Wave function and probability interpretation

From trajectory to probability cloud. Classical mechanics tracks a particle along one definite path $x(t)$. Quantum mechanics abandons that picture entirely: all the physical information about a particle in one dimension is packed into a single complex-valued function $\psi(x,t)$, and $\psi$ itself carries no direct physical meaning — you cannot measure it on a dial. What you can measure are probabilities, and Born's rule is the bridge between the mathematics and the lab.

Born's rule and probability density. The rule states that the squared magnitude of $\psi$ gives a probability density in space:

where: $\psi(x,t)$ is the (generally complex) wave function, $|\psi|^2=\psi^*\psi$ is real and never negative, and $dx$ is an infinitesimal length so that $|\psi|^2$ itself has units of probability per unit length — a density, not a probability. To get an actual probability you must integrate $|\psi|^2$ over a finite stretch of the line.

Normalization. Because the particle is somewhere on the line with certainty, the probabilities over all space must sum to exactly 1:

where: the integral runs over the whole region the particle can occupy. A $\psi$ obeying this is called normalized. If a solution of the wave equation instead integrates to some finite $N\ne 1$, rescaling $\psi_{\text{norm}}=\psi/\sqrt N$ restores normalization without changing the physics, since $|\psi_{\text{norm}}|^2 = |\psi|^2/N$ integrates to exactly 1.

Expectation values. Once $\psi$ is normalized, every measurable quantity is obtained by an "operator sandwich" — the appropriate operator placed between $\psi^*$ and $\psi$, then integrated over all space. Position is represented by plain multiplication by $x$, so

the mean position weighted by the probability density $|\psi|^2$. Momentum is represented instead by a derivative operator, $\hat p = -i\hbar\,\dfrac{d}{dx}$, so that

where: $\hbar = h/2\pi$ is the reduced Planck constant, $\psi^*$ is the complex conjugate of $\psi$, and the integral is taken over all $x$ for which $\psi$ is defined. This is the general pattern behind every expectation value in quantum mechanics: pick the operator for the quantity you want, sandwich it, integrate.

Good behaviour of $\psi$. Not every function is an acceptable wave function. Physically reasonable $\psi$ must be single-valued, and $\psi$ together with $d\psi/dx$ must stay continuous wherever the potential $V(x)$ is finite (the derivative can jump only across an infinite discontinuity in $V$). $\psi$ must also vanish, or stay bounded, as $x\to\pm\infty$, otherwise the normalization integral could not converge to a finite number. These conditions are not decoration — the next chapter uses exactly this continuity to match solutions of the Schrödinger equation across a potential step or well.

A particle confined to $0\le x\le L$ has, at some instant, the real wave function $\psi(x) = A\sin(\pi x/L)$ on that interval and $\psi(x)=0$ outside it. Find $A$, and then find the probability that the particle is found in the left half of the box, $0\le x\le L/2$.

Step 1. Normalize. We need $\displaystyle\int_0^L|\psi(x)|^2dx=1$, i.e. $A^2\displaystyle\int_0^L\sin^2(\pi x/L)\,dx=1$. Using $\sin^2\theta=\tfrac12(1-\cos2\theta)$, the integral evaluates to $L/2$. So $A^2(L/2)=1$, giving $A=\sqrt{2/L}$.

Step 2. Compute the left-half probability. $P(0\le x\le L/2) = \displaystyle\int_0^{L/2}\frac{2}{L}\sin^2(\pi x/L)\,dx = \dfrac{2}{L}\left[\dfrac{x}{2}-\dfrac{L}{4\pi}\sin\!\left(\dfrac{2\pi x}{L}\right)\right]_0^{L/2} = \dfrac{2}{L}\left(\dfrac{L}{4}-0\right) = \dfrac12$.

Step 3. Extend to the expectation value. With this same normalized $\psi$, $\langle x\rangle = \displaystyle\int_0^L\psi^*x\psi\,dx = \dfrac{2}{L}\int_0^Lx\sin^2(\pi x/L)\,dx$. Splitting via the same identity, the first term gives $\tfrac12\cdot\tfrac{L^2}{2}=\tfrac{L^2}{4}$ and the cosine term integrates to zero over the interval (a standard result for $\int_0^Lx\cos(2\pi x/L)dx$), so $\langle x\rangle = \dfrac{2}{L}\cdot\dfrac{L^2}{4} = \dfrac{L}{2}$.

Step 4. Interpret. Both results are exactly what symmetry demands: $|\psi|^2\propto\sin^2(\pi x/L)$ is symmetric about the midpoint $x=L/2$, so half the probability sits in each half of the box, and the mean position is precisely the centre. Final answers: $A=\sqrt{2/L}$, $P(\text{left half})=1/2$, and $\langle x\rangle = L/2$.

• Treating $\psi$ itself as a probability — it is $|\psi|^2$, a real non-negative quantity, that plays that role; $\psi$ can be complex or negative.
• Skipping normalization before computing an expectation value: if $\int|\psi|^2dx\ne 1$, the raw formula $\langle x\rangle=\int\psi^*x\psi\,dx$ gives the wrong number.
• Confusing "probability" with "probability density" — $|\psi(x)|^2$ has units of probability per unit length; you must integrate over an interval to get a number between 0 and 1.
• Believing $\langle x\rangle$ tells you where the particle actually is on any single measurement — it is a statistical mean over many identically prepared particles, not an outcome.

Ask the tutor: want to work through normalizing a Gaussian wave packet, or see how the continuity of $\psi$ and $d\psi/dx$ is actually applied at a potential step?