- #1

- 218

- 4

As we know, the normal Schrodinger equation is:

(-ħ

^{2}/2m) (∂

^{2}Ψ/∂x

^{2}) + v(x)Ψ = EΨ

For a step potential however, my book and web resource both say that for the boundary conditions:

v(x)= 0 for x < 0 and V

_{0}for x ≥ 0

the Schrodinger equations are:

(∂

^{2}Ψ

_{1}/∂x

^{2}) + K

_{1}

^{2}Ψ

_{1}(x) = 0

and

(∂

^{2}Ψ

_{2}/∂x

^{2}) + K

_{2}

^{2}Ψ

_{2}(x) = 0

where K

_{1}= squrt(2mE) / ħ and K

_{2}= squrt(2m(E- V

_{0})) / ħ

(the plus sign in the second one changes into a minus when the particle doesn't have enough energy to overcome the step).

Where/How exactly did Schrodinger get these step potential equations from the original one? The step potential equations don't even seem to have the (-ħ

^{2}/2m) term in front. Can someone please explain to me why step potentials seem to have different Schrodinger equations than the normal one?