# Does the Heisenberg uncertainty principle preclude moving in a straight path with certainty?

The uncertainty principle is σₓσₚ ≥ 0.5 ℏ where x is position and p is momentum.

Consider a 2d plane.

If one moves along a straight line along the plane (possibly backtracking or moving forwards but always along the same straight line) then along the dimension orthogonal to the line one is moving on one has a position of zero and a momentum of zero which contradicts the uncertainty principle.

Does this mean according to the Heisenberg uncertainty principle one cannot move along a straight line with certainty?

• The uncertainty principle is an expression that nothing in quantum mechanics "moves" by the standards of classical physics. – CuriousOne Jul 5 '15 at 0:05

## 1 Answer

When defining the uncertainty principle one has first of all to be very careful with the domain of definitions the operators have: in particular, if $A, B$ are the observables whose uncertainties we want to measure together, what needs to be calculated is the commutator $\left[A,B\right]$. In order this commutator to be well defined we must have that the states of your system $|\psi\rangle$ are such that $\left[A,B\right]|\psi\rangle$ is well defined, namely $$|\psi\rangle\in\mathcal{D}(A)\cap\mathcal{D}(B)\qquad\textrm{and}\qquad A|\psi\rangle\in\mathcal{D}(B),\quad B|\psi\rangle\in\mathcal{D}(A).$$ Whenever the above conditions do not hold the Heisenberg uncertainty principle simply does not hold true, or maybe it does only on the subset of states fulfilling that property. There is a very famous example when the above is not fulfilled, that is the point particle on the circle with conjugate operators $\varphi, p_{\varphi}$: the particle is confined to have $0\leq\varphi\leq 2\pi$ (which causes restrictions on its position uncertainty) and we can always choose the conjugate momentum $p_{\varphi}$ so that the product of the two is smaller than $\hbar$. This is because $p_{\varphi}|\psi\rangle$ does not belong to the domain of definition of the angular position operator $\varphi$.

This said alway note that, however, that in quantum mechanics there is no such thing as the trajectory of the particle and there is no way to tell that a particle is moving along a straight line (actually, there is no way to tell how a particle is moving anyway). You do not know what the trajectory of the particle is: the only thing you can do is to perform a measurement at a time $t$ and look at the position outcome $x(t)$; then you can perform another measurement at a later time $t'$ and have $x'(t')$ and so on and so forth. Eventually, after having performed infinite measurements, the positions $x(t)$ distribute according to a density functions proportional to $|\psi(x(t))|^2$.

In your specific example the statement

then along the dimension orthogonal to the line one is moving on one has a position of zero and a momentum of zero

is wrong, because you can have a measurement giving $y=0$ as outcome once, but if you perform another measurement the $y$ coordinate of the particle can be in principle anywhere in the universe (and so can its conjugate momentum). This unless you have a specific constraint that forces the particle to have $y=0$ always, but in that case the system would be one dimensional rather than two (two dimensions plus a constraint is equal to one dimension).

• I think a useful analogy is to think of a seismic wave. It isn't some point-particle. If it moves from A to B, it isn't just the houses on the AB line that shake. So it take "many paths". See this: "It has since become clear, however, that the uncertainty principle is inherent in the properties of all wave-like systems, and that it arises in quantum mechanics simply due to the matter wave nature of all quantum objects." – John Duffield Jul 4 '15 at 21:59