lecture12_sup1

# lecture12_sup1 - 5.61 Fall 2007 Page 1 12-15 Lecture...

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5.61 Fall 2007 12-15 Lecture Supplement Page 1 Harmonic Oscillator Energies and Wavefunctions via Raising and Lowering Operators We can rearrange the Schrödinger equation for the HO into an interesting form . .. 2 + ( m ω x ) 2 ψ = 1 2 m p 2 + ( m x ) 2 = E 1 ! d 2 m i dx with H = 1 p 2 + ( m x ) 2 2 m which has the same form as u 2 + v 2 = ( iu + v )( iu + v ) . We now define two operators a ± 1 2 ! m ip + m x ( ) that operate on the test function f(x) to yield ( a a + x 1 ( ip + m x ip + m x ) f ( ) ) f ( ) = x 2 ! m = 1 p 2 + ( m x ) 2 i m ( xp px ) f ( x ) 2 ! m ( p 2 + ( m x ) 2 ) [ x , p } f ( ) = { 2 ! 1 m 2 i ! ] x a a + = 1 ( p 2 + ( m x ) 2 ) + 2 1 = ! 1 2 1 H + 2 ! m Which leads to a new form of the Schrödinger equation in terms of a + and a - H = ! a a + 1 2 If we reverse the order of the operators-- a a + a + a -- we obtain …

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5.61 Fall 2007 12-15 Lecture Supplement Page 2 H ψ = ! ω a + a + 1 2 or ! a ± a ± 1 2 = E and the interesting relation a a + a + a = [ a a ] = 1 + A CLAIM: If satisfies the Schrödinger equation with energy E , then a + satisfies it with energy (E+ ! ω ) ! H ( a + ) = ! a + a + 2 1 ( a + ) = ! a + a a + + 2 1 a + = ! a + a a + + 1 2 = a + ! a + a + 1 + 1 2 ! a + a + 1 2 + ! = a + = a + ( H + ! ) = ( E + ! ) ( a + ) H ( a + ) = ( E + ! ) ( a + ) Likewise, - satisfies the Schrödinger equation with energy (E- ! ω ) … H ( a ) = ! a a + 1 2 ( a ) = ! a a + a 1 2 a = a ! a + a 1 2 = a ! a a + 1 2 1 = a ( H !
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## This note was uploaded on 09/24/2011 for the course MATH 1090 taught by Professor Greenwood during the Spring '08 term at MIT.

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lecture12_sup1 - 5.61 Fall 2007 Page 1 12-15 Lecture...

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