The Uncertainty Principle

The generalized statistical interpretation provides the complete prescription for extracting physical predictions from the quantum formalism. Given a state $|\psi\rangle$ and an observable $\hat{Q}$ with eigenvalues $\{q_n\}$ and orthonormal eigenstates $\{|q_n\rangle\}$, we have:

The possible outcomes of measuring $Q$ are the eigenvalues $q_n$. The probability of obtaining $q_n$ is $P(q_n) = |\langle q_n|\psi\rangle|^2$. After the measurement, the state collapses to $|q_n\rangle$.

For a continuous spectrum with eigenstates $|q\rangle$ delta-function normalized as $\langle q'|q\rangle = \delta(q'-q)$, the probability density is $\rho(q) = |\langle q|\psi\rangle|^2$, and the probability of finding the outcome in $[q, q+dq]$ is $\rho(q)\,dq$.

The expectation value of any function $f(Q)$ is $\langle f(Q)\rangle = \langle\psi|f(\hat{Q})|\psi\rangle = \int f(q)\rho(q)\,dq$. This unifies the discrete and continuous cases.

For any two observables $A$ and $B$, the product of their uncertainties satisfies the Robertson relation:

The proof uses the Cauchy-Schwarz inequality and the properties of Hermitian operators. Define $|f\rangle = (\hat{A} - \langle A\rangle)|\psi\rangle$ and $|g\rangle = (\hat{B} - \langle B\rangle)|\psi\rangle$. Then $(\Delta A)^2 = \langle f|f\rangle$ and $(\Delta B)^2 = \langle g|g\rangle$.

By Cauchy-Schwarz: $\langle f|f\rangle\langle g|g\rangle \ge |\langle f|g\rangle|^2$. Now write $\langle f|g\rangle = (1/2)\langle f|g\rangle + (1/2)\langle g|f\rangle^* = (1/2)(\langle f|g\rangle + \langle g|f\rangle) + (1/2)(\langle f|g\rangle - \langle g|f\rangle)$. The first term is real; the second is purely imaginary.

Since $|x+iy|^2 = x^2 + y^2 \ge y^2$, we have $|\langle f|g\rangle|^2 \ge (1/4)|\langle f|g\rangle - \langle g|f\rangle|^2 = (1/4)|[\hat{A}, \hat{B}]|^2$. Combining with Cauchy-Schwarz gives the Robertson relation.

The uncertainty principle sets a lower bound on $\Delta A\,\Delta B$. States that saturate this bound are called minimum-uncertainty states.

For position and momentum, the Cauchy-Schwarz inequality becomes an equality when $|g\rangle = c|f\rangle$ for some complex constant $c$. Substituting the definitions gives a differential equation for $\psi$.

For position and momentum, the condition becomes: $(\hat{p} - \langle p\rangle)|\psi\rangle = i\lambda (\hat{x} - \langle x\rangle)|\psi\rangle$ with $\lambda$ real. This is a first-order differential equation whose solution is a Gaussian:

The width parameter $a$ determines the uncertainties: $\Delta x = a$ and $\Delta p = \hbar/(2a)$. Their product is $\Delta x\,\Delta p = \hbar/2$, the minimum allowed.

The energy-time uncertainty principle is subtle because time is not an operator in quantum mechanics. The relation $\Delta E\,\Delta t \ge \hbar/2$ has a different interpretation than $\Delta x\,\Delta p \ge \hbar/2$.

One interpretation uses the rate of change of expectation values. For any observable $Q$, define $\Delta t_Q = \Delta Q / |d\langle Q\rangle/dt|$. Then $\Delta E\,\Delta t_Q \ge \hbar/2$. This says that if a state has a small energy uncertainty, the expectation values of observables change slowly.

Another interpretation concerns the lifetime of unstable states. If a state has energy uncertainty $\Delta E$, its lifetime is approximately $\tau \sim \hbar/\Delta E$. This explains why unstable particles have natural linewidths.

The energy-time uncertainty is not a statement about commuting operators (since there is no time operator). It reflects the fact that measuring energy precisely requires a long measurement time.

For the ground state of the infinite square well of width $a$, compute $\Delta x$, $\Delta p$, and verify the uncertainty principle.

Step 1 — $\langle x\rangle$. By symmetry, $\langle x\rangle = a/2$.

Step 2 — $\langle x^2\rangle$. Using $\int_0^a x^2 \sin^2(\pi x/a)\,dx = a^3/6 - a^3/(2\pi^2)$, we get $\langle x^2\rangle = a^2/3 - a^2/(2\pi^2)$.

Step 3 — $\Delta x$. $\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = \sqrt{a^2/12 - a^2/(2\pi^2)} = a\sqrt{1/12 - 1/(2\pi^2)} \approx 0.181a$.

Step 4 — $\langle p\rangle$. The integrand involves $\sin(\pi x/a)\cos(\pi x/a)$, which integrates to zero. So $\langle p\rangle = 0$.

Step 5 — $\langle p^2\rangle$. Since $\phi_1$ is an energy eigenstate, $\hat{p}^2\phi_1 = 2m(E_1 - V)\phi_1 = (\pi^2\hbar^2/a^2)\phi_1$. Thus $\langle p^2\rangle = \pi^2\hbar^2/a^2$.

Step 6 — $\Delta p$. $\Delta p = \sqrt{\langle p^2\rangle - \langle p\rangle^2} = \pi\hbar/a$.

Step 7 — Check. $\Delta x\,\Delta p = a\sqrt{1/12 - 1/(2\pi^2)} \cdot \pi\hbar/a = \pi\hbar\sqrt{1/12 - 1/(2\pi^2)} \approx 0.568\hbar > \hbar/2$. The uncertainty principle is satisfied.