Question

a. Suppose that at time ta the state function of a one particle system is Ψ = (2/πc2)3/4 e(exp [– (x2 + y2 + z2)/c2)] where c = 2 nm. Find the probability that a measurement of the particle’s position at ta will find the particle in the tiny cubic region with its center at x = 1.2 nm, y = -1.0 nm, z = 0 and with edges each of length 0.004 nm. Note that 1 nm = 10-9 m. b. At what point is the probability density a maximum for the Ψ above. Answer by simply looking at IΨI2. Compare the result when x is at the origin.

Answer 1

Ans-

This solution consists of a plane wave of unit amplitude traveling to the right [since the
time-dependent wavefunction is multiplied by exp(−i ωt), where ω = E/¯h > 0], and
a plane wave of complex amplitude R traveling to the left. We interpret the first plane
wave as an incoming particle (or, rather, a stream of incoming particles), and the second
as a particle (or stream of particles) reflected by the potential barrier. Hence, |R|
is the probability of reflection. This can be seen by calculating the probability current

In particular, it depends on the shape of the quantizer cell or Voronoi region centered at is fairly independent of the specific probability density of the input vector.acubic lattice with a fixed density of points and two possible support regions.