I got this question in one of my periodic test in coaching, I (seemingly) solved the question but later recognised what mess I did.

So, the question is :

Calculate the uncertainty in velocity of an electron. If the uncertainty in its position is ${10^ {-15}} \; \mathrm m$.

By simple calculations one can see that the answer would be ${5.79\times{10^{10}}} \; \mathrm{m/s}$. However, as one can see, the uncertainty in velocity is greater than the velocity of light then how is that possible or there is some other way to handle such problems.

P.S. I know the concept of dynamic mass but I am not very familiar with it. (I think that would work here, but don't know how?)

  • $\begingroup$ There is nothing wrong here, it demonstrates that the uncertainty is in effect, its impossible to know both to the precision in question 10^-15 m is a very small uncertainty which means the velocity uncertainty should be huge. $\endgroup$ – A. C. A. C. Sep 14 '17 at 17:31
  • $\begingroup$ What does the uncertainty greater than light logically mean, that the speed of electron may be greater than speed of light, well that seems illogical. What I want to say is that the range being encapsulated by that answer is quite wrong logically. It should have been limited (at max) upto speed of light. $\endgroup$ – geeky me Sep 14 '17 at 17:37
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    $\begingroup$ There isn't a prediction for electron velocities faster than light. There just isn't. There is a prediction for electrons having highly relativistic momenta and then you made a mistake by applying a classical approximation to find the velocity that goes with that momentum. $\endgroup$ – dmckee Sep 15 '17 at 3:55

The uncertainty relation that gets you to $$ \Delta p \gtrsim \frac{\hbar}{2\Delta x} \approx 5.3\times 10^{-20}\mathrm{\:kg \:m/s} \tag1 $$ is trustworthy enough. However, as you've noticed, that would require the electron to be relativistic, so the newtonian equivalent $p=mv$ cannot be used.

Frankly, I would take this as an indication that you're so deep in relativistic territory that the nonlinear relationship between momentum and velocity makes all the naive associations useless, and I would work exclusively in terms of momentum (since velocity is rarely a useful concept in quantum mechanics).

If, for whatever reason, you're required to translate that momentum into a velocity, then you should use the relativistic version, $$ p=\gamma mv = \frac{mv}{\sqrt{1-v^2/c^2}}, \tag 2 $$ which can be inverted to give $$ v=\frac{p}{\sqrt{m^2+p^2/c^2}}, $$ which produces $v=0.999987c$ when fed the momentum in $(1)$ ─ though again, that's a pretty useless number.

I should also mention that the concept of "relativistic mass" is a pedagogical crutch that's essentially been retired, and which should not be used. Use the dynamical relations as in $(2)$ instead.

  • $\begingroup$ To understand how relativistic this circumstance is, note that this electron would be confined by a volume about the size of the proton $\endgroup$ – Lewis Miller Sep 15 '17 at 1:27

how is that possible or there is some other way to handle such problems.

You're conflating uncertainty in a quantity for the quantity itself.

Calculate the uncertainty in velocity of an electron

By simple calculations one can see that the answer would be $5.79 \times 10^{10} m/s$

What does the uncertainty greater than light logically mean [?]

This means that the uncertainty in the velocity of an electron is 193x greater than the speed of light, not the velocity of the electron.

Logically this means that upon measuring the position of an electron to such high precision a simultaneous measurement of its velocity must be proportionately imprecise.

Fundamentally this means that if we were capable of calibrating a laser to measure the position of an electron to that precision then the laser would impart a massive amount of kinetic energy upon the electron. Arguably this would be enough energy to accelerate the electron to relativistic speeds.


Suppose this were an experiment

Electron Position

The position of an electron was measured to be 1067 femtometers ($1067 \times 10^{-15} m$) with an uncertainty of 1 femtometer ($10^{−15}m$). Assuming our experiment was well performed and we have confidence in this measurement, we could state with confidence and certainty that the electron's position is $1067 \pm 1$ femtometers. Therefore we can say with confidence that the actual position could be 1066-1068 femtometers. One would say this is an extremely precise measurement. For the benefit of readers in introductory physics or engineering classes, this would qualitatively sum up to a percent error of 0.0937%

Electron Velocity

Now let's invert the problem. The velocity of the electron was measured to be 6,843,651 m/s with an uncertainty of $5.79 \times 10^{10}$. Since the uncertainty in the measurement is far greater than the measurement, we know a priori that the measurement should probably not be trusted. We go ahead and claim that the electron's velocity is $6.843651 \times 10^{6} m/s \pm 10^{15} m/s $; but, that would mean that the range of possible values for the velocity is +1,000,000,006,843,651 m/s to -999,999,993,156,349 m/s. In SI units: 1 Pm/s to -1fm/s. With such a massive range of values, we can't say with any certainty what the velocity of the electron is. Therefore the measurement is extremely imprecise

The Uncertainty Principle.

Fundamentally, the physics behind the uncertainty principle are founded upon how we measure things at the atomic scale. If we need to know the position of particle A, shoot laser B or particle B at it and measure the response of B. With a laser our measurement error in the position is directly tied to the wavelength of our laser. The smaller the wavelength, the better our measurement. With a particle it'd be tied to speed. This is great until we realize that by impacting Particle A with a high velocity particle B or a highly energetic laser (wavelength is inversely proportional to energy remember?) we are imparting energy to it. Basic kinematics with momentum or kinetic energy tells us that the more energetic our particle or laser, the more likely Particle A will absorb that energy and fly in some direction at blazing fast velocities. The end result is an equal imprecision in any measurement of the particle's velocity when we measure its position.

  • $\begingroup$ I think the way you think of uncertainty principle is quite wrong? Its not like " if we shoot a laser....." Its a fundamental property of nature. $\endgroup$ – geeky me Sep 16 '17 at 14:35

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