# Forbidden region (Homework 7, Physics 325)

1. Consider the WKB solution for a ball of mass m with energy E in a gravitational potential $V (z) = mgz$,   $z > 0$, and $V = +\infty$,  z < 0, with a hard surface at z = 0.
(b) Explain the interpretation of the quantum mechanical probabilities $P(z) = |\Psi (z)|^2$ in terms of the time t spent in the region between $z$ and $z +\Delta$z, i.e., $P = |\psi (z)|^2\Delta$z.
(c) Estimate the probability of finding the ball in the classically forbidden region and it’s approximate width $z$, where $\int _R^{z0+\Delta z} p(z)dz \approx 3\hbar$.

b)
In simple words

$P(z)=|ψ(z) |^2~1/p(z) ~1/v(z)$

So that the bigger the probability to find the particle is space Δz  centered on z the smaller the speed of the particle is in that region of space and the longer the time spent by the particle in space Δz is.

c)

The width of the forbidden region (outside the well to the right) is Δz

$∫_{z_0}^{z_0+Δz} \sqrt{2m(mgz-E)}*dx=\sqrt{2m}*(2(mgz-E)^{3/2})/3mg (\text{from z0 to z0+Δz})=$

$=\sqrt{8/9m} (mgΔz)^{3/2}≈3ℏ$

$(8/9m)^{1/3}*mgΔz=(3ℏ)^{2/3}$

or $Δz=(3ℏ)^{2/3}/mg*(3^{2/3}*m^{1/3})/2=1/2g*(9ℏ/m)^{2/3}$

Inside the forbidden region the probability is ($p(z_0 )=0$ so at the denominator one has $p(z_0+Δz)$)

$P≈D^2/(|p(z_0+Δz)|)*exp⁡(-2*2π/T*z_0 )*Δz$

$P≈m/T*1/\sqrt{(2m^2 gΔz)}*exp⁡(-4π/T*E/mg)*Δz=$

$=1/(T√2g) √Δz*exp(-π*√(2E/m))$

$P≈1/T*\sqrt[3]{9ℏ/m}*exp⁡(-π*√(2E/m))$