# Does the Heisenberg's uncertainty equation holds when one of the observable have zero variance?

From this link Heisenberg uncertainty principle, It says:

Clearly, when $$\Delta p_x$$ shrinks, $$\Delta x$$ has to grow larger and larger in order to satisfy the Heisenberg inequality. For example, a plane wave $$\psi(x) = \exp[2\pi i px/h]$$ is an eigenfunction of $$p = -ih/(2\pi)\ d/dx$$, so that $$\Delta p_x = 0$$; a plane wave particle has a position $$x$$ that is completely undetermined. Conversely, if the position of the particle is very well determined, its momentum is very uncertain. The p-expansion (Fourier transform) of a well-localized wave packet ($$\Delta x \approx 0$$) requires eigenstates of many different eigenvalues $$p$$ and hence gives rise to a large spread in $$p$$.

so one of the variance from the left side of this equation could be zero (my calculation about measure spin in different basis also implies that one of the hermitian operator can have zero variance), this equation becomes to $$0\geq h/4\pi$$ which is clearly wrong. Does this mean the inequality does not work for 0 variance?

• Note that we use MathJax to typeset mathematics; you can find a good tutorial here. – Emilio Pisanty May 19 '20 at 11:59
• – Dvij D.C. May 19 '20 at 12:08

## 3 Answers

For the cases you've mentioned (plane-wave eigenstates of the momentum, and Dirac-delta eigenstates of the position), one of the position/momentum variances is zero and the other one is infinite, so the Heisenberg inequality formally reads $$\Delta p \, \Delta x = 0\times\infty \geq \frac h{4\pi}.$$ (As an obvious note: randomly ignoring infinities from your calculations, just because you don't know what to do with it, is obviously a recipe for trouble.)

This should be tempered with the fact that these are not physical states (it is impossible to have a true plane-wave, which occupies all of space, and it is impossible to localize a particle to infinite precision), and they do not fall in the class of wavefunctions (i.e. the Hilbert space) where the Heisenberg inequality is a theorem. For these states, the inequality should be understood as a limit: $$\Delta p \geq \frac1{\Delta x}\frac{h}{4\pi} \to \infty \quad \text{as}\quad \Delta x\to 0.$$

Zero variance means that the system is in the eigenstate of the corresponding operators. E.g., if we measure an eigenstate with a particular momentum, $$\psi_p(x) = e^{i\frac{px}{\hbar}},$$ The probability density is constant in space $$|\psi_p(x)|^2 = const,$$ which means that the position of the particle is completely undefined. (The mathematical ambiguity here is usually resolved by taking periodic boundary conditions in a regions $$[-L/2, L/2]$$, so that $$\psi_p(x) = e^{i\frac{px}{\hbar}}/\sqrt{L},$$ and $$|\psi_p(x)|^2 = 1/L$$, i.e. uniform everywhere in the region.)

From the point of view of the position-momentum uncertainty relation we have $$\Delta x \geq \frac{\hbar}{2\Delta p}\rightarrow +\infty \text{ as }{\Delta p \rightarrow 0}.$$

The derivation of the inequality holds under two conditions: the operators are self-adjoint AND the states on which they act are normalizable.

Plane waves are not normalizable. For a situation where the operators are not self-adjoint, see this question.

The inequality is otherwise airtight, and works all the time, including for $$0$$-variance. Indeed in the specific example of angular momentum one can use $$\Delta L_x\Delta L_y\ge \frac{1}{2}\vert \langle L_z\rangle\vert$$ and cyclic permutations to show that eigenstates of $$L_x$$ necessarily have $$0$$ average value of $$L_y$$ or $$L_z$$.