What a wavefunction tells us
The previous essay ended with an electron described by one complex number for every point along a line: its wavefunction, $\psi(x)$. What does the value of that function at a particular point tell us about where the electron will be found? And what rule decides how the function changes as time passes?
This essay answers the two questions in that order. First, what $\psi$ means: its size gives probabilities, and its phase carries the information that makes interference possible. Then the equation that moves it, written down by Erwin Schrödinger in 1926, read term by term. At the end, a special kind of solution appears whose probabilities do not change at all, and it turns out to be what an atom's energy levels are.
Size means probability
In the double-slit essay, $P(x)$ was the probability of landing in a small detector window of fixed width at position $x$, and it was the squared size of an amplitude. The wavefunction packages all those amplitudes into one function. The rule that connects it to experiment was proposed by Max Born in 1926: the probability of finding the particle in a small window of width $dx$ around $x$ is
$$P(\text{in } dx \text{ at } x) = |\psi(x)|^2\,dx .$$So $|\psi(x)|^2$ is a probability density: probability per unit length, in the way that a population density is people per square kilometre. It is not itself a probability, and it can be larger than 1. To get a probability you multiply by a width, or, for a region of any size, take the area under it:
$$P(a < x < b) = \int_a^b |\psi(x)|^2\,dx .$$The particle is certainly somewhere, so the area under the whole curve must be 1. A wavefunction that satisfies this is normalised. Any wavefunction can be normalised by multiplying it by a suitable constant; the shape of $|\psi|^2$ is the physics, and the constant only fixes the overall scale.
Born reached this reading by thinking about electrons scattered from atoms. The wave calculation gives a scattered wave spreading in all directions, but each electron is detected in one direction. The wave's intensity in a direction, he proposed, is the probability that an electron goes that way. This is what lets the wave and the particle coexist: the electron is always found whole, at one place, and the wave says how likely each place is.
Phase is invisible, until it isn't
Each value $\psi(x)$ is a complex number, an arrow with a length and a direction. Born's rule uses only the length. The direction, the phase, is invisible in $|\psi|^2$ at any single point, and multiplying the whole wavefunction by the same unit arrow $e^{i\alpha}$ changes no prediction at all.
But the phase differences between points, or between the parts of a superposition (a state formed by adding states together, as the two-slit amplitude added the two routes), are physical. They are what made the double-slit fringes. An electron moving to the right, for example, has a wavefunction whose phase turns steadily along $x$, like $e^{ikx}$. Its density $|e^{ikx}|^2 = 1$ is the same everywhere, but the rate at which the phase turns records how fast it is moving, as we will see in a moment. The figure shows a more realistic case, a wave packet: an electron located roughly in one region and moving.
Worked example: a wavefunction with numbers
Take an electron whose wavefunction falls off exponentially on both sides of a central point,
$$\psi(x) = A\,e^{-|x|/b},\qquad b = 0.1\ \text{nm}.$$This is the actual shape of an electron bound to a single very short-range attraction, but here it serves as a definite example to calculate with.
Normalise it. The density is $A^2 e^{-2|x|/b}$. The area under it is twice the area on the right-hand side:
$$\int_{-\infty}^{\infty} A^2 e^{-2|x|/b}\,dx = 2A^2\cdot\frac{b}{2} = A^2 b .$$Setting this to 1 gives $A = 1/\sqrt{b}$. With $b = 10^{-10}$ m, $A = 10^{5}$ m−1/2. The odd unit is right: $|\psi|^2$ must have units of one over length, so that multiplying by a length gives a pure number.
Probability in a region. The chance of finding the electron within $b$ of the centre:
$$\begin{aligned} P(|x| < b) &= 2\int_0^b \frac{1}{b}\,e^{-2x/b}\,dx\\[4pt] &= 1 - e^{-2} = 0.865 . \end{aligned}$$About 86 percent within 0.1 nm of the centre, 14 percent further out.
Probability in a small window. A detector window 0.001 nm wide at the centre catches the electron with probability about $|\psi(0)|^2 \times w = (10^{10}\ \text{m}^{-1})(10^{-12}\ \text{m}) = 0.01$. The same window placed at $x = b$ catches it with probability $e^{-2}$ times that, 0.0014.
Check. All the probabilities are between 0 and 1, the density has the right units, and the total is 1 by construction. Notice that $|\psi(0)|^2 = 10^{10}$ m−1 is an enormous number: a density is not a probability until it is multiplied by a width.
The equation that moves it
Now the second question: how does $\psi$ change in time? Schrödinger's equation can be motivated, though not derived, from two relations earlier essays established. De Broglie's wavelength, $\lambda = h/p$, and Planck's quantum, $E = h\nu$, can be written with $\hbar = h/2\pi$ and with the wavenumber $k = 2\pi/\lambda$ and angular frequency $\omega = 2\pi\nu$:
$$p = \hbar k, \qquad E = \hbar\omega .$$A particle with definite momentum and energy should then be described by a wave with definite $k$ and $\omega$, the travelling wave $\psi = e^{i(kx - \omega t)}$. Look at what differentiating this wave does:
$$\begin{aligned} -i\hbar\,\frac{\partial\psi}{\partial x} &= \hbar k\,\psi = p\,\psi,\\[4pt] i\hbar\,\frac{\partial\psi}{\partial t} &= \hbar\omega\,\psi = E\,\psi . \end{aligned}$$Differentiating in space picks out the momentum; differentiating in time picks out the energy. For a free particle, energy and momentum are related by $E = p^2/2m$. Applying each side of that relation to the wave, with the derivatives standing in for $E$ and $p$, gives an equation the wave satisfies. Adding a potential energy $V(x)$, the energy of the particle's position in a force field, gives Schrödinger's equation:
$$i\hbar\,\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\,\frac{\partial^2\psi}{\partial x^2} + V(x)\,\psi .$$Read it as the energy bookkeeping, total energy equals kinetic plus potential, turned into a rule for how the wave changes:
- Left side: the rate at which $\psi$ changes in time, with the factor $i\hbar$ that the travelling wave showed is tied to energy.
- First term on the right, the kinetic energy. It is the curvature of $\psi$. A tightly wiggling wavefunction has a short wavelength, a large momentum, and a large kinetic energy.
- Second term, the potential energy. It weights $\psi$ at each point by the potential energy there.
The argument from the travelling wave only fixes the free case. Adding $V(x)$ in the simplest way is a guess, and Schrödinger's justification was that it worked: applied to the electron in a hydrogen atom it produced Bohr's energy levels without orbits. Two further properties make it fit what we already know. The equation is linear: if $\psi_1$ and $\psi_2$ are solutions, so is any combination of them, which is what the double-slit rule of adding amplitudes needs. And it keeps the total probability equal to 1 as time runs: probability can flow from one place to another but is never created or lost.
States whose probabilities stand still
Some solutions have a special form: a fixed shape in space, multiplied by a phase that turns in time at a steady rate,
$$\psi(x,t) = \varphi(x)\,e^{-iEt/\hbar}.$$Because the time dependence is a single overall phase, the density $|\psi(x,t)|^2 = |\varphi(x)|^2$ does not change at all. Every probability for position, and in fact for any measurement, is the same at every moment. Such a state is called stationary, and it has a definite energy $E$. Putting this form into Schrödinger's equation, the time derivative gives $E$ times $\psi$, the phase cancels, and what is left is an equation for the shape alone:
$$-\frac{\hbar^2}{2m}\,\frac{d^2\varphi}{dx^2} + V(x)\,\varphi = E\,\varphi .$$Here is where energy levels come from. For a particle held in by a potential, as an electron in an atom is, this equation has acceptable solutions only for particular values of $E$. For other values, the solution runs off to infinity far from the atom and cannot be normalised. The allowed values are the energy levels. Bohr had to assume them; the wave equation produces them as the energies at which a bound wave fits. The essay on confinement will show this in the simplest case, a particle in a box.
Stationary states answer part of the objection that sank the classical atom. In the model used here, a single electron in a fixed potential, a stationary state's density does not move, so nothing is oscillating that could radiate, and within that model the state lasts forever. Change comes only from a superposition of states with different energies.
That model leaves something out: the electron's coupling to the electromagnetic field, which Schrödinger's equation as written here does not include. With the coupling taken into account, an excited state is no longer permanent. It decays to a lower level by emitting a photon, a process called spontaneous emission, typically within a few billionths of a second for atomic levels. The ground state has no lower level to decay to, and that is the state that stays. So the stability of atoms rests on two facts: the lowest energy level exists, and nothing lies below it.
Optional: how fast a wave packet spreads
A packet like the one in the figure does not keep its shape. Its parts carry a range of momenta, so they drift apart. For a Gaussian packet of initial width $\sigma_0$ (the standard deviation of position), Schrödinger's equation gives a width that grows as
$$\sigma(t) = \sigma_0\sqrt{1 + \left(\frac{t}{\tau}\right)^2},\qquad \tau = \frac{2m\sigma_0^2}{\hbar}.$$For times much shorter than $\tau$ the packet barely changes; for longer times it grows in proportion to $t$.
An electron located to within $\sigma_0 = 1$ nm has $\tau = 2(9.11\times10^{-31})(10^{-9})^2/(1.055\times10^{-34}) = 1.7\times10^{-14}$ s. After one picosecond its packet is about 60 nm wide; after a microsecond, about 6 cm. Free electrons do not stay localised.
A dust grain of one microgram, located to within one micrometre, has $\tau \approx 2\times10^{13}$ s, about 600,000 years. For all practical purposes it keeps its shape forever and moves like a classical particle. The difference is the mass, a factor of $10^{21}$, which enters $\tau$ directly.
Two questions
1. Far from the centre. For the electron in the worked example, what is the probability of finding it more than $2b$ from the centre?
Worked answer
The two tails beyond $\pm 2b$ together hold $2\int_{2b}^{\infty}\frac1b e^{-2x/b}\,dx = e^{-4} = 0.018$, a little under 2 percent.
The easy slip is to use $\psi$ instead of $|\psi|^2$: the wavefunction itself falls by $e^{-2}$ between the centre and $2b$, which might suggest an answer near $e^{-2} = 0.135$. Probabilities come from the density $|\psi|^2$, which falls twice as fast in the exponent.
2. Does the phase mean motion? In a stationary state, $\psi$ is multiplied by $e^{-iEt/\hbar}$, which keeps turning as time passes. A friend concludes that the electron in a stationary state must be moving back and forth, just too fast to see. What is wrong with this? What would have to be true of the state for something observable to change in time?
Worked answer
The turning factor is the same at every point, an overall phase, and Born's rule uses only $|\psi|^2$. Multiplying the whole wavefunction by one unit arrow changes no prediction. Every probability in a stationary state is constant; there is nothing moving back and forth, because nothing observable depends on time.
For something to change, the state must be a superposition of parts with different energies, say $E_1$ and $E_2$. Then the two parts' phases turn at different rates, their relative phase changes, and the cross term in $|\psi|^2$, the same kind of term that made the double-slit fringes, rises and falls with frequency $(E_2 - E_1)/h$. That is exactly the frequency of the photon emitted in a jump between the two levels: Bohr's rule reappears.
The tempting wrong turn is to treat the phase as a picture of motion, the way a rotating hand on a dial represents a moving part. An overall phase represents nothing that can be measured.
Further reading
- Feynman, Lectures on Physics, Vol. III, Ch. 16 "The Dependence of Amplitudes on Position", §16-2 "The wave function" and §16-5 "The Schrödinger equation": the equation reached from amplitudes for an electron hopping along a line of atoms, with a candid account of what is motivated and what is guessed.
- Landau & Lifshitz, Quantum Mechanics (2nd ed.), §10 "Stationary states" and §17 "Schrödinger's equation": a compact derivation of the equation's form from general principles.
- Born, Atomic Physics (Dover, 8th ed.), Ch. V §4 "Wave Mechanics" and §7 "The Statistical Interpretation of Wave Mechanics", the latter by the physicist who proposed it.
Where this leaves us
The value of the wavefunction at a point tells us, through its squared size, how likely the particle is to be found near there; its phase is invisible at a single point but decides how parts of the wave interfere. Schrödinger's equation moves it, with kinetic energy appearing as curvature, and its stationary solutions, whose probabilities never change, are the energy levels of bound systems.
We now have two descriptions of states in hand: short lists of amplitudes, like the photon's polarisation, and continuous functions, like $\psi(x)$. We also have questions that go beyond "where is it?": what is its energy, its momentum, its average position? The next essay gives a single compact language that handles all of these the same way, for lists and functions alike, and says exactly what a measurement returns.