16c. Spherical coordinates

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In order to solve now the hydrogen atom, the idea will be that of reformulating the Schrödinger equation in spherical coordinates. And for this purpose, we will need:

Theorem

The Laplace operator in spherical coordinates is

[[math]] \Delta=\frac{1}{r^2}\cdot\frac{d}{dr}\left(r^2\cdot\frac{d}{dr}\right) +\frac{1}{r^2\sin s}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d}{ds}\right) +\frac{1}{r^2\sin^2s}\cdot\frac{d^2}{dt^2} [[/math]]

with our standard conventions for these coordinates, in 3D.

Show Proof

There are several proofs here, a short, elementary one being as follows:

(1) Let us first see how [math]\Delta[/math] behaves under a change of coordinates [math]\{x_i\}\to\{y_i\}[/math], in arbitrary [math]N[/math] dimensions. Our starting point is the chain rule for derivatives:

[[math]] \frac{d}{dx_i}=\sum_j\frac{d}{dy_j}\cdot\frac{dy_j}{dx_i} [[/math]]

By using this rule, then Leibnitz for products, then again this rule, we obtain:

[[math]] \begin{eqnarray*} \frac{d^2f}{dx_i^2} &=&\sum_j\frac{d}{dx_i}\left(\frac{df}{dy_j}\cdot\frac{dy_j}{dx_i}\right)\\ &=&\sum_j\frac{d}{dx_i}\left(\frac{df}{dy_j}\right)\cdot\frac{dy_j}{dx_i}+\frac{df}{dy_j}\cdot\frac{d}{dx_i}\left(\frac{dy_j}{dx_i}\right)\\ &=&\sum_j\left(\sum_k\frac{d}{dy_k}\cdot\frac{dy_k}{dx_i}\right) \left(\frac{df}{dy_j}\right)\cdot\frac{dy_j}{dx_i}+\frac{df}{dy_j}\cdot\frac{d^2y_j}{dx_i^2}\\ &=&\sum_{jk}\frac{d^2f}{dy_kdy_j}\cdot\frac{dy_k}{dx_i}\cdot\frac{dy_j}{dx_i}+\sum_j\frac{df}{dy_j}\cdot\frac{d^2y_j}{dx_i^2} \end{eqnarray*} [[/math]]

(2) Now by summing over [math]i[/math], we obtain the following formula, with [math]A[/math] being the derivative of [math]x\to y[/math], that is to say, the matrix of partial derivatives [math]dy_i/dx_j[/math]:

[[math]] \begin{eqnarray*} \Delta f &=&\sum_{ijk}\frac{d^2f}{dy_kdy_j}\cdot\frac{dy_k}{dx_i}\cdot\frac{dy_j}{dx_i}+\sum_{ij}\frac{df}{dy_j}\cdot\frac{d^2y_j}{dx_i^2}\\ &=&\sum_{ijk}A_{ki}A_{ji}\frac{d^2f}{dy_kdy_j}+\sum_{ij}\frac{d^2y_j}{dx_i^2}\cdot\frac{df}{dy_j}\\ &=&\sum_{jk}(AA^t)_{jk}\frac{d^2f}{dy_kdy_j}+\sum_j\Delta(y_j)\frac{df}{dy_j} \end{eqnarray*} [[/math]]

(3) So, this will be the formula that we will need. Observe that this formula can be further compacted as follows, with all the notations being self-explanatory:

[[math]] \Delta f=Tr(AA^tH_y(f))+ \lt \Delta(y),\nabla_y(f) \gt [[/math]]

(4) Getting now to spherical coordinates, [math](x,y,z)\to(r,s,t)[/math], the derivative of the inverse, obtained by differentiating [math]x,y,z[/math] with respect to [math]r,s,t[/math], is given by:

[[math]] A^{-1}=\begin{pmatrix} \cos s&-r\sin s&0\\ \sin s\cos t&r\cos s\cos t&-r\sin s\sin t\\ \sin s\sin t&r\cos s\sin t&r\sin s\cos t \end{pmatrix} [[/math]]

The product [math](A^{-1})^tA^{-1}[/math] of the transpose of this matrix with itself is then:

[[math]] \begin{pmatrix} \cos s&\sin s\cos t&\sin s\sin t\\ -r\sin s&r\cos s\cos t&r\cos s\sin t\\ 0&-r\sin s\sin t&r\sin s\cos t \end{pmatrix} \begin{pmatrix} \cos s&-r\sin s&0\\ \sin s\cos t&r\cos s\cos t&-r\sin s\sin t\\ \sin s\sin t&r\cos s\sin t&r\sin s\cos t \end{pmatrix} [[/math]]

But everything simplifies here, and we have the following remarkable formula, which by the way is something very useful, worth to be memorized:

[[math]] (A^{-1})^tA^{-1}=\begin{pmatrix}1&0&0\\ 0&r^2&0\\ 0&0&r^2\sin^2s\end{pmatrix} [[/math]]

Now by inverting, we obtain the following formula, in relation with the above:

[[math]] AA^t=\begin{pmatrix}1&0&0\\ 0&1/r^2&0\\ 0&0&1/(r^2\sin^2s)\end{pmatrix} [[/math]]

(5) Let us compute now the Laplacian of [math]r,s,t[/math]. We first have the following formula, that we will use many times in what follows, and is worth to be memorized:

[[math]] \begin{eqnarray*} \frac{dr}{dx} &=&\frac{d}{dx}\sqrt{x^2+y^2+z^2}\\ &=&\frac{1}{2}\cdot\frac{2x}{\sqrt{x^2+y^2+z^2}}\\ &=&\frac{x}{r} \end{eqnarray*} [[/math]]

Of course the same computation works for [math]y,z[/math] too, and we therefore have:

[[math]] \frac{dr}{dx}=\frac{x}{r}\quad,\quad\frac{dr}{dy}=\frac{y}{r}\quad,\quad\frac{dr}{dz}=\frac{z}{r} [[/math]]

(6) By using the above formulae, twice, we can compute the Laplacian of [math]r[/math]:

[[math]] \begin{eqnarray*} \Delta(r) &=&\Delta\left(\sqrt{x^2+y^2+z^2}\right)\\ &=&\frac{d}{dx}\left(\frac{x}{r}\right)+\frac{d}{dy}\left(\frac{y}{r}\right)+\frac{d}{dz}\left(\frac{z}{r}\right)\\ &=&\frac{r^2-x^2}{r^3}+\frac{r^2-y^2}{r^3}+\frac{r^2-z^2}{r^3}\\ &=&\frac{2}{r} \end{eqnarray*} [[/math]]

(7) In what regards now [math]s[/math], the computation here goes as follows:

[[math]] \begin{eqnarray*} \Delta(s) &=&\Delta\left(\arccos\left(\frac{x}{r}\right)\right)\\ &=&\frac{d}{dx}\left(-\frac{\sqrt{r^2-x^2}}{r^2}\right) +\frac{d}{dy}\left(\frac{xy}{r^2\sqrt{r^2-x^2}}\right) +\frac{d}{dz}\left(\frac{xz}{r^2\sqrt{r^2-x^2}}\right)\\ &=&\frac{2x\sqrt{r^2-x^2}}{r^4} +\frac{r^2(z^2-2y^2)+2x^2y^2}{r^4\sqrt{r^2-x^2}} +\frac{r^2(y^2-2z^2)+2x^2z^2}{r^4\sqrt{r^2-x^2}}\\ &=&\frac{2x\sqrt{r^2-x^2}}{r^4} +\frac{x(2x^2-r^2)}{r^4\sqrt{r^2-x^2}}\\ &=&\frac{x}{r^2\sqrt{r^2-x^2}}\\ &=&\frac{\cos s}{r^2\sin s} \end{eqnarray*} [[/math]]

(8) Finally, in what regards [math]t[/math], the computation here goes as follows:

[[math]] \begin{eqnarray*} \Delta(t) &=&\Delta\left(\arctan\left(\frac{z}{y}\right)\right)\\ &=&\frac{d}{dx}(0)+\frac{d}{dy}\left(-\frac{z}{y^2+z^2}\right)+\frac{d}{dz}\left(\frac{y}{y^2+z^2}\right)\\ &=&0-\frac{2yz}{(y^2+z^2)^2}+\frac{2yz}{(y^2+z^2)^2}\\ &=&0 \end{eqnarray*} [[/math]]

(9) We can now plug the data from (4) and (6,7,8) in the general formula that we found in (2) above, and we obtain in this way:

[[math]] \begin{eqnarray*} \Delta f &=&\frac{d^2f}{dr^2}+\frac{1}{r^2}\cdot\frac{d^2f}{ds^2}+\frac{1}{r^2\sin^2s}\cdot\frac{d^2f}{dt^2}+\frac{2}{r}\cdot\frac{df}{dr}+\frac{\cos s}{r^2\sin s}\cdot\frac{df}{ds}\\ &=&\frac{2}{r}\cdot\frac{df}{dr}+\frac{d^2f}{dr^2}+\frac{\cos s}{r^2\sin s}\cdot\frac{df}{ds}+\frac{1}{r^2}\cdot\frac{d^2f}{ds^2}+\frac{1}{r^2\sin^2s}\cdot\frac{d^2f}{dt^2}\\ &=&\frac{1}{r^2}\cdot\frac{d}{dr}\left(r^2\cdot\frac{df}{dr}\right) +\frac{1}{r^2\sin s}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{df}{ds}\right) +\frac{1}{r^2\sin^2s}\cdot\frac{d^2f}{dt^2} \end{eqnarray*} [[/math]]

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

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Still with me, I hope, and do not worry, one day you will have such computations for breakfast. We can now reformulate the Schrödinger equation in spherical coordinates, and separate the variables, which leads to a radial and angular equation, as follows:

Theorem

The time-independent Schrödinger equation in spherical coordinates separates, for solutions of type [math]\phi=\rho(r)\alpha(s,t)[/math], into two equations, as follows,

[[math]] \frac{d}{dr}\left(r^2\cdot\frac{d\rho}{dr}\right) -\frac{2mr^2}{h^2}(V-E)\rho=K\rho [[/math]]

[[math]] \sin s\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\alpha}{ds}\right) +\frac{d^2\alpha}{dt^2}=-K\sin^2s\cdot\alpha [[/math]]

with [math]K[/math] being a constant, called radial equation, and angular equation.

Show Proof

By using the formula in Theorem 16.14, the time-independent Schrödinger equation reformulates in spherical coordinates as follows:

[[math]] (V-E)\phi=\frac{h^2}{2m}\left[\frac{1}{r^2}\cdot\frac{d}{dr}\left(r^2\cdot\frac{d\phi}{dr}\right) +\frac{1}{r^2\sin s}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\phi}{ds}\right) +\frac{1}{r^2\sin^2s}\cdot\frac{d^2\phi}{dt^2}\right] [[/math]]

Let us look now for separable solutions for this latter equation, consisting of a radial part and an angular part, as in the statement, namely:

[[math]] \phi(r,s,t)=\rho(r)\alpha(s,t) [[/math]]

By plugging this function into our equation, we obtain:

[[math]] (V-E)\rho\alpha=\frac{h^2}{2m}\left[\frac{\alpha}{r^2}\cdot\frac{d}{dr}\left(r^2\cdot\frac{d\rho}{dr}\right) +\frac{\rho}{r^2\sin s}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\alpha}{ds}\right) +\frac{\rho}{r^2\sin^2s}\cdot\frac{d^2\alpha}{dt^2}\right] [[/math]]

In order to solve this equation, we will do two manipulations. First, by multiplying everything by [math]2mr^2/(h^2\rho\alpha)[/math], this equation takes the following more convenient form:

[[math]] \frac{2mr^2}{h^2}(V-E)=\frac{1}{\rho}\cdot\frac{d}{dr}\left(r^2\cdot\frac{d\rho}{dr}\right) +\frac{1}{\alpha\sin s}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\alpha}{ds}\right) +\frac{1}{\alpha\sin^2s}\cdot\frac{d^2\alpha}{dt^2} [[/math]]

Now observe that by moving the radial terms to the left, and the angular terms to the right, this latter equation can be written as follows:

[[math]] \frac{2mr^2}{h^2}(V-E)-\frac{1}{\rho}\cdot\frac{d}{dr}\left(r^2\cdot\frac{d\rho}{dr}\right) =\frac{1}{\alpha\sin^2s}\left[\sin s\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\alpha}{ds}\right) +\frac{d^2\alpha}{dt^2}\right] [[/math]]

Since this latter equation is now separated between radial and angular variables, both sides must be equal to a certain constant [math]-K[/math], as follows:

[[math]] \frac{2mr^2}{h^2}(V-E)-\frac{1}{\rho}\cdot\frac{d}{dr}\left(r^2\cdot\frac{d\rho}{dr}\right) =-K [[/math]]

[[math]] \frac{1}{\alpha\sin^2s}\left[\sin s\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\alpha}{ds}\right) +\frac{d^2\alpha}{dt^2}\right]=-K [[/math]]

But this leads to the conclusion in the statement.

■

Let us first study the angular equation. We first have the following result:

Proposition

The angular equation that we found before, namely

[[math]] \sin s\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\alpha}{ds}\right) +\frac{d^2\alpha}{dt^2}=-K\sin^2s\cdot\alpha [[/math]]

separates, for solutions of type [math]\alpha=\sigma(s)\theta(t)[/math], into two equations, as follows,

[[math]] \frac{1}{\theta}\cdot\frac{d^2\theta}{dt^2}=-m^2 [[/math]]

[[math]] \frac{\sin s}{\sigma}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\sigma}{ds}\right) +K\sin^2s=m^2 [[/math]]

with [math]m[/math] being a constant, called azimuthal equation, and polar equation.

Show Proof

This is something elementary, the idea being as follows:

(1) Let us first recall that [math]r\in[0,\infty)[/math] is the radius, [math]s\in[0,\pi][/math] is the polar angle, and [math]t\in[0,2\pi][/math] is the azimuthal angle. Be said in passing, there are several conventions and notations here, and the above ones, that we use here, come from the general ones in [math]N[/math] dimensions, because further coordinates can be easily added, in the obvious way.

(2) Getting back now to our question, by plugging [math]\alpha=\sigma(s)\theta(t)[/math] into the angular equation, we obtain:

[[math]] \sin s\cdot\theta\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\sigma}{ds}\right) +\sigma\cdot\frac{d^2\theta}{dt^2}=-K\sin^2s\cdot\sigma\theta [[/math]]

By dividing everything by [math]\sigma\theta[/math], this equation can be written as follows:

[[math]] -\frac{1}{\theta}\cdot\frac{d^2\theta}{dt^2}=\frac{\sin s}{\sigma}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\sigma}{ds}\right) +K\sin^2s [[/math]]

Since the variables are separated, we must have, for a certain constant [math]m[/math]:

[[math]] \frac{1}{\theta}\cdot\frac{d^2\theta}{dt^2}=-m^2 [[/math]]

[[math]] \frac{\sin s}{\sigma}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\sigma}{ds}\right) +K\sin^2s=m^2 [[/math]]

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

■

Regarding the azimuthal equation, things here are quickly settled, as follows:

Proposition

The solutions of the azimuthal equation, namely

[[math]] \frac{1}{\theta}\cdot\frac{d^2\theta}{dt^2}=-m^2 [[/math]]

are the functions as follows, with [math]a,b\in\mathbb C[/math] being parameters,

[[math]] \theta(t)=ae^{imt}+be^{-imt} [[/math]]

and with only the case [math]m\in\mathbb Z[/math] being acceptable, on physical grounds.

Show Proof

The first assertion is clear, because we have a second order equation, and two obvious solutions for it, [math]e^{\pm imt}[/math], and then their linear combinations, and that's all. Regarding the last assertion, the point here is that by using [math]\theta(t)=\theta(t+2\pi)[/math], which is a natural physical assumption on the wave function, we are led to [math]m\in\mathbb Z[/math], as stated.

■

We are now about to solve the angular equation, with only the polar equation remaining to be studied. However, in practice, this polar equation is 10 times more difficult that everything what we did so far, so be patient. We first have:

Proposition

The polar equation that we found before, namely

[[math]] \frac{\sin s}{\sigma}\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\sigma}{ds}\right) +K\sin^2s=m^2 [[/math]]

with [math]m\in\mathbb Z[/math], translates via [math]\sigma(s)=f(\cos s)[/math] into the following equation,

[[math]] (1-x^2)f''(x)-2xf'(x)=\left(\frac{m^2}{1-x^2}-K\right)f(x) [[/math]]

where [math]x=\cos s[/math], called Legendre equation.

Show Proof

Let us first do a number of manipulations on our equation, before making the change of variables. By multiplying by [math]\sigma[/math], our equation becomes:

[[math]] \sin s\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\sigma}{ds}\right)=\left(m^2-K\sin^2s\right)\sigma [[/math]]

By differentiating at left, this equation becomes:

[[math]] \sin s\left(\cos s\cdot\sigma'+\sin s\cdot\sigma''\right)=\left(m^2-K\sin^2s\right)\sigma [[/math]]

Finally, by dividing everything by [math]\sin^2s[/math], our equation becomes:

[[math]] \sigma''+\frac{\cos s}{\sin s}\cdot\sigma'=\left(\frac{m^2}{\sin^2s}-K\right)\sigma [[/math]]

Now let us set [math]\sigma(s)=f(\cos s)[/math]. With this change of variables, we have:

[[math]] \sigma=f(\cos s) [[/math]]

[[math]] \sigma'=-\sin s\cdot f'(\cos s) [[/math]]

[[math]] \sigma''=-\cos s\cdot f'(\cos s)+\sin^2s\cdot f''(\cos s) [[/math]]

By plugging this data, our radial equation becomes:

[[math]] \sin^2s\cdot f''(\cos s)-2\cos s\cdot f'(\cos s)=\left(\frac{m^2}{\sin^2s}-K\right)f(\cos s) [[/math]]

Now with [math]x=\cos s[/math], which is our new variable, this equation reads:

[[math]] (1-x^2)f''(x)-2xf'(x)=\left(\frac{m^2}{1-x^2}-K\right)f(x) [[/math]]

But this is the Legendre equation, as stated.

■

Here comes now the difficult point. We have the following non-trivial result:

Theorem

The solutions of the Legendre equation, namely

[[math]] (1-x^2)f''(x)-2xf'(x)=\left(\frac{m^2}{1-x^2}-K\right)f(x) [[/math]]

can be explicitely computed, via complicated math, and only the case

[[math]] K=l(l+1)\quad:\quad l\in\mathbb N [[/math]]

is acceptable, on physical grounds.

Show Proof

The first part is something quite complicated, involving the hypergeometric functions [math]\!\!{\ }_2F_1[/math], that you don't want to hear about, believe me. As for the second part, analysis and physics, this is something not trivial either. See Griffiths .

■

In order to construct the solutions, we will need:

Theorem

The orthonormal basis of [math]L^2[-1,1][/math] obtained by starting with the Weierstrass basis [math]\{x^l\}[/math], and doing Gram-Schmidt, is the family of polynomials [math]\{P_l\}[/math], with each [math]P_l[/math] being of degree [math]l[/math], and with positive leading coefficient, subject to:

[[math]] \int_{-1}^1P_k(x)P_l(x)\,dx=\delta_{kl} [[/math]]

These polynomials, called Legendre polynomials, satisfy the equation

[[math]] (1-x^2)P_l''(x)-2xP_l'(x)+l(l+1)P_l(x)=0 [[/math]]

which is the Legendre equation at [math]m=0[/math], and with [math]K=l(l+1)[/math]. Moreover,

[[math]] P_l(x)=\frac{1}{2^ll!}\left(\frac{d}{dx}\right)^l(x^2-1)^l [[/math]]

which is called the Rodrigues formula for Legendre polynomials.

Show Proof

As a first observation, we are not lost somewhere in abstract math, because of the occurrence of the Legendre equation. As for the proof, this goes as follows:

(1) The first assertion is clear, because the Gram-Schmidt procedure applied to the Weierstrass basis [math]\{x^l\}[/math] can only lead to a certain family of polynomials [math]\{P_l\}[/math], with each [math]P_l[/math] being of degree [math]l[/math], and also unique, if we assume that it has positive leading coefficient, with this [math]\pm[/math] choice being needed, as usual, at each step of Gram-Schmidt.

(2) In order to have now an idea about these beasts, here are the first few of them, which can be obtained say via a straightforward application of Gram-Schmidt:

[[math]] \begin{eqnarray*} P_0&=&1\\ P_1&=&x\\ P_2&=&(3x^2-1)/2\\ P_3&=&(5x^3-3x)/2\\ P_4&=&(35x^4-30x^2+3)/8\\ P_5&=&(63x^5-70x^3+15x)/8 \end{eqnarray*} [[/math]]

(3) Now thinking about what Gram-Schmidt does, this is certainly something by recurrence. And examining the recurrence leads to the Legendre equation, as stated.

(4) As for the Rodrigues formula, by uniqueness no need to try to understand where this formula comes from, and we have two choices here, either by verifying that [math]\{P_l\}[/math] is orthonormal, or by verifying the Legendre equation. And both methods work.

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Going ahead now, we can solve in fact the Legendre equation at any [math]m[/math], as follows:

Proposition

The general Legendre equation, with parameters [math]m\in\mathbb N[/math] and [math]K=l(l+1)[/math] with [math]l\in\mathbb N[/math], namely

[[math]] (1-x^2)f''(x)-2xf'(x)=\left(\frac{m^2}{1-x^2}-l(l+1)\right)f(x) [[/math]]

is solved by the following functions, called Legendre functions,

[[math]] P_l^m(x)=(-1)^m(1-x^2)^{m/2}\left(\frac{d}{dx}\right)^mP_l(x) [[/math]]

where [math]P_l[/math] are as before the Legendre polynomials. Also, we have

[[math]] P_l^m(x)=(-1)^m\frac{(1-x^2)^{m/2}}{2^ll!}\left(\frac{d}{dx}\right)^{l+m}(x^2-1)^l [[/math]]

called Rodrigues formula for Legendre functions.

Show Proof

The first assertion is something elementary, coming by differentiating [math]m[/math] times the Legendre equation, which leads to the general Legendre equation. As for the second assertion, this follows from the Rodrigues formula for Legendre polynomials.

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And this is the end of our study. Eventually. By putting together all the above results, we are led to the following conclusion:

Theorem

The separated solutions [math]\alpha=\sigma(s)\theta(t)[/math] of the angular equation,

[[math]] \sin s\cdot\frac{d}{ds}\left(\sin s\cdot\frac{d\alpha}{ds}\right) +\frac{d^2\alpha}{dt^2}=-K\sin^2s\cdot\alpha [[/math]]

are given by the following formulae, where [math]l\in\mathbb N[/math] is such that [math]K=l(l+1)[/math],

[[math]] \sigma(s)=P_l^m(\cos s)\quad,\quad\theta(t)=e^{imt} [[/math]]

and where [math]m\in\mathbb Z[/math] is a constant, and with [math]P_l^m[/math] being the Legendre function,

[[math]] P_l^m(x)=(-1)^m(1-x^2)^{m/2}\left(\frac{d}{dx}\right)^mP_l(x) [[/math]]

where [math]P_l[/math] are the Legendre polynomials, given by the following formula:

[[math]] P_l(x)=\frac{1}{2^ll!}\left(\frac{d}{dx}\right)^l(x^2-1)^l [[/math]]

These solutions [math]\alpha=\sigma(s)\theta(t)[/math] are called spherical harmonics.

Show Proof

This follows indeed from all the above, and with the comment that everything is taken up to linear combinations. We will normalize the wave function later.

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General references

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