A (probably) simple Heisenberg Uncertainty Principle question

Number Six · 7th August 2011, 09:44 PM

As I understand it, part of it says that

x times y > z

where is the uncertainty in momentum and y is the uncertainty in position and z is greater than 0. So the more precisely we measure momentum the less precisely we measure position, and vice-versa.

But I think I've also heard reference to measuring momentum or position perfectly or exactly.

Those two contradict though. If the product of the uncertainties of momentum and position is greater than 0 then the uncertainty of momentum and of position must both always be greater than 0.

What gives?

Ziggurat · 7th August 2011, 10:22 PM

Originally Posted by Number Six

If the product of the uncertainties of momentum and position is greater than 0 then the uncertainty of momentum and of position must both always be greater than 0.

Not exactly. The theory allows for zero position (momentum) uncertainty if the corresponding momentum (position) uncertainty is infinite. These are not physically realizable scenarios, but they're mathematically consistent with the theory.

sol invictus · 8th August 2011, 02:44 AM

Originally Posted by Number Six

But I think I've also heard reference to measuring momentum or position perfectly or exactly.

As Zig says, it would require infinite uncertainty in the other quantity - and that's impossible physically. I don't completely agree with him that it's mathematically consistent, although it does depend on what conditions you decide to impose on the theory.

Do you recall where you heard that, or in what context?

Ziggurat · 8th August 2011, 07:18 AM

Originally Posted by sol invictus

As Zig says, it would require infinite uncertainty in the other quantity - and that's impossible physically. I don't completely agree with him that it's mathematically consistent, although it does depend on what conditions you decide to impose on the theory.

To make this more explicit, the problem with such states is that they aren't normalizable. Normalization is a requirement for any physically realizable state, but you can still play with such states in the theory if you don't mind that lack of normalization.

Almo · 8th August 2011, 01:05 PM

Think of it this way. The wavefunction that describes the particle is composed of many overlapping functions.

If you imagine you know its speed, it's like it's just a single sine wave. The particle could be at any of its bumps.

If you know its position, it's a delta function (infinite spike). That gives you no idea of its travel.

Vorpal · 8th August 2011, 01:06 PM

Originally Posted by Ziggurat

To make this more explicit, the problem with such states is that they aren't normalizable. Normalization is a requirement for any physically realizable state, but you can still play with such states in the theory if you don't mind that lack of normalization.

Are you talking about definite states or states of infinite uncertainty, as per sol's comment? For infinite uncertainties, it's not necessarily so. For example, a wavefunction that gives a position distribution in a Cauchy-Lorentz form, ψ(x) ∝ 1/√(1+x²), is normalizable but has infinite position uncertainty. It looks something like an instantaneously reflecting plane wave in momentum-space, so I guess it's still unphysical, though.

I'm not sure what mathematical inconsistencies occur for even for the non-normalizable states of definite position or momentum, if one thinks of states (kets) as antilinear functionals on some Hilbert space, or some appropriate subspace thereof. Over a whole Hilbert space, there's a simple correspondence: a state is normalizable iff it is a continuous functional.

Ziggurat · 8th August 2011, 01:22 PM

Originally Posted by Vorpal

Are you talking about definite states or states of infinite uncertainty, as per sol's comment?

Definite states, since what we wanted was not infinite uncertainty but zero uncertain, and infinite uncertainty in the other variable was part of the price we needed to pay to get zero uncertainty (definite states). But your example is still interesting since it demonstrates that infinite uncertainty in one variable is not sufficient to ensure zero uncertainty in the other variable.

Delvo · 9th August 2011, 09:55 PM

Zero uncertainty in anything has always been impossible. No matter how many digits of something's value you are able to determine, there's always the next digit after that.

Vorpal · 10th August 2011, 09:05 AM

Originally Posted by Delvo

Zero uncertainty in anything has always been impossible. No matter how many digits of something's value you are able to determine, there's always the next digit after that.

It's a fundamentally different barrier, though. In classical physics, the state contains the exact positions and momenta of particles. How precisely you can measure them is only a matter of your personal cleverness and technological ability: maybe in a century, experimentalists will get a dozen more digits, etc.

That's no longer the case in quantum physics. The state still contains all the information about positions and momenta of particles, but starkly defined positions and momenta aren't part of that any more.

7th August 2011, 09:44 PM	#1
Number Six JREF Kid Join Date: Sep 2001 Posts: 5,017	A (probably) simple Heisenberg Uncertainty Principle question As I understand it, part of it says that x times y > z where is the uncertainty in momentum and y is the uncertainty in position and z is greater than 0. So the more precisely we measure momentum the less precisely we measure position, and vice-versa. But I think I've also heard reference to measuring momentum or position perfectly or exactly. Those two contradict though. If the product of the uncertainties of momentum and position is greater than 0 then the uncertainty of momentum and of position must both always be greater than 0. What gives?
Number Six JREF Kid Join Date: Sep 2001 Posts: 5,017

7th August 2011, 10:22 PM	#2
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,227	Originally Posted by Number Six If the product of the uncertainties of momentum and position is greater than 0 then the uncertainty of momentum and of position must both always be greater than 0. Not exactly. The theory allows for zero position (momentum) uncertainty if the corresponding momentum (position) uncertainty is infinite. These are not physically realizable scenarios, but they're mathematically consistent with the theory.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

8th August 2011, 02:44 AM	#3
sol invictus Philosopher Join Date: Oct 2007 Location: Nova Roma Posts: 8,419	Originally Posted by Number Six But I think I've also heard reference to measuring momentum or position perfectly or exactly. As Zig says, it would require infinite uncertainty in the other quantity - and that's impossible physically. I don't completely agree with him that it's mathematically consistent, although it does depend on what conditions you decide to impose on the theory. Do you recall where you heard that, or in what context?
	Last edited by sol invictus; 8th August 2011 at 02:47 AM.

8th August 2011, 07:18 AM	#4
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,227	Originally Posted by sol invictus As Zig says, it would require infinite uncertainty in the other quantity - and that's impossible physically. I don't completely agree with him that it's mathematically consistent, although it does depend on what conditions you decide to impose on the theory. To make this more explicit, the problem with such states is that they aren't normalizable. Normalization is a requirement for any physically realizable state, but you can still play with such states in the theory if you don't mind that lack of normalization.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

8th August 2011, 01:05 PM	#5
Almo Masterblazer Join Date: Aug 2005 Location: Montreal, Quebec Posts: 6,409	Think of it this way. The wavefunction that describes the particle is composed of many overlapping functions. If you imagine you know its speed, it's like it's just a single sine wave. The particle could be at any of its bumps. If you know its position, it's a delta function (infinite spike). That gives you no idea of its travel.
	__________________ Almo! My Blog "No society ever collapsed because the poor had too much." — LeftySergeant "It may be that there is no body really at rest, to which the places and motions of others may be referred." –Issac Newton in the Principia

8th August 2011, 01:06 PM	#6
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,009	Originally Posted by Ziggurat To make this more explicit, the problem with such states is that they aren't normalizable. Normalization is a requirement for any physically realizable state, but you can still play with such states in the theory if you don't mind that lack of normalization. Are you talking about definite states or states of infinite uncertainty, as per sol's comment? For infinite uncertainties, it's not necessarily so. For example, a wavefunction that gives a position distribution in a Cauchy-Lorentz form, ψ(x) ∝ 1/√(1+x²), is normalizable but has infinite position uncertainty. It looks something like an instantaneously reflecting plane wave in momentum-space, so I guess it's still unphysical, though. I'm not sure what mathematical inconsistencies occur for even for the non-normalizable states of definite position or momentum, if one thinks of states (kets) as antilinear functionals on some Hilbert space, or some appropriate subspace thereof. Over a whole Hilbert space, there's a simple correspondence: a state is normalizable iff it is a continuous functional.
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,009	__________________ For every philosopher, there exists an equal and opposite philosopher. They're both wrong.

8th August 2011, 01:22 PM	#7
Ziggurat Penultimate Amazing Join Date: Jun 2003 Posts: 26,227	Originally Posted by Vorpal Are you talking about definite states or states of infinite uncertainty, as per sol's comment? Definite states, since what we wanted was not infinite uncertainty but zero uncertain, and infinite uncertainty in the other variable was part of the price we needed to pay to get zero uncertainty (definite states). But your example is still interesting since it demonstrates that infinite uncertainty in one variable is not sufficient to ensure zero uncertainty in the other variable.
	__________________ "As long as it is admitted that the law may be diverted from its true purpose -- that it may violate property instead of protecting it -- then everyone will want to participate in making the law, either to protect himself against plunder or to use it for plunder. Political questions will always be prejudicial, dominant, and all-absorbing. There will be fighting at the door of the Legislative Palace, and the struggle within will be no less furious." - Bastiat, The Law

9th August 2011, 09:55 PM	#8
Delvo Illuminator Join Date: Jul 2008 Location: Harrisburg, Pennsylvania Posts: 3,894	Zero uncertainty in anything has always been impossible. No matter how many digits of something's value you are able to determine, there's always the next digit after that.

10th August 2011, 09:05 AM	#9
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,009	Originally Posted by Delvo Zero uncertainty in anything has always been impossible. No matter how many digits of something's value you are able to determine, there's always the next digit after that. It's a fundamentally different barrier, though. In classical physics, the state contains the exact positions and momenta of particles. How precisely you can measure them is only a matter of your personal cleverness and technological ability: maybe in a century, experimentalists will get a dozen more digits, etc. That's no longer the case in quantum physics. The state still contains all the information about positions and momenta of particles, but starkly defined positions and momenta aren't part of that any more.
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,009	__________________ For every philosopher, there exists an equal and opposite philosopher. They're both wrong.