Hi, What is a good way to explain the uncertainty principle to a new freshman who has never had any quantum mechanics? That is, in plain English and without mentioning momentum space or anything.
First make the assertion that particles can be described using wave functions. Consider then a sine function, it has one distinct frequency, however it stretches to infinity in both directions, i.e. it is completely non-localised. Frequency is related to momentum, so one can make the assertion that, when the momentum is precisely known (one frequency), the uncertainity in position is infinite (completely non-localised wavefunction). Add a second frequency component. The wavefunction 'beats', making the particle more localised within the antinodes of the beats. In order to construct the wavefunction such that it only has one 'beat', there must be a frequency range, dF. The frequency range can be identified as the uncertainty in momentum, the 'beat width' can be identified as the uncertainty in space. It is then quite easy to show that the two are complementry, if one gets bigger, the other gets smaller. Include more frequencies, the beat width is reduced and vice-versa. Claude.
There is one famous description of the "quantum behavior".It's the one made by R.P.Feynman in his famous "Lectures on physics".Everybody trying to explain the Uncertainty Principle without quantum mechanics formalism inspired from this great book.Let it be known that i strongly doubt that Feyman hadn't inspired himself from other book.But at least he got a Nobel Prize and is considered alongside Dirac as the (theoretical) physicist of all time. So my warm reccomandation is the first chapter from the third volume from any edition of:Richard P.Feynman "Lectures on Physics". P.S.Hopefully you won't mind my assertion that:"Even a dumbass would understand physics if he was to learn it apud Feynman".
A very short explanation for absolute non-scientific people (like part of my family) is the following. Imagine an electron. If you want to know where it is, you'll have to "see" it, so it must shoot away at least a single photon. When you see the photon, you can know where it originated from (ea location = fixed). By shooting off a photon however, it's speed has changed, and hence it's momentum. So you know where it was, but you don't have a clue where it will be heading. It's a very rough explanation, which rambles on all sides for people knowledgeable, but it gives a hint of explanation. Greetz, Leo
And you can only measure the position of that electron to within one WAVELENGTH of the light you are using. The more precise you want that position, the smaller wavelength light you will need to use- and the shorter the wavelength the more energy the photon has and will "kick" your electron harder.
Does electron slowing produce photons? Hi Leo. Thanks for your effort to make the HUP (Heisenberg's uncertainty principle) more understandable to us real people. Your argument may be very possible under the conditions you have postulated. However, although it is true that Heisenberg's electrons were singular, their location was limited to those that were very near atomic kernels; i.e., the unpaired valence electrons of the elements. In the interest of simplification I invite you to open the simplist of highschool Chemistry texts where you will see those probability balloons protruding from an atom; such balloons epitomize the proximity of HUP electrons to the quantum field radiating from the element's nucleus. Now about the conditions whereby an electron can be responsible for the emission of a photon. Normally, when an electron collides with matter, its momentum is transferred to that matter; however, when it is considered that that electron is stopped abruptly by the momentum well of the quantum field, its momentum transfer does effect the production of a single photon. Once stopped in a radial way, the electron is swept, in an annular/circular way into a paired-electron orbital whose momentum is in the complete control of the quantum field; if the electron is stopped at the position of an empty orbital, then because it is still single, it remains a target of HUP. Thanks for your audience and patience; Cheers, Jim
The only qualm I have about explaining the HUP this way is that it makes it appears as if the HUP is merely the result of our inability to make a measurement. We need smaller and smaller wavelength to pinpoint the position, but by doing that, we destroy the momentum of the particle we want to measure. While this is true, it makes it appears as if the HUP is really a measurement limitation. This certainly isn't the case since the HUP is a more intrinsic property of QM. I would suggest using the diffraction from a single slit. This is the clearest, in-your-face manifestation of the HUP. I think I've described this at least a couple of times, so I won't do it here and bore a bunch of people. It clearly shows that the HUP isn't the inability to make a measurement of the position and momentum, but rather the inability to make accurate predictions of a momentum as the knowledge of the position improves. This is an extremely profound implication of the HUP that needs to be understood. Zz. Edit: I was responding to the responses BEFORE NeoClassic posting. I am not going to touch what he wrote even with a 20 mile "momentum well".
Certain notions and physical quantities routinely measured and analyzed in classical physics aren't even defined for quantum systems and so you cannot talk about micro-systems in their terms. Even then if you try to do that, you can never get a level of accuracy higher than a certain minimum amount in those measurements. That is HUP puts a limit on classical analysis of quantum systems. spacetime www.geocities.com/physics_all/index.html
I have this medical condition where I can't sleep at night if I don't point out the difference between the observer effect and the uncertainty principle when needed. Really, can you blame me?