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Uncertainty is due to the measurement techniques humans tend to use requiring photoelectric effect. The Planck constant is due to the photoelectric effect. If the standard deviation of measurement was zero, both sides of the equation would be zero and the uncertainty equation would be meaningless. So it has already been assumed for uncertainty to work.

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closed as unclear what you're asking by Aaron Stevens, Jon Custer, Dale, John Rennie, Kyle Kanos Nov 13 '18 at 22:33

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.

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    $\begingroup$ Can you expand on your reasoning here? What equation are you talking about? Why are you only focusing on the photoelectric effect? Also, realize that the uncertainty principle is independent of our measurement techniques. $\endgroup$ – Aaron Stevens Nov 13 '18 at 2:09
  • $\begingroup$ Then it doesn't mean anything because the standard deviation can be zero $\endgroup$ – Liu Nov 13 '18 at 2:24
  • $\begingroup$ I still don't understand what you mean. $\endgroup$ – Aaron Stevens Nov 13 '18 at 2:30
  • $\begingroup$ The uncertainty principle says that uncetainty exists if uncerainty exists $\endgroup$ – Liu Nov 13 '18 at 15:58
  • $\begingroup$ I understand that is what you are saying, but I don't follow your reasoning. Can you please expand on it. $\endgroup$ – Aaron Stevens Nov 13 '18 at 16:12
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Uncertainty is due to the measurement techniques humans tend to use requiring photoelectric effect.

I think you're confusing what uncertainty means in the theory of Quantum Mechanics. The famous "Uncertainty Principle" has nothing to do with what we measure. It a fundemental aspect regardless of whether we measure or not.

If you want to understand how we got to the uncertainty principle: $$\Delta x \Delta p \geq \frac \hbar 2,$$ there are lots of great resources to start with only a Google away.

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  • $\begingroup$ H is only relevant to measurement because the meausrment techniques humans tend to use require photoelectric effect. If you measure eg via wave interference, with the result being the resulting wave function then h is not involved $\endgroup$ – Liu Nov 13 '18 at 15:51

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