# How to use Heisenberg to explain atom cannot be at rest

In a Dutch national physics exam at the level just before entering university there was a question about a molecule HI: one atom hydrogen plus one atom iodine. Within the molecule, the hydrogen vibrates in one dimension with respect to the iodine. Pupils are asked to explain that the hydrogen atom (or nucleus) cannot be at rest, using Heisenberg's uncertainty principle written in the form $\Delta x\cdot\Delta p\geq h/(4\pi)$. (The statistical form $\sigma_x\cdot\sigma_p=\geq h/(4\pi)$ isn't part of the curriculum at this level.)

I'm comfortable with one explanation: if H were at rest, then $\Delta p$ would be 0, meaning that $\Delta x$ diverges. But H must remain within the space of the HI molecule (not considering tunneling) so we have a contradiction. It follows that the assumption of H at rest cannot be true. So far, no problem for me.

I'm puzzled by the second explanation in the official text that must be used by teachers like me to judge the pupil's anwers. The text assigns a score point for the "insight that being at rest means $\Delta x$ and/or $\Delta p$ is 0". This validates an explanation using "being at rest means $\Delta x$ is 0". This doesn't seem true to me. The exam makers have tried to clarify: "If succesive measurements of the H position produce the same $x$ value, then $\Delta x=0$ and H is at rest. This violates the uncertainty principle"

But that doesn't seem to be what $\Delta x$ is about: I've learnt that $\Delta x$ and $\Delta p$ are indications of the uncertainty in $x$ and $p$ at the same instant of time.

My question: is it possible to explain that H cannot be at rest within the HI molecule using $\Delta x=0$? I don't think that's possible.

"To be at rest" in classical mechanics means "to have definite position and zero momentum", the two properties being (again, in CM) equivalent: if something has definite position, then it must have zero velocity thus zero momentum, and if it has zero momentum, then it must have zero velocity thus definite position. Depending on which of the two properties above you decide is most suitable to define an object at rest, you have, at least classically (i.e. without talking about probabilities), $\Delta x=0$ and/or $p=0\ \Longrightarrow\ \Delta p=0$, the "and" being necessary due to the classical definition of "rest", the "or" to give (if necessary) the possibility to chose the direction of the implication ($\Delta x=0 \ \Longrightarrow\ \Delta p=0$ or $p=0\ \Longrightarrow\ \Delta x=0, \Delta p=0$). At an undergraduate level, students are taught to think in terms of classical mechanics even when dealing with quantum mechanics, so we can assume that this is the definition they are implying. I think that you are missing the fact that they don't want to go into the details of why the statement is true. They just want the student to show that he has an elementary understanding of the Uncertainty Principle (or rather, of its pop-science formulation) and of what it means to "be at rest". From this perspective, their answer is right, while yours is imprecise: if $p=0$ (hence $\Delta p=0$), then $\Delta x$ is not infinity, it is zero as well. Keep in mind that they are identifying $\Delta x$ and $\Delta p$ with classical observables, so that $p=0$ implies $\Delta x=0$ and vice versa. On the other hand, either $\Delta p$ or $\Delta x=0$ being zero - without talking about the other one - is sufficient to invalidate the Uncertainty Principle: regardless of the other you have $0\geq \hbar/2$. An undergraduate isn't usually required to speak in terms of divergences, so I guess that the argument "the other one should diverge" isn't taken into consideration from the very beginning. In the end, as they say, they don't want an argument for the statement to be true, they're just looking for the "insight that being at rest means [more correctly, "implies"] Δx and/or Δp is $0$". Then you plug it in the uncertainty relation, and you get the point.
On the other hand, if you are willing to go beyond the scope of the question, there is no argument to prove that a quantum object cannot be at rest. In fact, the very concept of "being at rest" is ill-defined in quantum mechanics. What one usually does in these situations is to take a classical concept and try to transpose it in quantum mechanical terms. But asking whether an object is at rest or not clearly implies that you are able to make simultaneous statements about the position and the momentum of an object, which you can't do in QM [this impossibility can be formulated in a way that does not make use of the uncertainty principle]. So the very concept of being at rest doesn't make sense in QM, and as such there is nothing to prove or to disprove. Nevertheless, the inequality $\sigma_{x}\sigma_{p}\geq \hbar/2$ is well-defined. The latter is a statistical statement, meaning that in order to see its physical consequences you should be able to make experiments on an ensamble of physical systems, not on just one object. Then the statement "the uncertainty relation is/is not verified" is a statement about the ensamble, not about the single object; so again, it doesn't imply any notion of rest. By the way, it is true that the relation $\sigma_{x}\sigma_{p}\geq \hbar/2$ implies that $x$ and $p$ are measured at the same instant of time. This is done as follows: you prepare $2N$ identical systems in the same way, so that you can say that at a time $t$ from the moment in which they have been prepared they all are in the same state. Then for $N$ of them you measure $x$ and for $N$ of them you measure $p$, in both cases at time $t$, and you compute the statistical $\sigma_{x}$ and $\sigma_{p}$ of the ensamble. According to QM, you'll find that $\sigma_{x}\sigma_{p}\geq \hbar/2$.
• $\Delta p=0$ is not enough to say that something is at rest. $\Delta p=0$ may mean $p=1\ \text{g}\cdot \text{m/s}$ with complete certainty, which for a body that has mass 1 gram means that the body is traveling at 1 m/s, so the body is not at rest. Something is at rest if $p=0$, i.e. ($p=mv$) if its velocity is zero, i.e. if its position doesn't change in time, i.e. if $\Delta x=0$. Then $p=0$ implies $\Delta p=0$, and it also implies $\Delta x=0$. – Giorgio Comitini Jul 26 '16 at 15:51
• I stated this in words at the beginning of my answer, then I got it mixed it up (I'm about to correct that). In your example, if you measure zero doppler shift, then you know that the particle is at rest. In classical mechanics, $v=0$ implies $x=const$, so this means that it has definite position, so $\Delta x=0$ regardless of whether you know or not the specific position. In classical mechanics, $\Delta x$ is not a statement about what you know of the system. $\sigma_{x}$ is, in quantum mechanics, but you explicitly asked for the classical picture. – Giorgio Comitini Jul 26 '16 at 15:59