# Extended Well (Perturbations)

Consider a particle in the quantum infinite well of width $2a$ centered on origin. The wavefunction of the particle is $\psi(x)=\sqrt{1/a}\cos(\pi x/2a)$ for $-a<x<a$ and zero for all other values of x. Suppose that at t=0 the width of the well is doubled so that the potential is zero for $-2a<x<2a$. the wavefunction at this instant remains the same.

a) Determine an expression for the wavefunction in terms of the energy eigenfunctions of the new potential (of width $4a$). Write the integral for the coefficients of your wave function.

b) Write the wavefunction for this particle at a latter time $t>0$.

c) determine an expression for the probability of measuring of each of the energies:

i) The n-th excited state of the original well.

ii) The n-th excited state of the new well.

Initial wavefunction is

$ψ(x)=\sqrt{1/a}*\cos(π/2a x)$

When the well extends the new energy wavefunctions are (primed)

$ψ_n’ (x)=A*\cos(nπ/4a x)$ so that $ψ(x)=A*\cos(2π/4a x)=B*ψ_2′ (x)$

Since initial wavefunctions are normalized we have

$1=∫ψ* ψdx=B^2*∫ ψ_2(‘*) ψ_2’ dx$

Explanation: When the well extends the shape of the initial wavefunctions is kept, however because the space available is bigger, the amplitude of the wavefunction needs to change.

c) At a later time the total wavefunction need to obey the time dependent Sch equation. That is is the wavefunction is $Ψ(x,t)$ then

$iℏ* dΨ(x,t)/dt=HΨ(x,t)$

The wavefunction that obeys this equation is of the form

$Ψ(x,t)=ψ(x)*f(t)$ with $f(t)=e^{-iω*t}=e^{-(iE_2′)/ℏ*t}$

Thus at a later time the wavefunction is

$Ψ(x,t)=\sqrt{(1/a)}*\cos(πx/2a)*exp(-(iE_2′)/ℏ*t)$ where $E_2’$ is second bounded energy in the $4a$ well

d) If given a wavefunction as a sum of mixed states (of bound energies $E_n$)

$ψ(x)=∑C_n*ϕ_n (n)$

Then the probability to find the particle in the n-th state having $E_n$ energy is

$P_n=|C_n |^2$

For the initial well the wavefunction ψ is exactly those that corresponds to the first energy level that is

$ψ(x)=1*ϕ_1 (x)$

Which means all probabilities of energies having $n≥2$ will be zero and the probability of $E_1$ will be 1.

For the case of the extended level since

$ψ(x)=B*ϕ_2′ (x)$

Then the probability to find the particle in the energy level $E_2’$ is

$P_2’=|B|^2$

and the probability to find the particle in all other energy level is zero.