Skip to main content

Posts

Showing posts with the label Uncertainty relations

Unusual position-momentum uncertainty relations

Consider a one-dimensional quantum particle, and let \(X\) and \(P\) represent its position and momentum operators, respectively. \(X\) and \(P\) are assumed to satisfy the canonical commutation relation, \[[X,P] = i \hbar\] with \(\hbar\) being the reduced Planck constant. For the following discussion, it is convenient to introduce dimensionless position and momentum as \[x = \frac{X}{L} \qquad \text{and} \qquad p = \frac{L P}{\hbar} \,,\] where \(L\) is some (chosen for convenience, but otherwise arbitrary) length scale. In terms of \(x\) and \(p\), the canonical commutation relation reads \[[x, p] = i \,.\] The operators \(x\) and \(p\) fulfill the Heisenberg uncertainty relation: \[\Delta x \Delta p \ge \frac{1}{2} \,,\] where \(\Delta x \ge 0\) and \(\Delta p \ge 0\) are uncertainties in the particle's position and momentum, respectively. (The Heisenberg uncertainty relation is a straightforward consequence of a more general Robertson-Schrödinger uncertainty rel

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