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WE all know that:

dE dt >= h/2pi

or

dS dt >= h/2pi

But... mathematically, what does that mean?

dE dt >= h/2pi

or

dS dt >= h/2pi

But... mathematically, what does that mean?

- Thread starter PrudensOptimus
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WE all know that:

dE dt >= h/2pi

or

dS dt >= h/2pi

But... mathematically, what does that mean?

dE dt >= h/2pi

or

dS dt >= h/2pi

But... mathematically, what does that mean?

- #2

jcsd

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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:

ΔxΔp_{x} ≥ ħ/2

ΔEΔt ≥ ħ/2

Δω_{x}Δω_{z} ≥ 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.

There are several sets of complementary pairs, for example:

ΔxΔp

ΔEΔt ≥ ħ/2

Δω

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 ∫x dp >= h/4pi??

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jcsd

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ΔxΔp

means that, for an elecrton (for example), uncertainty in postion along the x axis(Δx) mutplied by uncertainty in momentum along the x axis (Δp

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Integral

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While this may be true, it is not the point of the HUP, in fact to use the Δ 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.Originally posted by PrudensOptimus

so this means ∫x dp >= h/4pi??

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

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ΔA = √(<A

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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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- #9

pmb

To understand the uncertainty relation one has to first understand what an uncertainty is. For that seeOriginally posted by PrudensOptimus

WE all know that:

dE dt >= h/2pi

or

dS dt >= h/2pi

But... mathematically, what does that mean?

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

- #10

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.The point is that neither of those quanities can be zero.

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

Pete

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jcsd

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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:

ΔxΔp

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h or h bar??Originally posted by jcsd

Most of the time it can be assumed that:

ΔxΔp_{x}≈ h/2π

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Integral

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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 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

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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.Originally posted by Integral

Setting either of those quanities to zero violates the HUP.

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jcsd

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

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- #17

pmb

Setting either does not violate HUP. E.g. if the uncertainty in x is zero and the uncertainty in pOriginally 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?

Pete

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