3890ps7_10

# 3890ps7_10 - Chem 3890 Physical Chemistry I Fall 2010...

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Unformatted text preview: Chem 3890 Physical Chemistry I Fall 2010 Chemistry 3890 Problem Set 7 Fall 2010 Due: in class, Friday, October 29 Last revised: October 22, 2010 Problem 1 The Hermite polynomials H n [ ξ ] (Griffith, Sec. 2.3, see also http://mathworld.wolfram.com/ HermitePolynomial.html ) satisfy the following relations ( n a non-negative integer) H prime n = 2 nH n- 1 (1a) H n +1 = 2 ξH n- 2 nH n- 1 (1b) where H prime n ≡ d H n / d ξ . 1. Using these relations, show that the Hermite polynomials satisfy Hermite’s equation: H primeprime n- 2 ξH prime n + 2 nH n = 0 . (2) Normalized HO eigenstates are (McQ&S, § 5.6) φ n ( x ) = bracketleftbigg √ α 2 n n ! √ π bracketrightbigg 1 2 H n ( √ α x ) e- αx 2 / 2 , n = 0 , 1 , 2 , . . . (3) where α ≡ mω/ planckover2pi1 . 2. Use the above information to derive expressions for the matrix elements ( n prime | ˆ x | n ) ≡ integraldisplay + ∞-∞ d x φ * n prime ( x )ˆ xφ n ( x ) (4a) and ( n prime | ˆ p | n ) ≡ integraldisplay + ∞-∞ d x φ * n prime ( x )ˆ pφ n ( x ) (4b) for general n and n prime (non-negative integers). 3. Evaluate the expectation values ( ˆ x ) n ≡ ( φ n | ˆ x | φ n ) , ( ˆ x 2 ) n , ( ˆ p ) n , and ( ˆ p 2 ) n . 4. Show that ( ˆ T ) n = ( ˆ V ) n , where ˆ T is the kinetic energy operator and ˆ V is the HO potential operator....
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## This note was uploaded on 08/25/2011 for the course CHEM 3890 at Cornell.

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3890ps7_10 - Chem 3890 Physical Chemistry I Fall 2010...

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