11. Suppose a particle of mass m moves in one spatial dimension, subject to the

Question

11. Suppose a particle of mass m moves in one spatial dimension, subject to the potential energy V (x)m2r2. In a stationary state lW) of energy E: (a) Show that (X)0 and (P) 0 (b) Show that AXAP E How does this compare to the limit set by the uncertainty principle? (c) Show that ?K)-(V). Compare to the predictions of the virial theorem (proved on last week's problem set). (d) Show that ???? > ( )2, where ?? and ?? represent the standard deviations of the kinetic and potential energy, respectively

Explanation / Answer

11. given

potential Energy function, V(x) = 0.5mw^2*x^2

KE = 0.5mv^2

now, for a stationary state

total energy

E = KE + PE = 0.5mw^2x^2 + 0.5mv^2

also, v = x'

a. <v> = integral(v*dt)/integral(dt) from t = 0 to t = T

here T = 2*pi/w

E = 0.5mw^2x^2 + 0.5mv^2

dE/dt = 0

0 = mw^2*x*v + mvv'

w^2*x = - mv'

the solution to this is

x = Asin(wt)

hence

v = Awcos(wt)

<v> = A(sin(2pi) - sin(0))/(2*pi/w - 0) = 0

and

P = mv = mAw*cos(wt)

hence

<P> = integral(mv*dt)/integral(dt) = m<v> = 0

b. now, x = Asin(wt)

dx = Acos(wt)*w*dt

P = mAsin(wt)

dP = mAcos(wt)w*dt

dx*dP = mA^2*w^2cos^2(wt)*dt^2

now, E = 0.5mw^2*x^2 + 0.5mv^2 = 0.5mw^2A^2*sin^2(wt) + 0.5mA^2*w^2*cos^2(wt) = 0.5mw^2*A^2

E/w = 0.5mw*A^2

dx*dP/(E/w) = 2*wcos^2(wt)*dt^2

now, dX = dx/dt

dP = dp/dt

hence

dX*dP / (E/w) = 2*w*cos^2(wt)

the average value of the expression on the right is more than 1

hence

dX*dP > = E/w

From uncertianity principle

dX*dP > = h/2*pi

c. <K> = <0.5mv^2> = 0.5m<v^2> = 0.5m(integral(v^2*dt)/integral(dt)) = 0.5mw^2*A^2

similiarly

<V> = <0.5mw^2*A^2*sin^2(wt)> = 0.5mw^2A^2

hence <V> = <K>

d. dK = mv*dv

dV = mw^2*xdx

dK*dV = m^2w^2*vxdxdv = (dP*dx)^2*w^2 > (hw/2*pi)^2/4

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