Solution of the Schrodinger Equation for an Atom

Radial Schrodinger equation for central potentials

Stationary solutions of the time dependent Schrodinger equation

$i\hbar {\frac {\partial \psi ({\bf {{r},t)}}}{\partial t}}=H({\bf {{r})\psi ({\bf {r}},t)}}$

($H$ is the Hamiltonian operator) can be represented as

$\psi {({\bf {r}},t)}=e^{-iE_{n}t/\hbar }\psi _{n}({\bf {{r}),}}$

where $n$ stands for all quantum numbers necessary to label the state. This

leads to the time-independent Schrodinger equation

$[H({\bf {r}})-E_{n}]\psi _{n}({\bf {{r})=0}}$

A pervasive application of this equation in atomic physics is fo the case of a spherically symmetric one-particle system of mass $\mu$. In this case the Hamiltonian is

${\begin{aligned}H&\equiv {\rm {Kinetic~Energy}}~+~{\rm {Potential~Energy}}\\&=p^{2}/2\mu +V(r)={\frac {-\hbar ^{2}\nabla ^{2}}{2\mu }}+V(r)\\&=-{\frac {\hbar ^{2}}{2\mu }}{\left[{\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}{\left(r^{2}{\frac {\partial }{\partial r}}\right)}+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}{\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)}+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right]}+V(r)\end{aligned}}$

\noindent where the kinetic energy operator $\nabla ^{2}$ has been written in spherical coordinates. Because $V$ is spherically symmetric, the angular dependence of the solution is characteristic of spherically symmetric systems in general and may be factored out:

$\psi _{n\ell m}({\bf {r}})=R_{n\ell }(r)Y_{\ell m}(\theta ,\phi )$

\noindent $Y_{\ell m}$ are the {\it spherical harmonics} and $\ell$ is the eigenvalue of the operator for the orbital angular momentum, $L$,

$L^{2}Y_{\ell m}=\ell (\ell +1)\hbar ^{2}Y_{\ell m}$

and $m$ is the eigenvalue of the projection of $L$ on the quantization axis (which may be chosen at will)

$L_{z}Y_{\ell m}=m\hbar Y_{\ell m}$

With this substitution Eq.\ \ref{EQ_secpfive} the time independent radial Schrodinger equation becomes

${\frac {1}{r^{2}}}{\frac {d}{dr}}{\left(r^{2}{\frac {dR_{n\ell }}{dr}}\right)}+{\left[{\frac {2\mu }{\hbar ^{2}}}{\left[E_{n\ell }-V(r)\right]}-{\frac {\ell (\ell +1)}{r^{2}}}\right]}R_{n\ell }=0$

This is the equation which is customarily solved for the hydrogen atom's radial wave functions. For most applications (atoms, scattering by a central potential, diatomic molecules) it is more convenient to make a further substitution.

$R_{n\ell }(r)=y_{n\ell }(r)/r$

which leads to

${\frac {d^{2}y_{n\ell }(r)}{dr^{2}}}+{\left[{\frac {2\mu }{\hbar ^{2}}}[E_{n\ell }-V(r)]-{\frac {\ell (\ell +1)}{r^{2}}}\right]}y_{n\ell }(r)=0$

with the boundary condition $y_{n\ell }(0)=0$. This equation is identical with the time independent Schrodinger equation for a particle of mass $\mu$ in an effective one dimensional potential,

$V^{\rm {eff}}(r)=V(r)+L^{2}/(2\mu r^{2})=V(r)+{\frac {\hbar ^{2}\ell (\ell +1)}{2\mu r^{2}}}$

The term $\hbar ^{2}\ell (\ell +1)/2\mu r^{2}$ is called the {\it centrifugal potential}, and adds to the actual potential the kinetic energy of the circular motion that must be present to conserve angular momentum.

Radial equation for hydrogen

The quantum treatment of hydrogenic atoms or ions appears in many textbooks and we present only a summary\footnote{The most comprehensive treatment of hydrogen is the classic text of Bethe and Salpeter, { \it The Quantum Mechanics of One- and Two-Electron Atoms}, H. A. Bethe and E. E. Salpeter, Academic Press (1957). Messiah is also excellent.}. For hydrogen Eq.\ \ref{EQ_secpten} becomes

${\frac {d^{2}y_{n\ell }(r)}{dr^{2}}}+{\left[{\frac {2\mu }{\hbar ^{2}}}\left[E_{n}+{\frac {e^{2}}{r}}\right]-{\frac {\ell (\ell +1)}{r^{2}}}\right]}y_{n\ell }(r)=0$

\noindent First look at this as $r\rightarrow 0$, the dominant terms are

${\frac {d^{2}y}{dr^{2}}}-{\frac {\ell (\ell +1)}{r^{2}}}y=0$

for any value of $E$. It is easily verified that the two independent solutions are

$y\sim r^{\ell +1}~{\mbox{and}}~y\sim r^{-\ell }$

For $\ell \geq 1$ the only normalizable solution is $y\sim r^{\ell +1}$. Question: What happens to these arguments for $\ell =0$? What implications does this have for the final solution? Messiah has a good discussion of this. We look next at the solution for $r\rightarrow \infty$ where we may investigate a simpler equation if $\lim _{r\rightarrow \infty }rV(r)=0$. For large r:

${\frac {d^{2}y}{dr^{2}}}+{\frac {2\mu E}{\hbar ^{2}}}y=0$

If $E>0$, this equation has oscillating solutions corresponding to a free particle. For $E<0$ the equation has exponential solutions, but only the decaying exponential is physically acceptable (i.e.. normalizable)

$R(r)=y(r)/r={\frac {1}{r}}e^{(-2\mu E/\hbar ^{2})^{\frac {1}{2}}r}$

When $E<0$, it is possible to obtain physically reasonable solutions to Eq.\ \ref{EQ_secpten} (or indeed {\it any} bound state problem) only for certain discrete values of $E$, its eigenvalues. This situation arises from the requirement that the radial solution be normalizable, which requires that $\int _{o}^{\infty }y^{2}(r)dr=1$, or alternatively, that $y(r)$ vanishes sufficiently rapidly at large r.). Eq.\ref{EQ_rehone} is a prescription for generating a function $y(r)$ for arbitrary $E<0$ given $y$ and ${\mbox{d}}y/{\mbox{d}}r$ at any point. This can be solved exactly as hydrogen. For other central potentials, one can find the eigenvalues and eigenstates by computation. One proceeds as follows: Select a trial eigenvalue, $E<0$. Starting at large $r$ a "solution" of the form of Eq.\ \ref{EQ_rehfive} is selected and extended in to some intermediate value of $r,~r_{m}$. At the origin one must select the solution of the form $y_{\ell }\sim r^{\ell +1}$ Eq.\ (\ref{EQ_rehthree}); this "solution" is then extended out to $r_{m}$. The two "solutions" may be made to have the same value at $r_{m}$ by multiplying one by a constant; however, the resulting function is a valid solution only if the first derivative is continuous at $r_{m}$, and this occurs only for a discrete set of $E_{n\ell }$. The procedure described here is, in fact, the standard Numerov-Cooley algorithm for finding bound states. Its most elegant feature is a procedure for adjusting the trial eigenvalue using the discontinuity in the derivative that converges to the correct energy very rapidly. For the hydrogen atom, the eigenvalues can be determined analytically. The substitution

$y_{\ell }(r)=r^{\ell +1}e^{-(-2\mu E/\hbar ^{2})^{1/2}r}v_{\ell }(r)$

leads to a particularly simple equation. To make it dimensionless, one changes the variable from $r$ to $x$

$x=[2(-2\mu E)^{1/2}/\hbar ]r$

so the exponential in Eq.\ \ref{EQ_rehsix} becomes $e^{-1/2x})$, and defines

$v={\frac {\hbar }{(-2\mu E)^{1/2}a_{o}}}$

where $a_{o}$ is the Bohr radius:

${\left[x{\frac {d^{2}}{dx^{2}}}+(2\ell +2-x){\frac {d}{dx}}-(\ell +1-v)\right]}v_{\ell }=0$

This is a Laplace equation and its solution is a confluent hypergeometric series. To find the eigenvalues one now tries a Taylor series

$v_{\ell }(x)=1+a_{1}x+a_{2}x^{2}+\cdots$

for $v_{\ell }$. This satisfies the equation only if the coefficients of each power of x are satisfied, i.e..\

${\begin{array}{rcl}x^{o}:&&(2\ell +2)a_{1}-(\ell +1-v)=0\\x^{1}:&&2(2\ell +3)a_{2}-(\ell +2-v)a_{\ell }=0\\x^{p-1}:&&p(2\ell +1+p)a_{p}-(\ell +p-v)a_{p-1}=0\end{array}}$

The first line fixes $a_{1}$, the second then determines $a_{2}$, and in general

$a_{p}={\frac {(\ell +p-v)}{p(2\ell +1+p)}}a_{p-1}$

In general Eq.\ \ref{EQ_reheleven} will give a $p^{th}$ coefficient on the order of 1/$p$! so

$v_{\ell }(x)\sim \sum _{p=0}x^{p}/p!=e^{x}$

this spells disaster because it means $y=r^{\ell +1}e^{-x/2}v_{\ell }(r)$ diverges. The only way in which this can be avoided is if the series truncates, i.e. if v is an integer:

$v=n=n^{\prime }+\ell +1~~~~~~~~~~n=0,1,2,\cdots$

so that $a_{n}$, will be zero and $v_{\ell }$ will have $n^{\prime }$ nodes. Since $n^{\prime }\geq =0$, it is clear that you must look at energy level $n\geq \ell +1$ to find a state with angular momentum $\ell$ (e.g.. the 2$d$ configuration does not exist). This gives the eigenvalues of hydrogen (from Eq.\ \ref{EQ_reheight})

$E_{n}=-{\frac {1}{2}}\alpha ^{2}\mu c^{2}/(n^{\prime }+\ell +1)^{2}=-{\frac {1}{2}}{\frac {\mu (e^{2}/4\pi \epsilon _{o})^{2}}{\hbar ^{2}}}{\frac {1}{n^{2}}}=-{\frac {hcR_{H}}{n^{2}}},$

which agrees with the Bohr formula.