Basic factors on which uncertainity depends on.?

[aditya23456]

we know as per uncertainity, (px-xp)>=h/4(pie)...ie can px-xp take any value which is more than h/(4(pie))..can it be infinite also.?
On what factors do this extend of greatness depend on.?

 

[DrewD]

On what factors do this extend of greatness depend on.?

Are you asking what the uncertainty in a measurement depends on? The quick and unenlightening question is just about everything.

First, the uncertainty principle and the commutator relationships are not the same.
(px-xp)=i\hbar
\Delta x\Delta p\geq \frac{\hbar}{2}

The uncertainty principle is basically saying that there is a maximum precision. The tools that are used to make a measurement can make the uncertainty greater and certain preparations of particle will not even attain the minimum uncertainty. On a daily basis we make measurements that are not even near this level of precision, but no matter how hard we try, they will never be better.

 

[Naty1]

Here are some explanations I saved [and edited] from a very,very long discussion in these forums on Heisenberg uncertainty:

Course Lecture Notes, Dr. Donald Luttermoser,East Tennessee State University:

The HUP strikes at the heart of classical physics: the trajectory. Obviously, if we cannot know the position and momentum of a particle at t[0] we cannot specify the initial conditions of the particle and hence cannot calculate the trajectory...Due to quantum mechanics probabilistic nature, only statistical information about aggregates of identical systems can be obtained. QM can tell us nothing about the behavior of individual systems. ….QUOTE]

what the quotes means: unlike classical physics, quantum physics means future predications of state [like position, momentum] are NOT precise.] 'system' in the above means for example a particle like an electron or photon.

But why does this situation pop up in quantum mechanics? because of our mathematical representations. QM math is different than classical math.

[scattering, mentioned below, happens to be a convenient example of our limited ability to make measurements of arbitrary precision. ]The basic ideas are these: HUP [Heinsenberg uncertainty principle] is a result of nature, not of experimental based uncertainties. From the axioms of QM and the math that is used to build observables and states of systems, it turns out that position and velocity (and also momentum) are examples of what are called "canonical conjugates" [a function and its Fourier transform]; They cannot be both be "sharply localized". That is, they cannot be measured to an arbitrary level of precision. It is a mathematical fact that any function and its Fourier transform cannot both be made sharp. This not a matter of test instrument sensitivity but of nature.

The wave function describes not a single scattering particle but an ensemble of similarly accelerated particles. Physical systems [like particles] which have been subjected to the same state preparation will be similar in some of their properties but not all of them. The physical implication of the uncertainty principle is that NO STATE PREPARATION PROCEDURE IS POSSIBLE WHICH WOULD YIELD AN ENSEMBLE OF SYSTEMS IDENTICAL IN ALL OF THEIR OBSERVABLE PROPERTIES.

A few cornerstone mathematical underpinnings:
The wave function describes an ensemble of similarly prepared particles rather than a single scattering particle. A wave function with a well defined wavelength must have a large special extension, and conversely a wave function which is localized in a small region of space must be a Fourier synthesis of components with a wide range of wavelengths. We cannot measure them both to an arbitrary level of precision. A function and its Fourier transform cannot both be made sharp. This is a purely a mathematical fact and so has nothing to do with our ability to do experiments or our present-day technology. As long as QM is based on the present mathematical theory an arbitrary level of precision cannot be achieved. The HUP isn't about the knowledge of the conjugate observables of a single particle in a single measurement. The uncertainty theorem is about the statistical distribution of the results of future measurements. The theorem doesn't say anything about whether you can measure both at the same time. That is a separate issue. A single scattering experiment consists of shooting a single particle at a target and measuring its angle of scatter. Quantum theory does not deal with such an experiment but rather with the statistical distribution of the results of an ensemble of similar results. Quantum theory predicts the statistical frequencies of the various angles through which a an ensemble of similarly prepared particles may be scattered.

What we can't do is to prepare a state such that we would be able to make an accurate prediction about what the result of a position measurement would be, and an accurate prediction about what the result of a momentum measurement would be.

Physical systems which have been subjected to the same state preparation will be similar in some of their properties but not all of them. The physical implication of the uncertainty principle is that NO STATE PREPARATION PROCEDURE IS POSSIBLE WHICH WOULD YIELD AN ENSEMBLE OF SYSTEMS IDENTICAL IN ALL OF THEIR OBSERVABLE PROPERTIES.

 

[aditya23456]

"As long as QM is based on the present mathematical theory an arbitrary level of precision cannot be achieved."

Does this mean its not possible to control the inaccuracy(as I stated can there be an infinite inaccuracy)
And we've seen-
(px-xp)=ih/2(pie)
Where i is imaginary no.
How can a imaginary quantity enter into something real.?

 

[aditya23456]

can this be true---
Suppose a particle has position dx1 and momentum dp1 at an instant and later it has dx2 and dp2...now can
(dx1)(dp1)=(dx2)(dp2)

 

[Nugatory]

"As long as QM is based on the present mathematical theory an arbitrary level of precision cannot be achieved."

Does this mean its not possible to control the inaccuracy?

Not quite. The problem isn't that our measurements of position and momentum are inaccurate; it is that the position and momentum that we're measuring are naturally fuzzy from the start. Suppose I were to measure the position of a particle exactly. When I then say that after the position measurement the momentum is completely uncertain, I don't mean that the particle has some momentum that I can't measure accurately. I mean that if I now measure the momentum with complete accuracy the answer could be anything - but whatever it is, it is right.

And we've seen-
(px-xp)=ih/2(pie)
Where i is imaginary no.
How can a imaginary quantity enter into something real.?

The way the math of QM works, the \sqrt{-1} disappears in the final calculation of anything real.

 

[Nugatory]

can this be true---
Suppose a particle has position dx1 and momentum dp1 at an instant and later it has dx2 and dp2...now can
(dx1)(dp1)=(dx2)(dp2)

Careful... dx1 and dp1 aren't a position and a momentum, they're the uncertainty in these quantities... Could you restate your question more precisely?

 

[aditya23456]

ok this is my doubt..
(px- xp)>=h/4(pie)
(1) here we find it to be greater than h/4(pie)..how much greater than is it ?on what factor does this greatness depend on.?
Or let's say---its some k times more,so we get--
K(px- xp)=h/4(pie)
From this can't we state--
K(dp1)(dx1)=k(dp2)(dx2)

 

[DrewD]

First of all, again, (px-xp) does not represent the uncertainty. This is the commutator and it is different. Perhaps somewhere somebody uses this to represent the uncertainty, but I would be surprised.
The values \Delta p and \Delta x can have two different meanings. One is the intrinsic uncertainty which is calculated the same way that standard deviations are calculated for any probability distribution. What do they depend on? The wave function. I can't think of a good way to intuitively describe this. Naty1 did an excellent job, I thought, but if that isn't good enough, you might need to actually study QM.

The other way that [/itex]\Delta p[/itex] and \Delta x are used is to describe the uncertainty in a measurement. In this case the precision is affected by the tools that are used. Again, there are too many variable to describe.

To get to the new part of your question. I think you are asking if the uncertainty can be the same at a later time. Is that correct? So dx_1\equiv\Delta x(t_1),\ dx_2\equiv\Delta x(t_2) etc. Certainly that is possible. The deltas here are not change over time. The uncertainty can stay the same indefinitely. So if we have a wave function and calculate
\Delta x(t_1)\Delta p(t_1)= \hbar
it is possible (in fact, under many circumstance certain) that
\Delta x(t_2)\Delta p(t_2)=\hbar

 

[haael]

ok this is my doubt..
(px- xp)>=h/4(pie)
(1) here we find it to be greater than h/4(pie)..how much greater than is it ?on what factor does this greatness depend on.?

For the commutator relation, there is no inequality. The commutator does not depend on anything, it is exactly equal to the reduced Planck constant times i.

The commutator relation says that particles are in fact waves, not tiny balls.

Or let's say---its some k times more,so we get--
K(px- xp)=h/4(pie)
From this can't we state--
K(dp1)(dx1)=k(dp2)(dx2)

You make a mathematical error here.

First of all, there is little sense for defining Δp for a single particle. It's just like you toss a coin once, get a head and say: "the probability of getting a head is 100%". The meaning of Δp is the standard deviation of many measurements. Suppose you have dozens of particles, each produced in an identical process. When you try to measure the momentum of any of them, you can get just any number. They are not bounded by anything - perhaps this is what you mean by "infinite". But when you compute a mean value of all the measurements, you will see that the difference of each individual particle and the mean concentrates near zero with some standard deviation, called Δp.