Undergrad Can we quantize a static EM field?

[fxdung]

Can we have a quantization static EM field?If not, how can we interpret static EM field in stand point of QM?

 

[PeterDonis]

Can we have a quantization static EM field?

If you mean, can a static EM field be described using QM, of course it can. That's how QM models the hydrogen atom, for example, and derives all of the detailed data on energy levels that has been confirmed by many experiments.

 

[atyy]

The static EM field is treated as a potential in non-relativistic QM.

 

[Delta2]

If you mean, can a static EM field be described using QM, of course it can. That's how QM models the hydrogen atom, for example, and derives all of the detailed data on energy levels that has been confirmed by many experiments.

You treat the static E-field of the nucleus as a classical field. I think he means if we can do a quantization procedure (like that of assigning a quantum harmonic oscillator to each point of the field) for a static EM field which i think we cannot since there is nothing that varies or oscillates with time.

 

[PeterDonis]

You treat the static E-field of the nucleus as a classical field.

No, in non-relativistic QM, you treat it as a potential in the Schrodinger Equation, as @atyy said. That's still doing quantum mechanics, not classical mechanics.

I think he means if we can do a quantization procedure (like that of assigning a quantum harmonic oscillator to each point of the field) for a static EM field

You can do this in quantum field theory. QFT can describe any EM field state.

 

[Demystifier]

An analogous, but technically simpler, question would be can we quantize a static particle at rest? So let's try it! We first quantize a free non-static particle (for simplicity, in one dimension) and find that their energy eigenstates are plane waves proportional to $e^{-i\omega t}e^{ikx}$, where $\omega=E(k)/\hbar$ and $E(k)=\hbar^2k^2/2m$. The static particle corresponds to $k=0$, so the wave function of a static particle is just a constant. So yes, we can quantize it, but the wave function is very trivial.

 

[Delta2]

You can do this in quantum field theory. QFT can describe any EM field state.

Take a look at this wikipedia article

Quantization of the electromagnetic field - Wikipedia

I don't know if it is well written, I personally don't seem to understand it very well, but at some point it replaces the coefficients $\alpha_k^{(\mu)}(t)$ with operators and it says something about that there must be time dependence in order to do this. Maybe I 've misunderstood.

 

[PeterDonis]

Take a look at this wikipedia article

If you want to learn QFT, you should be looking at textbooks, not Wikipedia.

The short answer is that the time dependence the article is talking about is not the same as the one you're concerned about. The article is talking about the difference between the Schrodinger picture, in which all of the time dependence is in the wave function, and the Heisenberg picture, in which all of the time dependence is in the operators. That has nothing to do with whether observables are time dependent or not (even in a stationary state, where no observables change with time, there is still a time dependence of phase that has to be captured somewhere).

 

[hutchphd]

If you want to learn QFT, you should be looking at textbooks, not Wikipedia.

Of course this is the best idea, but I found this wikipedia article to be interesting and useful.
https://en.wikipedia.org/wiki/Stati...le_exchange#The_Coulomb_potential_in_a_vacuum

But caveat emptor

 

[Delta2]

If you want to learn QFT, you should be looking at textbooks, not Wikipedia.

Yes I think you are right on this, many Wikipedia articles are written in such a style that if you previously have read on the subject from a good textbook, then you can understand what the article says, however if you are completely new to the subject and just read the Wikipedia article then you ll probably make a mess of your understanding.

The short answer is that the time dependence the article is talking about is not the same as the one you're concerned about. The article is talking about the difference between the Schrodinger picture, in which all of the time dependence is in the wave function, and the Heisenberg picture, in which all of the time dependence is in the operators. That has nothing to do with whether observables are time dependent or not (even in a stationary state, where no observables change with time, there is still a time dependence of phase that has to be captured somewhere).

And before I learn QFT I have to learn QM cause I have little to no clue of what the Schrodinger and Heisenberg pictures are. What's a good book to learn QM starting from complete unknown, with only thing provided a good mathematical background. I am considering Griffiths, or is there a better option?

 

[hutchphd]

If you really are going full tilt on this I would recommend shopping for a used copy of Sakurai for your shelf. For learning Griffiths is OK but I personally prefer Gasiorowicz; but that may just be familiarity from having taught from it. I don't think the edition matters much, but there may be solution sets for some

 

[PeterDonis]

What's a good book to learn QM starting from complete unknown, with only thing provided a good mathematical background. I am considering Griffiths, or is there a better option?

I personally like Ballentine, but there are lots of different opinions about QM textbooks. One of the issues is that, since there are many different QM interpretations, every textbook author tends to write with their own preferred interpretation in mind, so it's very hard to find a textbook that just teaches basic QM without any interpretation used or intended. (Ballentine's preferred interpretation is the statistical interpretation, and he devotes a significant amount of space to explaining why. Which is fine as far as it goes, but can be distracting if all you're trying to learn is basic QM and you aren't even ready for an interpretation discussion.)

 

[PeterDonis]

I have little to no clue of what the Schrodinger and Heisenberg pictures are.

The really short version is:

(1) Schrodinger picture: operators representing observables stay the same and the wave function changes with time.

(2) Heisenberg picture: the wave function stays the same, and operators representing observables change with time.

The Schrodinger picture is easier to grasp intuitively because that's how things work in classical physics: all the changes are in the state (which in QM is the wave function), and we assume that observables don't change (e.g., "measure the spin-z of this electron" is the same observable whether we do it now or next year).

However, the Heisenberg picture can be made to seem intuitive if you think of the state (wave function) as representing a preparation process. In other words, it basically takes the viewpoint: we prepared this quantum system in a certain way, and that remains true no matter how long we wait until we measure it. The possibility of something else changing depending on how long we wait to make a measurement is best captured, on this view, by making "measure the spin-z of this electron today" and "measure the spin-z of this electron tomorrow" different observables, just as "measure the spin-z of this electron" and "measure the spin-x of this electron" are different observables.

Of course all this assumes that we don't do any other state preparation in between--but that is true for both pictures, since in the Schrodinger picture doing a new state preparation means we need to change the wave function anyway and recompute everything starting with the new wave function.

 

[atyy]

What's a good book to learn QM starting from complete unknown, with only thing provided a good mathematical background. I am considering Griffiths, or is there a better option?

I like French and Taylor for a quick introduction to calculations before something like Griffiths or Gasiorowicz. Shankar is nice because he puts the postulates in terms of quantum states as vectors in a vector space quite early (more properly they are unit vectors or directions, since the length of the vector does not change the state). I think Ballentine is plain wrong on fundamentals, so should not be recommended unless one is already an expert.

 

[fxdung]

If QFT can describe static EM field, then can we take the notion of photon for static EM field?

 

[PeterDonis]

If QFT can describe static EM field, then can we take the notion of photon for static EM field?

You can use the notion of a virtual photon, with limitations. But a static EM field state is not the kind of quantum field state that has a useful interpretation in terms of real photons, no. This is an example of the more general fact that not all states of quantum fields have useful particle interpretations.

 

[vanhees71]

The static EM field is treated as a potential in non-relativistic QM.

This is the semiclassical approximation, i.e., the em. field is treated as a classical field. Of course you can quantize the field also within a non-relativistic "1st quantization" treatment of the particles. As in full QED of course the static fields are included within the formalism. For details, see

S. Weinberg, Lectures on Quantum Mechanics

Of course, it's not possible to quantize only the static field, because to quantize you need some dynamics in the sense of the Hamiltonian formalism.

 

[fxdung]

What do you mean when saying "1st quantization treatment of the particle"? Does "the particle" mean "matter particle" or mean "virtue photon" of static EM field?Why above you say: static EM field are included in QED theory, but later you say :it is not possible to quantize static field?

 

[vanhees71]

"1st quantization" means that you deal with position, momentum, and spin operators for each particle. "2nd quantization" means field quantization. For non-relativistic physics in situations where particle numbers are conserved both formalisms are equivalent.

I never use the expression "virtual particles". There's no such thing in the theory. All there is are the mathematical expressions in perturbation theory most economically written down in terms of Feynman diagrams. Internal lines in Feynman diagrams do NOT represent particles (i.e., asymptotic free states) but propagators.

It doesn't make sense to quantize something static. Quantum theory is a theory of dynamical entities, and the electromagnetic field is such a dynamical entity and can be quantized but you cannot quantize special solutions of the dynamical entities. It just doesn't make sense.

 

[PeterDonis]

later you say :it is not possible to quantize static field?

That's not what he said. The point he was making is that your original question, "can we quantize a static EM field", doesn't make sense, because what you are calling a "static EM field" is a particular solution to the equations that you obtain when you have already done quantization. It makes no sense to ask if you can quantize particular solutions, since you have to have already done quantization to even know what they are.

Note that in my response to you in post #2, I said "If you mean, can a static EM field be described using QM"--in other words, I reworded your question so it made sense. I did not answer the question exactly as you asked it--I couldn't, since it didn't make sense exactly as you asked it.

 

[fxdung]

Then can we apply creation and annihilation operator formalism and Feymann diagram for static field?

 

[PeterDonis]

Then can we apply creation and annihilation operator formalism and Feymann diagram for static field?

Yes. Most QFT texts will derive the Coulomb potential for the static case using this formalism. The Wikipedia article linked to in post #9 is actually doing a sketchy version of this.

 

[vanhees71]

The Coulomb potential is an approximation of the full interaction. You get it when considering the scattering of a particle whose energy is much smaller than the mass of another particle such that you can assume that the heavy particle just stays at rest and thus that the light particle can be treated as moving in the corresponding classical Coulomb field. To derive this you need to sum over infinitely many Feynman diagrams ("soft-photon ladders").