Quantum That ClicksEpisode 2 / The wave function ── phase is S/ℏ, probability is |ψ|²

The true identity of the "one entity" glued together from particle and wave ── the complex wave ψ, whose phase is exactly "how many ℏ's of action"

The wave function ── phase is S/ℏ, probability is |ψ|² A quantum entity is represented by a complex wave ψ＝|ψ|e^(iS/ℏ). The phase is exactly how many ℏ's of action.
Only the magnitude squared |ψ|² (＝probability) is observable ── but the difference in phase shows up as interference.

Tools you'll need: Episode 1's S/ℏ, complex numbers (a turning arrow), addition This episode: ψ = |ψ| e^(iS/ℏ), probability = |ψ|²

In Episode 1 we said particle and wave are two faces of one quantum entity. The true identity of that "one entity" is this episode's star ── the wave function $\psi$. $\psi$ is not an ordinary real-valued wave but a complex wave. Think of it as carrying a turning arrow (a magnitude and a direction) at each point. Its direction (phase) is Episode 1's star, "how many ℏ's of action" ＝ $S/\hbar$. And what is observable is the square of the arrow's length, $|\psi|^2$ ── this is "the probability of being found at that place." The phase itself is not directly visible. But the difference in phase between two paths shows up in observation as reinforcement or cancellation (interference) ── and there lies the key to solving the riddle of particle-or-wave.

01The true identity of particle and wave ── one complex wave ψ

Send a single electron through a double slit, and it strikes the screen one point at a time, as a particle. Yet let tens of thousands pile up, and they form the interference fringes of a wave. "Each one a particle, yet the crowd a wave of fringes" ── the only tool that can tell this without contradiction is $\psi$. $\psi$ spreads as a wave and interferes, and at the instant of observation it is found at one point as a particle, with probability given by its $|\psi|^2$. Spread as a wave, detected as a particle ── $\psi$ takes on both single-handedly.

02Phase is S/ℏ, magnitude squared is probability

The wave function ── magnitude, and phase S/ℏ

$$\psi=|\psi|\,e^{i\,S/\hbar}\qquad(\text{probability} = |\psi|^2)$$

The complex number $\psi$ has a magnitude $|\psi|$ (the length of the turning arrow) and a phase $S/\hbar$ (the direction of the arrow). Phase ＝ the action $S$ divided by $\hbar$ ── Episode 1's ratio becomes the "direction" of the wave here. As the particle advances, the action $S$ builds up and the phase $S/\hbar$ turns round and round. Only the magnitude squared $|\psi|^2$ (the Born rule) ＝ the probability density is observable.

If the phase is $S/\hbar$, then for every $\hbar$'s worth the action increases, the arrow turns by exactly one radian. Episode 1's de Broglie relation $p=\hbar k$ is precisely this ── advancing through space turns the phase by $kx=px/\hbar$, and as time passes it turns by $\omega t=Et/\hbar$. The wave's wavelength and frequency were the very way the phase $S/\hbar$ turns.

03The overall phase is invisible ── but the "difference" is visible

Phase has an important property. Rotate all of $\psi$ by the same angle ($\psi\to e^{i\theta}\psi$), and the magnitude squared $|\psi|^2$ does not change ── the overall phase does not affect observation. Where we set the origin of the phase is merely a matter of our own bookkeeping.

A handshake with "Fields That Click" and "Cosmology That Clicks" "The overall phase does not affect observation" ── this is exactly that phase ambiguity we saw in "Fields That Click" Bonus ② (gauge) and "Cosmology That Clicks" Episode 8. Demanding that this ambiguity be freely choosable at each point gave birth to a field (the electromagnetic force) ── the phase $S/\hbar$ of the quantum $\psi$ is directly contiguous with the theory of forces. The $i$ (the complex number) of this episode is the same $i$ that generates temperature in Cosmology Episode 9.

But superpose two waves, and the difference in phase reveals itself. Add the two $\psi$'s that came via path A and path B, and ──

Superposition and interference ── what matters is the phase difference Δφ = ΔS/ℏ

$$|\psi_A+\psi_B|^2=|\psi_A|^2+|\psi_B|^2+\underbrace{2|\psi_A||\psi_B|\cos\Delta\varphi}_{\text{interference term}},\quad \Delta\varphi=\frac{S_B-S_A}{\hbar}$$

On top of the plain sum of probabilities (the first two terms), an interference term appears. Its size is set by the $\cos$ of the phase difference $\Delta\varphi=\Delta S/\hbar$ ── $\Delta\varphi=0$ means reinforcement (a bright fringe), $\Delta\varphi=\pi$ means cancellation (a dark fringe). This is the moment when the phase, which was invisible, shows its face in observation as a difference.

04Play with it ── adding the phases of two paths

In the figure below, we add up the wave functions that came via the two paths A and B, treated as turning arrows (phasors). On the left is the addition of the arrows (append B to the tip of A, from the origin to the resultant arrow); on the right is a graph of "the resultant's length² ＝ probability $|\psi|^2$" against the phase difference. Use the slider to vary the phase difference $\Delta\varphi=\Delta S/\hbar$.

At $\Delta\varphi=0$, the two point the same way and the resultant is longest ── reinforcement (four times the probability, a bright line). At $\Delta\varphi=\pi$ they point opposite and the resultant is zero ── cancellation (zero probability, a dark line). Neither path's own probability changes, yet by the phase difference alone the combined probability swings from 0 to maximum. This is interference, and it is proof that $\psi$ is a "wave." How many ℏ's the difference in action $\Delta S$ amounts to ── $S/\hbar$ is again the star here.

Figure: adding the wave functions of the two paths A and B as turning arrows (phasors). Left = addition of arrows, right = the resultant's length² (probability) vs. phase difference. Vary Δφ=ΔS/ℏ, and reinforcement (bright) ↔ cancellation (dark)

phase difference Δφ = ΔS/ℏ 0.00

arrows of paths A and B resultant (probability |ψ|²)

05So "particle or wave" vanishes

The wave function $\psi$ holds two pieces of information single-handedly ── the magnitude $|\psi|$ (where it is likely to be found ＝ the particle-like probability) and the phase $S/\hbar$ (how it interferes ＝ the wave-like property). The question "particle or wave?" was merely the difference between looking at only the magnitude and looking at only the phase of this one $\psi$. Detect it, and it is found at one point with probability $|\psi|^2$ (particle); while it travels, it carries a phase and interferes (wave) ── both are separate aspects of $\psi$. The true face beneath the faces that we called "two faces of particle and wave" in Episode 1 was the complex wave $\psi$.

◇ ◇ ◇

The honest line ── the reach of ψ and the probability interpretation

That the wave function is complex and its phase corresponds to the action $S/\hbar$ ($\psi\sim e^{iS/\hbar}$), that $|\psi|^2$ is the probability density (the Born rule), superposition and interference via the phase difference $\Delta\varphi=\Delta S/\hbar$ ($2|\psi_A||\psi_B|\cos\Delta\varphi$), and that the overall phase is unobservable (a gauge freedom), demonstrated as the double-slit interference of electron beams ── all of these are established physics.

However. ① In $\psi=|\psi|e^{iS/\hbar}$, "phase ＝ $S/\hbar$" is a correspondence in the semiclassical (WKB) approximation where the classical action $S$ is meaningful; strictly, the phase is a quantity given by quantum theory (made rigorous by Episode 3's sum over paths). ② The Born rule ($|\psi|^2$ ＝ probability) does not explain why it should be so; it is laid down as a postulate (part of the measurement problem, Episode 6). ③ When there are many particles, $\psi$ is not a wave in three-dimensional space but a wave in a high-dimensional "configuration space" gathering all the particles' coordinates (the stage of entanglement). The picture of "a turning arrow at each point" is the intuition for a single particle; for many particles it is far richer.

Practice problems (solvable with this episode's formulas alone)

When all of the wave function $\psi$ is multiplied by $e^{i\theta}$, how does the probability $|\psi|^2$ change? What is this freedom called?
See the answer
$|e^{i\theta}\psi|^2=|\psi|^2$, so it does not change. The overall phase does not affect observation ── the gauge (phase) freedom. The same as "Fields That Click" Bonus ②.
When the two paths' phase difference is $\Delta\varphi=\pi$, what happens to the combined probability, even though neither path's individual probability is zero?
See the answer
The interference term is $2|\psi_A||\psi_B|\cos\pi=-2|\psi_A||\psi_B|$. For equal amplitudes the total is $|\psi_A|^2+|\psi_B|^2-2|\psi_A||\psi_B|=0$. Complete cancellation (a dark line). The path exists, yet nothing arrives.
How does the de Broglie relation $p=\hbar k$ relate to the wave function's phase $S/\hbar$?
See the answer
Advancing $x$ through space turns the phase by $kx=px/\hbar$ ＝ the spatial part of the action $px$ divided by $\hbar$. The wavenumber $k$ is exactly the spatial way the phase $S/\hbar$ turns.
Why does the question "particle or wave?" vanish? Answer with ψ's two pieces of information.
See the answer
ψ carries at once the magnitude |ψ| (probability of being found ＝ particle-like) and the phase S/ℏ (interference ＝ wave-like). Particle-or-wave is a difference of which aspect of ψ you look at; the true face is one complex wave ψ.

Episode 2 summaryψ ── magnitude² is probability, phase is S/ℏ

The true identity of the "one entity" glued together from particle and wave is the complex wave, the wave function $\psi=|\psi|e^{iS/\hbar}$. At each point there is a "turning arrow"; the square of its length $|\psi|^2$ is the probability of being found (the Born rule), and its direction (phase) is Episode 1's $S/\hbar$. As action builds, the phase turns, and the way it turns is the de Broglie wavelength and frequency. The overall phase does not affect observation (＝ the gauge freedom of "Fields That Click," contiguous with cosmology's $i$).

But superpose two paths, and the difference in phase $\Delta\varphi=\Delta S/\hbar$ shows its face as the interference term $2|\psi_A||\psi_B|\cos\Delta\varphi$ ── reinforcement at $\Delta\varphi=0$, cancellation at $\pi$. Each path's probability is unchanged, yet by the phase difference alone the total swings from 0 to maximum. This is the proof that $\psi$ is a wave. "Particle or wave" is merely the difference between looking at $\psi$'s magnitude and looking at its phase. What counts is, again, the action ratio $S/\hbar$ ── the second step in the story of $\hbar$.

Next time ── Episode 3 "Summing over all paths ── least action and the emergence of the classical." We added the phases of two paths ── but what if there were countless paths? Feynman's answer is bold ── nature adds up the phase $e^{iS/\hbar}$ of "every possible path" from start to finish. Usually the phases are scattered and cancel, but only near the path where $S$ is stationary (＝ least action) do the phases line up and survive. And in the classical world of $S\gg\hbar$, that single one remains and ── Newton's trajectory appears. Here it connects with the least action of "Force That Clicks."

This document is Episode 2 of the "Quantum That Clicks" series, a reading for high-schoolers and undergrads who love physics. That the wave function is complex-valued and gives the probability amplitude, that $|\psi|^2$ is the probability density (the Born rule), the superposition principle and interference via the phase difference (the interference term $2\,\mathrm{Re}(\psi_A^*\psi_B)=2|\psi_A||\psi_B|\cos\Delta\varphi$), the unobservability of the global phase (the U(1) phase freedom), the correspondence of phase and action in the semiclassical regime $\psi\sim e^{iS/\hbar}$, and the experimental establishment of electron double-slit interference ── all are standard physics. That phase ＝ $S/\hbar$ is a correspondence in the semiclassical (WKB) approximation and the rigorous phase is set by quantum theory (made rigorous in Episode 3 via the path integral), that the Born rule is laid down as an axiom and involves interpretive aspects of the unresolved measurement problem (Episode 6), and that in many-particle systems $\psi$ is a function not on three-dimensional space but on high-dimensional configuration space and is the stage of entanglement, are all stated in the "honest line" in the body. The figure is a schematic of the phasor sum and the interference intensity $|\psi_A+\psi_B|^2=2(1+\cos\Delta\varphi)$. ── To print, use your browser's "Print" → "Save as PDF" (in the print version the slider and answers are frozen or hidden). Adjacent episodes: Episode 1: ℏ, the grain size / contents / sister series Fields That Click, Bonus ② (gauge).

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen, varying the phase difference Δφ makes the resultant of the two arrows (the probability) reinforce ↔ cancel. "See the answer" opens each solution.