# How is an electric field quantized?

1. Feb 14, 2015

I have asked this question before on various fora, but I never got a really satisfactory answer. Perhaps physicists simply don't know...

Anyway I'm asking again; an electric field emanating from, say a small pointed metal electrode, is for many practical purposes a continuous field that is "infinitely differentiable". To me this means that if you are at a distance of e.g. 3 cm from the electrode and then move 10 Å further from the electrode the field strength will diminish exactly in proportion to the extra distance. However I have an intuitive feeling that this is not really true; that there are really "steps" in the field. You move 5 Å and nothing happens to the field strength and then at 12 Å the field changes abruptly by an integer multiple of something.

So am I right or wrong? What are you comments? Is my intuition correct about quantization of fields? Is it possible to put numbers on this? Include Planck's constant in some formula?

S.A.

2. Feb 14, 2015

### Staff: Mentor

They know, but it takes years of study to understand quantum field theory.
There are not. At least there is absolutely no theory predicting such a thing and no experiment ever measured something like that. But this has nothing to do with quantization of fields as used in physics.

3. Feb 14, 2015

### Staff: Mentor

You are wrong. There is zero evidence for such a view.

The quantization of fields is no mystery. Its analogous to what goes on in classical field theory. You take a field and approximate it as a mechanical system where the value of the field is discretised (the values are treated as large number of 'lumps') with each discretised value treated analogously to a particle in a classical mechanical system (technically that means its described by a Lagrangian). But the field is not really discretised (which is your incorrect assumption) so one takes a limit to get a continuum and you end up with what's called Lagrangian Field Theory:
http://www.staff.science.uu.nl/~wit00103/ftip/Ch01.pdf

Quantum Firld Theory (QFT) does exactly the same thing - except its treated analogously to a quantum mechanical system (this is done by the Poisson bracket relations that Dirac worked out in the early days of QM):

There are other ways of proceeding as well - that's just one way - but all roads lead to Rome.

The interesting thing about QFT is, without going into the details, by doing that particles emerge in a natural way from the field - but you need to study the full theory to see why. Just as a hint if you do a Fourier transform on the field you end up with an integral where what is integrated mathematically is very similar to a Harmonic oscillator which is very interesting when quantised:
http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

Its behaviour is described by creation and annihilation operators. Its not generally talked about but QM can in fact be formulated in a number of equivalent ways:
http://math.bu.edu/people/mak/Styer Am J Phys 2002.pdf

One of those ways, the second quantisation formulation in the above link, is exactly what you get from QFT. In other words particles and fields are treated in a unified way.

Thanks
Bill

Last edited: Feb 14, 2015
4. Feb 20, 2015

Thanks for an interesting reply! I'm asking questions about matters that I don't really understand, yet. Hope to understand them one day when I have more mathematics. Eh, where was I, ah yes quantisation of electric fields. OK, the field itself is not "steplike", but would the particles, say protons experience a continuous acceleration when accelerated by that field? Or would acceleration be in "steps"?

5. Feb 20, 2015

### Staff: Mentor

In quantum mechanics, you cannot represent acceleration with a single number any more. But there are no steps involved.

6. Feb 21, 2015

### Staff: Mentor

QM is a theory about observations. What its doing, accelerating etc when not observed the theory is silent.

So the answer to your question is the theory doesn't say anything one way or the other.

Thanks
Bill

7. Feb 21, 2015

Acceleration should be observable! It's not metaphysics, how fast a particle travels! Perhaps, with a slightly different setup of QM and a carefully thought out experiment, step-like behavior in acceleration can be proven?

Best regards,

S.A.

8. Feb 21, 2015

### Staff: Mentor

Velocity can be observable, acceleration itself is not. There is no measurement that directly gives acceleration.
There is absolutely no reason to expect a step-like behavior. Why do you expect an experiment to find one?

9. Feb 21, 2015

### Staff: Mentor

Acceleration is not how fast a particle travels. Its the instantaneous change in how fast a particle travels.

To be specific one can find a velocity operator reasonably easily (see chapter 3 Ballentine) - but an acceleration operator is more difficult and dependant on the Hamiltonian - see page 56 - Landau - Quantum Mechanics

Thanks
Bill

Last edited: Feb 21, 2015
10. Feb 26, 2015

What would an acceleration operator look like and what would be the solution to the simplest of problems? Say a particle inside a box, with a or without a potential or a two body problem like the hydrogen atom?

Thanks BTW for taking the time to answer some strange questions from a novice!

Best regards,

11. Feb 26, 2015

### Staff: Mentor

12. Feb 26, 2015

### Spinnor

You might turn your question into an assumption and then ask how would physics in say the hydrogen atom be any different? Try and come up with a potential that is like 1/r but not quite, something that models your idea. Then solve for the spectrum of the hydrogen atom using your modified potential and compare with experiment. But theory already works really, really good with the 1/r potential. Experts here could tell you how good.