Say you have a particle in a one-dimensional box of length L.(adsbygoogle = window.adsbygoogle || []).push({});

The particle can only have momentum values of the form

[itex] p_{n} = \frac{nh}{2L} [/itex] according to the De Broglie standing wave condition.

Now say I don't measure the position of the particle, but I know for certain that it is in the box. Then the uncertainty in the position is [itex] Δx = \frac{L}{2} [/itex].

Therefore, the uncertainty of the momentum is [itex] Δp ≥ \frac{h}{\pi L} [/itex] (where I used Heisenberg's Uncertainty Principle that [itex] Δx Δp ≥ \frac{h}{2π} [/itex].

Now, suppose my instruments are good enough so that their only limit is that set by Heisenberg. Then [itex] Δp = \frac{h}{\pi L} [/itex]. But then, suppose I measured a momentum of [itex]\frac{2h}{2L} [/itex]. Then the uncertainty of my momentum measurement is less than the distance to the next integer-multiple momentum value. Therefore I know, for sure, that the momentum is exactly 2h/(2L) since assuming otherwise would mean accepting values in-between, contradicting quantization. But then the uncertainty is 0, contradicting Heisenberg.

I reach a contradiction. I assume Quantum mechanics are correct, so there must be a mistake in my reasoning. But I don't see it. Can you show me?

My only guess, for now, is that one of the formulas I used are not exactly correct, and a more advanced course in QM will explain it better. (By the way, I'm only in a college-level course so I don't know much about QM formalism)

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# Particle in a box and Heisenberg Uncertainty principle paradox?

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