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Module 02

The wavefunction and the Born rule

Turn "ψ is complex and |ψ|² is the probability density" from a slogan into a picture you can rotate.

What you will see

  • ψ as a curve writhing in the complex plane, with the real and imaginary parts as its two shadows
  • Normalisation is not mathematical fussiness: it is the sentence "the particle is somewhere"
  • One position measurement smashes the curve into a narrow spike, which then spreads again

Assumed background

  • Complex numbers: modulus, argument, Euler's formula
  • What an integral means (area)

The previous module showed that the landing points obey some probability distribution. This module introduces whatever it is that governs that distribution.

It is called the wavefunction, written ψ(x)\psi(x), and three things about it have to be said at once:

  1. ψ\psi is complex — at every point it has a modulus and a phase;
  2. what has physical meaning is ∣ψ∣2|\psi|^2, the probability density, not the probability itself;
  3. the phase of ψ\psi is not directly observable, yet differences of phase decide every interference effect.

A flat figure cannot show the first of these, so the first thing to do is stand the complex plane up.

The wavefunction: a curve writhing in the complex plane

The tube in the middle is ψ itself. The curves on the two walls are its real and imaginary shadows; the curtain on the right is |ψ|², the only one of the three that maps directly onto an experiment.

Loading 3D scene…

Prepare a wave packet
-8.0
3.2
0.70
Normalise
×1.00
∫|ψ|²dx = 1.0000

Drag the overall amplitude: the shape never changes, only the height scales. Press "Normalise" to pull ∫|ψ|²dx back to 1 — that equation is just "the particle is somewhere".

Measure position
1.20

After the measurement the packet is squeezed to width δ and immediately starts spreading — the smaller δ, the faster it spreads. That is not an instrument defect, it is Δx·Δp ≥ ħ/2.

Evolution and display
t = 0.0
Speed
ψ(x)=∣ψ(x)∣ eiθ(x),P(x) dx=∣ψ(x)∣2 dx\psi(x) = |\psi(x)|\,\ee^{\ii\theta(x)},\qquad P(x)\,\dd x = |\psi(x)|^2\,\dd x
⟨x⟩ = 0.000Δx = 1.000Δp ≈ 1.000ΔxΔp ≈ 1.000 (lower bound 0.5)
  • ψ (complex tube) and its real-part shadow
  • Imaginary-part shadow
  • |ψ|²

What to look for

  • Start with the "Along x" view. You are facing the complex plane, and at each x the value of ψ is one complex number — one little arrow. The twist of the tube is how the phase changes along x.
  • Now switch to "At the real part": you get exactly the oscillating curve textbooks print. It is only one shadow of the 3D curve — from this angle the imaginary part is entirely lost.
  • Drag k₀ to zero: the tube stops twisting (the phase is the same everywhere), the real part survives and the imaginary part vanishes. The packet no longer travels — it just spreads where it stands.
  • Drag "Overall amplitude": the shape is untouched while ∫|ψ|² drifts away from 1 and turns red. Press "Normalise" and the height snaps back. That is all normalisation ever does.
  • Press "Measure the position once" a few times: the outcome differs each time but clusters where |ψ|² is large. After each measurement the packet is narrow, then spreads fast. Set δ to its smallest value and measure again — it spreads much faster.

✕ "The real part is the wavefunction; the imaginary part is a mathematical trick"

The two are on exactly equal footing. Keeping only the real part throws away the phase, and the phase determines the probability current and every interference effect. That is why the imaginary shadow stays visible even in the "real part" view.

✕ "|ψ(x)|² is the probability of the particle being at x"

It is a probability **density**. The probability at a single point is zero; only an integral over an interval is a probability. Dimensionally |ψ|² is 1/length.

✕ "Measurement turns the wavefunction into a point"

Real measurements have finite resolution, so the packet is squeezed to that width, not to a delta function. The tighter the squeeze, the more uncertain the momentum and the faster it spreads.

Think it through

  1. Set k₀ positive and σ very small, then watch the Δp readout. Why does a narrower packet mean a more uncertain momentum? (Hint: the Fourier transform of a narrow Gaussian is a wide Gaussian.)
  2. Switch on the counter-propagating packet: fringes appear where the two blobs of probability meet. Could those fringes ever appear in a *sum* of two classical probability distributions?
  3. A free particle's |ψ|² spreads forever and never contracts. Does that contradict time-reversal symmetry? (Hint: what do you see if you run the film backwards?)

The Born rule: modulus squared, and nothing else

P(x∈[a,b])=∫ab∣ψ(x)∣2 dx.P(x \in [a,b]) = \int_a^b |\psi(x)|^2\,\dd x .

The left side is a probability (dimensionless); the integrand on the right is a probability density (units of 1/length). Asking “what is the probability of the particle being at x=0.3x = 0.3” is meaningless — that probability is zero. What is meaningful is the probability of landing in a small interval around 0.3.

Normalisation: one physical sentence, written as an equation

∫−∞∞∣ψ(x)∣2 dx=1.\int_{-\infty}^{\infty} |\psi(x)|^2\,\dd x = 1 .

All this says is: the particle is somewhere.

The “Normalise” button in the scene is worth pressing a few times. Drag the overall amplitude up and down: the shape never changes, only the height scales, while the area under ∣ψ∣2|\psi|^2 is pulled back to 1 every time. ψ\psi and 2ψ2\psi describe the same physical state — which will later become a stronger statement: neither the overall length nor the overall phase of a state vector means anything.

Probability conservation and the currentadvanced

Differentiate ρ=∣ψ∣2\rho = |\psi|^2 with respect to time, using the Schrödinger equation

iℏ ∂tψ=−ℏ22m∂x2ψ+Vψ\ii\hbar\,\partial_t \psi = -\frac{\hbar^2}{2m}\partial_x^2\psi + V\psi

and its complex conjugate. The result is

∂tρ+∂xj=0,j=ℏm Im ⁣(ψ∗ ∂xψ).\partial_t \rho + \partial_x j = 0,\qquad j = \frac{\hbar}{m}\,\mathrm{Im}\!\left(\psi^*\,\partial_x\psi\right).

This is a continuity equation, formally identical to the ones in fluid mechanics and electrodynamics. It guarantees that total probability is conserved: probability does not appear or vanish, it only flows.

Look at the expression for jj: it is proportional to the spatial gradient of the phase. The phase is not observable, but its gradient is the probability current — the first hard evidence that the phase is not decoration. Flatten the twist of ψ in the scene (constant phase everywhere) and the current goes to zero.

Measurement and collapse, stated honestly

The “Measure position” button smashes ψ\psi into a narrow spike. Note the wording carefully.

  • If your instrument has resolution δ\delta, the wavefunction afterwards is localised to a width of about δ\delta. It does not become a mathematical delta function.
  • Once localised, the momentum uncertainty grows (ΔxΔp≥ℏ/2\Delta x \Delta p \ge \hbar/2) and the packet immediately starts spreading fast. Release the button in the scene and you watch it blow up.
  • “Collapse” is a rule for how to continue calculating after a measurement, not a mechanism anyone has pinned down. Whether it is a physical process, or merely reflects our projecting the entangled system-plus-apparatus state, is the core dispute in quantum foundations. Module 07 and module 12 go into it.

Expectation values and uncertainty

Once you have a density, all the statistics follow:

⟨x⟩=∫x ∣ψ∣2 dx,⟨x2⟩=∫x2 ∣ψ∣2 dx,Δx=⟨x2⟩−⟨x⟩2.\langle x\rangle = \int x\,|\psi|^2\,\dd x,\qquad \langle x^2\rangle = \int x^2\,|\psi|^2\,\dd x,\qquad \Delta x = \sqrt{\langle x^2\rangle-\langle x\rangle^2}.

The readout shows ⟨x⟩\langle x\rangle and Δx\Delta x live. Squeeze the packet: Δx\Delta x shrinks while the momentum spread grows. The uncertainty relation is not an instrumental defect but a mathematical fact about Fourier transforms — a function and its transform cannot both be narrow.

Key formulas

  • Born rule

    P(a<x<b)=∫ab∣ψ∣2 dxP(a<x<b) = \int_a^b |\psi|^2\,\dd x

    Modulus squared is a density, not a probability

  • Normalisation

    ∫∣ψ∣2dx=1\int |\psi|^2 \dd x = 1

    "The particle is somewhere"

  • Probability current

    j=ℏmIm(ψ∗∂xψ)j = \frac{\hbar}{m}\mathrm{Im}(\psi^*\partial_x\psi)

    A phase gradient is a flow

  • Uncertainty relation

    Δx Δp≥ℏ/2\Delta x\,\Delta p \ge \hbar/2

    An inevitable consequence of Fourier duality

Think it through

  1. ψ\psi and eiαψ\ee^{\ii\alpha}\psi (constant α\alpha) give identical ∣ψ∣2|\psi|^2 and identical expectation values. Why does “the phase has no physical meaning” hold only for the global phase?
  2. A real wavefunction with constant phase has zero probability current jj — even where ∣ψ∣2|\psi|^2 has a steep slope. Does that clash with the intuition that stuff flows from high to low concentration? Why not?
  3. After a measurement squeezes the packet, what sets the rate at which it spreads? Guess first, then go back and shrink the instrument resolution.

Go deeper · matching textbook sections

The 3D scenes build the picture; the full derivations and exercises live in the textbook.

Having finished this module