# Why is heisenberg uncertainty not a limit of technology?

by danphan323
Tags: heisenberg, limit, technology, uncertainty
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 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?
 PF Gold P: 4,287 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.
 Mentor P: 11,749 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.
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Why is heisenberg uncertainty not a limit of technology?

 Quote by danphan323 How do we know that the uncertainty principle is a property of an electron and not a limit of our measuring ability?
We don't.

 Quote by danphan323 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?
Yes, it could. In fact, the Bohmian interpretation of quantum mechanics proposes a very precise value of both position and momentum at a given time. See e.g.
http://xxx.lanl.gov/abs/quant-ph/0611032
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PF Gold
P: 5,345
 Quote by danphan323 Couldn't it have both a position and a momentum at a given point in time?
Welcome to PhysicsForums, danphan323!

Demystifier made some comments which are accurate in a certain sense. However, the best answer is NO, particles do not have simultaneously well-defined values for non-commuting properties.

Notice that I said "non-commuting". Commuting properties CAN have simultaneously well-defined 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.
P: 25
 Quote by danphan323 Couldn't it have both a position and a momentum at a given point in time?
 Quote by Demystifier Yes, it could. In fact, the Bohmian interpretation of quantum mechanics proposes a very precise value of both position and momentum at a given time.
True. But maybe one should point out here also that the price you pay for this objectivity in Bohm-theory is that it is explicitly non-local, i.e. particles have to move (much!) faster than light.

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 non-local (i.e. allow super-luminal signaling), otherwise that model will contradict experimental results in the setups made with entangled particle pairs.
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PF Gold
P: 9,355
 Quote by DrChinese Demystifier made some comments which are accurate in a certain sense. However, the best answer is NO, particles do not have simultaneously well-defined values for non-commuting properties.
I was thinking that too. Demystifier's comments are correct, as far as I can tell, but only because to "have a position" can mean something different from to "be prepared in a state represented by a sharply peaked wavefunction".

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.
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PF Gold
P: 5,345
 Quote by Fredrik I was thinking that too. Demystifier's comments are correct, as far as I can tell, but only because to "have a position" can mean something different from to "be prepared in a state represented by a sharply peaked wavefunction".
I always love Demystifier's answers. Always sharp. Of course with that Bohmian edge as well.
P: 915
 Quote by Pythagorean 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.
are time and space non-commuting?

can all non-commuting properties can be broken down (reducible to or derived from) time-space?
PF Gold
P: 1,376
 Quote by jtbell 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.
You have made similar posts in number of thread about Heisenberg uncertainty. But is there discussion where this explanation has been discussed in more details?

Anyways I would like to understand to what extent this explanation works so let me ask some questions.
You can't make $f(x)$ and $\hat{f}(\xi)$ 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 $\hat{f}(\xi)$. 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?
PF Gold
P: 4,287
 Quote by San K are time and space non-commuting? can all non-commuting properties can be broken down (reducible to or derived from) time-space?
no (pretty sure time and space straight forward operators) and I don't know about all non-commuting properties. I know a lot of properties in classical physics can be reduced to time and space (but only if you forgive that mass is a ratio of distances).
 P: 150 Time is not an operator in QM.
PF Gold
P: 4,287
 Quote by Dead Boss Time is not an operator in QM.
hrm, so then it's not an observable?
 PF Gold P: 4,287 so then how do they commute time and energy for the uncertainty principle involving them? I thought commutation was for observable operators.
 P: 150 No, time is not an observable and the time-energy uncertainty is a different principle than the position-momentum 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).
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PF Gold
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 Quote by Pythagorean so then how do they commute time and energy for the uncertainty principle involving them? I thought commutation was for observable operators.
It's explained here.
PF Gold
P: 4,287
Apparently, that's exactly where it came from ("it kinda feels right")

 The energy-time uncertainty relation is not, however, an obvious consequence of the general Robertson–Schrödinger relation. Since energy bears the same relation to time as momentum does to space in special relativity, it was clear to many early founders, Niels Bohr among them, that the following relation should hold:[8][9]
http://en.wikipedia.org/wiki/Uncerta...inty_principle

addendum:

thanks Fredrik
Mentor
P: 11,749
 Quote by zonde You can't make $f(x)$ and $\hat{f}(\xi)$ 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 $\hat{f}(\xi)$.It takes as an argument frequency [...]
I think most practical applications of Fourier analysis (e.g. signal processing) use time and frequency as the conjugate variables. However, the same mathematics applies when you use position and wavenumber ##k = 2 \pi / \lambda## as the conjugate variables:

$$\psi(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} {A(k)e^{ikx} dk}$$
$$A(k) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} {\psi(x)e^{-ikx} dx}$$

Wavenumber and momentum are of course related by ##p = \hbar k##.

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