hw3akl - PHYS 507 Answers to Homework III 1. (a) The Taylor...

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PHYS 507 Answers to Homework III 1. (a) The Taylor series is V 0 ( x ) = V 0 ( a ) + V 00 ( a )( x - a ) + 1 2 V 000 ( a )( x - a ) 2 + ··· and the expectation value is h V 0 ( x ) i = V 0 ( a ) + 1 2 V 000 ( a x 2 + ··· Including only the lowest order difference we have h V 0 ( x ) i- V 0 ( a ) V 000 ( a x 2 / 2 and the fractional error made is V 000 ( a x 2 2 V 0 ( a ) . If this last expression is much smaller than 1, it is reasonable to use the classical equation of motion for h x i t and errors introduced in doing so will be small. We can write this condition (fractional error much smaller than 1) as a bound on the uncertainty in x as follows. Δ x ¿ s 2 V 0 ( a ) V 000 ( a ) = pot . We have written the same equation in the class, where we have said that pot is a typical length scale for potential (the typical length over which potential changes appreciably), without saying how to calculate that length scale. Here we have an estimate of that scale. To repeat, if Δ x is much smaller than pot , it is all right to use classical mechanics. But if Δ x pot , then we have to use quantum mechanics. (b) Here we have pot = 2 a a = h x i . As a result, if Δ x ¿ h x i , classical equation of motion can be used. Same result can be obtained for the 3D Coulomb potential. For example, for the Earth we use classical mechanics but for the electron in the Hydrogen atom we use quantum mechanics. 2. (a) h B i n = h ϕ n | B | ϕ n i = h ϕ n | HA - AH | ϕ n i = h ϕ n | HA | ϕ n i - h ϕ n | AH | ϕ n i = ( E n h ϕ n | ) A | ϕ n i - h ϕ n | A ( E n | ϕ n i ) = ( E n - E n ) h ϕ n | A | ϕ n i = 0 . 1
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(b) [ T,A ] = 1 2 m 1 2 [ p 2 ,px + xp ] = 1 2 m 1 2 ( p [ p 2 ,x ] + [ p 2 ,p ] x + x [ p 2 ,p ] + [ p 2 ,x ] p ) = 1 2 m 1 2 ± p 2 ¯ h i p + 0 + 0 + 2 ¯ h i p · p = 1 2 m 1 2 · 4 ¯ h i p 2 = 2 ¯ h i 1 2 m p 2 = 2 ¯ h i T [ V,A ] = 1 2 [ V ( x ) ,px + xp ] = 1 2 ( p [ V ( x ) ,x ] + [ V ( x ) ,p ] x + x [ V ( x ) ,p ] + [ V ( x ) ,x ] p ) = 1 2 (0 + i ¯ hV 0 ( x ) x + x · i ¯ hV 0 ( x ) + 0) = i ¯ hxV 0 ( x ) . h [ H,A ] i n = 0 2 ¯ h i h T i n + i ¯ h h xV 0 ( x ) i n = 0 2 h T i n = h xV 0 ( x ) i n . (c) For V ( x ) = Ax k we have xV 0 ( x ) = kAx k = kV ( x ). As a result, the virial theorem says 2 h T i n = k h V i n for any bound eigenstate of the Hamiltonian. Since E n = h H i n = h T i n + h V i n we get h T i n = k k + 2 E n h V i n = 2 k + 2 E n . (i) For the harmonic oscillator we have
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This note was uploaded on 02/11/2012 for the course MATH 435 taught by Professor Starg during the Spring '11 term at Al Ahliyya Amman University.

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hw3akl - PHYS 507 Answers to Homework III 1. (a) The Taylor...

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