Uncertainty Principle/Talk

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Are you quite sure it's h/4π? I've always seen the Uncertainty Principle written as Δx Δph-bar, which is equal to h/2π. -- Xaonon

I'm not sure, and neither are the experts:

I just checked Encyclopedia Britannica, and they give both 2π and 4π in different articles. A case for Wikipedia commentary/Making fun of Britannica :-)

I think we should leave it at 4π -- at least we are on the safe side. --AxelBoldt

I think it's 4π, though it's a while since I did any QM. Doesn't it come from:

(ΔA)(ΔB) >= (1/2) |[A,B]|

and

[x^,p^] = h_bar / i ,

so

(Δx^)(Δp^) >= h_bar / 2 .

-- DrBob

Ok, I will then happily make fun of Britannica now... --AxelBoldt


I think to remember that DrBob is right. The thing is that when you use it to estimate a value, sometimes you say

(Δx^)(Δp^) ~ h_bar

that is compatible with the previous--AN


Sadly, the source is a bit old, but in Feynman's lectures, Feynman defines the measurements Δx and Δp to be the width of a gaussian distribution, which may be the source of the confusion. He does, however, state precisely:

ΔxΔp ≥ h/2π

It was originally published 1963-1965, so I guess it may have changed since then.

BlackGriffen

Ah, that must be it. They never clearly say what they mean with "uncertainty". Feynman takes the "width", which is probably twice the standard deviation. We take the standard deviation itself, that's why we get half his number. So we are fine, but EB is still screwed :-) --AxelBoldt

Not quite Axel. Check the math. If Δx' and Δp' are the standard deviations, equal to half the width, then:
(2Δx')(2Δp') ≥ h/2π
=> Δx'Δp' ≥ h/8π
That might be the source of the confusion if the competitors were:
ΔxΔp ≥ h_bar
ΔxΔp ≥ h_bar/4
I'm not familiar enough with statistical data analysis to say much more, however. Does the width of a gaussian divided by √2 mean anything? --BlackGriffen
Yup, √2 times standard deviation is the width of the range where 50% of the values will be. But it only works for a gaussian distribution, and there is no reason to assume that all observables are normally distributed, in fact they're most definitely not, so our use of the standard deviation is much cleaner. --AxelBoldt
So, the width over √2 corresponds to the 50% of observations. Doesn't that mean that the formula:
ΔxΔp ≥ h/2
is the incorrect one? the correct ones being:
ΔxΔp ≥ h
σxσph/4
lower case sigma is the standard character for a standard deviation, right?--BlackGriffen
In stats, they use sigma, but it seems that physisists use Δx both for standard deviation and for the 50% range, and call both "uncertainty". If we used Δx for the 50% range and σx for the standard deviation, then the correct formulas would be
ΔxΔp ≥ h
σxσph/2
But the first of those is really pointless since it makes the unjustified assumption that the variables are normally distributed. --AxelBoldt

If there are two differing definitions of Δx and Δp we should note this, and that the uncertainity principle takes different forms depending on what definition is chosen. -- SJK