# Basic factors on which uncertainity depends on.?

1. Jun 8, 2012

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.?

2. Jun 9, 2012

### DrewD

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.

3. Jun 9, 2012

### 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:

4. Jun 9, 2012

"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.?

5. Jun 11, 2012

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)

6. Jun 12, 2012

### Staff: Mentor

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.

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

7. Jun 12, 2012

### Staff: Mentor

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?

8. Jun 12, 2012

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 lets 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)

9. Jun 12, 2012

### 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
$\Delta x(t_2)\Delta p(t_2)=\hbar$

10. Jun 12, 2012

### haael

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.

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.