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The question is

By choosing reasonable numerical values for mass and velocity, show that $\Delta x \Delta p >=\frac{ \hbar}{2}$ doesn't impose any limitations on the precision with which the position and momentum of a macroscopic body can be determined.

Now I think we can choose 1kg and maybe 1m/s. But $v = \frac{\omega}{k}$. So $\omega = k$. Now I don't know what to do from here.

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  • $\begingroup$ what are v, w and k? $\endgroup$ – danimal Mar 16 '15 at 15:23
  • $\begingroup$ Sorry, in the wave function, v velocity, k wave number, w frequency $\endgroup$ – Zeeshan Ahmad Mar 16 '15 at 15:28
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The $\Delta$ values in the uncertainty relation tell you the errors (or uncertainties) in the position and momentum, not the actual/measured values of these. For a "normal" sized body, with momentum 1kgm/s, there may be an error of say $\Delta p =\pm0.01$kgm/s for example. Follow through with this value in the error for position and see what you get...

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  • $\begingroup$ I get $\delta x$ very small, so that means that the minimum error I have in position is very small, 5.3e-33m. But it could be greater than this, as it is an inequality, no? $\endgroup$ – Zeeshan Ahmad Mar 16 '15 at 15:31
  • $\begingroup$ Oh I think I get it. Uncertainty relationship limits the minimal error of position (in this case). But for macroscopic bodies, that limit becomes very low, ie more precision possible. Correct? $\endgroup$ – Zeeshan Ahmad Mar 16 '15 at 15:38
  • $\begingroup$ yes, the uncertainty principle tells you the lowest the error (ie the max precision) $\endgroup$ – danimal Mar 16 '15 at 16:13
  • $\begingroup$ I'm not sure about your language, but I think you've got it. The point is that, for any "reasonable" "every-day" measurements, the product of uncertainties is enormously greater than that specified by the Uncertainty Principle", so the principle doesn't become relevant. $\endgroup$ – WhatRoughBeast Mar 16 '15 at 19:05
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The relevant physics argument here is at heart a scaling argument, which is why it would be more instructive to do this exercise in natural units where $\hbar = 1$. If we consider objects of uniform density then to study what happens if these objects become extremely large, we can define a new scale to measure their positions and momenta. We define new variables such that as the limit to infinite size is taken, in terms of these new variables the size and momenta remain constant.

So, we consider objects of sizes $r_0t$ where $r_0$ is kept fixed and $t$ becomes arbitrarily large. The uncertainty relation in natural units is:

$$\Delta x \Delta p \geq \frac{1}{2}$$

We now define new variables such that for $t\to \infty$ the values of the size and momenta of the object stay the same. A macroscopic observer who lives at a scale corresponding to very large values of $t$ is bound to use these variables to describe its macroscopic world. This means that we need to scale the position by defining:

$$y(t) = \frac{x}{t}$$

To define the new momentum variable, we note that macroscopic motion is typically the result of conversion of part of the internal energy to macroscopic work. The internal energy scales as $t^3$, the mass also scales as $t^3$, the typical momentum of objects should thus scale as the square root of the product of the mass and the internal energy, therefore as $t^3$. We thus need to definine:

$$q(t) = \frac{p}{t^3}$$

This then yieds:

$$\Delta y \Delta q = \frac{\Delta x\Delta p}{t^4} \geq \frac{1}{2 t^4}$$

So, we see that in the limit of $t\to \infty$ the product of the uncertainty and the momentum tends to zero, as measured in terms of the rescaled macroscopic variables. Now, you can ask if the SI units we use in practice are consistent with the argument given here.

The effective radius of a hydrogen atom is about $0.5\times 10^{-10}$ meters. But we would like to measure things in terms of meters, so the value for $t$ appropriate for to define our practical unit system should be of the order of $10^{10}$ which leads to a factor of order $10^{-40}$ in the uncertainty relations. This is indeed consistent with what you get if you put back $\hbar$ in the uncertainty relation and use the SI value for this quantity.

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  • $\begingroup$ Why must the size and momenta remain the same? Have they not increased at a macroscopic level? $\endgroup$ – Zeeshan Ahmad Mar 16 '15 at 16:56
  • $\begingroup$ They do increase, but we measure them using units that also scale relative to the microscopic scale so that typical values for macroscopic objects have values of order unity. $\endgroup$ – Count Iblis Mar 16 '15 at 17:08
  • $\begingroup$ Yes, after some pondering, it all does make sense. I applaud your imagination $\endgroup$ – Zeeshan Ahmad Mar 16 '15 at 18:04

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