The Wave Function

The fundamental equation of quantum mechanics is the time-dependent Schrödinger equation:

Here $\psi(x,t)$ is the wave function, a complex-valued function of position $x$ and time $t$. The Hamiltonian operator $\hat{H}$ contains the kinetic and potential energy of the particle. For a single particle moving in one dimension under the influence of a potential $V(x,t)$:

The Schrödinger equation is a postulate of quantum mechanics. It cannot be derived from classical mechanics. Its validity rests entirely on its agreement with experiment. The equation is first-order in time, which means that given the wave function $\psi(x,0)$ at $t=0$, the Schrödinger equation determines $\psi(x,t)$ for all future times.

A crucial feature of the Schrödinger equation is that it preserves normalization. If the wave function is normalized at $t=0$, it remains normalized for all time. This follows from the Hermiticity of the Hamiltonian.

Max Born proposed in 1926 that $|\psi(x,t)|^2$ gives the probability density for finding the particle at position $x$ at time $t$. This is the statistical interpretation of the wave function.

The probability of finding the particle between $a$ and $b$ at time $t$ is the integral of the probability density over that interval:

This interpretation is fundamentally different from classical physics. In classical mechanics, a particle has a definite trajectory $x(t)$. In quantum mechanics, the particle does not have a definite position until a measurement is performed. The wave function only provides probabilities.

The wave function itself is not directly observable. Only probabilities derived from it can be measured. This is one of the deepest features of quantum mechanics and the source of much philosophical debate.

For a discrete variable $j$ that can take values $j_1, j_2, \ldots$, the probability of obtaining value $j_n$ is $P(j_n)$. The probabilities satisfy:

The expectation value (average) of any function $f(j)$ is:

For a continuous variable $x$, the probability density $\rho(x)$ is defined so that $\rho(x)\,dx$ is the probability of finding $x$ in an infinitesimal interval $[x, x+dx]$. The normalization condition becomes:

The expectation value of a function $f(x)$ is:

The variance measures the spread around the mean: $\sigma^2 = \langle j^2 \rangle - \langle j \rangle^2$ for discrete variables, or $\sigma^2 = \langle x^2 \rangle - \langle x \rangle^2$ for continuous variables. The standard deviation is $\sigma = \sqrt{\text{variance}}$.

Since the particle must exist somewhere in space, the total probability of finding it must equal 1. This gives the normalization condition:

A wave function satisfying this condition is said to be normalized. If a wave function is not normalized, we can normalize it by dividing by the square root of the integral of $|\psi|^2$ over all space.

Normalization is preserved by the Schrödinger equation for physically realistic potentials. This means that if we prepare the system in a normalized state at $t=0$, it will remain normalized for all time. The proof relies on the Hermiticity of the Hamiltonian and the fact that the probability current satisfies a continuity equation.

In quantum mechanics, momentum is represented by an operator rather than a number. The momentum operator in the position representation is:

The expectation value of momentum is computed by sandwiching the momentum operator between $\psi^*$ and $\psi$:

For a plane wave $\psi_k(x) = A e^{ikx}$, applying the momentum operator gives $\hat{p}\psi_k = \hbar k \psi_k$. Thus plane waves are momentum eigenstates with eigenvalue $p = \hbar k$. This is the de Broglie relation made precise.

The expectation value of any function of momentum $Q(p)$ can be written as:

This formula, together with the corresponding formula for functions of position, allows us to compute the expectation value of any observable that can be expressed in terms of position and momentum.

The Heisenberg uncertainty principle states that the product of the uncertainties in position and momentum satisfies:

This is not a statement about measurement precision or technological limitations. It is a fundamental property of nature. A quantum particle simply does not have simultaneously well-defined position and momentum.

The uncertainty principle arises from the wave nature of quantum particles. A narrow wave packet in position space (small $\Delta x$) requires a broad superposition of momentum eigenstates (large $\Delta p$), because position and momentum are related by Fourier transformation.

The minimum uncertainty product $\Delta x\,\Delta p = \hbar/2$ is achieved by Gaussian wave packets. Any other wave shape gives a strictly larger product.

Consider the normalized Gaussian wave packet at $t=0$: $\psi(x) = (2a/\pi)^{1/4} e^{-a x^2}$, where $a > 0$.

Step 1 — Verify normalization. We need $\int |\psi|^2 \, dx = (2a/\pi)^{1/2} \int e^{-2ax^2} \, dx = (2a/\pi)^{1/2} \cdot \sqrt{\pi/2a} = 1$.

Step 2 — Compute $\langle x \rangle$. The integrand $x|\psi|^2$ is an odd function, so $\langle x \rangle = 0$.

Step 3 — Compute $\langle x^2 \rangle$. Using $\int x^2 e^{-2ax^2} \, dx = \sqrt{\pi}/(2(2a)^{3/2})$, we get $\langle x^2 \rangle = 1/(4a)$, so $\Delta x = 1/(2\sqrt{a})$.

Step 4 — Compute $\langle p \rangle$. Since $\psi$ is real and even, $\langle p \rangle = -i\hbar \int \psi \, \frac{d\psi}{dx} \, dx = -\frac{i\hbar}{2}[\psi^2]_{-\infty}^{+\infty} = 0$.

Step 5 — Compute $\langle p^2 \rangle$. After computing $d^2\psi/dx^2$ and integrating, $\langle p^2 \rangle = a\hbar^2$, so $\Delta p = \hbar\sqrt{a}$.

Step 6 — Check uncertainty principle. $\Delta x\,\Delta p = (1/(2\sqrt{a})) \cdot (\hbar\sqrt{a}) = \hbar/2$. This Gaussian wave packet is a minimum-uncertainty state.