# A Is the Klein-Gordon equation a quantization of classical particles?

#### jordi

The Schrödinger equation can be derived from the path integral quantization of the Lagrangian of classical, non-relativistic particles.

Can the Klein-Gordon (and maybe the Dirac) equation be derived from the path integral quantization of a given classical (supposedly relativistic) Lagrangian of particles? If so, which Lagrangian?

Usually, the Klein-Gordon equation is derived by taking the energy dispersion equation of special relativity, and promoting the energy and the momentum to their respective differential operators, in the same way the Schrödinger equation is derived from the energy dispersion equation of non-relativistic classical mechanics.

But as said, there is a second derivation of the Schrödinger equation, coming from the path integral quantization. I have not seen the same for the Klein-Gordon equation. Even worse for the Dirac equation, which is not derived from any energy dispersion equation (granted, it satisfies the energy dispersion equation when applied twice).

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#### hilbert2

Gold Member
The KG equation isn't quantized at all, as it is. It's like the classical equation of waves on a string, except that the solution can be complex-valued. After quantization it describes a system with a possibly uncertain number of boson particles. The "uncertain number" means here that, e.g. there's a 50% chance to find 1 boson from the system and 50% chance to find two bosons.

#### jordi

Well, let me put it in another way: the Schrödinger equation can be considered also as a classical field equation. And it continues being true that, after path integral quantization of classical particles, the Schrödinger equation "appears there".

In the same way, is there any classical Lagrangian, such that after path integral quantization, it results into the K-G equation?

#### PeterDonis

Mentor
is there any classical Lagrangian, such that after path integral quantization, it results into the K-G equation?
Wouldn't that just be the Klein-Gordon Lagrangian?

#### jordi

Wouldn't that just be the Klein-Gordon Lagrangian?

No. Of course, the KG equation is derived, "classically" (via the Euler-Lagrange field equations) from the KG Lagrangian.

What I am saying is that quantization can be understood, loosely, as "going one level up": a classical, non-relativistic mechanical Lagrangian is defined on paths. The Euler-Lagrange equations "choose" a path (the classical path). But by quantizing this classical system, one goes one level up. You can see it from two ways: by path integral, since you are summing over all paths, in fact you are going one level up geometrically, i.e. you are going to the whole space (all paths in the space are "the same" as the space itself). By the canonical quantization, you are going one level up since your end result is a field (which is "the same" as "gluing" all paths in the same space).

As a consequence, it makes all the sense in the world that quantizing a classical particle results in a classical field theory, the Schrödinger equation in this case.

The same happens with electromagnetism. The wave equation can be understood as the quantization of geometrical optics. Here we find exactly the same situation: "classically", light rays are paths. By quantizing them, one sums over all paths, i.e. one fills the space. It makes all the sense in the world that the quantization of light rays results in a field equation, and this is what the wave equation is.

This idea of "quantization is going one level up" also allows us to understand second quantization / field theory: relativistic field theories (e.g. KG, Dirac ...) have a problem, even being "one level up" already: a relativistic field "contains" multiparticle states (you can see this from the path integral: all classical field configurations contribute "a little" in the sum of the path integral, irrespective of how many particles there are in the bra/ket which is calculated, since by relativity, if a particle has enough energy, it can create new particles ex-nihilo).

As a consequence, if we want to calculate empirical numbers, we need to calculate S-matrix components, which have a definite particle definition. So, we cannot just keep one field (the solution of a field equation), we need to sum over all possible fields, such that the contributions of all fields result in a definite particle amount. So, in QFT we are "two levels up" (we have quantized "twice").

But in order this picture outlined above works, it would be necessary that both the KG and the Dirac equations can be derived as field equations of a classical (possibly non-relativistic) Lagrangian, via path integral quantization (in the same way as the Schrödinger equation and the wave equation are field equations result from the quantization of paths, via the analogy "all paths in the space = space itself").

I assume the KG should not be too difficult, since the KG is just the wave equation with a mass term (maybe trying the action being the relativistic length?), but for the Dirac equation, I have no idea where to start.

#### PeterDonis

Mentor
The same happens with electromagnetism.
The K-G equation is quantized the same way that Maxwell's Equations are quantized (in fact it's easier because spin-0 is simpler than spin-1). So if you accept that the path integral works for electromagnetism, why wouldn't it work for the K-G equation? I'm afraid I don't understand what the issue is here.

#### jordi

The K-G equation is quantized the same way that Maxwell's Equations are quantized (in fact it's easier because spin-0 is simpler than spin-1). So if you accept that the path integral works for electromagnetism, why wouldn't it work for the K-G equation? I'm afraid I don't understand what the issue is here.
Well, I need to find the equivalent of light rays (the classical Lagrangian which is quantized to result in classical electromagnetism) for KG and Dirac. I do not know what these classical particles are (i.e. which classical particles result, via quantization, in the KG and Dirac field equations).

#### PeterDonis

Mentor
I need to find the equivalent of light rays (the classical Lagrangian which is quantized to result in classical electromagnetism)
Which classical Lagrangian do you think this is?

#### jordi

The Lagrangian of geometrical optics. You can see this in Zeidler vol. 2 of the QFT series, and the quantization of geometrical optics (resulting in the wave equation) in Marcuse's Light Transmission Optics.

#### PeterDonis

Mentor
The Lagrangian of geometrical optics.
Please write it down explicitly. I don't have Zeidler's textbook, and the PF LaTeX feature is easy to use.

#### jordi

$$L(x,y,y') := \frac{n(x,y)}{c} \sqrt{1 + y'^2}$$

#### jordi

Some random idea: could the Dirac equation be the quantization of the Pauli equation? Edit: no, it cannot be, since the Dirac equation is relativistic and the Pauli equation is not. But maybe a relativistic generalization of the Pauli equation could work.

#### PeterDonis

Mentor
$$L(x,y,y') := \frac{n(x,y)}{c} \sqrt{1 + y'^2}$$
This just looks like the Lagrangian used in Fermat's principle; i.e., you're doing a path integral in space, not spacetime. I thought we were talking about path integrals in spacetime; that's not the same thing.

#### jordi

This just looks like the Lagrangian used in Fermat's principle; i.e., you're doing a path integral in space, not spacetime. I thought we were talking about path integrals in spacetime; that's not the same thing.
The Schrödinger equation also results from a path integral in space. The difference between space and spacetime is a relativistic one, not a quantum one per se. Path integrals work in both cases.

#### PeterDonis

Mentor
The Schrödinger equation also results from a path integral in space.
No, it doesn't. You're integrating a Lagrangian that is a function of time (though the free particle Lagrangian turns out not to be explicitly a function of time), position, and the time rate of change of position, over an interval of time. That's a spacetime integral, even though it's non-relativistic.

Sure. And?

#### PeterDonis

Mentor
And that means that the Lagrangian you gave for geometric optics, as far as I can tell, has nothing to do with how electromagnetism is quantized. The path integral you refer to for geometric optics is a classical path integral, not a quantum one. It's a very useful formulation of classical electromagnetism in the geometric optics approximation; but it does not lead to a path integral formulation of quantum electrodynamics, at least not that I'm aware of. (As I said, I don't have the textbooks you refer to; from what I can gather about them online, they cover geometric optics classically, and rely on the derivation of geometric optics as an approximation in classical electrodynamics, but don't quantize geometric optics directly.)

The Lagrangian that leads to the photon part (i.e., the part that describes light by itself) of quantum electrodynamics is the Maxwell Lagrangian:

$$L = - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}$$

This is the Lagrangian of a spin-1 field, not of any kind of particle.

#### PeterDonis

Mentor
I do not know what these classical particles are (i.e. which classical particles result, via quantization, in the KG and Dirac field equations).
As far as I know, there aren't any. The K-G equation describes, quantum mechanically, a spin-0 field (or, classically, something like waves on a string, as @hilbert2 said earlier). the Dirac equation describes a spin-1/2 field, something for which there is no classical analogue that I'm aware of.

#### jordi

What I am saying is that quantizing geometrical optics results in a ("classical") wave equation. Classical electromagnetism can be proved to result in a wave equation, so this is why I argue that classical electromagnetism is the quantization of geometrical optics.

Here, quantization is "going one level up", not necessarily having "quantum effects". For example, quantizing a classical particle is going one level up (we use all paths in the space, i.e. we use all space), and also it implies having quantum effects.

The quantum effects of light result from the second quantization (here really understood as second, after a first quantization!).

#### PeterDonis

Mentor
What I am saying is that quantizing geometrical optics results in a ("classical") wave equation.
What wave equation?

#### PeterDonis

Mentor
The wave equation
I know what a wave equation is. I'm asking you what specific wave equation you think you get from the Lagrangian of geometric optics.

Whichever one it is, it isn't Maxwell's Equations (since you get those from the Lagrangian I gave earlier), which invalidates this argument you give:

Classical electromagnetism can be proved to result in a wave equation, so this is why I argue that classical electromagnetism is the quantization of geometrical optics.

#### jordi

Sure, my statement was an abuse of language. Maxwell's equations imply the wave equation, but they are "more than only" a wave equation. This is self-understood.

What I state is that quantizing geometrical optics results in the wave equation. Of course, it is not clear what the function what is solution of the wave equation is (in the same way it is not clear what the function which is solution of the Schrödinger equation is; one needs further arguments), for the specific case of light (it is clear what it is for a string, for example).

Then, to interpret the solution of the wave equation in terms of light, one needs Maxwell's equations, sure. It is not "automatic" and self-implied by the wave equation.

But again: once we understand (with other means) that Maxwell's equations imply a wave equation for the electromagnetic fields, then everything makes sense: light rays in geometrical optics have been quantized, i.e. now we do not have a single ray, but all rays in space. And this is Huygens' principle, basically. So, quantizing geometrical optics "is" Huygens' principle, i.e. wave behaviour.

#### jordi

Maybe a simpler question: which is the classical, non-relativistic Lagrangian, which results after quantization in the Pauli equation? Here the analogy with the Schrödinger equation is clear: a classical, non-relativistic Lagrangian for classical particles results, after quantization, in the Schrödinger equation. The Pauli equation is the equivalent of the Schrödinger equation, but with spin one half particles. Is there a classical Lagrangian that results in the Pauli equation after quantization? Grassmann variables maybe?

#### PeterDonis

Mentor
What I state is that quantizing geometrical optics results in the wave equation.
And again I ask, what wave equation? It isn't the usual one that you derive from Maxwell's Equations, because that one, as I've already said, comes (via Maxwell's Equations) from the Lagrangian I gave earlier.

In geometric optics presentations I've seen, there are equations like the eikonal equation, which is an approximation to the wave equation that is derived from Maxwell's Equations. But that equation, like all equations in geometric optics, describes the spatial paths of light rays. It does not describe the rays in spacetime, which is what the wave equation derived from Maxwell's Equations does.

So I do not agree that quantizing geometric optics results in "the wave equation", because to get the wave equation that is derived from Maxwell's Equations, you have to use the Maxwell Lagrangian that I wrote down earlier, not the one you wrote down. The one you wrote down can only give you things like the eikonal equation, and while you can derive the Eikonal equation from Maxwell's Equations, you can't go the other way: you can't get electromagnetic wave optics from geometric optics, you can only get geometric optics from EM wave optics.

"Is the Klein-Gordon equation a quantization of classical particles?"

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