- #1

mrrelativistic

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Why is it ψ(x) = Ae^ikx + Be^-ikx (x < -a)

ψ(x) = Fe^ikx (x > a)

ψ(x) = Ce^μx + De^-μx (-a < x < a)

for k^2 = (2mE)/(hbarr), μ^2 = (2m(V - E))/(hbarr) and E < V

The potential barrier is 0 at x < -a and x > +a and V at -a < x < a.

I know that there are two cases (E < V and E > V). What I really don't know is how do you get the solution form of the Schrodinger equation.

And for a potential well:

Why is it ψ(x) = Ae^kx + Be^-kx (x < -a)

ψ(x) = Fe^kx + Ge^-kx (x > a)

ψ(x) = Csin(μx) + Dcos(μx) (-a < x < a)

The potential well is 0 at x < -a and x > +a and -V at -a < x < a.