quantum mechanics - Why is Heisenberg's uncertainty principle not an experimental error since it is the error created by photons striking on elementary particles?

Why is Heisenberg's uncertainty principle not an experimental error since it is the error created by photons striking on elementary particles?

Answer

it is the error created by photons striking on elementary particles

It's not. Heisenberg's uncertainty principle actually has nothing to do with any particular experiment, or any particular interaction. It's a purely mathematical statement about waves.

Its true meaning is explained in detail on the Wikipedia page, but the gist is that if you have a wave, you can express it as a function of position, $\psi(x)$, or of momentum, $\phi(p)$. These two functions are Fourier transforms of each other. You can then calculate the variance of each function, $\sigma_x^2$ and $\sigma_p^2$ respectively, using formulas given on Wikipedia, and you will find that these two quantities obey the relationship

$$\sigma_x \sigma_p \ge \frac{\hbar}{2}$$

Since $\sigma$ is a measure of how tightly concentrated the wave is around one particular point, this tells you that a wave which is tightly concentrated in position must be fairly spread out in momentum, and vice versa. (For the proper definitions of "concentrated" and "spread out," of course.)

The only way this connects to measurement is that, if you make a series of position measurements on objects with the same quantum state, the variance of those measurements will tend to the variance of the wavefunction. And similarly for momentum. So with a large number of measurements of both position and momentum, if you compute their variances, you'll find that they have to satisfy that inequality. In a sense, it's a statement about the particle's state before it gets hit with a photon (or something else), not some effect of the photon hitting it.