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Cauchy-Schwarz inequality

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 becomes \[\langle u | A | u \rangle |z|^2 + \langle u | A | v \rangle z^* + \langle u | A | v \rangle^* z + \langle v | A | v \rangle \ge 0 \,.\] Let \(z = x + i y\), where \(x\) and \(y\) are real. Then, \[\langle u | A | u \rangle \left( x^2 + y^2 \right) + 2 \operatorname{Re} \langle u | A | v \rangle x + 2 \operatorname{Im} \langle u | A | v \rangle y + \langle v | A | v \rangle \ge 0 \,.\] Completing the squares in \(x\) and \(y\), we get \[\left( x + \frac{\operatorname{Re} \langle u | A | v \rangle}{\langle u | A | u \rangle} \right)^2 + \left( y + \frac{ \operatorname{Im} \langle u | A | v \rangle}{\langle u | A | u \rangle} \right)^2 + \frac{\langle u | A | u \rangle \langle v | A | v \rangle - \big| \langle u | A | v \rangle \big|^2}{\langle u | A | u \rangle^2} \ge 0 \,.\] The last inequality is fulfilled for all \(x\) and \(y\) if and only if \[\langle u | A | u \rangle \langle v | A | v \rangle - \big| \langle u | A | v \rangle \big|^2 \ge 0 \,,\] which concludes the proof.