[Physics] Is it true that we can’t measure position and momentum together

commutatorheisenberg-uncertainty-principlemomentumquantum mechanicsquantum-measurements

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.

Best Answer

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.