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The uncertainty principle states that there always will be mean variance if we measure position or momentum. It does not state that the measurement is wrong. It only states that there always will be a deviation from the mean value of position/momentum or $<x>,<p>$. The closer the position measurement is to the mean, the momentum measurement is further from the mean. In short each experiment will give a different result. Then why do so many sources give the wrong idea that we can't measure them precisely. According to me we can measure them precisely

https://www.theguardian.com/science/2013/nov/10/what-is-heisenbergs-uncertainty-principle

The uncertainty principle says that we cannot measure the position(x) and the momentum(p)of a particle with absolute precision.

Instead of 'measure' shouldn't it be 'predict'?

Griffiths is correct although,

Please understand what uncertainty means-Like position measurements, momentum measurements yield precise answers- the "spread" here refers to the fact that measurements on identical systems do not yield consistent results

Am I wrong anywhere?

EDIT: Suppose someone prepares an ensemble of N identical systems and makes position measurements.Let $N \rightarrow \infty$

$$\bar{x} = \frac{1}{N}\sum_{i=1}^Nx_i,\\$$ $$(\Delta x)^2 = \frac{1}{N-1}\sum_{i=1}^N(x_i - \bar{x})^2.$$

Similarly $$\bar{p} = \frac{1}{N}\sum_{i=1}^Np_i,\\$$ $$(\Delta p)^2 = \frac{1}{N-1}\sum_{i=1}^N(p_i - \bar{p})^2.$$

I'm confused that, if $x_i$ deviates very little from $\bar{x}$, $p_i$ will deviate a lot from $\bar{p}$, does that make the measurement $p_i$ less accurate ? As the momentum literally could take up any value, so the measurement $p_i$ is useless. Was the measurement $p_i$ merely coincidental ? Was the particle in a superposition of widely varying momentum eigenstates.

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  • $\begingroup$ We can – and do – measure them with arbitrary accuracy and simultaneously. What's important is to make clear what do we mean by "measurement", "uncertainty", "position", and "momentum". I agree with you that "predict" would be in many instances a better choice than "measure". A detailed explanation with references is given in the answer to this question. $\endgroup$
    – pglpm
    Feb 24 '21 at 7:34
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Here is what you are missing.

You cannot measure both position and momentum simultaneously with arbitrary precision for a quantum (very very small) object. The more precisely you pin down its location, the more uncertain its momentum becomes, and vice versa.

The classic example is an electron, which we place in a trapping apparatus by which we can progressively "squeeze down" its location by confining it to a smaller and smaller bit of space as we squeeze.

The electron responds to this state of affairs by bouncing around inside that shrinking space with greater and greater violence, exhibiting increasingly uncertain momentum. At some point, the energy it possesses at any given instant exceeds the rest energy equivalent of its mass and if we clamp down more on its position, we begin generating electron-antielectron pairs in that space.

This means that if we try to pin down the electron's position with greater and greater precision, we eventually come to a place where we no longer know for certain how many electrons we have inside that space.

Alternatively, we could try shooting particles at the electron so as to measure its momentum by determining how it recoils from the collision. In this case, to get more precision in the momentum measurement requires us to strike the electron with higher and higher energy particles with shorter and shorter wavelengths- and the electron recoils with greater and greater violence and its location becomes more and more uncertain.

This means that the more precisely we try to pin down the electron's momentum, we will eventually reach the point where we have no idea where the electron is located.

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Yes, you are wrong, and so is the Guardian article. The uncertainty principle doesn't actually say that we can't measure both position and momentum. It says that a particle can't have both well-defined position and well-defined momentum at a given time.

In order to measure a particle, you have to interact with it somehow. When you measure $x$, the associated interaction changes $p$. However, $p$ doesn't change from one well-defined value to another well-defined value. Instead, $p$ ends up indefinite.

Now, what Griffiths is referring to is this: If you later measure $p$, the associated interaction will cause $p$ to have a definite value (immediately after the second measurement.) This doesn't mean that $p$ had a definite value before you measured it, however.

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No, you can indeed measure position and momentum together. What you can't do is measure them together beyond a certain point. In particular, we can't measure them down beyond

$$\Delta x \Delta p = \frac{1}{2}\hbar$$

. It does not say that they cannot be measured simultaneously at all - only up to a certain "resolution limit", at which point you begin trading off accuracy in one for accuracy in the other.

The "measurements are always exact" part is an idealization. You can define a so-called "weak measurement" that only measures the position or momentum to some minimum scale (this should be distinguished from a faulty measurement, which returns a value that is notionally at some accuracy, but has an error to it that isn't just limited resolution. You can carry out a faithful measurement to no more than a set resolution; mathematically, the two behave differently.). In this case, there is no problem if that scale is suitably large. For example, we can define a weak measurement down to, say, $1\ \mathrm{mm}$ in length, and $1\ \mathrm{kg \cdot \frac{mm}{s}}$ (equiv. $1\ \mathrm{mN \cdot s}$) in momentum. That's an uncertainty product of $10^{-6}\ \mathrm{kg \cdot \frac{m^2}{s}}$, while Planck's constant is on the order of $10^{-34}$, so there is no difficulty here in simultaneously measuring the two if we only ask for that much information. Basically, "don't overdraw the bit bank".

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Quantum mechanics involves a lot of probability so you have to be prepared to come across such problems. As you mention, if we prepare an ensemble of N identical objects whose $\Delta x$ is calculated to be very less and consequently $\Delta p$ is large.Others have provided good reasons for this fundamental axiom. You would get(just to give an idea) $$x_i = {2, 2.2, 2.3, 3, 2.4, 2.33333, 2.45.......}nm $$ $$p_i = {2, 5, 100, 2500, 50, 1000, 45, 450, .......}mkg^2/s$$ Each of your measurement is accurate(given that there is no error in the apparatus) but this behavior is rather hard to model. Quantum mechanics assumes that a particle is the superposition of all these states so it is easier to model it's behavior. And indeed the modelling is very useful.

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Yes, Heisenberg's uncertainty principle tells us that two canonically conjugate quantities can't be measured simultaneously with 100% accuracy that means if you try to measure position of a particle, there will be uncertainty in momentum and vice versa.. This is only applicable to microscopic scale not for macroscopic.. Classically, both parameters can be measured simultaneously with100% accuracy.

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