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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 relation.)

In fact, there are (infinitely) many other position-momentum uncertainty relations. For instance, \(\Delta x\) and \(\Delta p\) satisfy the following two simple, symmetric inequalities: \[\Delta x + \Delta p \ge \sqrt{2} \,,\] \[(\Delta x)^2 + (\Delta p)^2 \ge 1 \,.\] These inequalities however are less precise than the Heisenberg one (which, in turn, is less precise than the Robertson-Schrödinger inequality).

The validity of the above uncertainty relations can be easily established from the following figure.

All points of the \(\Delta x\)-\(\Delta p\) plane that lie above the (blue) minimal uncertainty curve corresponding to the Heisenberg inequality also lie above the (red and green) minimal uncertainty curves corresponding to the other two inequalities.