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Why is heisenberg uncertainty not a limit of technology? 
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#1
May412, 12:42 AM

P: 3

How do we know that the uncertainty principle is a property of an electron and not a limit of our measuring ability? I understand that photons striking an electron alter its momentum, but imagine an electron that is not being observed. Couldn't it have both a position and a momentum at a given point in time?



#2
May412, 12:48 AM

PF Gold
P: 4,262

Because it's not a problem of measurement, it's a problem of definition of time and space as we know it. It's a theoretical, mathematical problem that can be experimentally verified; it's not an experimental nuisance that we gave a name.



#3
May412, 01:07 AM

Mentor
P: 11,630

There is a very general uncertainty principle that applies to all kinds of "wave packets", and can be derived from the mathematics of Fourier analysis:
$$\Delta x \Delta k \ge \frac{1}{2}$$ It applies to electromagnetic waves, sound waves, electrical signals in wires, etc. The HUP is simply the application of this principle to the wavelike behavior of particles. It is no more a reflection of technological limitations on measurement, than is the case with sound waves, electrical signal pulses, etc. 


#4
May412, 02:04 AM

Sci Advisor
P: 4,575

Why is heisenberg uncertainty not a limit of technology?
http://xxx.lanl.gov/abs/quantph/0611032 


#5
May412, 09:07 AM

Sci Advisor
PF Gold
P: 5,299

Demystifier made some comments which are accurate in a certain sense. However, the best answer is NO, particles do not have simultaneously welldefined values for noncommuting properties. Notice that I said "noncommuting". Commuting properties CAN have simultaneously welldefined values. So for example: spin and momentum can both be known, but not position and momentum. In addition, a well known paper from 1935 referred to as EPR (Einstein is the E) tackled this issue from your perspective. A series of works over nearly 50 years answered the question in the negative. See EPR, Bell (1965), Aspect (1981) for more on this. 


#6
May412, 09:17 AM

P: 25

We also know this is true in general from Bell's theorem which shows that any model insisting on particles "having" propertied even when they are not measured (i.e. objective theories) will have to be nonlocal (i.e. allow superluminal signaling), otherwise that model will contradict experimental results in the setups made with entangled particle pairs. 


#7
May412, 11:56 AM

Emeritus
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PF Gold
P: 9,274

I started writing a much longer explanation, but it's taking too long. So this short comment will have to do, at least for now. 


#8
May412, 02:25 PM

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PF Gold
P: 5,299




#9
May412, 02:38 PM

P: 915

can all noncommuting properties can be broken down (reducible to or derived from) timespace? 


#10
May612, 02:16 AM

PF Gold
P: 1,376

Anyways I would like to understand to what extent this explanation works so let me ask some questions. You can't make [itex]f(x)[/itex] and [itex]\hat{f}(\xi)[/itex] peak sharply at the same time so it seems very elegant explanation for uncertainty principle in QM. Now the question I have is what would be physical interpretation of functions [itex]\hat{f}(\xi)[/itex]. It takes as an argument frequency and produces amplitude and phase for particular frequency. It seems like dimension of frequency can't span real space or time so it should be something more complex and indirect, right? 


#11
May612, 12:19 PM

PF Gold
P: 4,262




#12
May612, 12:46 PM

P: 150

Time is not an operator in QM.



#14
May612, 12:50 PM

PF Gold
P: 4,262

so then how do they commute time and energy for the uncertainty principle involving them? I thought commutation was for observable operators.



#15
May612, 01:13 PM

P: 150

No, time is not an observable and the timeenergy uncertainty is a different principle than the positionmomentum relation (it doesn't help that they look the same).
Unfortunately I don't really know how it's derived or why it should be true (but it kinda feels right in the light of special relativity). 


#16
May612, 01:24 PM

Emeritus
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PF Gold
P: 9,274




#17
May612, 01:27 PM

PF Gold
P: 4,262

Apparently, that's exactly where it came from ("it kinda feels right")
addendum: thanks Fredrik 


#18
May612, 09:33 PM

Mentor
P: 11,630

$$\psi(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{\infty} {A(k)e^{ikx} dk}$$ $$A(k) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{\infty} {\psi(x)e^{ikx} dx}$$ Wavenumber and momentum are of course related by ##p = \hbar k##. 


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