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In the proof of Heisenberg's uncertainty principle (HUP), it doesn't constrict the observables in only one object(one particle for example). So if I can have two particles have the same observable, name it $O_1$ and $O_2$, then I can measure $O_1$ from object 1 and measure $O_2$ from object 2. Finally, I get the precise values of $O_1$ and $O_2$! The problem that might occur in the reasoning I can think is that the two objects have the same $O_1$ and $O_2$ is impossible.

If we can create this kind of objects, the value of $O_1$ and the value of $O_2$ will have uncertainty when we create it even though we want to create all the objects to have the same values of $O_1$ and $O_2$(might be an instrument limit). So the variance of the $O_1$ variable and the variance of observable $O_2$ will have the HUP?

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You are wrong in thinking you can get the precise values of O1 and O2 by measurements on two particles if you are assuming they correspond to the values of O1 and O2 that a single particle would exhibit. The HUP does not stop you from measuring, say, an electron's exact position and exact momentum- it just stops you from being able to say that the electron has those attributes at the same time, because it doesn't. When it has a well-defined momentum, its position becomes very ill-defined, and when it has well defined position its momentum becomes very ill-defined.

That all follows directly from the wavelike character of an electron. The spatial extent of an electron's wave function represents, broadly, the degree of uncertainty about its position. To pin down the position almost exactly, the wave function needs to approximate a delta function, becoming an extremely narrow spike. Such a wave has no single associated frequency, and instead is a superposition of waves with a very wide range of frequencies- broadly speaking, the narrower the spike of the very localised wave, the greater the spread of frequencies in its components. Since the momentum of an electron is related to the frequency of its wave function, when the electron is in a very localised state, its wave function is a superposition of components with a very wide range of associated momentum values. If you measure the momentum of a localised electron, its wave-function changes to become any one of the component momentum states that had previously been superimposed to create the narrow peak. You cannot predict for certain which moment state it will 'jump' into.

So, while an electron has a very well-defined position, or, in other words, a tightly localised wave-function, that wave function is a superposition of components with a wide range of momentum values. The more tightly you contain the wave function (ie the more precisely you nail down the electron's position) the wider becomes the range of momentum values in the components of its wave function. The HUP quantifies the trade-off between how far you can pin down one observable at the expense of increasing the potential spread of the other.

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  • $\begingroup$ While I agree with everything you write I do not think it explains anything, really, because of this assumption "Since the momentum of an electron is related to the frequency of its wave function, etc.". The association of "frequency" with momentum is what makes the commutation relations, get us HUP, and makes an electron a "wave" but it does not add to intuition of the "why". Or, does it? $\endgroup$
    – hyportnex
    Commented Jun 3, 2021 at 0:27
  • $\begingroup$ That depends upon which 'why' you mean. If you accept that the electron has an associated wave function, then my explanation shows how uncertainty arises straightforwardly from that. If the 'why' is why does an electron has a wave function or why momentum is related to frequency, then they would be different questions, and I would have answered them accordingly. $\endgroup$ Commented Jun 3, 2021 at 5:47

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