### 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 $$A = x$$.