University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

468

Pages:

5
Author:

Jabin F.

#Hamilton EOM These equations are obtained from the Legendre transformation of Lagrange or by extremization of action. Canonical equations of motion are Hamilton's EOM. Lagrange's EOM and Hamilton's EOM give the same final result. P-68: Lagrangian of a system L(x,ẋ) = (1/2)mẋ² + (1/2)kx² - (1/2)λe^(x²/λ) Q-14: Equation of motion is: d/dt(∂L/∂ẋ) - ∂L/∂x = 0 Solution: d/dt(∂L/∂ẋ) - ∂L/∂x = 0 d/dt(mx - λxe^(x²/λ)) - (kx - λxe^(x²/λ)) = 0 mẍ - λ(e^(x²/λ) + 2x²e^(x²/λ)) - kx + λxe^(x²/λ) = 0 mẍ - λe^(x²/λ) - 2λx²e^(x²/λ) - kx + λxe^(x²/λ) = 0 mẍ - λe^(x²/λ) - λx²e^(x²/λ) - kx = 0 ẍ - (λ/m)e^(x²/λ) - (λ/m)x²e^(x²/λ) - (k/m)x = 0 ẍ + (λ/m)e^(x²/λ) + (λ/m)x²e^(x²/λ) + (k/m)x = 0 ẍ + (λ/m)(1 + x²)e^(x²/λ) + (k/m)x = 0 P-102 Let q and p be canonical coordinate and momentum of a dynamical system. Which of the following transformation is canonical? a) Q₁ = (1/√2)(q + p), P₁ = (1/√2)(q - p) b) Q₂ = (1/√2)(q + ip), P₂ = (1/√2)(p - iq) Solution: a) {Q₁, P₁} = (∂Q₁/∂q)(∂P₁/∂p) - (∂Q₁/∂p)(∂P₁/∂q) = (1/2)(1/2) - (1/2)(1/2) = 0, not canonical transformation. b) {Q₂, P₂} = (∂Q₂/∂q)(∂P₂/∂p) - (∂Q₂/∂p)(∂P₂/∂q) = (1/√2)(1/√2) - (i/√2)(-i/√2) = 1, canonical transformation. #Hamiltonian of a quantum mechanical SHO: H = (p²/2m) + (1/2)mω²x² What is the eigen energy of the system? Solution: Making perfect square: H = (1/2m)(p² + m²ω²x²) H = (1/2m)(p + imωx)(p - imωx) Canonical transformation: Put p = imωx H = (1/2m)(imωx + imωx)(imωx - imωx) E = (n + 1/2)ħω How canonical transformation? Put p = imωx H = (p²/2m) + (1/2)mω²x² E = (n + 1/2)ħω V(x,y,z) = (1/2)k(x² + y² + z²) is 3D SHO in QM Then E = ħω(n₁ + n₂ + n₃) + (3/2)ħω where (ω) → frequency of normal mode