# If all particles are fields, why does first quantization work for some particles?

After a lot of Google and asking professors about the two quantization methods, I have learned that first quantization is what you use to quantize classical particles, while second quantization is what you use to quantize classical fields or a large number of classical particles.

If, according to QFT, every particle is an excitation of a field, then why does the first quantization method work for some particles? Is it an approximation that works because the wavelengths involved are much smaller for, say, an electron than a photon?

If that's not the right question because I'm wrongly assuming that you shouldn't be able to use 1st quantization for fields, then why can't you use it for the electromagnetic field? I mean, you can to an extent, but it feels forced, the whole reason being Gauss' law forbidding a position eigenstate, but why does something similar not happen for other particles' fields?

I'm interested in why the localization problem applies to the photon and not to, say, the electron: Why can you (in principle) exactly locate an electron in space, if you can't do it for the photon? They are both fields, shouldn't they follow similar rules? Obviously they don't, so what is the difference?

• You're asking us to explain quantum field theory, and that's far too broad a question to be usefully answered here. It is very hard to get an intuitive understanding of how QFT works because it's complicated, so I'm not surprised your Googling has left you mystified. If you can get your rep score up to 20 or more you can join the Physics chat room and ask there. Jan 21, 2017 at 7:21
• I think my question is specific enough for someone to give a rough explanation. Maybe the electron field is scalar and thus there is no flux to be maintained to zero (I heard something about it being a 2-spinor, but that doesn't mean much to me). Even if the answer is too technical for me to understand completely, at least I could get a flavour of why they are such different fields. Jan 21, 2017 at 7:41
• I know that in some cases second quantization is just a change of notation, but if that was always the case, then first quantization should be able to describe every single thing that second quantization can, which I don't think is true. And the other answers agree, saying that first quantization is a low-energy non-relativistic approximation. Jan 21, 2017 at 15:04
• Hi Pato. Your edit is an interesting question but I would ask it seperately . Jan 22, 2017 at 6:40
• @JohnRennie, That would make this question a bit incomplete. Should I just edit the whole thing? Jan 22, 2017 at 9:38

First quantisation cannot accommodate pair-production, which is non-negligible at energies comparable to the mass of the particle. Therefore,

First quantisation only works for massive particles, and only in the range of energies that make motion non-relativistic.

Of course, this doesn't mean that a massive particle can always be modelled in a first quantisation scheme: for example, another requirement is that the particle is stable, and unaffected by confinement, etc.

Towards the edit, let me stress that the localisation of particles in a relativistic context is a very subtle issue (cf. this and this PSE questions). A nice reference for this problem is No place for particles in relativistic quantum theories?, by H. Halvorson and R. Clifton.

• @PatoRaimundo at high energies you have to use fields to describe the electron. But there are no weird effects, just QFT. It's a complicated subject but perfectly logical. Jan 21, 2017 at 10:54
• I'm mainly interested in the fact that photons can't have a well-defined position, while low-energy electrons can. What is the fundamental cause of that? Why are both fields so different in that respect? I know Gauss' law forbids an electric field delta function, but why doesn't something similar happen to the electron? Jan 21, 2017 at 11:03
• @PatoRaimundo But generic electrons don't have a well-defined position. Even in an introductory quantum mechanics course, the wavefunction is not a delta function. Jan 21, 2017 at 19:44
• @knzhou Not always, but in theory it could be a delta function. Especially after a position measurement. That's not the case with photons. Jan 21, 2017 at 20:12
• @PatoRaimundo A perfect position measurement doesn't exist. If you want to just consider an incredibly precise position measurement, it would be impart so much energy to the electron that the nonrelativistic limit wouldn't even make sense anymore. Jan 21, 2017 at 20:13

AccidentalFourierTransform gives you physical reasons but first quantization also came about partially for both historical and pedagogical reasons.

If you take the electron, a hundred years ago it was cutting edge science to match up the predictions of the "weird" quantum world to what was "obvious" in the classical world.

In something similiar to the simple (but misleading) Bohr semi-classical picture of the atom, first quantization makes certain simplifying assumptions and mixes classical and quantum descriptions.

In first quantization, some physical properties, such as electric or magnetic fields, and the potential wells associated with them, are treated classically, but their effects are seen as changes in the quantum description (utilising wave functions and matrices) of particles, especially the electron.

Actually, imo there is no other real alternative to learning QM basics, except by treating quantum systems as they are done in the first quantization method. As John says in the comments, otherwise you would need to know more about QED and QFT. If you were treating a system of say, an electron in a magnetic field without the use of first quantization, it would be a confusing chicken and egg like affair, with the need to explain the $E$ and $B$ fields in field terms.

Its much easier to learn the basics of QM this way, and allows for the particle's motion and spin to be dealt with in the first "explanation" of how a particle behaves quantum mechanically and without the distraction of having to explain non essential "external" effects.

Second quantization then appears after you upgrade from the Schroedinger equation to the Dirac equation, with it's built in description of spin, particle production and the later introduction of the field concept as the basis for modern particle physics.

• Well, I like to make sense of what I learn. If I "just use it", I will probably forget it in a couple of months. I want to know what am I using and why it works, otherwise it's just "monkey see, monkey do". And while that's inevitable when first learning something, I want to get out of it as quickly as possible. Jan 21, 2017 at 11:17
• Sorry, my first paragraph is badly worded, I didn't really mean ........anyway, I will change it, the more I try to explain what I meant, the the worse it gets:) You are dead right in what you say. There is also canonical quantization, which converts classical equations to QM versions.
– user140606
Jan 21, 2017 at 11:50

Localisation of electrons and photons are based on very different concepts. In fact, it is possible to construct a wave-function for an electron (under the caveats mentioned below) because even with spinors (what is used to represent electrons) you can construct a quantity which behaves under the Lorentz transformation as the zeroth component of a 4-vector (a necessary condition to fulfill the continuity equation).

But for a photon, which is described by a 4-vector potential $A_\mu$, such a quantity cannot be found. Partly because they are bosons, but mainly because they always move at the speed of light (you cannot choose an system of reference in which the photon is at rest, while for the electron you always can), which makes their localisation impossible (except perhaps in geometrical optics, but that's a very coarse theory, so the localisation is also very coarse). Not to mention that photons can only be described in a relativistic quantum theory, where the concept of wave function has lost its meaning because it violates causality. That problem does not occur in non-relativistic QM where instant interaction is assumed.

Finally, the quick summary of the difference between the quantization methods is that in 1st quantization you replace the real 4-vector $p^{\mu}$ of a particle with a differential operator $i\nabla^{\mu}$, which acts on the one-particle wave function; whereas in 2nd quantisation, the momentum and energy of your system are contained in operators on the Fock space (suited for dealing with many particles). They consist of creation and annihilation operators, so there aren't any more differential operators.

{\bf Edit:} I've a couple of precisions to make. Actually, being a boson does not prevent the particle from being localized, that's not fully true. In a limited sense they can be still localized. Charged pions, for instance, are bosons, and a probability current $j^{\mu}$ can be constructed which fulfills the continuity equation $\partial_\mu j^\mu=0$. But already in this case appear problems, the zeroth component $j^0$ is not positive definite. Whereas in case of the photon field this current cannot be constructed at all, shortly there is no photon current at all.

• Wow, thanks for taking the time to respond, but I'm afraid it has been lost in me. I feel like your answer is the most specific one, but also the most obscure. If you don't mind I'll try to edit it a bit for clarity (you can change it back if you don't like it), but I don't think I have enough background to understand what you are explaining Jan 25, 2017 at 0:35
• So almost 5 years later, I just randomly remembered this. Now I find the opposite problem: I would like more details. I see now that you're saying that there is a conserved current associated to electrons, but not to photons, which means that photons have no conserved quantities (which is why photon number is not conserved). I also see now that you stated the result "If there is no rest frame, localization is impossible", but not how to reach that conclusion. Could you please share it? I'd very much like to revisit this. (Also, I can make a new question if you want to give a full answer) Oct 18, 2021 at 8:20
• I don't have proof in the proper sense. But in wellknown experiments like the double-slit or in an interferometer where the photon beam is splitted up in 2 parts we don' know the way the photons takes. Attempts to localize it destroys the experiment. 2) some particles, like electrons or neutrons and may be muons as well can be held in traps but for a photon it is impossible. Such a trap should move with the photons and this is simply not possible due to SR. A photon always moves with c WRT such a trap. I think mathematically this is expressed by the non-existence of a photon current. Oct 18, 2021 at 21:26
• But the double slit experiment isn't exclusive to photons, just easier (can be done with a laser pointer and tin foil). I assume the same applies to interferometers. I understand photons can't be trapped (at rest), but how do you reach the statement "localizability implies trapability"? It's the same statement as before, worded differently. And if I'm reading you correctly, you're also saying that objects travelling at c can't have conserved currents. So massless particles can't have conserved charges. Now, I don't know QCD, but afaik gluons are massless and color charge is conserved. Oct 25, 2021 at 6:12
• If you want a full-fledged proof, I invite to post a new question. Then you get the opportunity of answers of other people. As I already said I don't have a proof in the proper sense. But what gluons concerns, a gluon current on its own is not conserved. One has to add a quark current contribution. Oct 25, 2021 at 8:19