Why is heisenberg uncertainty not a limit of technology?

danphan323

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?

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.

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.

Demystifier

We don't.

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

DrChinese

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.

Dali

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.

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

DrChinese

I always love Demystifier's answers. Always sharp. Of course with that Bohmian edge as well.

San K

are time and space non-commuting?

can all non-commuting properties can be broken down (reducible to or derived from) time-space?

zonde

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 [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?

Pythagorean

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).

Dead Boss

Time is not an operator in QM.

Pythagorean

hrm, so then it's not an observable?

Pythagorean

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

Dead Boss

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).

Fredrik

It's explained here.

Pythagorean

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

http://en.wikipedia.org/wiki/Uncerta...inty_principle

addendum:

thanks Fredrik