Quantum Physics Corner

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

Let \(A\) be a positive-definite Hermitian operator. The following inequality holds for any pair of states \(| u \rangle\) and \(| v \rangle\): \[\big| \langle u | A | v \rangle \big|^2 \le \langle u | A | u \rangle \langle v | A | v \rangle \,.\] This is the Cauchy-Schwarz inequality (in a generalized form). In the case when \(A\) is the identity operator, the inequality reads \[\big| \langle u | v \rangle \big|^2 \le \langle u | u \rangle \langle v | v \rangle \,.\] Proof: Consider the state \[| \psi \rangle = z | u \rangle + | v \rangle \,,\] where \(z\) is a complex number. Since \(A\) is positive-definite, we have \[\langle \psi | A | \psi \rangle \ge 0 \,.\] This can be rewritten as \[\Big( \langle u | z^* + \langle v | \Big) A \Big( z | u \rangle + | v \rangle \Big) \ge 0 \,,\] or \[\langle u | A | u \rangle |z|^2 + \langle u | A | v \rangle z^* + \langle v | A | u \rangle z + \langle v | A | v \rangle \ge 0 \,.\] Since \(A^{\dagger} = A\), the last inequality becom