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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 of change of the expectation value of \(A\) is \[\frac{d \langle A \rangle}{d t} = \frac{i}{\hbar} \langle H A - A H \rangle \,,\] the uncertainty relation reads \[\Delta E \Delta A \ge \frac{\hbar}{2} \left| \frac{d \langle A \rangle}{dt} \right| \,.\] Defining \[\Delta t_A = \left| \frac{d \langle A \rangle}{dt} \right|^{-1} \Delta A \,,\] one arrives at the Mandelstam-Tamm uncertainty relation. It follows from the above definition that \(\Delta t_A\) is the time required for \(\langle A \rangle\) to change by an amount of the order of \(\Delta A\).

Example

Consider a quantum particle of mass \(m\) moving freely along a line, taken to be the \(x\) axis. Let the particle's state \(| \psi \rangle\) be represented by a free-particle Gaussian wave packet: \[\langle x | \psi \rangle = \left( \frac{2 \operatorname{Re} \alpha_t}{\pi} \right)^{1/4} \exp \left( -\alpha_t (x - x_t)^2 + \frac{i}{\hbar} p_0 (x - x_t) + i \frac{p_0^2 t}{2 \hbar m} + \frac{i}{2} \arg \frac{\alpha_t}{\alpha_0} \right) \,,\] where \(t\) is time, and \[x_t = x_0 + \frac{p_0 t}{m} \,, \qquad \alpha_t = \frac{\alpha_0}{1 + i \frac{2 \hbar t}{m} \alpha_0} \,.\] Here, \(x_0\) and \(p_0\) are the initial mean position and momentum of the particle, respectively, and complex-valued parameter \(\alpha_0\) (satisfying \(\operatorname{Re} \alpha_0 > 0\)) quantifies the wave packet spread and position-momentum correlation at \(t=0\).

The uncertainty in energy of the particle is given by \[\Delta E = \frac{(\Delta p)^2}{m} \sqrt{\frac{1}{2} + \left( \frac{p_0}{\Delta p} \right)^2} \,,\] where \[\Delta p = \frac{\hbar |\alpha_0|}{\sqrt{\operatorname{Re} \alpha_0}}\] is the momentum uncertainty.

The uncertainty in position is \[\Delta x = \frac{1}{2 \sqrt{ \operatorname{Re} \alpha_t} } \,,\] and the rate of change of the mean position, i.e. the mean velocity of the particle, is \[\frac{d \langle x \rangle}{d t} = \frac{p_0}{m} \,.\] Thus, the characteristic time associated with the change of particle's position is \[\Delta t_x = \left| \frac{d \langle x \rangle}{dt} \right|^{-1} \Delta x = \frac{m \Delta x}{|p_0|} \,.\]

Hence, \[\begin{align} \Delta E \Delta t_x &= \frac{(\Delta p)^2}{m} \sqrt{\frac{1}{2} + \left( \frac{p_0}{\Delta p} \right)^2} \frac{m \Delta x}{|p_0|} \\ &= \Delta x \Delta p \frac{\Delta p}{|p_0|} \sqrt{\frac{1}{2} + \left( \frac{p_0}{\Delta p} \right)^2} \\ &\ge \Delta x \Delta p \end{align} \,.\] Finally, the Heisenberg uncertainty relation, \[\Delta x \Delta p \ge \frac{\hbar}{2} \,,\] (that is easily verified from the formulas above) implies \[\Delta E \Delta t_x \ge \frac{\hbar}{2} \,.\] This is the Mandelstam-Tamm uncertainty relation in the case of \(A = x\).