MathLabs
TheoremProved

Robertson–Heisenberg uncertainty principle

Statement

For any two self-adjoint operators A^=A^∗\hat A=\hat A^*, B^=B^∗\hat B=\hat B^* on H\mathcal H and any unit vector ψ\psi in the domain of A^B^\hat A\hat B and B^A^\hat B\hat A, the standard deviations σA=⟨ψ∣(A^−⟨A^⟩)2∣ψ⟩\sigma_A=\sqrt{\langle\psi|(\hat A-\langle\hat A\rangle)^2|\psi\rangle} and σB\sigma_B (expectations taken in state ψ\psi) satisfy σAσB≥12∣⟨[A^,B^]⟩∣\sigma_A \sigma_B \ge \tfrac{1}{2}\left|\langle [\hat{A},\hat{B}]\rangle\right|, where [A^,B^]=A^B^−B^A^[\hat A,\hat B]=\hat A\hat B-\hat B\hat A.

Why is it true?

Two observables can only be measured with simultaneous perfect precision if their operators commute; the size of the commutator is a direct measure of how incompatible they are. Cauchy–Schwarz turns "these two vectors cannot both be short" into a hard numerical bound, and separating the inner product into its real and imaginary parts is exactly what isolates the commutator (the genuinely quantum part) from the anticommutator (a classical-looking correlation term).

Proof sketch

Step 1 (centered operators). Let ΔA^=A^−⟨A^⟩I\Delta\hat A=\hat A-\langle\hat A\rangle I and ΔB^=B^−⟨B^⟩I\Delta\hat B=\hat B-\langle\hat B\rangle I, both still self-adjoint since ⟨A^⟩,⟨B^⟩\langle\hat A\rangle,\langle\hat B\rangle are real scalars. By definition σA2=⟨ψ∣ΔA^2∣ψ⟩=∥ΔA^ ψ∥2\sigma_A^2=\langle\psi|\Delta\hat A^2|\psi\rangle=\|\Delta\hat A\,\psi\|^2 and likewise σB2=∥ΔB^ ψ∥2\sigma_B^2=\|\Delta\hat B\,\psi\|^2.

Step 2 (Cauchy–Schwarz). Apply the Cauchy–Schwarz inequality ∣⟨f∣g⟩∣2≤⟨f∣f⟩⟨g∣g⟩|\langle f|g\rangle|^2\le\langle f|f\rangle\langle g|g\rangle to f=ΔA^ ψf=\Delta\hat A\,\psi and g=ΔB^ ψg=\Delta\hat B\,\psi: ∣⟨ΔA^ ψ∣ΔB^ ψ⟩∣2≤σA2 σB2.|\langle\Delta\hat A\,\psi|\Delta\hat B\,\psi\rangle|^2 \le \sigma_A^2\,\sigma_B^2.

Step 3 (split into real and imaginary parts). Write ⟨ΔA^ ψ∣ΔB^ ψ⟩=⟨ψ∣ΔA^ΔB^∣ψ⟩\langle\Delta\hat A\,\psi|\Delta\hat B\,\psi\rangle=\langle\psi|\Delta\hat A\Delta\hat B|\psi\rangle. Since ΔA^,ΔB^\Delta\hat A,\Delta\hat B are self-adjoint, complex-conjugating flips the operator order: ⟨ΔA^ΔB^⟩∗=⟨ΔB^ΔA^⟩\langle\Delta\hat A\Delta\hat B\rangle^*=\langle\Delta\hat B\Delta\hat A\rangle. Hence the anticommutator expectation ⟨{ΔA^,ΔB^}⟩=⟨ΔA^ΔB^⟩+⟨ΔB^ΔA^⟩=2 Re⟨ΔA^ΔB^⟩\langle\{\Delta\hat A,\Delta\hat B\}\rangle=\langle\Delta\hat A\Delta\hat B\rangle+\langle\Delta\hat B\Delta\hat A\rangle=2\,\mathrm{Re}\langle\Delta\hat A\Delta\hat B\rangle is real, while the commutator expectation ⟨[ΔA^,ΔB^]⟩=⟨ΔA^ΔB^⟩−⟨ΔB^ΔA^⟩=2i Im⟨ΔA^ΔB^⟩\langle[\Delta\hat A,\Delta\hat B]\rangle=\langle\Delta\hat A\Delta\hat B\rangle-\langle\Delta\hat B\Delta\hat A\rangle=2i\,\mathrm{Im}\langle\Delta\hat A\Delta\hat B\rangle is purely imaginary. Also [ΔA^,ΔB^]=[A^,B^][\Delta\hat A,\Delta\hat B]=[\hat A,\hat B], since the constant shifts cancel in the commutator.

Step 4 (recombine). By Pythagoras applied to real and imaginary parts, ∣⟨ΔA^ΔB^⟩∣2=(Re⟨ΔA^ΔB^⟩)2+(Im⟨ΔA^ΔB^⟩)2=14∣⟨{ΔA^,ΔB^}⟩∣2+14∣⟨[A^,B^]⟩∣2≥14∣⟨[A^,B^]⟩∣2,|\langle\Delta\hat A\Delta\hat B\rangle|^2=\big(\mathrm{Re}\langle\Delta\hat A\Delta\hat B\rangle\big)^2+\big(\mathrm{Im}\langle\Delta\hat A\Delta\hat B\rangle\big)^2=\tfrac14|\langle\{\Delta\hat A,\Delta\hat B\}\rangle|^2+\tfrac14|\langle[\hat A,\hat B]\rangle|^2 \ge \tfrac14|\langle[\hat A,\hat B]\rangle|^2, dropping the manifestly nonnegative anticommutator term. Combining with Step 2, σA2σB2≥14∣⟨[A^,B^]⟩∣2\sigma_A^2\sigma_B^2\ge\tfrac14|\langle[\hat A,\hat B]\rangle|^2, and taking square roots (both sides nonnegative) gives σAσB≥12∣⟨[A^,B^]⟩∣\sigma_A \sigma_B \ge \tfrac{1}{2}\left|\langle [\hat{A},\hat{B}]\rangle\right|.

Topics that use this theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

  1. John von Neumann (1955). Mathematical Foundations of Quantum Mechanics
  2. Howard P. Robertson (1929). The Uncertainty Principle · DOI:10.1103/PhysRev.34.163
  3. Marshall H. Stone (1932). On One-Parameter Unitary Groups in Hilbert Space · DOI:10.2307/1968538