# Is it possible to find the hydrogen atom's radial wavefunctions?

Is there a way to actually find the equation of $R(r)$ without looking at a table with these equations already given? I'm given $n$, $\ell$, and $m$.

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"Is there a way to actually find the equation of $R(r)$?" Yes... plug the separation of variables ansatz into the Schrodinger equation. Alternatively, you can use the angular momentum operators to construct a Schrodinger equation satisfied by a particular state. – JamalS May 9 '14 at 17:24
@JamalS that should probably be an answer – David Z May 9 '14 at 17:28
Hi jerry2144: Welcome to Phys.SE. Is your question essentially How do I compute the radial wavefunctions $R(r)$ of the Hydrogen atom from first principles? – Qmechanic May 9 '14 at 20:00

An electron with total orbital angular momentum of $L^2 = \hbar^2 l(l+1)$ will experience a centrifugal force in addition to the Coulomb force from the nucleus. The result is that, in the frame rotating with the electron (don't read too much into this), the electron will see an effective potential energy given by:

$$V_l(r) = - \frac{e^2}{4 \pi \epsilon_o r} + \frac{\hbar^2}{2 m} \frac{l(l+1)}{r^2}$$ where $r$ is the distance between the electron and the nucleus. I put the subscript $l$ on the potential $V_l(r)$ to stress that this effective potential will change as the orbital quantum number $l$ of the electron changes.

The radial wave function $U_{ln}(r)$ obeys the ordinary 1D Schrodinger equation for a particle in a potential $V_l(r)$:

$$-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}U_{ln}(r) + V_l(r)U_{ln}(r)=E_{ln}U_{ln}(r)$$

The energies $E_{ln}$, as you can see, are the energy eigenvalues for an electron in an effective potential given by $V_l(r)$, one for each value of $n$. By convention we arrange these eigenvalues so that higher $n$ quantum numbers correspond to higher energies${}^*$.

There are fancy techniques for finding the $U_{ln}(r)$ in terms of familiar mathematical functions, but there is also nothing preventing you from writing a computer program to calculate these functions to arbitrary accuracy.

Once you find these $U_{ln}(r)$, the to calculate the probability $P(r_o,r^{\prime})$ that an electron with quantum numbers $l$ and $n$ (independent of $m$) is found in some spherical shell between $r_o$ and $r^{\prime}$, we only need to calculate the familiar integral: $$P(r_o,r^{\prime})=\int_{r_o}^{r^{\prime}}dr \left|U_{ln}(r)\right|^2$$ This works provided the function is properly normalized: $$P(0,\infty)=\int_{0}^{\infty}dr |U_{ln}(r)|^2=1$$ Typically the radial wave function is expressed in the form $R_{ln}(r)$ so that the probability $P(r,\theta,\phi)d^3x$ to be found in a small cube of volume $d^3x$ at the point $(r,\theta,\phi)$ is given by: $$P(r,\theta,\phi)d^3x=\left|R_{ln}(r)Y^m_l(\theta,\phi)\right|^2d^3x$$ It is easy to show that the two functions are related by the equation: $$R_{ln}(r) = \frac{U_{ln}(r)}{r}$$

${}^*$ sidenote : It turns out by some mysterious coincidence that the lowest energy eigenvalue for an electron with a given $l$ is equal to the second lowest energy eigenvalue for an electron with an angular momentum quantum number of $l-1$. Because of this, it is convention that for each $l$ we begin counting from $n=l,l+1,l+2,\ldots$ so that it is sufficient to only specify $n$ (independent of $l$) to know the energy of the eigenstate.

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