The Linear Potential Plotted with Airy Function Solution

Remove["Global`*"]

Pick a value for the (scaled) energy

ϵ=3;

A series of plots to show the potential, the energy level, and the classical turning point

potential=Plot[Abs[y],{y,-5,5},PlotStyle->Black]

energy=Plot[ϵ,{y,-5,5},PlotStyle->Red];
Show[potential,energy]

turning=ListLinePlot[{{ϵ,0},{ϵ,1.5ϵ}},PlotStyle->Dashed];
Show[potential,energy,turning]

Demonstrate a plot of the Airy function (Figure 2.3 in the textbook) and find the zeros

Plot[AiryAi[z],{z,-10,5}]

rootsA=FindRoot[AiryAi[z]==0,{z,{-2,-4,-5.5,-7}}][[1]]

rootsD=FindRoot[D[AiryAi[z],z]==0,{z,{-1,-3,-5,-6}}][[1]]

Define the Airy function argument in terms of the scaled position

zFunc=2^(1/3)(y-ϵ);

Prepare the potential plot for plotting eigenfunctions

potential=Plot[Abs[y],{y,-1,8},PlotStyle->Black,
PlotRange->{-8,8}];
all=Show[potential,energy,turning];

Find and plot the eigenfunction for the ground state energy

scale=-(z/.rootsD)[[1]]/(2^(1/3)ϵ);
ϵ0Plot=Plot[8 AiryAi[scale zFunc],{y,0,8}];
Show[all,ϵ0Plot]

Now the first excited state

scale=-(z/.rootsA)[[1]]/(2^(1/3)ϵ);
ϵ0Plot=Plot[8 AiryAi[scale zFunc],{y,0,8}];
Show[all,ϵ0Plot]

And next the second excited state

scale=-(z/.rootsD)[[2]]/(2^(1/3)ϵ);
ϵ0Plot=Plot[8 AiryAi[scale zFunc],{y,0,8}];
Show[all,ϵ0Plot]

Finally the third excited state

scale=-(z/.rootsA)[[2]]/(2^(1/3)ϵ);
ϵ0Plot=Plot[8 AiryAi[scale zFunc],{y,0,8}];
Show[all,ϵ0Plot]