16d. The hydrogen atom

[math] \newcommand{\mathds}{\mathbb}[/math]

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Hydrogen, eventually. In order now to finish our study, and eventually get to conclusions about hydrogen, it remains to solve the radial equation, for the Coulomb potential [math]V[/math] of the proton. Let us begin with some generalities, valid for any time-independent potential [math]V[/math]. As a first manipulation on the radial equation, we have:

Proposition

The radial equation, written with [math]K=l(l+1)[/math],

[[math]] (r^2\rho')'-\frac{2mr^2}{h^2}(V-E)\rho=l(l+1)\rho [[/math]]

takes with [math]\rho=u/r[/math] the following form, called modified radial equation,

[[math]] Eu=-\frac{h^2}{2m}\cdot u''+\left(V+\frac{h^2l(l+1)}{2mr^2}\right)u [[/math]]

which is a time-independent 1D Schrödinger equation.

Show Proof

With [math]\rho=u/r[/math] as in the statement, we have:

[[math]] \rho=\frac{u}{r}\quad,\quad \rho'=\frac{u'r-u}{r^2}\quad,\quad (r^2\rho')'=u''r [[/math]]

By plugging this data into the radial equation, this becomes:

[[math]] u''r-\frac{2mr}{h^2}(V-E)u=\frac{l(l+1)}{r}\cdot u [[/math]]

By multiplying everything by [math]h^2/(2mr)[/math], this latter equation becomes:

[[math]] \frac{h^2}{2m}\cdot u''-(V-E)u=\frac{h^2l(l+1)}{2mr^2}\cdot u [[/math]]

But this gives the formula in the statement. As for the interpretation, as time-independent 1D Schrödinger equation, this is clear as well, and with the comment here that the term added to the potential [math]V[/math] is some sort of centrifugal term.

■

Getting back now to the Coulomb potential of the proton, we have here: \begin{fact} The Coulomb potential of the hydrogen atom proton, acting on the electron by attraction, is given according to the Coulomb law by

[[math]] V=-\frac{Kep}{r} [[/math]]

where [math]p[/math] is the charge of the proton, and [math]K[/math] is the Coulomb constant. In practice however we have [math]p\simeq e[/math] up to order [math]10^{-7}[/math], and so our formula can be written as

[[math]] V\simeq-\frac{Ke^2}{r} [[/math]]

and we will use this latter formula, and with [math]=[/math] sign, for simplifying. \end{fact} Getting back now to math, it remains to solve the modified radial equation, for the above potential [math]V[/math]. And we have here the following result, which does not exactly solve this radial equation, but provides us instead with something far better, namely the proof of the original claim by Bohr, which was at the origin of everything:

Theorem (Schrödinger)

In the case of the hydrogen atom, where [math]V[/math] is the Coulomb potential of the proton, the modified radial equation, which reads

[[math]] Eu=-\frac{h^2}{2m}\cdot u''+\left(-\frac{Ke^2}{r}+\frac{h^2l(l+1)}{2mr^2}\right)u [[/math]]

leads to the Bohr formula for allowed energies,

[[math]] E_n=-\frac{m}{2}\left(\frac{Ke^2}{h}\right)^2\cdot\frac{1}{n^2} [[/math]]

with [math]n\in\mathbb N[/math], the binding energy being

[[math]] E_1\simeq-2.177\times 10^{-18} [[/math]]

with means [math]E_1\simeq-13.591\ {\rm eV}[/math].

Show Proof

This is again something non-trivial, and we will be following Griffiths , with some details missing. The idea is as follows:

(1) By dividing our modified radial equation by [math]E[/math], this becomes:

[[math]] -\frac{h^2}{2mE}\cdot u''=\left(1+\frac{Ke^2}{Er}-\frac{h^2l(l+1)}{2mEr^2}\right)u [[/math]]

In terms of [math]\alpha=\sqrt{-2mE}/h[/math], this equation takes the following form:

[[math]] \frac{u''}{\alpha^2}=\left(1+\frac{Ke^2}{Er}+\frac{l(l+1)}{(\alpha r)^2}\right)u [[/math]]

In terms of the new variable [math]p=\alpha r[/math], this latter equation reads:

[[math]] u''=\left(1+\frac{\alpha Ke^2}{Ep}+\frac{l(l+1)}{p^2}\right)u [[/math]]

Now let us introduce a new constant [math]S[/math] for our problem, as follows:

[[math]] S=-\frac{\alpha Ke^2}{E} [[/math]]

In terms of this new constant, our equation reads:

[[math]] u''=\left(1-\frac{S}{p}+\frac{l(l+1)}{p^2}\right)u [[/math]]

(2) The idea will be that of looking for a solution written as a power series, but before that, we must “peel off” the asymptotic behavior. Which is something that can be done, of course, heuristically. With [math]p\to\infty[/math] we are led to [math]u''=u[/math], and ignoring the solution [math]u=e^p[/math] which blows up, our approximate asymptotic solution is:

[[math]] u\sim e^{-p} [[/math]]

Similarly, with [math]p\to0[/math] we are led to [math]u''=l(l+1)u/p^2[/math], and ignoring the solution [math]u=p^{-l}[/math] which blows up, our approximate asymptotic solution is:

[[math]] u\sim p^{l+1} [[/math]]

(3) The above heuristic considerations suggest writing our function [math]u[/math] as follows:

[[math]] u=p^{l+1}e^{-p}v [[/math]]

So, let us do this. In terms of [math]v[/math], we have the following formula:

[[math]] u'=p^le^{-p}\left[(l+1-p)v+pv'\right] [[/math]]

Differentiating a second time gives the following formula:

[[math]] u''=p^le^{-p}\left[\left(\frac{l(l+1)}{p}-2l-2+p\right)v+2(l+1-p)v'+pv''\right] [[/math]]

Thus the radial equation, as modified in (1) above, reads:

[[math]] pv''+2(l+1-p)v'+(S-2(l+1))v=0 [[/math]]

(4) We will be looking for a solution [math]v[/math] appearing as a power series:

[[math]] v=\sum_{j=0}^\infty c_jp^j [[/math]]

But our equation leads to the following recurrence formula for the coefficients:

[[math]] c_{j+1}=\frac{2(j+l+1)-S}{(j+1)(j+2l+2)}\cdot c_j [[/math]]

(5) We are in principle done, but we still must check that, with this choice for the coefficients [math]c_j[/math], our solution [math]v[/math], or rather our solution [math]u[/math], does not blow up. And the whole point is here. Indeed, at [math]j \gt \gt 0[/math] our recurrence formula reads, approximately:

[[math]] c_{j+1}\simeq\frac{2c_j}{j} [[/math]]

But, surprisingly, this leads to [math]v\simeq c_0e^{2p}[/math], and so to [math]u\simeq c_0p^{l+1}e^p[/math], which blows up.

(6) As a conclusion, the only possibility for [math]u[/math] not to blow up is that where the series defining [math]v[/math] terminates at some point. Thus, we must have for a certain index [math]j[/math]:

[[math]] 2(j+l+1)=S [[/math]]

In other words, we must have, for a certain integer [math]n \gt l[/math]:

[[math]] S=2n [[/math]]

(7) We are almost there. Recall from (1) above that [math]S[/math] was defined as follows:

[[math]] S=-\frac{\alpha Ke^2}{E}\quad:\quad\alpha=\frac{\sqrt{-2mE}}{h} [[/math]]

Thus, we have the following formula for the square of [math]S[/math]:

[[math]] S^2=\frac{\alpha^2K^2e^4}{E^2}=-\frac{2mE}{h^2}\cdot\frac{K^2e^4}{E^2}=-\frac{2mK^2e^4}{h^2E} [[/math]]

Now by using the formula [math]S=2n[/math] from (6), the energy [math]E[/math] must be of the form:

[[math]] E=-\frac{2mK^2e^4}{h^2S^2}=-\frac{mK^2e^4}{2h^2n^2} [[/math]]

Calling this energy [math]E_n[/math], depending on [math]n\in\mathbb N[/math], we have, as claimed:

[[math]] E_n=-\frac{m}{2}\left(\frac{Ke^2}{h}\right)^2\cdot\frac{1}{n^2} [[/math]]

(8) Thus, we proved the Bohr formula. Regarding numerics, the data is as follows:

[[math]] K=8.988\times10^9\quad,\quad e=1.602\times10^{-19} [[/math]]

[[math]] h=1.055\times10^{-34}\quad,\quad m=9.109\times10^{-31} [[/math]]

But this gives the formula of [math]E_1[/math] in the statement.

■

As a first remark, all this agrees with the Rydberg formula, due to:

Theorem

The Rydberg constant for hydrogen is given by

[[math]] R=-\frac{E_1}{h_0c} [[/math]]

where [math]E_1[/math] is the Bohr binding energy, and the Rydberg formula itself, namely

[[math]] \frac{1}{\lambda_{n_1n_2}}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) [[/math]]

simply reads, via the energy formula in Theorem 16.25,

[[math]] \frac{1}{\lambda_{n_1n_2}}=\frac{E_{n_2}-E_{n_1}}{h_0c} [[/math]]

which is in agreement with the Planck formula [math]E=h_0c/\lambda[/math].

Show Proof

Here the first assertion is something numeric, coming from the fact that the formula in the statement gives, when evaluated, the Rydberg constant:

[[math]] R=\frac{-E_1}{h_0c}=\frac{2.177\times10^{-18}}{6.626\times10^{-34}\times 2.998\times 10^8}=1.096\times10^7 [[/math]]

As a consequence, and passed now what the experiments exactly say, we can define the Rydberg constant of hydrogen abstractly, by the following formula:

[[math]] R=\frac{m}{2h_0c}\left(\frac{Ke^2}{h}\right)^2 [[/math]]

Regarding now the second assertion, by dividing [math]R=-E_1/(h_0c)[/math] by any number of type [math]n^2[/math] we obtain, according to the energy convention in Theorem 16.25:

[[math]] \frac{R}{n^2}=-\frac{E_n}{h_0c} [[/math]]

But these are exactly the numbers which are subject to substraction in the Rydberg formula, and so we are led to the conclusion in the statement.

■

Let us go back now to our study of the Schrödinger equation. Our conclusions are:

Theorem

The wave functions of the hydrogen atom are the following functions, labelled by three quantum numbers, [math]n,l,m[/math],

[[math]] \phi_{nlm}(r,s,t)=\rho_{nl}(r)\alpha_l^m(s,t) [[/math]]

where [math]\rho_{nl}(r)=p^{l+1}e^{-p}v(p)/r[/math] with [math]p=\alpha r[/math] as before, with the coefficients of [math]v[/math] subject to

[[math]] c_{j+1}=\frac{2(j+l+1-n)}{(j+1)(j+2l+2)}\cdot c_j [[/math]]

and [math]\alpha_l^m(s,t)[/math] being the spherical harmonics found before.

Show Proof

This follows indeed by putting together all the results obtained so far, and with the remark that everything is up to the normalization of the wave function.

■

In what regards the main wave function, that of the ground state, we have:

Theorem

With the hydrogen atom in its ground state, the wave function is

[[math]] \phi_{100}(r,s,t)=\frac{1}{\sqrt{\pi a^3}}\,e^{-r/a} [[/math]]

where [math]a=1/\alpha[/math] is the inverse of the parameter appearing in our computations above,

[[math]] \alpha=\frac{\sqrt{-2mE}}{h} [[/math]]

called Bohr radius of the hydrogen atom. This Bohr radius is the mean distance between the electron and the proton, in the ground state, and is given by the formula

[[math]] a=\frac{h^2}{mKe^2} [[/math]]

which numerically means [math]a\simeq5.291\times10^{-11}[/math].

Show Proof

There are several things going on here, as follows:

(1) According to the various formulae in the proof of Theorem 16.25, taken at [math]n=1[/math], the parameter [math]\alpha[/math] appearing in the computations there is given by:

[[math]] \alpha=\frac{\sqrt{-2mE}}{h}=\frac{1}{h}\cdot m\cdot\frac{Ke^2}{h}=\frac{mKe^2}{h^2} [[/math]]

Thus, the inverse [math]\alpha=1/a[/math] is indeed given by the formula in the statement.

(2) Regarding the wave function, according to Theorem 16.27 this consists of:

[[math]] \rho_{10}(r)=\frac{2e^{-r/a}}{\sqrt{a^3}}\quad,\quad \alpha_0^0(s,t)=\frac{1}{2\sqrt{\pi}} [[/math]]

By making the product, we obtain the formula of [math]\phi_{100}[/math] in the statement.

(3) But this formula of [math]\phi_{100}[/math] shows in particular that the Bohr radius [math]a[/math] is indeed the mean distance between the electron and the proton, in the ground state.

(4) Finally, in what regards the numerics, these are as follows:

[[math]] a=\frac{1.055^2\times10^{-68}}{9.109\times10^{-31}\times8.988\times10^9\times1.602^2\times10^{-38}}=5.297\times10^{-11} [[/math]]

Thus, we are led to the conclusions in the statement.

■

Getting back now to the general setting of Theorem 16.25, the point is that the polynomials [math]v(p)[/math] appearing there are well-known objects in mathematics, as follows:

Proposition

The polynomials [math]v(p)[/math] are given by the formula

[[math]] v(p)=L_{n-l-1}^{2l+1}(p) [[/math]]

where the polynomials on the right, called associated Laguerre polynomials, are given by

[[math]] L_q^p(x)=(-1)^p\left(\frac{d}{dx}\right)^pL_{p+q}(x) [[/math]]

with [math]L_{p+q}[/math] being the Laguerre polynomials, given by the following formula:

[[math]] L_q(x)=\frac{e^x}{q!}\left(\frac{d}{dx}\right)^q(e^{-x}x^q) [[/math]]

Show Proof

The story here is very similar to that of the Legendre polynomials. Consider the Hilbert space [math]H=L^2[0,\infty)[/math], with the following scalar product on it:

[[math]] \lt f,g \gt =\int_0^\infty f(x)g(x)e^{-x}\,dx [[/math]]

(1) The orthogonal basis obtained by applying Gram-Schmidt to the Weierstrass basis [math]\{x^q\}[/math] is then the basis formed by the Laguerre polynomials [math]\{L_q\}[/math].

(2) We have the explicit formula for [math]L_q[/math] in the statement, which is analogous to the Rodrigues formula for the Legendre polynomials.

(3) The first assertion follows from the fact that the coefficients of the associated Laguerre polynomials satisfy the equation for the coefficients of [math]v(p)[/math].

(4) Alternatively, the first assertion follows as well by using an equation for the Laguerre polynomials, which is very similar to the Legendre equation.

■

With the above result in hand, we can now improve Theorem 16.25, as follows:

Theorem

The wave functions of the hydrogen atom are given by

[[math]] \phi_{nlm}(r,s,t)=\sqrt{\left(\frac{2}{na}\right)^3\frac{(n-l-1)!}{2n(n+l)!}}e^{-r/na}\left(\frac{2r}{na}\right)^lL_{n-l-1}^{2l+1}\left(\frac{2r}{na}\right)\alpha_l^m(s,t) [[/math]]

with [math]\alpha_l^m(s,t)[/math] being the spherical harmonics found before.

Show Proof

This follows indeed by putting together what we have, namely Theorem 16.25 and Proposition 16.29, and then doing some remaining work, concerning the normalization of the wave function, which leads to the normalization factor appearing above.

■

And good news, that is all. The above formula is all you need, in everyday life. \begin{exercises} Congratulations for having read this book, and no exercises for this final chapter. However, if you enjoyed this book, and looking for more to read, have a look at the various books referenced below. Normally these are all good books, which all sort of go to the point, without bothering much with annoying details, and normally you should like them too. So, have a look at them, and start with the one that you like the most.

Before everything, however, learn more calculus, and especially do many exercises, as many as needed, first in order to be at ease with calculus, and then in order to really love calculus. In fact, and in the hope that you got it, the present book was just an introduction to calculus, not calculus itself. Just a beginning. \begin{thebibliography}{99} \baselineskip=18.2pt \bibitem{ar1}V.I. Arnold, Ordinary differential equations, Springer (1973). \bibitem{ar2}V.I. Arnold, Mathematical methods of classical mechanics, Springer (1974). \bibitem{ar3}V.I. Arnold, Catastrophe theory, Springer (1984). \bibitem{ar4}V.I. Arnold, Lectures on partial differential equations, Springer (1997). \bibitem{akh}V.I. Arnold and B.A. Khesin, Topological methods in hydrodynamics, Springer (1998). \bibitem{at1}M.F. Atiyah, K-theory, CRC Press (1964). \bibitem{at2}M.F. Atiyah, The geometry and physics of knots, Cambridge Univ. Press (1990). \bibitem{ama}M.F. Atiyah and I.G. 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General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].