Why is the uncertainty in position equal to the wavelength?

[HermitOfThebes]

Why is the uncertainty in position of an object which is determined by a wave equal to the wavelength of the object? I don't get it. Isn't the wave diffracted at the surface of the object? we can then calculated the distance from velocity and time accurately, can't we? is it because we must receive a whole wave cycle by our devices before we signal the return of the wave? that would make sense, but I am not sure if it's correct.

 

[mfb]

Why is the uncertainty in position of an object which is determined by a wave equal to the wavelength of the object?

It is not. The wavelength just gives a lower limit - an object cannot be more localized than "a single blob" (something similar to a single half-wave). If you compress the wavefunction more you also reduce the wavelength.

 

[HermitOfThebes]

It is not. The wavelength just gives a lower limit - an object cannot be more localized than "a single blob" (something similar to a single half-wave). If you compress the wavefunction more you also reduce the wavelength.

If it was more localised than that the wave wouldn't reflect off it right?

 

[mfb]

Reflect off what, when, where?

 

[HermitOfThebes]

Reflect off what, when, where?

The wavelength of a wave "radiowave for instance" determines the minimum size of the object that it can detect, as it can't detect object smaller than half it's wavelength. is this statement true? and from your earlier comment I understand that the uncertainty in the position of the object is not equal to the wavelength of the wave used. What is the uncertainty in an object's position equal to then? and how does using a wave of shorter wavelength helps us pin down the object's position more accurately?

 

[HallsofIvy]

If it was more localised than that the wave wouldn't reflect off it right?

You appear to be very confused. You seem to be thinking of a wave of some sort hitting an object. But we are talking about a wave that is the "object" or (depending on your philosophical position) is a "guide wave" for the object or gives the probability of finding the object at a particular place. In none of those is the wave striking the object.

 

[mfb]

The wavelength of a wave "radiowave for instance" determines the minimum size of the object that it can detect, as it can't detect object smaller than half it's wavelength.

There is reflection of electromagnetic waves from smaller objects as well. The angular dependence of the reflected intensity depends a lot on the size, but that is a different topic.
That has nothing to do with your previous question, however.

What is the uncertainty in an object's position equal to then?

In quantum mechanics, <x²> - <x>², but I don't think that helps now.

 

[HermitOfThebes]

There is reflection of electromagnetic waves from smaller objects as well. The angular dependence of the reflected intensity depends a lot on the size, but that is a different topic.
That has nothing to do with your previous question, however.
In quantum mechanics, <x²> - <x>², but I don't think that helps now.

I am talking about uncertainty in the macroscopic world. Like the uncertainty in the position of an object that a bat detects using its echo. What is it equal to?

 

[phinds]

I am talking about uncertainty in the macroscopic world. Like the uncertainty in the position of an object that a bat detects using its echo. What is it equal to?

I think uncertainty in the macro world is orders of magnitude greater than in the quantum world and is due to different effects. In other words, in the macro world, you're not even measuring the quantum uncertainty at all, you are measuring uncertainty due to other issues (irregular shapes, movement of the target object, etc).

 

[HermitOfThebes]

I think uncertainty in the macro world is orders of magnitude greater than in the quantum world and is due to different effects. In other words, in the macro world, you're not even measuring the quantum uncertainty at all, you are measuring uncertainty due to other issues (irregular shapes, movement of the target object, etc).

right so using a wave of wavelength > 10m can still give me a precision of 1m (i.e. an answer to the nearest metre) ?

 

[mfb]

A precision on what? Distance, position, size, shape in some way? Something else?
Independent of that, the answer is probably yes, if your measurement is precise enough.

 

[jeremyfiennes]

I'm also confused by position uncertainty. The interaction between electrons and photons has the nature of colliding elastic spheres (Compton scattering). A low energy photon forming a point on a microscope screen gives an electron's position as exactly as a high energy photon does. I've seen explanations in terms of the resolving power of the lens. But this uses a classical wave model, whereas here we are thinking in particle terms.

 

[jtbell]

The interaction between electrons and photons has the nature of colliding elastic spheres (Compton scattering).

No, it doesn't. Energy and momentum are conserved in Compton scattering, and in collisions between elastic spheres in classical physics. This does not mean that the two processes have the same "nature."

Introductory textbooks and other presentations commonly have diagrams of Compton scattering that look like colliding billiard balls, but this simply provides an illustrative "hook" to display the quantities involved in the analysis (momenta, energies, angles).

 

[jeremyfiennes]

Thanks. I get your point. Even so, experiment (e.g. 2-slit eraser) shows that if you treat something as waves, it behaves as waves; and if you treat it as particles it behaves as particles. This is the wave-particle duality. There being no rational relation between the two sides, one cannot mix them. Waves don't interact: they superimpose. Particles do: they collide and bounce off each other. An interaction experiment is therefore a particle experiment, and its results should be explicable in particle terms. To bring in the wavelength is side-mixing.
In spite of which: what then is the relation between the wavelength and the position uncertainty? And what reasoning is it based on?

 

[Nugatory]

Even so, experiment (e.g. 2-slit eraser) shows that if you treat something as waves, it behaves as waves; and if you treat it as particles it behaves as particles. This is the wave-particle duality. This is the wave-particle duality. There being no rational relation between the two sides, one cannot mix them. Waves don't interact: they superimpose. Particles do: they collide and bounce off each other. An interaction experiment is therefore a particle experiment, and its results should be explicable in particle terms. To bring in the wavelength is side-mixing.

The concept of wave-particle duality, at least as you're understanding it, was abandoned more than 75 years ago when the modern formulation of quantum mechanics was hammered out. You will save yourself a fair amount of grief if you let go of it too.

A more accurate way of stating the conclusion of the double-slit experiment might be: After you measure an observable (in this case, whether a particle is present at a slit) in a quantum system the system will evolve from there in a manner consistent with the observed result. The interesting and important thing about the double-slit experiment is that this evolution leads to results that don't match our classical intuition about how things should turn out.

 

[Nugatory]

n spite of which: what then is the relation between the wavelength and the position uncertainty? And what reasoning is it based on?

In principle we can measure the position of a particle with arbitrary (but not infinite) accuracy. What we cannot do is predict ahead of time what our arbitrarily precise measurement will be. Instead, the mathematical Rules of Quantum Mechanics let us calculate the probability of getting any given position result when we perform the measurement. We do this calculation, and for the situations we're discussing here, we get a bell curve; if the curve is wide and flat there is more spread in the range of possible results than if it is very narrow and sharply peaked.

When you do the math, it turns out that it requires shorter wavelengths to form a sharply peaked bell curve than a wide flat one. (This result should also make intuitive sense, if you consider that the bell curve cannot be much narrower that the shortest wavelength in it. And if you are not familiar with Fourier transforms, this would be a really good time to start looking at them).

 

[jeremyfiennes]

The concept of wave-particle duality, at least as you're understanding it, was abandoned more than 75 years ago when the modern formulation of quantum mechanics was hammered out.

I'm not sure where my idea of the wave-particle duality came from, but the probability of its having been more than 75 years ago appears to be minimal. Can you give a reference to the hammering out?

 

[Nugatory]

I'm not sure where my idea of the wave-particle duality came from, but the probability of its having been more than 75 years ago appears to be minimal. Can you give a reference to the hammering out?

Wave-particle duality was part of the "old quantum mechanics"; to quote wikipedia: "The old quantum theory is a collection of results from the years 1900–1925 which predate modern quantum mechanics."

And what changed in 1925 is that the hammering out of modern quantum mechanics started: Heisenberg's matrix mechanics paper was published in in 1925. Schrödinger's wave mechanics paper was published in 1926. Dirac demonstrated the equivalence of the two formulations in the same year, and that's also when Max Born proposed the Born rule. The process was pretty much complete when von Neumann consolidated the mathematical formalism of modern QM in 1932.

So yes, a bit more than 75 years.

You have almost certainly read about "wave-particle duality" in more recent publications, but they won't be serious textbooks. The idea is so intriguing that once it leaked out into the popular imagination it took root like an urban legend, and populizers have been repeating it without checking it ever since.

 

[jeremyfiennes]

Ok.