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beginner · Physics · The Wavefunction & Born Rule

Probability Density |ψ|²

From wavefunction to probability

In quantum mechanics a particle in one dimension is described by its wavefunction ψ(x,t)\psi(x, t) — a complex-valued function of position and time. The wavefunction itself is not directly measurable, but its squared modulus

ρ(x,t)=ψ(x,t)2\rho(x, t) = |\psi(x, t)|^2

is the probability density: the probability of finding the particle in a small interval [x,x+dx][x,\, x + dx] at time tt is

dP=ψ(x,t)2dx.dP = |\psi(x, t)|^2\, dx.

This is Max Born's 1926 interpretation of the wavefunction, now called the Born rule. Because ψ2|\psi|^2 is a probability density, integrating over all space must give 1:

ψ(x,t)2dx=1.\int_{-\infty}^{\infty} |\psi(x, t)|^2\, dx = 1.

A wavefunction that satisfies this condition is said to be normalized.

Computing ψ2|\psi|^2 for a complex amplitude

For a general complex number z=a+biz = a + bi,

z2=zz=a2+b2,|z|^2 = z^* z = a^2 + b^2,

where zz^* is the complex conjugate. When ψ\psi is real-valued (as in many textbook examples) this simplifies to ψ2=ψ2|\psi|^2 = \psi^2. The squared modulus is always real and non-negative, which is necessary for it to be a probability density.

Example: Gaussian wavefunction

One of the most important wavefunctions is the Gaussian (also called a coherent state in the position basis):

ψ(x)=Aexp ⁣(x22σ2),\psi(x) = A \exp\!\left(-\frac{x^2}{2\sigma^2}\right),

where σ\sigma controls the spatial spread and AA is determined by normalization. Inserting into the normalization condition and using the Gaussian integral eu2/σ2du=σπ\int_{-\infty}^{\infty} e^{-u^2/\sigma^2}\, du = \sigma\sqrt{\pi}:

A2ex2/σ2dx=A2σπ=1    A2=1σπ.\int_{-\infty}^{\infty} |A|^2 e^{-x^2/\sigma^2}\, dx = |A|^2 \sigma\sqrt{\pi} = 1 \implies |A|^2 = \frac{1}{\sigma\sqrt{\pi}}.

For the standard choice σ=1\sigma = 1 this gives A2=1/π|A|^2 = 1/\sqrt{\pi}. It is conventional to write the normalization constant as

A=1π1/4(when σ=1),A = \frac{1}{\pi^{1/4}} \quad (\text{when } \sigma = 1),

because then A2=1/π|A|^2 = 1/\sqrt{\pi} and one can verify directly: π1/2ex2dx=π1/2π=1\int_{-\infty}^{\infty} \pi^{-1/2} e^{-x^2}\, dx = \pi^{-1/2} \cdot \sqrt{\pi} = 1.

The probability density is

ψ(x)2=1πex2,|\psi(x)|^2 = \frac{1}{\sqrt{\pi}} e^{-x^2},

which is a normalized Gaussian with peak 1/π0.5641/\sqrt{\pi} \approx 0.564 at x=0x = 0.

Key points

Try it

This is a numerical exercise — your code should return a number.

For the Gaussian wavefunction with σ=1\sigma = 1 and normalization constant A=π1/4A = \pi^{-1/4}, compute the probability density at x=0x = 0, that is ψ(0)2|\psi(0)|^2.

Run your code to see the quantum state.

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