Heisenberg uncertainty principle (simple stuff)

AI Thread Summary
The discussion centers on the energy-time variation of the Heisenberg uncertainty principle and its implications for energy conservation. The original poster seeks clarification on why a small delta t is necessary for the principle to hold, suggesting that a larger delta t would seem more consistent with the equation. Responses explain that a shorter measurement time leads to higher uncertainty in energy, which aligns with the principle's assertion that energy conservation can appear violated temporarily. The key takeaway is that increased measurement accuracy over longer time intervals reduces uncertainty, reinforcing the conservation of energy. This highlights the relationship between measurement duration and the perceived violation of energy conservation.
Kenny Lee
Messages
75
Reaction score
0
I'm having a little trouble with my textbook's explanation. This is regarding the energy - time variation of the uncertainty principle; very basic stuff, so I hope helping me out won't take too much. Let me quote exactly:

"... the energy conservation can appear to be violated by an amount delta E, as long as it is only for a short time interval, delta t, consistent with the equation (delta E) * (delta t) >= (h / 4 pi )"

I understand what is being said - and it makes sense. Otherwise we'd all be noticing these 'violations' of energy conservation.
But I don't see how it is 'consistent' with the equation. Why must delta t be small? To me, it seems more 'consistent' with the equation that delta t is large, so that the equation becomes greater than h/4pi.

Someone please help clarify. And thanks for your time.
 
Physics news on Phys.org
(deltaE)*(delta t) is always greater than " ". Think about it this way: If you take more measurements (greater accuracy-much data) over a longer time, say for a single body whose energy is conserved, then the uncertainty in energy tends toward zero. So, if you make a quick (inacurate-few data points) measurement the uncertainty in energy is high. This could lead you to believe that the energy was not conserved. Atleast, I think that was the point the book was trying to make.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
4
Views
2K
Replies
2
Views
2K
Replies
23
Views
5K
Replies
2
Views
2K
Replies
4
Views
3K
Replies
2
Views
1K
Back
Top