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410HW8 - dG = m X i =1 G u i du i m X i =1 G u i du i G x...

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1 Homework #8 Consider the Hamiltonian system for the harmonic oscillator. The Hamiltonian is H ( x, π ) = π 2 2 m + kx 2 2 The corresponding system of differential equations is dx dt = ∂H ∂π = x m (1) dt = - ∂H ∂x = - kx (2) (3) 1. x * = x + π, π * = π . dx * dt = dx dt + dt = π m - kx = π * m - k ( x * - π * ) (4) * dt = dt = - kx = k ( π * - x * ) (5) 2. H * ( x * , π * ) = ( π * ) 2 2 m + k ( x * - π * ) 2 2 ∂H * ∂π * = π * m - k ( x * - π * ) = dx * dt (6) ∂H * ∂x * = k ( x * - π * ) = - * dt (7) Therefore, we get the same set of differential equations as in (1), so the transformation preserves the form of the Hamiltonian equation. 3. x * = e x , π * = π . dx * dt = e x dx dt = x * π m = x * π * m (8) * dt = dt = - kx = - k ln( x * ) (9) 1
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4. H * ( x * , π * ) = ( π * ) 2 2 m + k (ln( x * )) 2 2 ∂H * ∂π * = π * m = dx * dt (10) ∂H * ∂x * = k ln( x * ) x * = - * dt (11) which are different than the equations in (3). Therefore the transformation is not canonical. 5. x * = e x , we can set ˜ G = ˜ G ( x, π * ). By case 2 in class, we have x * = - ˜ G π * = e x (12) ˜ G = - e x π * (13) π = - ˜ G x = e x π * (14) π * = πe - x (15) Thus, with x * = e x , π * = πe - x is a canonical transformation.
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Unformatted text preview: dG = m X i =1 G u i du i + m X i =1 G u * i du * i + G x dx (16) But since G x = 0 = ⇒ H * = H . Thus, dG = m X i =1 π * i du * i-m X i =1 π i du i (17) = m X i =1 π * i du * i-m X i =1 π i du i + ( H *-H ) (18) which gives us the transformation is canonical. 7. u * = √ u cos(2 π ) , π * = √ u sin(2 π ). π * du *-πdu = ( 1 4 sin(4 π )-π ) du-2 u sin 2 (2 π ) dπ (19) := F 1 du + F 2 dπ (20) 2 We can check that ∂F 1 ∂π = ∂F 2 ∂u , therefore this is an exact differential and we can solve for G such that dG = π * du *-πdu . Specifically, G = u 4 sin(4 π )-uπ. Thus, by problem (6), this is a canonical transformation. 3...
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