851HW8_09Solutions - PHYS851 Quantum Mechanics I Fall 2009...

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PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 8: SOLUTIONS Topics Covered: Algebraic approach to the quantized harmonic oscillator, coherent states. Some Key Concepts: Oscillator length, creation and annihilation operators, the phonon number oper- ator. 1. Start from the harmonic oscillator Hamiltonian H = 1 2 M P 2 + 1 2 2 X 2 . Make the change of variables X λ ¯ X , P planckover2pi1 λ ¯ P , and H planckover2pi1 2 2 ¯ H . Find the value of λ for which ¯ H = 1 2 ( ¯ X 2 + ¯ P 2 ) . We find planckover2pi1 2 2 ¯ H = planckover2pi1 2 2 2 ¯ P 2 + 1 2 2 λ 2 ¯ X 2 (1) For all of the constants to cancel requires 2 λ 2 = planckover2pi1 2 2 (2) solving for λ gives λ = radicalbigg planckover2pi1 (3) 2. Write down the harmonic oscillator Hamiltonian in terms of ω , A , and A , and then write the com- mutation relation between A and A . Use these to derive the equation of motion for the expectation value a ( t ) = ( ψ ( t ) | A | ψ ( t ) ) . Solve this equation for the general case a (0) = a 0 . Prove that a ( t ) := ( A ) = [ a ( t )] . The Hamiltonian is H = planckover2pi1 ω ( A A + 1 / 2) (4) and the commutator is bracketleftBig A,A bracketrightBig = 1 (5) The equation of motion for the expectation value is d dt ( A ) = i planckover2pi1 ( [ A,H ] ) = ( A ) (6) With a ( t ) = ( A ) , the solution is a ( t ) = a (0) e iωt (7) For A , we find a ( t ) = ( A ) = ( ψ ( t ) | A | ψ ( t ) ) = ( ψ ( t ) | A | ψ ( t ) ) = ( A ) = [ a ( t )] (8) The point here is that we can say a ( t ) = [ a (0)] e iωt (9) without deriving a separate equation of motion for A . 1
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3. Starting from ( x | X | n 1 ) = n 1 ( x ), express X in terms of A and A , to derive a recursion relation of the form: φ n ( x ) = f n ( x ) φ n 1 ( x ) + g n ( x ) φ n 2 . (10) Starting from φ 0 ( x ) = [ πλ ] 1 / 2 e 1 2 ( x/λ ) 2 , use your recursion relation to compute φ 2 ( x ), φ 3 ( x ), and φ 4 ( x ). n 1 ( x ) = ( x | X | n 1 ) = λ 2 parenleftBig ( x | A | n 1 ) + ( x | A | n 1 ) parenrightBig = λ 2 ( n 1 ( x | n 2 ) + n ( x | n ) ) = λ n 1 2 φ n 2 ( x ) + λ n 2 φ n ( x ) (11) solving for φ n ( x ) gives φ n ( x ) = radicalbigg 2 n x λ φ n 1 ( x ) radicalbigg n 1 n φ n 2 ( x ) (12) with n = 1 , 2 , 3 , 4 this gives φ 1 ( x ) = 2 x λ φ 0 ( x ) = bracketleftbig π 2 λ bracketrightbig 1 / 2 2 parenleftBig x λ parenrightBig e 1 2 ( x/λ ) 2 (13) ψ 2 ( x ) = x λ ψ 1 ( x ) radicalbigg 1 2 ψ 0 ( x ) = 2 [ πλ ] 1 / 2 x 2 λ 2 e x 2 2 λ 2 1 [2 πλ ] 1 / 2 e x 2 2 λ 2 = 1 [2 2 2! πλ ] 1 / 2 parenleftbigg 4 x 2 λ 2 2 parenrightbigg e x 2 2 λ 2 ψ 3 ( x ) = radicalbigg 2 3 x λ ψ 2 ( x ) radicalbigg 2 3 ψ 1 ( x ) = radicalBigg 2 24 πλ parenleftbigg 4 x 2 λ 2 2 parenrightbigg x λ e x 2 2 λ 2 radicalBigg 4 3 πλ x λ e x 2 2 λ 2 = 1 [2 3 3!
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