The Schrödinger equation on the cheap

April 17, 2021. A quick post explaining how matter waves lead to natural definitions of momentum and energy operators, and hence the Schrödinger equation. There is nothing new, just (I hope) a clear development of the topic.

Light and matter

We learnt in a previous post that Einstein’s famous formula $E = mc^2$ can also be written

\[E^2 = m_0^2 c^4 + p^2 c^2,\]

where $m_0$ is the rest mass of the object we’re considering and $p$ its momentum. If the object happens to be a massless particle, say a photon of light, then $m_0 = 0$ and we end up with the curious identity:

\[E = pc.\]

Now, it just so happens we have a different expression for the energy of a photon, also proposed by Einstein in 1905. Turns out you can make a battery from a light bulb and a lump of metal, but only when the light is blue enough; in this case, the light shines the electrons out of the metal and into a current. But if light delivered energy continuously, like we would expect for a continuous electromagnetic wave, then only the wattage should matter, not the colour. What gives?

Einstein’s brilliant explanation was that light isn’t interacting with electrons as a continuous wave, but as a discrete particle whose energy is determined by colour! From the experimental results, he deduced that the energy per particle of light is

\[E = hf,\]

where $f$ is the frequency (oscillations per second) and $h = 6.63 \times 10^{-34} \text{ J s}$ is Planck’s constant. Using our two expressions for the energy, we find that the momentum of a photon is

\[p = \frac{hf}{c}.\]

You can use this to compute how many laser pointers you would need to repel an incoming asteroid before it hits the earth (exercise left to the reader).

Plane waves

If we zoom out from the photon, or rather, measure it differently, say by getting it to diffract through some slits rather than dislodging individual electrons from a metal, it will be described by a wave. For simplicity, let’s consider a single spatial dimension $x$. A plane wave is a simple sinusoidal displacement of some medium (e.g. air pressure, the surface of a body of water, electromagnetic fields) with amplitude

\[\Psi (x, y) = \Psi_0 \sin \left[k(x - vt)\right].\]

The constant $ \Psi_0$ is just the maximum size of this displacement, but $k$ and $v$ require a bit more explanation. Here, $v$ is the speed of the wave, since a point of fixed $C = x - vt$ in time $\Delta t$ must move $\Delta x$ obeying

\[C = x - vt = (x + \Delta x) - v (t + \Delta t) = C + \Delta x -v \Delta t \quad \Longrightarrow \quad \frac{\Delta x}{\Delta t} = v.\]

If we take a snapshot of the wave at fixed time $t$, then it will repeat itself when the argument of the sine function is increased by $2\pi$, or

\[k(x - vt) \mapsto k(x - vt) + 2\pi = k\left(x + \frac{2\pi}{k} - vt\right).\]

Since $t$ is fixed, it follows that when we increment $x$ by $2\pi/k$, the wave repeats itself. In other words, $\lambda = 2\pi/k$ is the wavelength. By the same reasoning, if we freeze $x$ the wave repeats itself with a period in time,

\[T = \frac{2\pi}{vk}.\]

Since the frequency $f = 1/T$ is the inverse of the period, we find that frequency times wavelength equals speed:

\[f\lambda = \frac{vk}{2\pi}\cdot \frac{2\pi}{k} = v.\]

At this point, it will simplify things dramatically to use the exponential instead of the sine wave. Since Euler’s marvellous formula tells us that

\[e^{i\theta} = \cos\theta + i \sin\theta,\]

we can replace the sinusoid with

\[\Psi (x, t) = \Psi_0 e^{i k(x - vt)}.\]

If we really want the sine, we just take the imaginary part.

The momentum operator

Let’s now consider a photon, moving at speed $v =c$ and with momentum obeying

\[p = \frac{hf}{c} = \frac{hf}{f \lambda} = \frac{h}{\lambda} = \frac{h}{2\pi}\cdot \frac{2\pi}{\lambda} = \hbar k,\]

where we used $c = f\lambda$, $k = 2\pi/\lambda$, and defined a new constant $\hbar = h/2\pi$, called Planck’s reduced constant. This lets us rewrite the plane wave as

\[\Psi(x, t) = \Psi_0 e^{i (p/\hbar)(x - ct)} = \Psi_0 e^{i(px - Et)/\hbar},\]

since $E = pc$. In 1924, Louis de Broglie made the bold suggestion that matter could also be described as a wave, with momentum and energy obeying the same relation. If measured properly, these matter waves could exhibit wavelike-behaviour, like bending around obsctacles and interfering. This has been experimentally observed, not only with electrons, but giant molecules called buckyballs!

We won’t be concerned with this, however. Instead, we would like to think about how to extract the momentum from a plane wave. One way is to rearrange $\Psi(x, t)$ to find $p$. Instead, we will define a procedure which simply pulls $p$ out. It is, as you have probably already guessed, simply the derivative with respect to $x$:

\[\frac{\partial}{\partial x}\Psi(x, t) = \Psi_0 \frac{\partial}{\partial x}e^{i (px - Et)/\hbar} = \frac{ip}{\hbar} \Psi(x, t).\]

The momentum operator $\hat{p}$ is simply defined as the operation which gives us $p$ (in front of the original function) without the extra constants $i$ and $\hbar$. More precisely,

\[\hat{p} = -i\hbar \frac{\partial}{\partial x},.\]

Often, we treat $t$ as fixed, so that the partial derivative becomes $d/dx$. One of the distinctive characteristics of waves is that the displacements add and subtract nicely. This is called the superposition principle. What happens with momentum then? A cute thing about this operator, as opposed to the algebraic expression which isolates $p$, is that we can apply it to any combination of plane waves. For instance, if we consider

\[\Psi(x, t) = \Psi_1(x, t) + \Psi_2(x, t) = \Psi_{0}^{(1)} e^{i (p_1x - E_1t)/\hbar} + \Psi_0^{(2)}e^{i (p_2x - E_2t)/\hbar},\]

then

\[\hat{p}\Psi(x, t) = p_1 \Psi_1(x, t) + p_2 \Psi_2(x, t).\]

No single nice momentum sits out the front of the whole wave (unless $p_1 = p_2$), but this tells us something important: such a combination doesn’t have a well-defined momentum! The momentum will be uncertain. We might guess that the uncertainty depends on the relative size of the amplitudes $\Psi_{0}^{(1)}$ and $\Psi_{0}^{(2)}$, and this guess is correct. We won’t pursue the generalisations (the uncertainty principle and the Born rule) here.

The energy operator

Let’s end, briefly, with the Schrödinger equation. In 1925, Erwin Schrödinger went on vacation in the Swiss alps, taking only de Broglie’s thesis with him. By the end of his getaway, he had derived the fundamental equation of quantum mechanics. How did he do it? He was guided by many subtleties we won’t care about, but the basic observation is simple: define an energy operator in the same way we defined the momentum operator. For a plane wave $\Psi(x, t) = \Psi_0 e^{i(px - Et)/\hbar}$, we have

\[i\hbar \frac{\partial}{\partial t} \Psi(x, t) = E \Psi(x, t),\]

suggesting we define the energy operator

\[\hat{H} = i\hbar \frac{\partial}{\partial t}.\]

For historical reasons, we use $\hat{H}$ for “Hamiltonian” instead of $\hat{E}$ for “energy”. This is more or less it, but we will say a few more words to connect it to other versions of the Schrödinger equation you might encounter. The kinetic energy and momentum of a classical particle are related by

\[E = \frac{1}{2}mv^2 = \frac{(mv)^2}{2m} = \frac{p^2}{2m}.\]

Schrödinger simply guessed this held as an operator equation. Substituting the actual expressions in terms of partial derivatives:

\[i\hbar \frac{\partial}{\partial t}\Psi = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi.\]

Of course, a particle can get energy from other places. If it rolls around on a slope, for instance, there will be some potential energy $V$, and the classical energy is

\[E = \frac{p^2}{2m} + V.\]

Once again, we promote this to an operator. Schrödinger’s guess means that this is still identified with the time derivative, so

\[i\hbar \frac{\partial}{\partial t}\Psi = \left(-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \hat{V}\right)\Psi,\]

where $\hat{V}$ is now promoted as well. Actually doing this promotion is technical and ambiguous, so we won’t worry about it. Anyway, the simplest and most elegant way to write the Schrödinger equation is just

\[\hat{H}\Psi = i\hbar \frac{\partial}{\partial t}\Psi. \tag{1} \label{H}\]

Although I’ve drawn a slightly facile analogy between momentum and energy, there is a fundamental difference. The momentum operator is defined as the derivative with respect to $x$, so applying the momentum operator will always be the same as differentiating. The energy operator is defined in a totally different way! For instance, we often take a classical expression for energy and promote stuff to operators. There is no need for this to equal the time derivative, so Schrödinger’s equation is telling us that not any old wavefunction $\Psi$ is allowed, just the ones that evolve according to (\ref{H}). Still, the whole shebang drops out of the plane wave!

Acknowledgments

Thanks to J.A. for motivating discussion as usual!

Written on April 17, 2021
Physics   Hacks