Question

Quantum mechanics.

A Hamiltonian of the form $\dpi{100} \small H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2x^2-\epsilon x$ , is equivalent to the Hamiltonian of a harmonic oscillator with its equilibrium point displaced$\dpi{100} \small H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2(x-a)^2+C$ where $\dpi{100} \small a$ and C are constant, find them. With the previous result, find the exact spectrum of H. Calculate the same spectrum using the theory of disturbances to second order with $\dpi{100} \small H_l=-\epsilon x$ . Compare your results. Calculate the wave functions up to first order using $\dpi{100} \small H_l$ as a perturbation.

Answer 1

Hamiltorian of harmonic oscillator in presence of some field H = P^2/2m + 1/2 mw^2 x^2 - E x H = P^2/2m + 1/2 mw^2 (x^2 = 2E/mw^2 x) H = P^2/2m + 1/2 mw^2 (x - E/mw^2)^2 - 1/2 mw^2 times E^2/m^2 w^4 H = P^2/2m + 1/2 mw^2 (x - E/mw^2)^2 - E^2/2mw^2 comparison with H = P^2/2m + 1/2 mw^2 (x - a)62 + c a = E/mw^2 and C = -E^2/2 mw^2 solution of one dimensional harmonic oscillator E_n = (n + 1/2) planckconstantovertwopi w = E^2/2mw^2 wave function up to first order using perturbation Vertical-bar n_1 > = summation�_m < m_0 Vertical-bar H_1 Vertical-bar m > /E^(0)_m - E^(0)_m - (1) < m_0 Vertical-bar H_1 Vertical-bar n_0 > = -E < m_0 Vertical-bar x Vertical-bar n_0 > \r\n = -E squareroot planckconstantovertwopi/2mw < m_0 Vertical-bar a + a^t Vertical-bar n_0 > = -E squareroot planckconstantovertwopi/2mw [squareroot n < m_0 Vertical-bar n_0 - 1 > + squareroot n + 1 < m_0 Vertical-bar n_0 + 1 >] < m_0 Vertical-bar H_1 Vertical-bar n_0 > = -E squareroot planckconstantovertwopi /2mw [squareroot n delta_m_0, n_0 - 1 + squareroot n + 1 delta_m_0, n_0 + 1] - (2)