The Heisenberg Uncertainty Principle suggests that the more precisely the position of a particle is measured, the less precisely its momentum can be known, and vice versa.

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

What I've read is that in order to measure the position of a particle accurately, we need a very energetic quantum of light to measure it but the more energetic the photon is, the greater it will disturbe the velocity of the particle, thus letting us unable to obtain its momentum accurately.

This sounds just like an observer effect to me. However, I'm told that it shouldn't be thought of as an observer effect and instead, measurement only gives the the particle's momentum and location corresponding to one of its possible states (and so before the measurement it exists in all the possible states, in others words, it is in a superposition).

And this confuses me. Why can't it be interpreted as an observer effect and why do we have to think of the measurement as just obtaining one of the particle's possible states? And why it exists in all its possible state before measurement? Because it exhibits both particle properties and wave properties simultaneously? If that is the reason, how does its wave-particle duality imply that it exists in all states?


1 Answer 1


The argument that one can only detect positions precisely using high-energy photons is very often used to justify the uncertainty principle but it its not its ultimate justification. The argument's place in physics is curious: it essentially states that, even if quantum mechanics were wrong, these kind of observer effects would also prevent us from measuring position and momentum simultaneously.

The real ground the uncertainty principle stands on is de Broglie's postulate: particles of matter are also waves (of some sort) and these waves have wavelength $$\lambda=\frac hp.$$ If you substitute this expression for the particle's momentum into the uncertainty principle, you get it in the form $\Delta x \cdot\Delta\frac1\lambda\geq\frac1{2\pi}$, or in terms of the wavenumber $k=2\pi/\lambda$, $$\Delta x \cdot\Delta k\geq1.$$ This now expresses a fundamental fact of wave physics: you can have a localized wavepacket, but you can only localize it to - roughly - a region the size of the wavelength. More precisely, the more you want to localize a wavepacket, the more waves with different wavelengths you will need to superpose to create it.

  • $\begingroup$ Great answer. Note that you can know the position of a particle without touching it, sometimes. Just make a small hole in a wall, and send some particles through it. If they make it through the wall, then you know that the particle was in the region of the whole at some point in the past. This of course does not violate the HUP: there will be a corresponding uncertainty in the momentum of the particle, which comes from the wave aspects of the particles. $\endgroup$
    – Andrea
    Commented Jun 30, 2021 at 9:59

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