I found a problem that says:

Show by direct substitution that $R_{10}$ is a solution of Schrödinger's radial equation.

AFAIK Schrödinger's radial equation is


and $R_{10}$ in this case is:


where $\rho$ is

$\rho = \frac{2Zr}{na_0}$

and $E_p(r)$ is:

$E_p(r) = -\frac{Ze^2}{4\pi\epsilon_0r}$

The thing is I don't understand very well what $E$ is in this case and (I think) I need it to solve the problem. It's the total energy but in the book (Alonso y Finn) they never talk about the kinetic energy. I suppose it depends on $r$, $n$ and $l$ because once I do the substitution on the left side of the equation and I factorize it I get $R_{10}$ multiplied by a term that's full of those variables.

  • $\begingroup$ This isn't really a physics question, but seems to be a question about a less than ideally written textbook. $\endgroup$
    – DarenW
    Sep 29, 2011 at 5:45
  • $\begingroup$ Corrections to v1: $x$ should be $r$ (in two places), and $\phi$ should be $\pi$. $\endgroup$
    – Qmechanic
    Sep 29, 2011 at 12:21

1 Answer 1


$E$ is the total energy. Since the Hamiltonian $\hat{H}$ is

$\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V$

in a time-independent Schrodinger equation

$\hat{H}\psi = E\psi$

in a spherical coordinates, you can write the wave function as:

$\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$

The Laplace operator $\nabla^2$ in the spherical coordinates is:

$\nabla^2 = \frac{1}{r}\frac{\partial}{\partial r}\left( r\frac{\partial}{\partial r}\right) + \frac{1}{r^2sin\theta}\frac{\partial}{\partial \theta}\left( sin\theta\frac{\partial}{\partial\theta} \right)+\frac{1}{r^2sin^2\theta}\frac{\partial^2}{\partial\phi^2}$

Put the $\nabla^2$ into the Schrodinger equation and separate the variables, you can get the radial equation in your question.

$-\frac{\hbar^2}{2m}\left[\frac{d^2}{dr^2} +\frac{2}{r}\frac{d}{dr}-\frac{l(l+1)}{r^2}\right]R + E_p(r)R = ER$

This equation is the Legendre's differential equation, where $l$ is the eigen value of the colatitude equation (equation of variable $\theta$) and $l$ can only be non-negative integers. This $l$ is also called the orbital quantum number. By solving this differential equation, you can solve the radial part $R_{n,l}(r)$ and the principle quantum number $n$, which yields the the state energy $E_n$.

Since the $-\frac{\hbar^2}{2m}\nabla^2$ is the kinetic energy part in Hamiltonian, so I think this solves your second question.

  • $\begingroup$ B.T.W. you can refer to the Legendre polynomials to know how to solve this differential equation to get $E$ $\endgroup$
    – Jing
    Sep 29, 2011 at 4:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.