# On the Heisenberg uncertainty relation

Are there fundamental limits on the accuracy for measuring both position $q$ at time $t$ and momentum $p$ at time $t+\Delta t$, with tiny $\Delta t$?

If yes, why?

If no, why can't one then measure (in principle) both $q$ and $p$ arbitrarily well at the same time $t$ (which is not allowed by Heisenberg's uncertainty relation), by taking $\Delta t$ sufficiently small and noting that any measurement takes time?

• – AccidentalFourierTransform Jan 22 '17 at 18:35
• @AccidentalFourierTransform you mean the comments there? – anna v Jan 22 '17 at 18:47
• @annav yes, specially Emilio's comment. – AccidentalFourierTransform Jan 22 '17 at 18:54
• I think both this and my question is more interesting as being about the so-called statistical interpretation in which the $|\psi \rangle$ is not a distribution of a single system but instead it is referring to a similarly prepared statistical ensemble of identical QM systems. While I did not state it in my question as such then I was actually interested if one could use this non-simultaneous uncertainty lower limit to distinguish between the statistical and the orthodox (Copenhagen) interpretation of QM. I still do not know but would like to... – hyportnex Jan 22 '17 at 21:48
• Hmm .. the width of the non-zero part of the position PDF can grow no faster than $2c$, so we can presumably just apply the usual HEP to a width of $\Delta q + 2c \Delta t$ as a first semi-classical approximation, no? – dmckee Jan 22 '17 at 21:58

There should be fundamental limitations, but not in a way you'd expect. The uncertainty principle is essentially a reflection of the non-commutativity of the two quantities $x$ and $p$.

According to the Copenhagen interpretation of QM, once you make an exact measurement of the position $x$ at time $t$ (the first measurement can, in itself, be made arbitrarily exact), the wave function collapses to a particular position state, which here would be a dirac delta at $x = x_0$ (say) $$|\Psi(t_+)\rangle = |x_0\rangle$$

Assuming we chose the $\Delta t$ to be arbitrarily small so that the state on which one makes the next observation (of $p$) is the same as the one $|\Psi(t_+)\rangle$. In the momentum eigen-basis, this wave function can be written as \begin{align} |\Psi\rangle &= |x_0\rangle,\\ &= \int dp ~|p\rangle \langle p| x_0\rangle, \\ &= \int dp ~|p\rangle e^{-ipx_0/\hbar}. \end{align} Since the number $e^{-ipx_0/\hbar}$ is a just a phase factor, the probability for the particle to be in any momentum state (or to possess any momentum value) is uniform over all $p$ values. One can as well say that momentum is not well defined right after measurement of position. I hope this answers your question.

If $\Delta t$ is non-zero yet small, uncertainty still shows up and can be seen through either evolving the state (Schrodinger picture) or the operators (Heisenberg picture). In the latter approach, we have a time dependent operator given as $$p(\Delta t) = e^{iH \Delta t/\hbar} p(0) e^{-iH \Delta t/\hbar}$$ For small $\Delta t$, $p(\Delta t)$ can be approximated through a truncated Hausdorff expansion. We can quantify the uncertainty through the commutator \begin{align} C &= [x(0),p(\Delta t)],\\ &= [x(0),e^{iH \Delta t/\hbar} p(0) e^{-iH \Delta t/\hbar}] , \end{align} which can be evaluated approximately or accurately depending on the complexity of the Hamiltonian $H$.

• A rather more interesting scenario is - what if the position measurement does not lead to a complete collapse? Such Weak Measurements are topics of current research. Here's a blog article readers may find interesting. – the boy who believed May 8 '18 at 6:57
• You assume an infinitesimal time delay. Your answer implies nothing for a finite but tiny delay, which was my question. Moreover, you assume collapse to an unnormalizable state, which is impossible. Note that Landau & Lifschitz (Vol. 3, Section 7) explicitly remark that the state after the measurement is in general not an eigenstate - though this is often claimed in less careful expositions of quantum measurement. – Arnold Neumaier May 8 '18 at 10:10
• Agreed! I assume infinitesimal time to show the extreme case scenario. In general, after the measurement, the state would evolve - one should then look the non-equal time commutator - edits included in the answer. – the boy who believed May 8 '18 at 15:06

No, there are no particular limits, but this doesn't cheat the uncertainty principle in the way it might seem. First, it's very useful to remember that the uncertainty principle is not a statement about our knowledge of a quantum state or the nature of measurement, it's a statement about the nature of quantum states. A wavefunction with a well-defined position simply does not have a well-defined momentum.

Let me try to analyze your example. First, you measure the position of the particle extremely accurately. Now it's in a quantum state that has a very well-defined position and a very poorly defined momentum (like a delta function). Then, at a time $$\Delta t$$ later, you measure the momentum very accurately. Now it's in a state that has a very well-defined momentum but very poorly defined position (like a sine wave). Other than more or less standard limits on measuring things and state evolution, nothing prevents us from doing exactly this. Importantly, at no point in the above procedure does the state violate Heisenberg's principle.

I'm not sure exactly what you mean by phrases like "any measurement takes time" (and I'm not sure I agree) but if you view the above process that the wavefunction must evolve smoothly from $$t$$ to $$t+\Delta t$$ and in no sense do we know the position and momentum arbitrarily well at the same time, namely because at no point in time are they (nor can they be) well defined for the particle.

Aside: I find it most intuitive to think about these things and the measurement process in the many-worlds picture rather than starting to think about exactly at what point in time the wave function "collapses" or anything like that. Hope this helps!

• Would you mind sharing how this looks to you in the Many Worlds picture? – electronpusher Jun 18 at 20:24

After you measured position $q$ at time $t$, even the probability distribution of the momentum you will measure at time $t+\Delta t$ will be "mostly unrelated" to the probability distribution of the momentum you would have measured at time $t-\Delta t$. And if you would measure position again at time $t+2\Delta t$, it will be "mostly unrelated" to both the actual measurement outcome and the probability distribution of the position measurement at time $t$.

You might still ask whether there is any limitation to how accurate you can measure momentum at time $t+\Delta t$, but if you measured position from time $t-\Delta T$ to time $t$ and momentum from time $t$ till time $t+\Delta t$, then this probably just boils down to how accurate you can make a momentum measurement in a finite time $\Delta t$. For a quantum object with mass $m$, this is probably given via the relation $E=\frac{p^2}{2m}$ by $\Delta E=\frac{p+\Delta p/2}{m}\Delta p$. Assuming $\Delta E\Delta t \geq\frac{\hbar}{2}$, then $\frac{p+\Delta p/2}{m}\Delta p\Delta t \geq\frac{\hbar}{2}$.

Measuring arbitrarily well means observing eigenstates of the observables, so the answer depends on whether the state can evolve from a position eigenstate to a momentum eigenstate in time $\Delta t$.

In my experimentalist's view, if the HUP is about both

$$\Delta x\Delta p\ge\frac{\hbar}{2}$$

$$\Delta E\Delta t\ge\frac{\hbar}{2}$$

then a very small $\Delta t$ would imply a very large $\Delta E$ and given that, for a particle, measuring the energy constrains the magnitude of the momentum, making $\Delta t$ very small gives a large uncertainty to $\Delta p$.

True, the the energy/time uncertainty cannot be derived from commutators, and in this link I found some arguments that imply one has to define well what $t$ in $\Delta t$ is.

does it refer to the accuracy of measurement, to the duration of measurement or perhaps to the lifetime of a decaying state.

In general, I think that paradoxes appear when one mixes frames of reference and this might be one of the cases where one has to keep careful track of the measurement context.

• I wish that people who down vote would state their objection. One of the reasons I follow this site is because I am still learning physics, and a downvote without comment is not enlightening. – anna v Jan 23 '17 at 4:53
• I agree. I’d take it a step further and say a comment should be required for a down vote. – Lambda Jan 26 '18 at 15:15
• Are you implying that if we make a "quick" measurement, or two measurements in quick succession, it greatly increases the energy of the particle? I am no expert, but I'm suspecting this may be a naive application of the energy-time uncertainty relation. – electronpusher Jun 18 at 20:28
• @electronpusher No, only that if we know with great accuracy t, then E is known incaccurately. – anna v Jun 19 at 4:09
• Thank you, my mistake. – electronpusher Jun 19 at 8:29