- #1
maria clara
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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?
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