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If two similar systems are created and In the first system the position is measured with accuracy and in the second one the momentum is measured with accuracy can this avoid the uncertainty in the measurements.
Say two square potentials where $V_1 = V_2$, $m_1 = m_2$, and the width is 2a for both of them and thus the energy is equal. $p_1$ is measured with accuracy and $x_2$ is measured with accuracy. So wont we be able to find 'x_1' without any uncertainty in the result as $x_1 = x_2$ because both the systems are same

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closed as unclear what you're asking by Ben Crowell, StephenG, ZeroTheHero, user191954, John Rennie Oct 28 '18 at 6:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I don't understand the question. You describe preparing two systems and measuring some things. Why is this of interest? Why would it violate "a major law?" $\endgroup$ – Ben Crowell Oct 27 '18 at 18:16
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    $\begingroup$ Okay, so now you know the position of one electron and the momentum of a different electron. How is the uncertainty principle supposed to be violated? $\endgroup$ – knzhou Oct 27 '18 at 18:19
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    $\begingroup$ Pressure and temperature are undefined for a single atom. $\endgroup$ – my2cts Oct 27 '18 at 21:26
  • $\begingroup$ @my2cts Why? It is a typical example to consider an atom in a heat bath. Each energy level has some degeneracy (to a first-approximation). Thus, you can compute the derivative of energy wrt entropy. It might be debatable if this statistical quantity coincides with thermodynamic temperature though. $\endgroup$ – jinawee Oct 28 '18 at 17:48
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First, the uncertainty principle isn't a problem. It is an important principle of physics that is universally valid, according to all the evidence we have, and there is no way to avoid it.

Your situation doesn't avoid it, either. If you measure the position $x_1$ of the first atom and the momentum $p_2$ of the second atom, the uncertainty principle predicts that the minimum product of uncertainties is zero because $$[x_1,p_2]=0$$ because these act on different degrees of freedom (those describing the first or the second particle, respectively). So in this case, when you do two truly independent measurements, the uncertainty principle only says that the product that was non-negative to start with is non-negative. The uncertainty principle is completely vacuous in your case which means that it is mathematically impossible to violate it.

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