Question

Uncertainty Principle

Evaluate, by considering Eq. (9.2-3), the limitation on the simultaneous specification of the following observables for a one-dimensional system (a) the kinetic energy and the position (b) the kinetic energy and the potential energy (c) the electric dipole moment and the total energy

Answer 1

(Delta A) (Delta B) greaterthanorequalto 1/2 Vertical-bar < [A^cap, B^cap] > Vertical-bar A = K = p^2/2m, B = x (Delta K) (Delta X) greaterthanorequalto 1/2 Vertical-bar < [K^cap, X^cap] > Vertical-bar = 1/2 Vertical-bar < [p^cap^2/2m, x^cap] > Vertical-bar = 1/2 1/2m Vertical-bar < p^cap[p^cap, x^cap] + [p^cap, x^cap] + [p^cap, x^cap] p^cap > Vertical-bar = 1/4m Vertical-bar p^cap(-i planck) + (-i planck) p^cap Vertical-bar = 1/4m Vertical-bar p^cap (-i planck) + (-i planck) p^cap Vertical-bar = 1/4m Vertical-bar -2 i planck p^cap Vertical-bar Vertical-bar 0 - i 2 planck Vertical-bar = squareroot 0^2 + (-2 planck)^2 = 2 planck Therefore (Delta K)(Delta X) greaterthanorequalto planck /2m p A = K = p^2/2m, B = V (Delta K)(Delta V) = 1/2 Vertical-bar < p^cap^2/2m, V^cap] > = 1/4m Vertical-bar p^cap [p^cap, v] - [p^cap, v] p^cap = 0 Therefore (Delta K)(Delta V) lessthanorequalto 0 A = p = qx, B = E = p^2/2m + v (Delta p)(Delta x) = 1/2 Vertical-bar < qx, p^2/2m + V] > Vertical-bar = q/2 [X, p^2/2m] + [X, V] 2q planck p/4m = q planck p/2m Therefore (Delta p)(Delta x) greaterthanorequalto q planck p/2m