*Michael Fowler, University of Virginia*

The aim of this section is to give a fairly brief review of waves in various shaped elastic media$\u2014$beginning with a taut string, then going on to an elastic sheet, a drumhead, first of rectangular shape then circular, and finally considering elastic waves on a spherical surface, like a balloon.

The reason we look at this material here is that these are “real waves”, hopefully not too difficult to think about, and yet mathematically they are the solutions of the same wave equation the Schrödinger wave function obeys in various contexts, so should be helpful in visualizing solutions to that equation, in particular for the hydrogen atom.

We begin with the stretched string, then go on to the rectangular and circular drumheads. We derive the wave equation from $F=ma$ for a little bit of string or sheet. The equation corresponds exactly to the Schrödinger equation for a free particle with the given boundary conditions.

The most important section here is the one on waves on a
sphere. We find the first few standing
wave solutions. These waves correspond
to Schrödinger’s wave function for a free particle on the surface of a
sphere. This is what we need to analyze to
understand the hydrogen atom, because using separation of variables we split
the electron’s motion into radial motion and motion on the surface of a
sphere. The potential *only* affects the *radial* motion, so the motion on the sphere is free particle motion,
described by the same waves we find for vibrations of a balloon. (There *is* the generalization to
complex non-standing waves, parallel to the one-dimensional extension from $\mathrm{sin}kx$ and $\mathrm{cos}kx$ to ${e}^{ikx}$ and ${e}^{-ikx},$ but
this does not affect the structure of the equations.)

Let’s begin by reminding ourselves of the wave equation for waves on a taut string, stretched between $x=0$ and $x=L,$ tension $T$ newtons, density $\rho $ kg/meter. Assuming the string’s equilibrium position is a straight horizontal line (and, therefore, ignoring gravity), and assuming it oscillates in a vertical plane, we use $f\left(x,t\right)$ to denote its shape at instant $t,$ so $f\left(x,t\right)$ is the instantaneous upward displacement of the string at position $x.$ We assume the amplitude of oscillation remains small enough that the string tension can be taken constant throughout.

The *wave equation*
is derived by applying $F=ma$ to an infinitesimal length $dx$ of string (see the diagram below). We picture our little length of string as
bobbing up and down in simple harmonic motion, which we can verify by finding
the net force on it as follows.

At the left hand end of the string fragment, point $x,$ say, the tension $T$ is at a small angle $df\left(x\right)/dx$ to the horizontal, since the tension acts
necessarily along the line of the string.
Since it is pulling to the left, there is a *downward* force component $Tdf\left(x\right)/dx.$ At the
right hand end of the string fragment there is an *upward* force $Tdf\left(x+dx\right)/dx.$

Putting $f\left(x+dx\right)=f\left(x\right)+\left(df/dx\right)dx,$ and adding the almost canceling upwards and
downwards forces together, we find a net force $T\left({d}^{2}f/d{x}^{2}\right)dx$ *T*(*d*^{2}*f*/*dx*^{2})*dx* on the bit of string. The string mass is $\rho dx,$ so $F=ma$ becomes

$$T\frac{{\partial}^{2}f(x,t)}{\partial {x}^{2}}dx=\rho dx\frac{{\partial}^{2}f(x,t)}{\partial {t}^{2}}$$

giving the standard wave equation

$$\frac{{\partial}^{2}f(x,t)}{\partial {x}^{2}}=\frac{1}{{c}^{2}}\frac{{\partial}^{2}f(x,t)}{\partial {t}^{2}}$$

with wave velocity given by ${c}^{2}=T/\rho .$ (A more detailed discussion is given in my Physics 152 Course, plus an animation here.)

This equation can of course be solved by separation of
variables, $f\left(x,t\right)=f\left(x\right)g\left(t\right),$ and the
equation for $f\left(x\right)$ is identical to the time independent
Schrödinger equation for a particle confined to $\left(0,L\right)$ by infinitely high walls at the two ends. This is why the eigenfunctions (states of
definite energy) for a Schrödinger particle confined to $\left(0,L\right)$ are identical to the modes of vibration of a
string held between those points.
(However, it should be realized that the *time dependence* of the string wave equation and the Schrödinger *time-dependent* equation are quite
different, so a nonstationary state, one corresponding to a sum of waves of
different energies, will develop differently in the two systems.)

Let us now move up to two dimensions, and consider the analogue to the taut string problem, which is waves in a taut horizontal elastic sheet, like, say, a drumhead. Let us assume a rectangular drumhead to begin with. Then, parallel to the argument above, we would apply $F=ma$ to a small square of elastic with sides parallel to the $x$ and $y$ axes. The tension from the rest of the sheet tugs along all four sides of the little square, and we realize that tension in a sheet of this kind must be defined in newtons per meter, so the force on one side of the little square is given by multiplying this “tension per unit length” by the length of the side.

Following the string analysis, we take the vertical
displacement of the sheet at instant $t$ to be given by $f\left(x,y,t\right).$ We assume this displacement is quite small, so
the tension itself doesn’t vary, and that each bit of the sheet oscillates up
and down (the sheet is not tugged to one side).
Suppose the bottom left-hand corner (so to speak) of the square is $\left(x,y\right),$ the top right-hand corner $\left(x+dx,\text{\hspace{0.17em}}y+dy\right).$ Then
the left and right edges of the square have lengths $dy.$ Now, what is the total force on the left
edge? The force is $Tdy,$ in the
local plane of the sheet, perpendicular to the edge $dy.$ Factoring in the slope of the sheet in the
direction of the force, the vertically downward component of the force must be $Tdy\partial f\left(x,y,t\right)/\partial x.$ By the
same argument, the force on the right hand edge has to have an *upward* component $Tdy\partial f\left(x+dx,y,t\right)/\partial x.$ *Tdy* ∂*f*(*x+dx*,
*y*, *t*)/∂*x*.

Thus the net upward force on the little square from the sheet tension tugging on its left and right sides is

_{$$Tdy\left(\frac{\partial f(x+dx,y,t)}{\partial x}-\frac{\partial f(x,y,t)}{\partial x}\right)=Tdydx\left(\frac{{\partial}^{2}f}{\partial {x}^{2}}\right).$$}

The net vertical force from the sheet tension on the other two sides is the same with $x$ and $y$ interchanged.

The mass of the little square of elastic sheet is $\rho dxdy,$ and its upward acceleration is ${\partial}^{2}f/\partial {t}^{2}.$ Thus $F=ma$ becomes:

$$Tdydx\left(\frac{{\partial}^{2}f}{\partial {x}^{2}}\right)+Tdydx\left(\frac{{\partial}^{2}f}{\partial {y}^{2}}\right)=\rho dxdy\left(\frac{{\partial}^{2}f}{\partial {t}^{2}}\right),$$

giving

$$\left(\frac{{\partial}^{2}f}{\partial {x}^{2}}\right)+\left(\frac{{\partial}^{2}f}{\partial {y}^{2}}\right)=\frac{1}{{c}^{2}}\left(\frac{{\partial}^{2}f}{\partial {t}^{2}}\right).$$

with ${c}^{2}=T/\rho .$

This equation can be solved by separation of variables, and the time independent part is identical to the Schrödinger time independent equation for a free particle confined to a rectangular box.

A similar argument gives the wave equation for a circular drumhead, this time in $\left(r,\phi \right)$ coordinates (we use $\phi $ rather than $\theta $ here because of its parallel role in the spherical case, to be discussed shortly).

This time, instead of a tiny square of elastic, we take the small area $rdrd\phi $ bounded by the circles of radius $r$ and $r+dr$ and lines through the origin at angles $\phi $ and $\phi +d\phi .$ Now, the downward force from the tension $T$ in the sheet on the inward curved edge, which has length $rd\phi ,$ is $Trdrd\phi \partial f\left(r.\phi .t\right)/\partial r.$ On putting this together with the upward force from the other curved edge, it is important to realize that the $r$ in $Trd\phi $ varies as well as $\partial f/\partial r$ on going from $r$ to $r+dr,$ so the sum of the two terms is $Td\phi \left(\partial /\partial r\right)\left(r\partial f/\partial r\right)dr.$ To find the vertical elastic forces from the straight sides, we need to find how the sheet slopes in the direction perpendicular to those sides. The measure of length in that direction is not $\phi ,$ but $r\phi ,$ so the slope is $\left(1/r\right)\left(\partial f/\partial \phi \right),$ and the net upward elastic force contribution from those sides (which have length $dr$ ) is $Tdrd\phi \left(\partial /\partial \phi \right)\left(1/r\right)\left(\partial f/\partial \phi \right).$

Writing $F=ma$ for this small area of elastic sheet, of mass $\rho rdrd\phi ,$ gives then

$$Td\phi \frac{\partial}{\partial r}r\frac{\partial f}{\partial r}dr+Tdrd\phi \frac{\partial}{\partial \phi}\frac{1}{r}\frac{\partial f}{\partial \phi}=\rho rdrd\phi \frac{{\partial}^{2}f}{\partial {t}^{2}}$$

which can be written

$$\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial f}{\partial r}+\frac{1}{{r}^{2}}\frac{{\partial}^{2}f}{\partial {\phi}^{2}}=\frac{1}{{c}^{2}}\frac{{\partial}^{2}f}{\partial {t}^{2}}.$$

This is the wave equation in polar coordinates. Separation of variables gives a radial equation called Bessel’s equation, the solutions are called Bessel functions. The corresponding electron standing waves have actually been observed for an electron captured in a circular corral on a surface.

Finally, let us consider elastic waves on the surface of a sphere, such as an inflated spherical balloon. The natural coordinate system here is spherical polar coordinates, with $\theta $ measuring latitude, but counting the north pole as zero, the south pole as $\pi .$ The angle $\phi $ measures longitude from some agreed origin.

We take a small elastic element bounded by longitude lines $\phi $ and $\phi +d\phi $ and latitudes $\theta $ and $\theta +d\theta .$ For a sphere of radius $r,$ the sides of the element have lengths $r\mathrm{sin}\theta d\phi ,\text{\hspace{0.17em}}rd\theta .$ Beginning with one of the longitude sides, length $rd\theta ,$ tension $T,$ the only slightly tricky point is figuring its deviation from the local horizontal, which is $\left(1/r\mathrm{sin}\theta \right)\left(\partial f/\partial \phi \right),$ since increasing $\phi $ by $d\phi $ means moving an actual distance $r\mathrm{sin}\theta d\phi $ on the surface, just analogous with the circular case above. Hence, by the usual method, the actual "vertical" force from tension on the two longitude sides is

$$Trd\theta d\phi \frac{\partial}{\partial \phi}\frac{1}{r\mathrm{sin}\theta}\frac{\partial f}{\partial \phi}.$$

To find the force on the latitude sides, taking the top one first, the slope is given by $\left(1/r\right)\left(\partial f/\partial \theta \right),$ so the force is just $Tr\mathrm{sin}\theta d\phi \left(1/r\right)\left(\partial f/\partial \theta \right).$ On putting this together with the opposite side, it is necessary to recall that $\mathrm{sin}\theta $ as well as $f$ varies with $\theta .$ so the sum is given by: $Trd\phi d\theta \left(\partial /\partial \theta \right)\mathrm{sin}\theta \left(1/r\right)\left(\partial f/\partial \theta \right).$ We are now ready to write down $F=ma$ once more, the mass of the element is $\rho {r}^{2}\mathrm{sin}\theta d\theta d\phi .$ Canceling out elements common to both sides of the equation, we find:

$$\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\mathrm{sin}\theta \frac{\partial f}{\partial \theta}+\frac{1}{{\mathrm{sin}}^{2}\theta}\frac{{\partial}^{2}f}{\partial {\phi}^{2}}=\frac{{r}^{2}}{{c}^{2}}\frac{{\partial}^{2}f}{\partial {t}^{2}}$$.

Again, this wave equation is solved by separation of
variables. The time-independent
solutions are called the *Legendre*
functions. They are the basis for
analyzing the vibrations of any object with spherical symmetry, for example a
planet struck by an asteroid, or vibrations in the sun generated by large solar
flares, or the cosmic background microwave radiation.

Recall that for the two dimensional circular case, after
separation of variables the angular dependence was all in the solution to ∂^{2}*f*/∂^{$\phi $}^{2}
= −*λf*, and the physical solutions must fit smoothly around
the circle (no kinks, or it would not satisfy the wave equation at the kink),
leading to solutions sin*m*$\phi $ and cos*m*$\phi $ (or e^{im}$\phi $) with *m* an integer, and *λ* = *m*^{2} (this
is why we took *λ* with a minus sign in
the first equation).

For the spherical case, the equation containing all the angular dependence is

$$\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\mathrm{sin}\theta \frac{\partial f}{\partial \theta}+\frac{1}{{\mathrm{sin}}^{2}\theta}\frac{{\partial}^{2}f}{\partial {\phi}^{2}}=-\lambda f$$

The standard approach here is, again, separation of
variables. Taking the first term on the left hand side over to the right, and
multiplying throughout by sin^{2}*θ*
isolates the $\phi $ term:

$$\frac{{\partial}^{2}f}{\partial {\phi}^{2}}=-{\mathrm{sin}}^{2}\theta \left(\lambda f+\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\mathrm{sin}\theta \frac{\partial f}{\partial \theta}\right)$$

Writing now

$f\left(\theta ,\phi \right)={f}_{\theta}\left(\theta \right){f}_{\phi}\left(\phi \right)$

in the above equation, and dividing throughout by *f*, we find as usual that the left hand
side depends only on $\phi $, the
right hand side only on *θ*, so both
sides must be constants. Taking the
constant as *$\u2013$**m*^{2}, the $\phi $ solution is e^{±im}$\phi $, and one
can insert that in the *θ* equation to
give

$$\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\mathrm{sin}\theta \frac{\partial {f}_{\theta}}{\partial \theta}-\frac{{m}^{2}}{{\mathrm{sin}}^{2}\theta}{f}_{\theta}=-\lambda {f}_{\theta}$$

What about possible solutions that don’t depend on $\phi $? The equation would be the simpler

$$\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\mathrm{sin}\theta \frac{\partial f}{\partial \theta}=-\lambda f$$

Obviously, *f* =
constant is a solution (for *m* = 0)
with eigenvalue *λ* = 0.

Try *f* = cos*θ*.
It is easy to check that this is a solution with *λ* = 2.

Try *f* = sin*θ*.
This is *not* a solution. In
fact, we should have realized it cannot be a solution to the wave equation by
visualizing the shape of the elastic sheet near the north pole. If *f*
= sin*θ*, *f* =
0 at the pole, but rises linearly (for small *θ *) going away from the pole. Thus the pole is at the bottom of a
conical valley. But this conical valley amounts to a kink in the elastic sheet$\u2014$the slope
of the sheet has a discontinuity if one moves along a line passing through the
pole, so the shape of the sheet cannot satisfy the wave equation at that point.
This is somewhat obscured by working in spherical coordinated centered there,
but locally the north pole is no different from any other point on the sphere,
we could just switch to local (*x*,*y*) coordinates, and the cone
configuration would clearly *not*
satisfy the wave equation.

However, *f* = sin*θ *sin$\phi $ *is* a solution to the equation. It is a worthwhile exercise to see how the $\phi $ term gets rid of the conical point at the
north pole by considering the value of *f*
as the north pole is approached for various values of $\phi $: $\phi $ = 0, *π*/2, *π*, 3*π*/2 say. The sheet is now smooth at the pole!

We find *f* = sin*θ *cos$\phi $, sin*θ *sin$\phi $ (and so
sin*θ *e^{i}$\phi $) are
solutions with *λ* = 2.

It is straightforward to verify that *f* = cos^{2}*θ* $\u2013$ 1/3 is a solution with *λ* = 6.

Finally, we mention that other *λ* = 6 solutions are
sin*θ *cos*θ *sin$\phi $ and sin^{2}*θ *sin2$\phi $.

We do not attempt to find the general case here, but we have
done enough to see the beginnings of the pattern. We have found the series of eigenvalues 0, 2,
6, … . It turns out that the complete
series is given by *λ* = *l*(*l*
+ 1), with *l* = 0, 1, 2, … . This integer *l* is the analogue of the integer *m* in the wave on a circle case.
Recall that for the wave on the *circle*,
if we chose real wave functions (cos*m*$\phi $, sin*m*$\phi $, not e^{im}$\phi $) then 2*m* gave the number of nodes the wave had
(that is, *m* complete wavelengths
fitted around the circle). It turns out
that on the sphere *l* gives the number
of nodal *lines* (or circles) on the
surface. This assumes that we again
choose the $\phi $-component
of the wave function to be real, so that there will be *m* nodal circles passing through the two poles corresponding to the
zeros of the cos*m*$\phi $
term. We find that there are *l* $\u2013$ *m*
nodal *latitude* circles corresponding
to zeros of the function of *θ*.

*λ* *l* *m* *form of solution (unnormalized)*

0 0 0 constant

2 1 0 cos*θ*

2 1 1 sin*θ
e ^{i}*$\phi $

2 1 -1 sin*θ*
*e ^{-i}*$\phi $

6 2 0 cos^{2}*θ *$\u2013$ 1/3

6 2 ±1 cos*θ *sin*θ*
*e** ^{±i}*$\phi $

6 2 ±2 sin^{2}*θ*
*e*^{±2i}$\phi $

In three dimensions, the Schrödinger equation for an electron in a potential can be written:

$$-\frac{{\hslash}^{2}}{2m}\left(\frac{{\partial}^{2}\psi}{\partial {x}^{2}}+\frac{{\partial}^{2}\psi}{\partial {y}^{2}}+\frac{{\partial}^{2}\psi}{\partial {z}^{2}}\right)+V(x,y,z)\psi =E\psi $$

This is the obvious generalization of our previous two-dimensional discussion, and we will later be using the equation in the above form to discuss electron wave functions in metals, where the standard approach is to work with standing waves in a rectangular box.

Recall that in our original “derivation” of the Schrödinger equation, by analogy with the Maxwell wave equation for light waves, we argued that the differential wave operators arose from the energy-momentum relationship for the particle, that is

$$\frac{{p}_{x}^{2}+{p}_{y}^{2}+{p}_{z}^{2}}{2m}\psi \equiv -\frac{{\hslash}^{2}}{2m}\left(\frac{{\partial}^{2}\psi}{\partial {x}^{2}}+\frac{{\partial}^{2}\psi}{\partial {y}^{2}}+\frac{{\partial}^{2}\psi}{\partial {z}^{2}}\right)=-\frac{{\hslash}^{2}{\nabla}^{2}\psi}{2m}$$

so that the time-independent Schrödinger wave equation is
nothing but the statement that *E* = *K*.*E*. + *P*.*E*.
with the kinetic energy expressed as the equivalent operator.

To make further progress in solving the equation, the only trick we know is separation of variables. Unfortunately, this won’t work with the equation as given above in $\left(x,y,z\right)$ coordinates, because the potential energy term is a function of $x,y$ and $z$ in a nonseparable form. The solution is, however, fairly obvious: the potential is a function of radial distance from the origin, independent of direction. Therefore, we need to take as our coordinates the radial distance $r$ and two parameters fixing direction, $\theta $ and $\phi .$ We should then be able to separate the variables, because the potential only affects radial motion. No potential term will appear in the equations for $\theta ,\phi $ motion, which will describe free particle motion on the surface of a sphere.

It is worth thinking about what are the natural momentum
components for describing motion in spherical polar coordinates $\left(r,\theta ,\phi \right).$ The radial component of momentum, ${p}_{r},$ points
along the radius, of course. The $\theta $ -component ${p}_{\theta}$ points along a line of longitude, away from
the north pole if positive (remember $\theta $ itself measures *latitude*, counting the north pole as zero). The $\phi $-momentum
component, ${p}_{\phi},$ points along a line of latitude.

It will be important in understanding the hydrogen atom to
connect these momentum components $\left({p}_{r},{p}_{\theta},{p}_{\phi}\right)$ with the *angular
momentum* components of the atom.
Evidently, momentum in the $r$ -direction, which passes directly through the
center of the atom, contributes nothing to the angular momentum.

Consider now a particle for which ${p}_{r}={p}_{\theta}=0,$ only ${p}_{\phi}$ being nonzero. Classically, such a particle is circling the north pole at constant latitude $\theta ,$ say, so it is moving in space in a circle or radius $r\mathrm{sin}\theta $ in a plane perpendicular to the north-south axis of the sphere. Therefore, it has an angular momentum about that axis

${p}_{\phi}r\mathrm{sin}\theta ={L}_{z},\text{say}\text{.}$

(The standard transformation from $\left(x,y,z\right)$ coordinates to $\left(r,\theta ,\phi \right)$ coordinates is to take the north pole of the $\theta ,\phi $ sphere to be on the $z$ -axis.)

As we shall see in detail below, the wave equation
describing the $\phi $ motion is a simple one, with solutions of the
form ${e}^{im\phi}$ with integer $m,$ just as in the two-dimensional circular well.
This just means that *the component of
angular momentum along the **$z$** -axis
is quantized, **${L}_{z}=m\hslash ,$** with **$m$** an
integer*.

The* total* angular
momentum is $L=r{p}_{\perp},$ where ${p}_{\perp}$ is the component of the particle’s momentum
perpendicular to the radius, so

${p}_{\perp}^{2}={p}_{\phi}^{2}+{p}_{\theta}^{2}.$

Thus the square of the total angular momentum is (apart from
a constant factor) the kinetic energy of a particle moving freely on the
surface of a sphere. The equivalent
Schrödinger equation for such a particle is the wave equation given in the last
section for waves on a balloon. (This can be established by the standard change
of variables routine on the differential operators). Therefore, the solutions we found for elastic
waves on a sphere actually describe the angular momentum wave function of the
hydrogen atom. We conclude that *the total angular momentum is quantized, **${L}^{2}=l\left(l+1\right){\hslash}^{2},$** with **$l$** an
integer.*

The conclusions of our above waves on a sphere analysis of the angular momentum of a quantum mechanical particle are a little strange. We found that the component of angular momentum in the $z$ -direction must be a whole number of $\hslash $ units, yet the square of the total angular momentum ${L}^{2}=l\left(l+1\right){\hslash}^{2}$ is not a perfect square! One might wonder if the component of angular momentum in the $x$ -direction isn’t also a whole number of $\hslash $ units as well, and if not, why not?

The key is that in questions of this type we are forgetting
the essentially wavelike nature of the particle’s motion, or, equivalently, the
uncertainty principle. Recall first that
the $z$ -component of angular momentum, that is, the
angular momentum ${L}_{z}$ about the $z$ -axis, is the product of the particle’s
momentum in the $xy$ -plane and the distance of the line of that
motion from the origin. There is no contradiction
in specifying that momentum and that position simultaneously, because they are
in perpendicular directions. However, we
cannot at the same time specify either of the other components ${L}_{x},{L}_{y}$ of the angular momentum, because that would
involve measuring some component of momentum in a direction in which we have
just specified a position measurement.
We *can* measure the *total* angular momentum, that involves
additionally only the component ${p}_{\theta}$ of momentum perpendicular to the ${p}_{\phi}$ needed for the $z$ -component.

Thus the uncertainty principle limits us to measuring at the same time only the total angular momentum and the component in one direction. Note also that if we knew the $z$ -component of angular momentum to be $m\hslash ,$ and the total angular momentum were ${L}^{2}={l}^{2}{\hslash}^{2},$ with $l=m,$ then we would also know that the $x$ and $y$ components of the angular momentum were exactly zero. Thus we would know all three components, in contradiction to our uncertainly principle arguments. This is the essential reason why the square of the total angular momentum is greater than the maximum square of any one component. It is as if there were a “zero point motion” fuzzing out the direction.

Another point related to the uncertainty principle concerns measuring just where in its circular (say) orbit the electron is at any given moment. How well can that be pinned down? There is an obvious resemblance here to measuring the position and momentum of a particle at the same time, where we know the fuzziness of the two measurements is related by $\Delta x\cdot \Delta p\sim h.$ Naïvely, for a circular orbit of radius $r$ in the $xy$ -plane, $rp={L}_{z},$ and distance measured around the circle is $r\theta ,$ so $\Delta x\cdot \Delta p\sim h$ suggests $\Delta \theta \cdot \Delta {L}_{z}\sim h.$ That is to say, precise knowledge of ${L}_{z}$ implies no knowledge of where on the circle the particle is. This is not surprising, because we have found that for ${L}_{z}=m\hslash $ the wave has the form ${e}^{im\phi},$ and so ${\left|\psi \right|}^{2},$ the relative probability of finding the particle, is the same anywhere in the circle. On the other hand, if we have a time-dependent wave function describing a particle orbiting the nucleus, so that the probability of finding the particle at a particular place varies with time, the particle cannot be in a definite angular momentum state. This is just the same as saying that a particle described by a wave packet cannot have a definite momentum.

It is worth writing first the energy equation for a classical particle in the Coulomb potential:

$$\frac{1}{2m}\left({p}_{r}^{2}+{p}_{\theta}^{2}+{p}_{\phi}^{2}\right)-\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}=E.$$

This makes it possible to see, term by term, what the various parts of the Schrödinger equation signify. In spherical polar coordinates, Schrödinger’s equation is:

$$-\frac{{\hslash}^{2}}{2m}\left(\frac{1}{r}\frac{{\partial}^{2}}{\partial {r}^{2}}(r\psi )+\frac{1}{{r}^{2}}\left\{\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\left(\mathrm{sin}\theta \frac{\partial \psi}{\partial \theta}\right)+\frac{1}{{\mathrm{sin}}^{2}\theta}\frac{{\partial}^{2}\psi}{\partial {\phi}^{2}}\right\}\right)-\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}\psi =E\psi .$$

We look for separable solutions of the form

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

We now follow the standard technique. That is to say, we substitute $R\Theta \Phi $ for $\psi $ in each term in the above equation. We then observe that the differential operators only actually operate on one of the factors in any given part of the expression, so we put the other two factors to the left of these operators. We then divide the entire equation by $R\Theta \Phi ,$ to get

$$-\frac{{\hslash}^{2}}{2m}\left(\frac{1}{R}\frac{1}{r}\frac{{\partial}^{2}}{\partial {r}^{2}}(rR)+\frac{1}{{r}^{2}}\left\{\frac{1}{\Theta}\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\left(\mathrm{sin}\theta \frac{\partial \Theta}{\partial \theta}\right)+\frac{1}{{\mathrm{sin}}^{2}\theta}\frac{1}{\Phi}\frac{{\partial}^{2}\Phi}{\partial {\phi}^{2}}\right\}\right)-\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}=E.$$

The above equation can be rearranged to give:

$$\left(\frac{1}{R}\frac{1}{r}\frac{{\partial}^{2}}{\partial {r}^{2}}(rR)+\frac{1}{{r}^{2}}\left\{\frac{1}{\Theta}\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\left(\mathrm{sin}\theta \frac{\partial \Theta}{\partial \theta}\right)+\frac{1}{{\mathrm{sin}}^{2}\theta}\frac{1}{\Phi}\frac{{\partial}^{2}\Phi}{\partial {\phi}^{2}}\right\}\right)=\frac{2m}{{\hslash}^{2}}\left(E+\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}\right).$$

Further rearrangement leads to:

$$\frac{1}{\Phi}\frac{{\partial}^{2}\Phi}{\partial {\phi}^{2}}={\mathrm{sin}}^{2}\theta \left[{r}^{2}\left\{\frac{2m}{{\hslash}^{2}}\left(E+\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}\right)-\frac{1}{R}\frac{1}{r}\frac{{\partial}^{2}}{\partial {r}^{2}}(rR)\right\}-\frac{1}{\Theta}\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\left(\mathrm{sin}\theta \frac{\partial \Theta}{\partial \theta}\right)\right].$$

At this point, we have achieved the separation of
variables! The left hand side of this
equation is a function *only of **$\phi ,$** *the
right hand side is a function *only of **$r$** and **$\theta .$** * The only way this can make sense is if *both sides of the equation are in fact
constant* (and of course equal to each other).

Taking the left hand side to be equal to a constant we denote for later convenience by $-{m}^{2},$

$$\frac{{\partial}^{2}\Phi (\phi )}{\partial {\phi}^{2}}=-{m}^{2}\Phi (\phi ).$$

We write the constant $-{m}^{2}$ because we know that as a factor in a wave function $\Phi \left(\phi \right)$ must be single valued as $\phi $ increases through $2\pi ,$ so an integer number of oscillations must fit around the circle, meaning $\Phi $ is $\mathrm{sin}m\phi ,\text{\hspace{0.17em}}\mathrm{cos}m\phi $ or ${e}^{im\phi}$ with $m$ an integer. These are the solutions of the above equation. Of course, this is very similar to the particle in the circle in two dimensions, $m$ signifies units of angular momentum about the $z$ -axis.

Backing up now to the equation in the form

$\left(\frac{1}{R}\frac{1}{r}\frac{{\partial}^{2}}{\partial {r}^{2}}(rR)+\frac{1}{{r}^{2}}\left\{\frac{1}{\Theta}\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\left(\mathrm{sin}\theta \frac{\partial \Theta}{\partial \theta}\right)+\frac{1}{{\mathrm{sin}}^{2}\theta}\frac{1}{\Phi}\frac{{\partial}^{2}\Phi}{\partial {\phi}^{2}}\right\}\right)=\frac{2m}{{\hslash}^{2}}\left(E+\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}\right)$

we can replace the $\frac{1}{\Phi}\frac{{\partial}^{2}\Phi}{\partial {\Phi}^{2}}$ term by $-{m}^{2},$ and move the $r$ term over to the right, to give

$$\frac{1}{\Theta}\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\left(\mathrm{sin}\theta \frac{\partial \Theta}{\partial \theta}\right)-\frac{{m}^{2}}{{\mathrm{sin}}^{2}\theta}={r}^{2}\left\{\frac{2m}{{\hslash}^{2}}\left(E+\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}\right)-\frac{1}{R}\frac{1}{r}\frac{{\partial}^{2}}{\partial {r}^{2}}(rR)\right\}.$$

We have again managed to *separate
the variables*$\u2014$the left
hand side is a function *only *of $\theta ,$ the right hand side a function of $r.$ Therefore both must be equal to the same
constant, which we set equal to $-\lambda .$

This gives the $\Theta \left(\theta \right)$ equation:

$$\frac{1}{\mathrm{sin}\theta}\frac{\partial}{\partial \theta}\mathrm{sin}\theta \frac{\partial \Theta (\theta )}{\partial \theta}-\frac{{m}^{2}}{{\mathrm{sin}}^{2}\theta}\Theta (\theta )=-\lambda \Theta (\theta ).$$

This is exactly the wave equation we discussed above for the elastic sphere, and the allowed eigenvalues $\lambda $ are $l\left(l+1\right)$ where $l=0,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}2,\dots $ . with $l\ge m.$

Replacing the $\theta ,\phi $ operator with the value found just above in the original Schrödinger equation gives the equation for the radial wave function:

$$-\frac{{\hslash}^{2}}{2m}\left(\frac{1}{r}\frac{{\partial}^{2}}{\partial {r}^{2}}(rR(r))-\frac{l(l+1)}{{r}^{2}}R(r)\right)-\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}R(r)=ER(r).$$

The first term in this radial equation is the usual radial kinetic energy term, equivalent to ${p}_{r}^{2}/2m$ in the classical picture. The third term is the Coulomb potential energy. The second term is an effective potential representing the centrifugal force. This is clarified by reconsidering the energy equation for the classical case,

$$\frac{1}{2m}\left({p}_{r}^{2}+{p}_{\theta}^{2}+{p}_{\phi}^{2}\right)-\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}=E.$$

The angular momentum squared

${L}^{2}={r}^{2}\left({p}_{\theta}^{2}+{p}_{\phi}^{2}\right)=l\left(l+1\right){\hslash}^{2}.$

Thus for fixed angular momentum, we can write the above “classical” equation as

$$\frac{1}{2m}\left({p}_{r}^{2}+\frac{l(l+1){\hslash}^{2}}{{r}^{2}}\right)-\frac{1}{4\pi {\epsilon}_{0}}\frac{{e}^{2}}{r}=E.$$

The parallel to the radial Schrödinger equation is then clear.

We must find the solutions of the radial Schrödinger equation that decay for large $r.$ These will be the bound states of the hydrogen atom. In natural units, measuring lengths in units of the first Bohr radius, and energies in Rydberg units

$$r={a}_{0}\rho =\frac{4\pi {\epsilon}_{0}{\hslash}^{2}}{m{e}^{2}}\rho \text{,}E={E}_{R}\epsilon =\frac{m{e}^{4}}{2{\hslash}^{2}{(4\pi {\epsilon}_{0})}^{2}}\epsilon $$.

Finally, taking $u\left(r\right)=rR\left(r\right),$ the radial equation becomes

$$-\frac{{d}^{2}u(\rho )}{d{\rho}^{2}}+\frac{l(l+1)}{{\rho}^{2}}u(\rho )-\frac{2}{\rho}u(\rho )=\epsilon u(\rho )$$.