The Time-Identity Tradeoff

Abstract

Distinguishability plays a major role in quantum and statistical physics. When particles are identical their wave function must be either symmetric or antisymmetric under permutations and the number of microscopic states, which determines entropy, is counted up to permutations. When the particles are distinguishable, wavefunctions have no symmetry and each permutation is a different microstate. This binary and discontinuous classification raises a few questions: one may wonder what happens if particles are almost identical, or when the property that distinguishes between them is irrelevant to the physical interactions in a given system. Here I sketch a general answer to these questions. For any pair of non-identical particles there is a timescale, $\tau _d$, required for a measurement to resolve the differences between them. Below $\tau _d$, particles seem identical, above it - different, and the uncertainty principle provides a lower bound for $\tau _d$. Thermal systems admit a conjugate temperature scale, $T_d$. Above this temperature the system appears to equilibrate before it resolves the differences between particles, below this temperature the system identifies these differences before equilibration. As the physical differences between particles decline towards zero, $\tau _d \rightarrow \infty$ and $T_d \rightarrow 0$.

Appendix: Detecting Mass Differences

In the main text (the discussion surrounding Fig. 2) we considered the ability of a physical system to detect charge differences, e.g., to distinguish between electric charge q and another charge $q+\delta q$. Basically, we assumed the existence of a field (in that case, electric field $\mathcal{E}$) that couples to the corresponding charge, so in a fixed field the forces on different charges (q and $q+\delta q$) differ by $\delta q \mathcal{E}$. When the force is applied through a period $\Delta t$, its contributions to the momenta of the two particles (in the direction of the applied force, that was chosen in Fig. 2 to be the y-direction) differ by $\Delta P_\mathcal{E} = \delta q \mathcal{E} \Delta t$. To resolve between the two charges $\Delta P_\mathcal{E}$ must be larger than the quantum mechanical uncertainty $\Delta P_{qm} = \hbar /\Delta y$, where $\Delta y$ is the width of the entrance slit. Therefore, $\Delta t$ must be greater from, or equal to, $\hbar /\delta q \mathcal{E} \Delta y$.

These considerations suggest that and charge difference $\delta q$ may be resolved in an arbitrary short time $\delta t$, provided that $\Delta y$ is large enough, $\Delta y > \hbar /\delta q \mathcal{E} \delta t$. The restriction $\Delta t \ge \hbar /\Delta E$ is related to energy, not to charge. In the deflection experiment the energy difference is $\delta q \mathcal{E} \Delta y$, so it grows linearly with $\Delta y$, and by taking the slit width to infinity one may reduce the measurement time to zero.

The same considerations apply to any other physical property or charge, with the exception of mass.

Let us try to distinguish between two particles, one of mass m and the other of mass $m+\delta m$, using the deflection setups of Fig. 2, where the field is a fixed gravitational field g pointing in the y direction. The above discussion appears to suggest that one may resolve between m and $m+\delta m$ as long as,

$$\begin{aligned} \Delta t_{qm} \ge \frac{\hbar }{g \, \delta m \, \Delta y} \end{aligned}$$

(2)

so the measurement time, again, goes to zero when $\Delta y$ diverges.

This conclusion must be wrong, since a mass scale is inherently related to an energy scale via $\Delta E = \delta m \, c^2$. On the contrary, Eq. (2) implies that by increasing $\Delta y$ one may break the limitation imposed by the energy-time uncertainty relationships.

To address this question the principles of general relativity must be invoked. The lapse of time changes along the gradient of the gravitational potential. When a weak gravitational field g is applied in the y direction,

$$\begin{aligned} \Delta t(y) = \left( 1+ \frac{g y}{c^2}\right) \Delta t(y=0). \end{aligned}$$

(3)

Therefore, the uncertainty in $\Delta y$ is translated to an uncertainty in the duration of the experiment. Without the gravitational effect on time, the minimum duration of the measurement, $\Delta t_{qm}$, is dictated by Eq. (2). If this is the measurement time for a particle that enters through the lower part of the slit, at $y=0$, then the measurement time for a particle that enters in the higher part of the slit must be greater than,

$$\begin{aligned} \Delta t_{qm+g} = \frac{g \Delta y}{c^2} \frac{\hbar }{g \, \delta m \, \Delta y} = \frac{\hbar }{\delta m \, c^2}, \end{aligned}$$

(4)

as requested.