How do you write the Schrödinger equation for the hydrogen atom?

Well, the Schrodinger equation in spherical coordinates is:

#hatHpsi_(nlm_l)(r,theta,phi) = Epsi_(nlm_l)(r,theta,phi)#

where #hatH# is the Hamiltonian operator for the system and #psi_(nlm_l)(r,theta,phi)# is the wave function.

The hydrogen atom contains a proton and an electron, so the Hamiltonian contains the kinetic energy #hatK# of the electron and proton, and the potential energy #hatV# for coulombic attraction with the nucleus.

[Since the nucleus is much more massive than the electron, the Born-Oppenheimer approximation allows us to ignore the nuclear kinetic energy relative to the electron.]

This is one common way to write this.

#hatH = hatK_(n) + hatK_(e) + hatV_(n e)#

#~~ hatK_e + hatV_(n e)#

#= -ℏ^2/(2mu)grad^2 - e^2/(4piepsilon_0vecr)#

where:

• #ℏ = h//2pi# is the reduced Planck's constant.
• #mu = (m_p m_e)/(m_p + m_e) ~~ m_e# is the reduced mass of the electron and proton.
• #nabla^2 = 1/r^2 (del)/(delr) (r^2 (del)/(delr)) + 1/(r^2 sintheta) (del)/(del theta)(sintheta (del)/(del theta)) + 1/(r^2sin^2theta) (del^2)/(delphi^2)# is the Laplacian operator in spherical coordinates.
• #e# is the elementary charge, #1.602 xx 10^(-19)# #"C"#, e.g. positive for a proton, negative for an electron.
• #epsilon_0 = 8.854187817 xx 10^(-12) "F"cdot"m"^(-1)# is the vacuum permittivity.
• #vecr# is the radial position. Under the Born-Oppenheimer approximation, we assume the nucleus is fixed so that #vecr# becomes the radial distance of the electron from the nucleus.

And so, assuming the wave function is square-integrable, the Schrodinger equation for the hydrogen atom in full is:

#[-ℏ^2/(2mu)[1/r^2 del/(delr)(r^2 (del)/(delr)) + 1/(r^2sintheta) (del)/(deltheta)(sintheta (del)/(del theta)) + 1/(r^2sin^2theta) (del^2)/(delphi^2)] - e^2/(4piepsilon_0vecr)]R_(nl)(r)Y_(l)^(m_l)(theta,phi) = E_(nlm_l)R_(nl)(r)Y_(l)^(m_l)(theta,phi)#

This has been solved in full before.