# How does Uncertainty Principle work?

WE all know that:

dE dt >= h/2pi

or

dS dt >= h/2pi

But... mathematically, what does that mean?

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jcsd
Gold Member
What's dS? That to me is change in entropy. Also the uncertainty is gretae than or equal to h/4pi

There are several sets of complementary pairs, for example:

&Delta;x&Delta;px &ge; ħ/2

&Delta;E&Delta;t &ge; ħ/2

&Delta;&omega;x&Delta;&omega;z &ge; ħ/2

etc.

What this does is place a definte limit on how accurately the results of an experiment can be predicted, note: this is not due experimental error, the uncertainity is intrinsic.

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so this means &int;x dp >= h/4pi??

jcsd
Gold Member
No, that's why using d instead of &Delta; may lead to confusion.

&Delta;x&Delta;px &ge; ħ/2

means that, for an elecrton (for example), uncertainty in postion along the x axis(&Delta;x) mutplied by uncertainty in momentum along the x axis (&Delta;px) is always greater than or equal to the rationalised Planck's constant (syn. with Dirac's constant and h/2pi) over 2

Integral
Staff Emeritus
Gold Member
Originally posted by PrudensOptimus
so this means &int;x dp >= h/4pi??
While this may be true, it is not the point of the HUP, in fact to use the &Delta; of the HUP to a differential is going against the entire meaning the HUP. The point is that neither of those quanities can be zero. To form the integral you must be have ture differentials, this means that they can be arbitrally small.

Edit, the more I think about this the less true is seems. Where is you differential on the RHS? Your relationship simply does not work.

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jcsd
Gold Member
Infact is worth noting that when a 'd' appears infront of a quantity in physics it does not mean you should neccesarily start thniking about intergration. For example in the formula for a change in entropy dS = dq/T, the dS and dq are actually non-integral and to avoid such confusion are sometimes annoted with a 'd-bar' instaed of a 'd'.

The &Delta; in the HUP actually refers to a standard deviation, i.e. if A is an observable, then

It's not the HUP dx*dp>=~(h/(2*pi))? At least is how it appears in my book "Quantum theory" of David Bohm

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pmb
Originally posted by PrudensOptimus
WE all know that:

dE dt >= h/2pi

or

dS dt >= h/2pi

But... mathematically, what does that mean?
To understand the uncertainty relation one has to first understand what an uncertainty is. For that see

http://www.geocities.com/physics_world/qm/probability.htm

But note that dt above is not an uncertainty, in fact dE dt >= h/2pi is not a true uncertainty relation.

Pete

pmb
The point is that neither of those quanities can be zero.
That is incorrect. There is no reason that an uncertainty can't be zero. If dx is zero that means that there is zero knowledge of p.

Of course dx is not an infinitesimal x - its an uncertainty in x. Big difference.

Pete

jcsd
Gold Member
Originally posted by meteor
It's not the HUP dx*dp>=~(h/(2*pi))? At least is how it appears in my book "Quantum theory" of David Bohm

Most of the time it can be assumed that:

&Delta;x&Delta;px &asymp; h/2&pi;

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Originally posted by jcsd
Most of the time it can be assumed that:

&Delta;x&Delta;px &asymp; h/2&pi;
h or h bar??

Integral
Staff Emeritus
Gold Member
Originally posted by pmb
That is incorrect. There is no reason that an uncertainty can't be zero. If dx is zero that means that there is zero knowledge of p.

Of course dx is not an infinitesimal x - its an uncertainty in x. Big difference.

Pete
Setting either of those quanities to zero violates the HUP. Isn't this a good protion of the argument for zero point energy? The fact that 0 energy is forbidden by the HUP?

Originally posted by Integral
Setting either of those quanities to zero violates the HUP.
Well ... if we have zero uncertainty in position, then the wavefunction isn't really a function, it's a Dirac delta distribution. The Fourier transform in momentum space is a constant function of infinite uncertainty. So the left-hand side of the uncertainty relation is an ill-defined zero times infinity. But you can define it using limits: take the wavefunction in position space to be an increasingly sharply peaked Gaussian, and thus in momentum space to be an increasingly broad Gaussian. In the limit as the position Gaussian becomes infinitely narrow (a Dirac delta), we can say that the limit of the product of the uncertainties is well-defined, finite, and does not violate the HUP.

jcsd
Gold Member
Prudens Optimus: In that equation h, but remember h-bar = h/2pi, so I could of simply said "h-bar".

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pmb
Originally posted by Integral
Setting either of those quanities to zero violates the HUP. Isn't this a good protion of the argument for zero point energy? The fact that 0 energy is forbidden by the HUP?
Setting either does not violate HUP. E.g. if the uncertainty in x is zero and the uncertainty in px is infinite then the equality is satisfied. Why do you think 0 energy is forbidden?

Pete