*Michael Fowler, University of Virginia*

We have seen that electrons and photons behave in a very
similar fashion—
both
exhibit diffraction effects, as in the double slit experiment, both have
particle like or quantum behavior. We
can in fact give a complete analysis of photon behavior—we can figure
out how the electromagnetic wave propagates, using Maxwell’s equations, then
find the probability that a photon is in a given small volume of space $dxdydz,$ is proportional to ${\left|E\right|}^{2}dxdydz,$ the energy density. On the other hand, our analysis of the
electron’s behavior is incomplete—we know
that it must also be described by a wave function $\psi (x,y,z,t)$ analogous to $E,$ such that $|\psi (x,y,z,t){|}^{2}dxdydz$ gives the probability of finding the electron
in a small volume $dxdydz$ around the point $\left(x,y,z\right)$ at the time $t.$ *However, we do not yet have the analog of
Maxwell’s equations to tell us how **$\psi $** varies
in time and space*. The purpose of
this section is to give a plausible derivation of such an equation by examining
how the Maxwell wave equation works for a single-particle (photon) wave, and
constructing parallel equations for particles which, unlike photons, have
nonzero rest mass.

Let us examine what Maxwell’s equations tell us about the motion of the simplest type of electromagnetic wave— a monochromatic wave in empty space, with no currents or charges present. First, we briefly review the derivation of the wave equation from Maxwell’s equations in empty space:

$$\begin{array}{l}\text{div}\overrightarrow{B}=0\\ \text{div}\overrightarrow{E}=0\\ \text{curl}\text{\hspace{0.05em}}\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}\\ \text{curl}\overrightarrow{B}=\frac{1}{{c}^{2}}\frac{\partial \overrightarrow{E}}{\partial t}\end{array}$$

To derive the wave equation, we take the curl of the third equation:

$$\text{curlcurl}\overrightarrow{E}=-\frac{\partial}{\partial t}\text{curl}\overrightarrow{B}=-\frac{1}{{c}^{2}}\frac{{\partial}^{2}\stackrel{\rightharpoonup}{E}}{\partial {t}^{2}}$$

together with the vector operator identity

$\text{curlcurl=grad(div)-}{\nabla}^{\text{2}}$

to give

$${\nabla}^{2}\overrightarrow{E}-\frac{1}{{c}^{2}}\frac{{\partial}^{2}\overrightarrow{E}}{\partial {t}^{2}}=0$$.

For a plane wave moving in the $x$ -direction this reduces to

*$$\frac{{\partial}^{2}\overrightarrow{E}}{\partial {x}^{2}}-\frac{1}{{c}^{2}}\frac{{\partial}^{2}\overrightarrow{E}}{\partial {t}^{2}}=0.$$*

The monochromatic solution to this wave equation has the form

$\overrightarrow{E}(x,t)={\overrightarrow{E}}_{0}{e}^{i(kx-\omega t)}$.

(Another possible solution is proportional to $\mathrm{cos}\left(kx-\omega t\right).$ We shall find that the exponential form, although a complex number, proves more convenient. The physical electric field can be taken to be the real part of the exponential for the classical case.)

Applying the wave equation differential operator to our plane wave solution

$$\left(\frac{{\partial}^{2}}{\partial {x}^{2}}-\frac{1}{{c}^{2}}\frac{{\partial}^{2}}{\partial {t}^{2}}\right){\overrightarrow{E}}_{0}{e}^{i(kx-\omega t)}=\left({k}^{2}-\frac{{\omega}^{2}}{{c}^{2}}\right){\overrightarrow{E}}_{0}{e}^{i(kx-\omega t)}=0.$$

If the plane wave is a solution to the wave equation, this must be true for all $x$ and $t,$ so we must have

$\omega =ck.$

This is just the familiar statement that the wave must travel at $c.$

We know from the photoelectric effect and

$\begin{array}{l}E=h\nu =\hslash \omega ,\\ p=\frac{h}{\lambda}=\hslash k.\end{array}$

Notice, then, that the wave equation tells us that $\omega =ck$ and hence $E=cp.$

To put it another way, if we think of ${e}^{i(kx-\omega t)}$ as describing a particle (photon) it would be more natural to write the plane wave as

${\overrightarrow{E}}_{0}{e}^{i(px-Et)/\hslash}.$

that is, in terms of the energy and momentum of the particle.

In these terms, applying the (Maxwell) wave equation operator to the plane wave yields

$$\left(\frac{{\partial}^{2}}{\partial {x}^{2}}-\frac{1}{{c}^{2}}\frac{{\partial}^{2}}{\partial {t}^{2}}\right){\overrightarrow{E}}_{0}{e}^{i(px-Et)/\hslash}=\left({p}^{2}-\frac{{E}^{2}}{{c}^{2}}\right){\overrightarrow{E}}_{0}{e}^{i(px-Et)/\hslash}=0$$

or

${E}^{2}={c}^{2}{p}^{2}.$

*The wave equation
operator applied to the plane wave describing the particle propagation yields
the energy-momentum relationship for the particle*.

The discussion above suggests how we might extend the wave equation operator from the photon case (zero rest mass) to a particle having rest mass $m.$ We need a wave equation operator that, when it operates on a plane wave, yields

${E}^{2}={c}^{2}{p}^{2}+{m}^{2}{c}^{4}.$

Writing the plane wave function

$\phi \text{\hspace{0.17em}}\text{\hspace{0.05em}}(x,t)=A{e}^{i(px-Et)/\hslash}$

where $A$ is a constant, we find we can get ${E}^{2}={c}^{2}{p}^{2}+{m}^{2}{c}^{4}$ by adding a constant (mass) term to the differentiation terms in the wave operator:

$$\left(\frac{{\partial}^{2}}{\partial {x}^{2}}-\frac{1}{{c}^{2}}\frac{{\partial}^{2}}{\partial {t}^{2}}-\frac{{m}^{2}{c}^{2}}{{\hslash}^{2}}\right)A{e}^{\frac{i}{\hslash}(px-Et)}=-\frac{1}{{\hslash}^{2}}\left({p}^{2}-\frac{{E}^{2}}{{c}^{2}}+{m}^{2}{c}^{2}\right)A{e}^{\frac{i}{\hslash}(px-Et)}=0.$$

This wave equation is called the *Klein-Gordon* equation and correctly describes the propagation of
relativistic particles of mass $m.$ However, it’s a bit inconvenient for
nonrelativistic particles, like the electron in the hydrogen atom, just as ${E}^{2}={c}^{2}{p}^{2}+{m}^{2}{c}^{4}$ is less
useful than $E={p}^{2}/2m$ for this case.

Continuing along the same lines, let us assume that a nonrelativistic electron in free space (no potentials, so no forces) is described by a plane wave:

$\psi (x,t)=A{e}^{i(px-Et)/\hslash}.$

We need to construct a wave equation operator which, applied
to this wave function, just gives us the ordinary nonrelativistic
energy-momentum relationship, $E={p}^{2}/2m.$ The ${p}^{2}$ obviously comes as usual from differentiating
twice with respect to $x,$ but the only way we can get $E$ is by having a *single *differentiation with respect to time, so this looks
different from previous wave equations:

$$i\hslash \frac{\partial \psi (x,t)}{\partial t}=-\frac{{\hslash}^{2}}{2m}\frac{{\partial}^{2}\psi (x,t)}{\partial {x}^{2}}.$$

This is *Schrödinger’s
equation* for a free particle. It is
easy to check that if $\psi (x,t)$ has the plane wave form given above, the
condition for it to be a solution of this wave equation is just $E={p}^{2}/2m.$

Notice one remarkable feature of the above equation—
the $i$ on the left means that $\psi $ *cannot*
be a real function.

The effect of a potential on a de Broglie wave was considered by Sommerfeld in an attempt to generalize the rather restrictive conditions in Bohr’s model of the atom. Since the electron was orbiting in an inverse square force, just like the planets around the sun, Sommerfeld couldn’t understand why Bohr’s atom had only circular orbits, no Kepler-like ellipses. (Recall that all the observed spectral lines of hydrogen were accounted for by energy differences between these circular orbits.)

De Broglie’s analysis of the allowed circular orbits can be formulated by assuming that at some instant in time the spatial variation of the wave function on going around the orbit includes a phase term of the form ${e}^{ipq/\hslash},$ where here the parameter $q$ measures distance around the orbit. Now for an acceptable wave function, the total phase change on going around the orbit must be $2n\pi ,$ where $n$ is an integer. For the usual Bohr circular orbit, $p$ is constant on going around, $q$ changes by $2\pi r,$ where $r$ is the radius of the orbit, giving

$\frac{1}{\hslash}p2\pi r=2n\pi $

so

$$pr=n\hslash ,$$

the usual angular momentum quantization.

What Sommerfeld did was to consider a general Kepler ellipse orbit, and visualize the wave going around such an orbit. Assuming the usual relationship $p=h/\lambda ,$ the wavelength will vary as the particle moves around the orbit, being shortest where the particle moves fastest, at its closest approach to the nucleus. Nevertheless, the phase change on moving a short distance $\Delta q$ should still be $p\Delta q/\hslash ,$, and requiring the wave function to link up smoothly on going once around the orbit gives

$\int pdq=nh}.$

Thus only certain elliptical orbits are allowed. The
mathematics is nontrivial, but it turns out that every allowed elliptical orbit
has the same energy as one of the allowed circular orbits. This is why Bohr’s
theory gave all the energy levels. Actually, this whole analysis is old
fashioned (it’s called the “old quantum theory”) but we’ve gone over it to
introduce the idea of* a wave with
variable wavelength, changing with the momentum as the particle moves through a
varying potential*.

Let us consider first the one-dimensional situation of a
particle going in the *x*-direction
subject to a “roller coaster” potential.
What do we expect the wave function to look like? We would expect the wavelength to be shortest
where the potential is lowest, in the valleys, because that’s where the
particle is going fastest—
maximum momentum. Perhaps slightly less obvious is that the
amplitude of the wave would be largest at the tops of the hills (provided the
particle has enough energy to get there) because that’s where the particle is
moving slowest, and therefore is most likely to be found.

With a nonzero potential present, the energy-momentum relationship for the particle becomes the energy equation

$$E=\frac{{p}^{2}}{2m}+V(x).$$

We need to construct a wave equation which leads naturally to this relationship. In contrast to the free particle cases discussed above, the relevant wave function here will no longer be a plane wave, since the wavelength varies with the potential. However, at a given $x,$ the momentum is determined by the “local wavelength”, that is,

$$p=-i\hslash \frac{\partial \psi}{\partial x}.$$

It follows that the appropriate wave equation is:

$$i\hslash \frac{\partial \psi (x,t)}{\partial t}=-\frac{{\hslash}^{2}}{2m}\frac{{\partial}^{2}\psi (x,t)}{\partial {x}^{2}}+V(x)\psi (x,t).$$

This is the standard one-dimensional Schrödinger equation.

In three dimensions, the argument is precisely analogous. The only difference is that the square of the momentum is now a sum of three squared components, for the $x,y,z$ directions, so $\frac{{\partial}^{2}}{\partial {x}^{2}}\text{becomes}\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{{\partial}^{2}}{\partial {y}^{2}}+\frac{{\partial}^{2}}{\partial {z}^{2}}={\nabla}^{2},$ and the equation is:

$$i\hslash \frac{\partial \psi (x,y,z,t)}{\partial t}=-\frac{{\hslash}^{2}}{2m}{\nabla}^{2}\psi (x,y,z,t)+V(x,y,z)\psi (x,y,z,t).$$

This is the complete Schrödinger equation.