851HW8_09Solutions

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 Mω 2 X 2 . Make the change of variables X → λ ¯ X , P → planckover2pi1 λ ¯ P , and H → planckover2pi1 2 Mλ 2 ¯ H . Find the value of λ for which ¯ H = 1 2 ( ¯ X 2 + ¯ P 2 ) . We find planckover2pi1 2 Mλ 2 ¯ H = planckover2pi1 2 2 Mλ 2 ¯ P 2 + 1 2 Mω 2 λ 2 ¯ X 2 (1) For all of the constants to cancel requires Mω 2 λ 2 = planckover2pi1 2 Mλ 2 (2) solving for λ gives λ = radicalbigg planckover2pi1 Mλ (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 . 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 ] ) = − iω ( 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 3. Starting from ( x | X | n − 1 ) = xφ 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 ( 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 ). xφ 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 ( 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 ( 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!...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online