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Showing posts with the label Uncertainty relations

Mandelstam-Tamm uncertainty relation

The Mandelstam-Tamm enerty-time uncertainty relation states that \[\Delta E \Delta t_A \ge \frac{\hbar}{2} \,,\] where \(\Delta E\) is the energy uncertainty of a quantum system, and \(\Delta t_A\) is the time required for a significant change of the expectation value of an observable \(A\). Derivation Consider a quantum system with a Hamiltonian \(H\). Let \(| \psi \rangle\) be the time-dependent state of the system, and let \(A\) be some observable. The uncertainty in the system's energy and the uncertainty in \(A\) are defined, respectively, as \[\begin{align} &\Delta E = \sqrt{\langle H^2 \rangle - \langle H \rangle^2} \,, \\ &\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2} \,, \end{align}\] where \(\langle \cdot \rangle = \langle \psi | \cdot | \psi \rangle\) denotes the expectation value. \(\Delta E\) and \(\Delta A\) satisfy the uncertainty relation \[\Delta E \Delta A \ge \frac{\big| \langle HA - AH \rangle \big|}{2} \,.\] Since the rate

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

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 becom