Past exam of the physics A course of the University of Cambridge / 2018 / Paper 1 / Section B / B9

The commutator is $\boxed{[A,B]=AB-BA}$. To evaluate it for differential operators, apply it to a smooth test wavefunction $f$:Thus the canonical commutation relation is $\boxed{[x,p_x]=i\hslash I}$. The associated Heisenberg uncertainty relation is $\Delta x\Delta p_x\ge \hslash /2$. Position and momentum cannot both have arbitrarily sharp outcome distributions in the same prepared state, even with ideal measuring apparatus. This concerns quantum-state preparation, not merely a calibration error in a particular instrument.

Expand the proposed product rule without interchanging any operator factors:Applying this commutator identity with $A=x$ and $B=C=p_x$ givesFor a differentiable multiplication potential, apply the other commutator to a test function:ThereforeThese operator equalities are understood on a common domain where the indicated products and derivatives are defined, with the boundary conditions appropriate to the state.

For a normalized state, the expectation value isFor a time-independent operator, differentiation givesThe Schrödinger equation says $|\dot{\psi }⟩=-iH|\psi ⟩/\hslash$. The self-adjointness of $H$ gives $⟨\dot{\psi }|=i⟨\psi |H/\hslash$, soIn the integral derivation, the same self-adjointness step is integration by parts with vanishing boundary terms. An explicitly time-dependent observable would also contribute $⟨\partial Q/\partial t⟩$. These are the expectation-value equations underlying the Heisenberg equation of motion.

The Hamiltonian operator isSince multiplication by $x$ commutes with multiplication by $V(x)$, part (b) gives $[x,H]=i\hslash p_x/m$. Similarly $[p_x,H]=-i\hslash V^{\prime }(x)$. Substituting into part (c) gives Ehrenfest theorem:Thus the mean momentum changes according to the mean force, and the derivative of mean position is mean velocity. These exact quantum equations resemble Hamilton's classical equations, but $⟨V^{\prime }(x)⟩$ generally differs from $V^{\prime }(⟨x⟩)$. The classical point-particle equation for the center is exact for potentials at most quadratic and is an approximation for a localized packet when the force varies little across it. A spatially broad packet in a nonlinear potential need not follow a classical trajectory. Even where its center follows classical motion, its width and measurement uncertainties remain quantum properties.