Use the commutator results, [ŵ, ô] = iħ, [x²,p] = 2iħâ, and [ÂÂ, Ĉ] = Â[B, Ĉ] + [‚Ĉ]Â, to find the commutators given below. (a) [x, p²] (b) [x³,p] (c) [x², p²]
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A: we can use the properties of commutator brackets.
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- Using the result from part (a), evaluate = (^²+x²) y(x). (choose one) ( 1 + x²) √(x) y (x) X ( 121 - 12/27) 1 (X) y(x) ○ (1-x²) y(x) ○ (1+x²) y(x) 2 (1+77) w(0) y(x) (1-27) (0) y(x) 1 2 + X 2 2 y(x)Show that the power law relationship P(Q) = kQr, for Q > 0 and r # 0, has an inverse that is also a power law, Q(P) = mP s, where m = k - l/r and s = 1/ r.Calculate the divergences of cach of the following vectors: (a) v = 3k (b) v = r (c) = (4xz + y?) î + (12a² – 2²) ĵ + (xy – yz) k (d) ở = -yî+ x} (c) v = . -
- If | jm > is an eigenstate of 1², then its eigenvalue is j(j+1)ħ², Z True FalseLet n 1 be an integer, let to to is given by a stationary path of the Lagrangian functional C: L[x] = 1 dt L(t,x,x), x(0) =x0, x(t)=X1, where LT - V and T is the total kinetic energy T = n 1 k=1 2 mark. Using the above first-integral, show that, if V is independent of t, the total energy E=T+V of the particle is a constant of the motion.