- #1

maria clara

- 58

- 0

the Hamiltonian for the hydrogen atom appears as

H= (Pr)^2/2m+(L^2)/2mr^2-e^2/r

The Schroedinger equation is

((Pr)^2/2m+(L^2)/2mr^2-Ze^2/r+|E|)PHI=0

|E| because we're looking for bound states.

m - reduced mass.

the radial equation is

{(-h/2m)[(1/r)(d^2/dr^2)r]+h^2l(l+1)/2mr^2-e^2/r+|E|}R(r)=0

(h denotes h-bar)

we change the the dependent variable to u=rR(r)

and we get

(-d^2/dr^2+l(l+1)/r^2-(2m/h^2)(e^2/r)+2m|E|/h^2)u=0

I don't understand how exactly the last equation is obtained. By the change u=rR, shouldn't we get the factor 1/r inside the brackets?

H= (Pr)^2/2m+(L^2)/2mr^2-e^2/r

The Schroedinger equation is

((Pr)^2/2m+(L^2)/2mr^2-Ze^2/r+|E|)PHI=0

|E| because we're looking for bound states.

m - reduced mass.

the radial equation is

{(-h/2m)[(1/r)(d^2/dr^2)r]+h^2l(l+1)/2mr^2-e^2/r+|E|}R(r)=0

(h denotes h-bar)

we change the the dependent variable to u=rR(r)

and we get

(-d^2/dr^2+l(l+1)/r^2-(2m/h^2)(e^2/r)+2m|E|/h^2)u=0

I don't understand how exactly the last equation is obtained. By the change u=rR, shouldn't we get the factor 1/r inside the brackets?

Last edited: