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Robertson-Schrödinger uncertainty relation

Consider a quantum system in a state \(| \psi \rangle\), and let \(A\) and \(B\) be Hermitian operators representing a pair of observables. One can choose to perform a measurement of \(A\) or a measurement of \(B\) on the system. Generally, the outcomes of these measurements cannot be predicted with certainty. The outcomes are statistical in nature and characterized by the respective expectation values \(\langle A \rangle\) and \(\langle B \rangle\). Hereinafter, \[\langle \cdot \rangle = \langle \psi | \cdot | \psi \rangle \,.\] The corresponding uncertainties are defined as \[\sigma_A = \sqrt{ \langle \big( A - \langle A \rangle \big)^2 \rangle } = \sqrt{\langle A^2 \rangle - \langle A \rangle^2} \,,\] \[\sigma_B = \sqrt{ \langle \big( B - \langle B \rangle \big)^2 \rangle } = \sqrt{\langle B^2 \rangle - \langle B \rangle^2} \,.\] The Robertson-Schrödinger uncertainty relation sates that \[\sigma_A^2 \sigma_B^2 \ge \langle \tfrac{i}{2} [A,B] \rangle^2 + \Big( \langle \tfrac{1}{2} \{ A, B \} \rangle - \langle A \rangle \langle B \rangle \Big)^2 \,,\] where \[[A,B] = A B - B A \,,\] \[\{ A, B \} = A B + B A \,.\] As the operators \(A\) and \(B\) are Hermitian, so are the operators \(i[A,B]\) and \(\{A,B\}\), implying that the right-hand side of the Robertson-Schrödinger relation is non-negative.

Proof: The relation follows from the Cauchy-Schwarz inequality: \[\big| \langle u | v \rangle \big|^2 \le \langle u | u \rangle \langle v | v \rangle \,,\] for any two states \(| u \rangle\) and \(| v \rangle\). Let us take \[| u \rangle = \big( A - \langle A \rangle \big) | \psi \rangle \,,\] \[| v \rangle = \big( B - \langle B \rangle \big) | \psi \rangle \,.\] Then, \[\langle u | u \rangle = \langle \big( A - \langle A \rangle \big)^2 \rangle = \sigma_A^2 \,,\] \[\langle v | v \rangle = \langle \big( B - \langle B \rangle \big)^2 \rangle = \sigma_B^2 \,,\] \[\langle u | v \rangle = \langle \big( A - \langle A \rangle \big) \big( B - \langle B \rangle \big) \rangle = \langle A B \rangle - \langle A \rangle \langle B \rangle \,.\] Substituting these expressions into the Cauchy-Schwarz inequality, we obtain \[\sigma_A^2 \sigma_B^2 \ge \big| \langle A B \rangle - \langle A \rangle \langle B \rangle \big|^2 \,.\] It remains to show that \(\big| \langle A B \rangle - \langle A \rangle \langle B \rangle \big|^2\) can be rewritten as the right-hand side of the Robertson-Schrödinger uncertainty relation. To this end, we define two Hermitian operators \(C_+\) and \(C_-\) as \[C_+ \equiv \{A,B\} = A B + B A \,,\] \[C_- \equiv i [A,B] = i (A B - B A) \,.\] Then, \[A B = \tfrac{1}{2} C_+ - \tfrac{i}{2} C_- \,,\] \[B A = \tfrac{1}{2} C_+ + \tfrac{i}{2} C_- \,,\] and \[\big| \langle A B \rangle - \langle A \rangle \langle B \rangle \big|^2 = \left| \langle \tfrac{1}{2} C_+ \rangle - \langle A \rangle \langle B \rangle - i \langle \tfrac{1}{2} C_- \rangle \right|^2 \,.\] Finally, since \(\langle C_+ \rangle\), \(\langle C_- \rangle\), and \(\langle A \rangle \langle B \rangle\) are all real, we obtain \[\big| \langle A B \rangle - \langle A \rangle \langle B \rangle \big|^2 = \Big( \langle \tfrac{1}{2} C_+ \rangle - \langle A \rangle \langle B \rangle \Big)^2 + \langle \tfrac{1}{2} C_- \rangle^2 \,,\] which concludes the proof.