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).

 

[hilbert2]

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]

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?

https://en.wikipedia.org/wiki/Klein–Gordon_equation#Action

 

[jordi]

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

https://en.wikipedia.org/wiki/Klein–Gordon_equation#Action

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

But I am not asking about this issue.

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]

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]

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]

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]

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

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]

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.

 

[jordi]

Sure. And?

 

[PeterDonis]

And?

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]

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]

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

What wave equation?

 

[jordi]

What wave equation?

The wave equation: https://en.wikipedia.org/wiki/Wave_equation

 

[PeterDonis]

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]

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.

 

[jordi]

You are wrong. Look at Marcuse book and you will see it. For sure, Maxwell's equations imply a (the!) wave equation. But quantizing geometrical optics too!

 

[jordi]

In a sense, quantizing is "giving superpowers" (or giving the capacity to generalize behaviour outside of the regime applicable to the system to be quantized). For this reason, quantizing light rays results in the wave equation. It is the same in mechanics: the quantization of a classical particle results in "superpowers", i.e. being able to explore the particle at a regime unheard of before.

 

[PeterDonis]

which is the classical, non-relativistic Lagrangian, which results after quantization in the Pauli equation?

I don't think there is one; non-relativistic spin-1/2 particles do not have any classical analogue that I'm aware of, as I said before. Note the word "particles", though; spin-1/2 fields can be written down classically, and you would indeed use Grassmann variables for a path integral involving spin-1/2 fields, as described, for example, in Zee's book Quantum Field Theory in a Nutshell. But doing this is inherently relativistic, in the sense that you are working with representations of the Poincare algebra; I'm not aware of any non-relativistic way to do it.

 

[PeterDonis]

Look at Marcuse book

I don't have it. Do you know of any online reference?

 

[jordi]

Unfortunately not, I found the geometrical optics quantization in Marcuse book just by serendipity. I have never ever found this derivation anywhere else.

But you can see it indirectly, but beautifully, in:

https://arxiv.org/abs/math-ph/0405032
This paper describes the general solutions of a wide variety of PDEs (including the wave equation) as path integrals. So, if the general solution of the wave equation is a path integral (i.e. summing over all paths), this means the wave equation is the quantization of "something".

Since Marcuse book gives what is this something, now everything fits: the wave equation is the quantization of light rays.

But even without having Marcuse result, and only having the paper I have given to you, you can see that it is clear that the wave equation is the quantization of "something".

 

[PeterDonis]

This paper describes the general solutions of a wide variety of PDEs (including the wave equation) as path integrals.

Very interesting! I was not aware of this result.

Marcuse book gives what is this something

Can you describe how this result is obtained? What I'm still having a problem with is that (as the paper you linked to shows for the particular example of the wave equation on $\mathbb{E}^{(1, 3)}$) the wave equation is a spacetime equation--it relates the second derivative with respect to time to the second derivative(s) with respect to space. But the geometric optics Lagrangian is a Lagrangian in space; time does not appear. And, as I've said, its usage in geometric optics is to describe the spatial paths of light rays. So how can a path integral using this Lagrangian lead to a spacetime equation?

 

[jordi]

With the Lagrangian I have given to you, if you compute its Hamiltonian, you see the Hamiltonian is the square root of a number minus the sum of the squares of the momenta.

By taking the usual quantization rules (promoting the momenta and the Hamiltonian to operators, both position and time, resp.), you immediately get the wave equation with a mass term (the K-G equation).

With this argument, I realize it is not exactly the wave equation which results from the quantization of geometrical optics, but K-G equation. This is somewhat mysterious.

 

[PeterDonis]

With the Lagrangian I have given to you, if you compute its Hamiltonian

How can I do that if there's no time appearing anywhere?

 

[PeterDonis]

How can I do that if there's no time appearing anywhere?

Unless you mean $x$ to play the role of time?

 

[Demystifier]

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?

What you look for is the path integral for a relativistic first-quantized particle. It can be found in some string theory textbooks, but also in the short paper attached here.

 

[jordi]

Unless you mean $x$ to play the role of time?

Yes, that is the solution to the riddle. The Hamiltonian corresponds to the z coordinate (not time!), so the resulting equation is not the K-G (wave equation plus mass) but the Helmholtz equation (Laplace equation plus mass).

And the argument is the Helmholtz equation is just the wave equation once the time part of the wave equation is variable-separated (in the same way the Schrödinger equation becomes the energy eigenvalue problem).

 

[PeterDonis]

that is the solution to the riddle

Here's what I get for the Hamiltonian, if I rewrite your Lagrangian using $t$ instead of $x$ and $q$ instead of $y$ (with $y'$ becoming $\dot{q}$), and using units where $c = 1$ to reduce clutter:

$$
L = n(t, q) \sqrt{1 + \dot{q}^2}
$$

First obtain the conjugate momentum:

$$
p = \frac{\partial L}{\partial \dot{q}} = n(t, q) \frac{\dot{q}}{\sqrt{1 + \dot{q}^2}}
$$

Invert this formula to obtain $\dot{q}$ in terms of $p$ (I'll stop showing the functional dependence of $n$ explicitly at this point):

$$
\dot{q} = \frac{p}{\sqrt{n^2 - p^2}}
$$

Rewrite the Lagrangian in terms of $p$ instead of $\dot{q}$ to facilitate deriving the Hamiltonian:

$$
L = n \sqrt{1 + \frac{p^2}{n^2 - p^2}} = \frac{n^2}{\sqrt{n^2 - p^2}}
$$

Now derive the Hamiltonian:

$$
H = \dot{q} p - L = \frac{p^2}{\sqrt{n^2 - p^2}} - \frac{n^2}{\sqrt{n^2 - p^2}} = - \sqrt{n^2 - p^2}
$$

 

[PeterDonis]

the Helmholtz equation is just the wave equation once the time part of the wave equation is variable-separated (in the same way the Schrödinger equation becomes the energy eigenvalue problem).

But you can't run this backwards; you can't derive the full time-dependent Schrodinger equation from the energy eigenvalue problem, and you can't derive the wave equation from the Helmholtz equation. You can only go in the other direction. By the logic you're giving here, the Lagrangian for a classical free particle should give us the Schrodinger energy eigenvalue problem, not the time-dependent Schrodinger Equation.

 

[jordi]

Exactly. Now square the Hamiltonian, and perform the usual substitutions (number to operator) for quantization.

 

[PeterDonis]

Now square the Hamiltonian, and perform the usual substitutions (number to operator) for quantization.

Yes, and when I do that, I get (in the position representation) the operator $\nabla^2 + n^2$. Which, as you have said, is the Helmholtz equation, not the K-G equation. There's no time derivative anywhere.

 

[jordi]

What you look for is the path integral for a relativistic first-quantized particle. It can be found in some string theory textbooks, but also in the short paper attached here.

Polyakov's Gauge Fields and Strings, chapter 9.11, seems to give an answer in (9.325): the Dirac equation can be obtained by quantizing a supersymmetric particle action. Really surprising (to me).

 

[jordi]

And in fact, I like this idea of using the action given by the "path distance", since this is equivalent to trying to get as much as we can from Fermat's principle (principle of minimization of time).

In a sense, the field equations (K-G, Dirac, wave equation, Schrödinger equation ...) are taking the Fermat's principle, and extract everything possible, including using the method of quantization (i.e. superpowers).

Once it is clear, by relativistic arguments, that a field is not enough, one needs to abandon particles and go into fields (and then quantize again!).

 

[jordi]

From Green, Schwarz and Witten:

"The innocent idea of a light ray in Minkowski space should be viewed as a hint of Yang-Mills theory and general relativity. The 'proper' theory of the massless point particle - namely the supersymmetric theory - gives rise upon quantization to the Maxwell and linearized Einstein equations, and these should then be generalized to the nonlinear Yang-Mills and Einstein equations."

Amazing! This is even more than I dreamt it could be true. By taking supersymmetric theories, one can get, after quantizing them (i.e. after giving them superpowers), all the classical field theories (not only Yang-Mills, Dirac and K-G, but also Einstein)!

 

[jordi]

"The innocent idea of a light ray in Minkowski space should be viewed as a hint of Yang-Mills theory (...)": mmm, if to the Lagrangian of geometrical optics one adds, by hand, Minkowski space, i.e. a minus derivative squared, this would result in the wave equation including time, right?

 

[jordi]

An interesting consequence of this view is that once particles "have given everything they could" (basically, the field equations), and field equations are experimentally proven to be not enough to describe reality, one has to move away from the particle viewpoint. But it is good to know that good ol' Fermat's principle has been so useful.

But since we want to keep as much as we can from the past, there are basically two ways to go ahead:

- Give superpowers to what already you have given superpowers before, i.e. quantize field theory
- Add one dimension to the particle, i.e. create a string, and give superpowers to it, i.e. first quantize string theory

Of course, one could give superpowers to the first-quantized string theory, i.e. do string field theory.

A fourth option would be to give superpowers to quantum field theory (i.e. to give superpowers to the superpowers of the superpowers), but I do not know how this could be done.

 

[HomogenousCow]

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.

In principle all boson fields (Higgs, W, Z) can have classical states (coherent states) the same way the electromagnetic field does but I think they're just strongly suppressed by their masses. However the "classical field" of the pre-SSB Higgs field is DEFINITELY present and very measurable since it gives mass to fermions.