# Energies of Asymetric Well

You are given the asymmetric well in the figure ($2V_0$ on one side and $5V_0$ on the other side). Sketch the first excited state, solve the Schrodinger equation inside the well. How less likely is for a particle to be found near $x=2L$ than near $x=L$?

The sketch is above. The central node will be always towards the side of the well where the wall has a lower height. (To clearly see this fact the right-left well walls height difference is amplified in the figure). This happens because the wave function will penetrate more into that wall than in the higher wall.  (If the left wall would be infinite, the wave function will have a zero exactly at the left wall. Since the right wall is finite, the wave function will penetrate this wall and will be displaced towards right with a small amount). The first excited state (n=2) has one node inside the well (the ground state has no node inside the well).

b)

In region I (in the left wall) one has

$-ℏ^2/2m*(d^2 ψ)/(d x^2 )+5V_0 ψ=Eψ$

or$-(d^2 ψ)/(d x^2 )=2m/ℏ^2 *(E-5V_0 )ψ$

That has a general solution

$ψ_I (x)=A e^{-i k x}+Be^{i k x}$   with $k=sqrt{(2m(E-5V_0))}/ℏ$  with $E<5V_0$

So that

$ψ_I (x)=A e^{K x}+Be^{-K x}$    with $K=sqrt{(2m(5V_0-E))}/ℏ$   real

Since when $x→-∞$,one has $ψ_I (x)→0$ it follows that:
$ψ_I (x)=A*e^K x$

c)

The probability of finding particle in space $d x$ at position x is

$P(x)=|ψ(x) |^2$

So that

$P(-L)/P(-2L) =|ψ_I (-L)|^2/|ψ_I (-2L)|^2 =(e^(-k L))^2/(e^(-2kL) )^2 =e^2KL$