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Is there a mathematically somewhat rigorous argument why you cannot do these to things at the same time?

  1. Observe the particle in a clasically forbidden region.
  2. Measure the particle having an energy less than e.g. $V_0$ (for a step-potential for example).

The only attempt of this I've seen is in a MIT 8.04 Lecture by Barton Zwiebach, but I was also slightly confused/dissatisfied with the derivation there, as it for example leaves out the factor 1/2 in the uncertainty principle and uses the "qualitative" uncertainty principle (that was first formulated by Heisenberg, as far as I know), instead of one of the more precise versions that were formulated later.

If anyone knows or has heard of such an approach, I would be interested in it!

Qmechanic
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user410662
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1 Answers1

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The reason this is impossible is that you can never measure the position and energy simultaneously. No observables corresponding to non-commuting operators may be measured simultaneously, and position and the kinetic contribution to the energy never commute.

Concretely, if you measure the position of a particle and find it at $x_{0}$, where the potential is $V(x_{0})$, this amounts to a measurement that $E>V(x_{0})$, so that the particle has enough total energy to be found at that location. You might think that if you prepared the particle in an eigenstate of the Hamiltonian, with energy $E_{1}<V_{0}$, then you should not subsequently be able to measure a different value of $E$—but this is incorrect. If you just measure $E$ after preparing the system this way, you will always get $E_{1}$. However, if you measure a non-commuting observable $x$, the system will collapse into a state that is no longer an energy eigenstate; this is fine, because the measurement process involved having new interactions of the system with the outside world, so it could not be expected that the energy of the system after it was probed would be the same as before.

Buzz
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