# Heisenberg Uncertainty Principle

1. Feb 25, 2012

### StevieTNZ

From what I've gathered reading the scientific literature, the more precise we know a quantum system's position, the more uncertain the momentum becomes.

Does the uncertainty principle place a limit on how well we can know a system's position when we measure that observable? I've read elsewhere on this forum if the system's position was made infinitely precise, the momentum uncertainty would be infinite - by the way it was stated, does that signal an issue if it happened?

2. Feb 25, 2012

### lugita15

In practice it's pretty much impossible to make an infinitely many precise measurement, so you don't see infinite uncertainties in real life.

3. Feb 25, 2012

### DaveC426913

But yes in principle, as the certainty in position increases without bound, so will the uncertainty in momentum increase without bound.

Simplistically, this is the source of the infinities than pop up when one attempts to combine QM and GR.

4. Feb 25, 2012

### lugita15

Could you elaborate on this? I thought the infinities in QFT were due to field self-energy problems. And I thought that the particular issues with quantum gravity is that an infinite number of independent coupling constants have to be experimentally determined for renormalization, and since it's hard to do an infinite number of experiments, we're stuck with self-energy divergences if we try doing QG perturbatively.

5. Feb 26, 2012

Steve, do you know about fourier transforms? If I show you a long plane wave with lots of cycles, you can easily measure the wavelength, right? But if I show you a short wobble that barely gets going at all, you're not quite sure where to put your ruler. That's about it really. The momentum is associated with the wavelength, but you can't measure that very well if the whole phenomenon starts and stops within a space that's comparable with or shorter than that wavelength.

More precisely, the position is a probability distribution and so is the momentum. The latter is the fourier transform of the former. So you can plot probability for a range of positions or probability for a range of momenta and the one distribution is the fourier transform of the other.

More precisely still, both are complex functions of space (ignoring time for now) where the magnitude of the complex number (squared) is the probability distribution. You can still do the fourier trick to transform one complex field into the other, but now I think you have to call it a laplace transform instead of a fourier transform. But the idea is the same.

Let's take both extremes as examples: the long plane wave in space has a precise wavelength and that precise value tells you the momentum. The fourier (laplace if you're feeling pedantic) transform is a spike. The other extreme is that the position distribution is a spike, and the FT will be an extended wave in momentum space, i.e., lots of equally probable values for momentum.

Between the two extremes, we can consider a gaussian distribution of probability vs position. You'll find that the FT is also a gaussian within a limited range. That's when you know a bit about both position and momentum.

Quantum observations operate on these wave functions and change them. For instance, if you accurately measure the position of an electron that formerly was somewhere, anywhere in a box, you'd be changing the first extreme into the second. The position distribution would go from a wave to a spike, so the momentum distribution would go from a spike to a wave.

Hope that helps,

6. Feb 26, 2012

### StevieTNZ

Does the position "spike" contain a variety of positions the system can be in? Can we narrow it down to one (obviously to a certain degree - not infinity).

7. Feb 26, 2012

### lugita15

A true position spike means that the wave function is a Dirac delta function, so zero position uncertainty and infinite momentum uncertainty. That is, of course, pretty much unrealizable in actual experiments.

8. Feb 26, 2012

When I said "spike" I was talking about a probability distribution that's very tall and narrow, e.g., it's extremely likely to be near 2cm from the end, but it's very unlikely to be at 2.001 or 1.999 and not at all likely to be near 2.002 or 1.998.

9. Feb 26, 2012

### lugita15

If by "spike" you just meant something tall and narrow, than by wave you cannot mean sine wave. The Fourier transform of a sine wave is a Dirac delta function, i.e. a spike with 0 width and infinite height. The fourier transform of a finite width, finitely tall spike is a wave packet with a large but finite bandwidth.

10. Feb 26, 2012