Normalisation and probability density · Quantum Mechanics

Intuition

The modulus squared of the wavefunction is a probability density: integrate it over a stretch of the line and you get the probability of finding the particle there. Over the whole line the probability must be one, which fixes the size of the wavefunction; multiplying it by the right constant is called normalising. The probability of any single exact point is zero, just as a single point of a coastline has no length.

Rainfall per square metre is a density; how much fell on a field is the density integrated over the field. The wavefunction squared is rainfall of probability, and over the whole map exactly one unit falls.

The density $|\psi|^{2}=e^{-2|x|}$ of the normalised state $\psi(x)=e^{-|x|}$ . The probability of finding the particle between the dashed lines, $-\tfrac12\le x\le\tfrac12$ , is the area under the curve there: $1-e^{-1}\approx0.63$ .

Probability density

By the Born rule, $|\psi(x)|^{2}$ is the probability density for position: the probability of finding the particle between $a$ and $b$ is the integral of the density over that stretch. Normalisation says the particle is somewhere.

\begin{aligned}P(a\le x\le b)&=\int_{a}^{b}|\psi(x)|^{2}\,dx\\ \int_{-\infty}^{\infty}|\psi(x)|^{2}\,dx&=1\end{aligned}

Working with densities

To normalise $\psi$ , divide it by $\sqrt{\int|\psi|^{2}dx}$ . A global phase is left free.

Normalising a decaying exponential

\psi(x)=A\,e^{-|x|/a}\ \ \text{is normalised when}\ \ |A|^{2}=\frac{1}{a}

The density $|A|^{2}e^{-2|x|/a}$ is symmetric, so its integral over the line is twice the integral over the positive half, which is an elementary exponential integral. Setting the result equal to one fixes the modulus of $A$ , and leaves its phase free.

Proof steps

|\psi|^{2}=|A|^{2}e^{-2|x|/a}

Square the modulus of the wavefunction.

\int_{-\infty}^{\infty}|\psi|^{2}dx=2|A|^{2}\int_{0}^{\infty}e^{-2x/a}dx

The density is even, so integrate over half the line and double.

2|A|^{2}\cdot\frac{a}{2}=|A|^{2}a

$\int_{0}^{\infty}e^{-2x/a}dx=a/2$ .

|A|^{2}a=1\ \Rightarrow\ |A|=\frac{1}{\sqrt a}

Set the total probability to one.

Applications

Practice

Area Under the Density

The probability of finding the particle in a stretch of the line is the area under the probability density over that stretch.

P(a\le x\le b)=\int_{a}^{b}|\psi(x)|^{2}dx

Try it

A particle has the normalised wavefunction $\psi(x)=\sqrt2\sin(\pi x)$ on $0\le x\le1$ and zero elsewhere. What is the probability of finding it in $0\le x\le\tfrac12$ ?