πmω four = 3m ω . 2 (b) n = zero: σx = σx σ p = x2 − x 2 = 2mω p2 − p ; σp = 2 = m ω ; 2 mω = . (Right on the uncertainty restrict. ) 2 2 2mω n = 1: σx = three ; 2mω σp = 3m ω ; 2 σx σp = three 2 > 2 . (c) T = 1 2 p = 2m 1 four three four T + V = H = ω (n = zero) ω (n = 1) ; V = 1 2 ω (n = zero) = E0 three 2 ω (n = 1) = E1 1 mω 2 x2 = 2 1 four ω (n = zero) three four ω (n = 1) . , as anticipated. challenge 2. 12 From Eq. 2. sixty nine, x= 2mω (a+ + a− ), p=i mω (a+ − a− ), 2 so x = 2mω ψn∗ (a+ + a− )ψn dx. c 2005 Pearson schooling, Inc. , top Saddle River, NJ. All rights reserved. This fabric is safe lower than all copyright legislation as they at present exist. No element of this fabric might be reproduced, in any shape or in any way, with out permission in writing from the writer. ¨ bankruptcy 2. THE TIME-INDEPENDENT SCHRODINGER EQUATION 22 yet (Eq. 2. sixty six) a+ ψn = So √ x = p =m x2 = 2mω dx = zero. dt 2mω 2mω n + 1ψn+1 , ψn∗ ψn+1 dx + n+1 x2 = √ (a+ + a− )2 = √ a− ψn = nψn−1 . ψn∗ ψn−1 dx = zero (by orthogonality). n 2mω √ a2+ + a+ a− + a− a+ + a2− . ψn∗ a2+ + a+ a− + a− a+ + a2− ψn . yet 2 a ψn = a+ a+ a ψ = a + − n + a a ψ = a + n − − a2− ψn = a− √ √ + 1 n + 2ψn+2 = √ n√ = n nψn √ = n + 1) n + 1ψn √ √ = n n − 1ψn−2 √ √n + 1ψn+1 nψ √ n−1 n + 1ψn+1 √ nψn−1 = (n + 1)(n + 2)ψn+2 . = nψn . = (n + 1)ψn . = (n − 1)nψn−2 . So x2 = p2 = − 2mω |ψn |2 dx + zero = 2mω (2n + 1) = n+ 1 . 2 mω mω mω 2 (a+ − a− )2 = − a+ − a+ a− − a− a+ + a2− ⇒ 2 2 p2 = − mω mω [0 − n − (n + 1) + zero] = (2n + 1) = 2 2 1 1 n+ 2 2 T = p2 /2m = σx = |ψn |2 dx + (n + 1) 0+n x2 − x 2 = n+ n+ 1 m ω.