The problem statement:

Measurement detects a position of a proton with accuracy of $\pm10pm$. How much is the position uncertainty $1s$ later? Assume the speed of a proton $v\ll c$.

What i understand:

I know that in general it holds that:

\begin{align} \Delta x \Delta p \geq \frac{\hbar}{2} \end{align}

This means i can calculate momentum uncertainty for the first measurement:

\begin{align} \Delta x_1\Delta p_1 &\geq \frac{\hbar}{2}\\ \Delta p_1 &\geq \frac{\hbar}{2\Delta x_1}\\ \Delta p_1 &\geq \frac{1.055\times10^{-34}Js}{2\cdot 10\times10^{-12}m \rlap{~\longleftarrow \substack{\text{should I insert 20pm}\\\text{instead of 10pm?}}}}\\ \Delta p_1 &\geq 5.275\times10^{-24} \frac{kg m}{s}\\ \Delta p_1 &\geq 9.845 keV/c \end{align}


Using the position uncertainty $\Delta x_1$ I calculated the momentum uncertainty in first measurement $\Delta p_1$.

How do I calculate the position uncertainty $\Delta x_2$ after $1s$? I am not sure what happens $1s$ later but is a momentum uncertainty conserved so that it holds $\Delta p_1 = \Delta p_2$? I know that the problem expects me to use the clasic relation $p=m_ev$ somehow but how?

  • $\begingroup$ Perhaps you could use $\Delta E\Delta t\geq\hbar/2$ somehow.... $\endgroup$
    – Kyle Kanos
    Commented Jul 27, 2013 at 11:19
  • $\begingroup$ Is it possible that the velocity of the proton is constant and therefore velocity uncertainty is constant too??? $\endgroup$
    – 71GA
    Commented Jul 27, 2013 at 12:23
  • $\begingroup$ One thing to note is, one usually does not expect that the uncertainty just decreases by itself. So roughly speaking you can assume the uncertainty $\Delta x$ will increase by the uncertainty in speed times $\Delta t$. $\endgroup$
    – Ali
    Commented Jul 27, 2013 at 13:34

2 Answers 2


You are on the right track in calculating the uncertainty in momentum using the uncertainty principle.

The new position will be

$$x_2 = x_1 + \frac{p}{m} t$$

There is a well known technique of error propagation which works like

$\delta(f(x_1, x_2, x_i, ...) = \sqrt{\Sigma\left(\frac{\partial f }{\partial x_i} \delta x_i\right)^2 } $,

where $\delta x_i$ means the uncertainty in $x_i$, which is an independent coordinate (including momenta and times) of the motion. You sum over every measurement that has an uncertainty.

This comes from the Taylor series.

Applying this, you will get $$\delta x_2 = \sqrt{\delta x_1^2 + \left(\frac{t}{m} \delta p\right)^2}$$


I thought about this a little more and I think that the addition in quadrature is not so appropriate here. Usually you use this for measurement uncertainties, where you look for one-sigma intervals, but for quantum mechanics, where you look for complete uncertainty, it might be more correct to add the components directly.

$$\delta x_2 = \delta x_1 + \left(\frac{t}{m} \delta p\right)$$

  • $\begingroup$ Can i ask you for an external link so i can read more about this error propagation. $\endgroup$
    – 71GA
    Commented Jul 27, 2013 at 23:32
  • $\begingroup$ The wikipedia page is pretty good -- en.wikipedia.org/wiki/Propagation_of_uncertainty $\endgroup$
    – Gremlin
    Commented Jul 28, 2013 at 15:09
  • $\begingroup$ Can i ask you why did you change your original sugestion in an EDIT? $\endgroup$
    – 71GA
    Commented Jul 28, 2013 at 18:42
  • $\begingroup$ It's because I thought about it some more and realised that in the uncertainty principle, it's the complete uncertainty interval that matters more. The first formula is more appropriate when you are trying to come up with the uncertainty on a quantity that depends on two independent measurements. It accounts for the fact that sometimes random fluctuations cancel each other out rather than always adding up. $\endgroup$
    – Gremlin
    Commented Jul 29, 2013 at 8:38

Let's suppose you did the measurement in the rest frame of the proton. You've worked out that $\Delta p$ is $5.275 \times 10^{-24}$ kg.m/s. This means the probability distribution of the momentum will be some curve (usually taken to be a gaussian) with a half width of $5.275 \times 10^{-24}$. To convert this to a velocity probability distribution just divide by the proton mass, $1.673 \times 10^{-27}$ kg, and you'll find the velocity probability distribution half width is $3,150$ m/s. Therefore after 1 second the probability distribution of the proton position will have a half width of $3,150$ m.


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