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

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Unformatted text preview: 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 M 2 X 2 . Make the change of variables X X , P planckover2pi1 P , and H planckover2pi1 2 M 2 H . Find the value of for which H = 1 2 ( X 2 + P 2 ) . We find planckover2pi1 2 M 2 H = planckover2pi1 2 2 M 2 P 2 + 1 2 M 2 2 X 2 (1) For all of the constants to cancel requires M 2 2 = planckover2pi1 2 M 2 (2) solving for gives = radicalbigg planckover2pi1 M (3) 2. Write down the harmonic oscillator Hamiltonian in terms of , 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 ) = ( ( t ) | A | ( t ) ) . Solve this equation for the general case a (0) = a . Prove that a ( t ) := ( A ) = [ a ( t )] . The Hamiltonian is H = planckover2pi1 ( 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 ] ) = i ( A ) (6) With a ( t ) = ( A ) , the solution is a ( t ) = a (0) e it (7) For A , we find a ( t ) = ( A ) = ( ( t ) | A | ( t ) ) = ( ( t ) | A | ( t ) ) = ( A ) = [ a ( t )] (8) The point here is that we can say a ( t ) = [ a (0)] e it (9) without deriving a separate equation of motion for A . 1 3. Starting from ( x | X | n 1 ) = x n 1 ( x ), express X in terms of A and A , to derive a recursion relation of the form: n ( x ) = f n ( x ) n 1 ( x ) + g n ( x ) n 2 . (10) Starting from ( x ) = [ ] 1 / 2 e 1 2 ( x/ ) 2 , use your recursion relation to compute 2 ( x ), 3 ( x ), and 4 ( x ). x n 1 ( x ) = ( x | X | n 1 ) = 2 parenleftBig ( x | A | n 1 ) + ( x | A | n 1 ) parenrightBig = 2 ( n 1 ( x | n 2 ) + n ( x | n ) ) = n 1 2 n 2 ( x ) + n 2 n ( x ) (11) solving for n ( x ) gives n ( x ) = radicalbigg 2 n x n 1 ( x ) radicalbigg n 1 n n 2 ( x ) (12) with n = 1 , 2 , 3 , 4 this gives 1 ( x ) = 2 x ( x ) = bracketleftbig 2 bracketrightbig 1 / 2 2 parenleftBig x parenrightBig e 1 2 ( x/ ) 2 (13) 2 ( x ) = x 1 ( x ) radicalbigg 1 2 ( x ) = 2 [ ] 1 / 2 x 2 2 e x 2 2 2 1 [2 ] 1 / 2 e x 2 2 2 = 1 [2 2 2!...
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851HW8_09Solutions - PHYS851 Quantum Mechanics I, Fall 2009...

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