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

[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 Schrödinger 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 Schrödinger energy eigenvalue problem, not the time-dependent Schrödinger 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.