# How precisely the Klein-Gordon equation is derived?

In various articles (and books) such as the wiki article of the Klein-Gordon equation wrote:

"The Klein-Gordon equation is a "quantized" version of the relativistic energy-momentum relation";

In the article of "canonical quantization" has been written:

"The Klein-Gordon equation is the classical equation of motion for a free massive scalar field, but also the "quantum" equation for a scalar massive particle wave-function.";

I'd like to know,

1) What is exact name of the procedure of derivation of Klein-Gordon equation from the energy-momentum relation? (based on the above articles, i.e. by replacing quantities energy and momentum and so on with their corresponding quantum operators)?

2) Finally, the Klein-Gordon equation is a classical massive field equation, or a quantum mechanical massive wave-function equation, or both?

This question may sound elementary, however, I would appreciate if somebody clearly and simply answer it.

• my apologiee, I did comment thinking my books covered it in the way you asked about, but as you say, lots of authors just deal with the mechanics of it – user108787 Oct 18 '16 at 7:48

In Quantum Mechanics, the Schrödinger equation is just the statement that energy is the generator of time evolution. In the QM framework this is written as

$$H|\psi(t)\rangle=i\hbar\dfrac{d|\psi(t)\rangle}{dt}.$$

Now, if we have the position representation $\mathbf{r}$ we can form the wavefunction $\Psi(\mathbf{r},t)=\langle \mathbf{r}|\psi(t)\rangle$ and this becomes

$$\langle \mathbf{r}|H|\psi(t)\rangle=i\hbar \dfrac{\partial\Psi}{\partial t}.$$

The usual Schrödinger equation is found when we replace $H$ by the quantized classical hamiltonian:

$$H=\dfrac{P^2}{2m}+V.$$

The question is that the equation you get for $\Psi(\mathbf{r},t)$ is not Lorentz invariant. And indeed, we used the non relativistic energy when we quantized.

Now, the canonical way to do it, is to try quantizing the relativistic version

$$E^2=p^2+m^2,$$

in units where $c=1$. To quantize this we insist that energy is the generator of time translations. This suggests that $E\mapsto i\hbar \partial_t$ while we insist that $p$ is the generator of spatial translations so that $p\mapsto -i\hbar \nabla$. This leads to

$$-\hbar^2\dfrac{\partial^2\Psi}{\partial t^2}=-\hbar^2\nabla^2\Psi+m^2\Psi,$$

or also choosing units where $\hbar =1$

$$(\square+m^2)\Psi=0.$$

Here, $\Psi$ is a wave function, hence $\Psi:\mathbb{R}^3\times \mathbb{R}\to \mathbb{C}$ and hence, despite this strange terminology, $\Psi$ is a classical field.

So for $(1)$, we just quantized the energy momentum relation, by requiring that the same relation holds in the quantum version and imposing that energy is the generator of time translations and momentum the generator of spatial translations.

Now for $(2)$, the Klein-Gordon is a wave function equation. You are just rewriting Schrödinger's equation with a particular Hamiltonian. In the same way, it is a classical field. It is a classical field because it is not operator valued. A quantum field is one operator valued field. Now, talking about making it into a quantum field, that is, dealing with the quantization of this field is another story.

• Thank you, Nice! So here we have two different steps of quantizations. First we quantize a simple algebraic relation (containing quantities p, E and so on) just by replacing corresponding quantum operators (i.e. by insisting that energy and momentum are the generators of time and spatial translations, respectively); then after deriving the the wave-functional equation(s) which are a sort of classical field equation(s) by this way, we should quantize these classical fields to obtain their corresponding quantum fields (as the next step), right? – user129968 Oct 17 '16 at 22:13
• When people talk about "quantization" in the context of Quantum Mechanics what they mean is exactly promote position and momentum to operators, so that the canonical commutation relation is verified. That's what one means to quantize a Hamiltonian. It is that traditonal procedure used in QM. When one picks a field and makes it "operator-valued", this can also be seems as a type of quantization. Indeed, people sometimes refer to this as "second quantization". For instance, the Dirac equation from Relativistic QM has problems. When you quantize and make the Dirac field the problems are solved. – user1620696 Oct 17 '16 at 22:37
• Fine, so what you have done in your argument above is precisely the 1st quantization of energy-momentum relation? Apologize I don't have enough reputation to vote your answer. – user129968 Oct 17 '16 at 22:49
• The argument is a way to obtain the Klein-Gordon equation in the context of Relativistic QM. It was a way to try to make QM and SR work well together. It is based in the idea of quantization: turn the dynamical variables into the correct observables. Now, this terminology of 1st and 2nd quantization, I admit I don't know if it's very used these days. – user1620696 Oct 18 '16 at 0:59
• @user1620696 The terminology of first and second quantization seems to be falling out of favor. – David Z Oct 18 '16 at 7:38

The starting point is the representation theory of the Poincaré group (or actually the induced representation theory, and in particular the little group method of Wigner).

For a massive particle of spin zero and mass $m$, the spectrum of the momentum operator is the hyperboloid $p^2 = m^2$, with the energy condition $p^0 > 0$, sometimes denoted by $\Omega_m^+$. One of the advantages of this description is that one gains a genuine invariant measure rather than a just quasi-invariant one, on $\Omega_m^+$, given by $$\text d\Omega_m^+(p) = \delta(p^2-m^2)\theta(p^0)\text d^4p,\qquad\forall p\in\Omega_m^+.$$ The physical Hilbert space is then $H=L^2(\Omega_m^+,\text d\Omega_m^+)$ and clearly any element of this space satisfies the equation $$(p^2 - m^2)\phi(p) = 0,\qquad\forall\phi\in H.$$ The Fourier transform of this equation gives the Klein-Gordon equation $$(\square + m^2)\psi(x) = 0.$$

• @Phoenix87 Thank you. Is this procedure which you wrote, practically, equivalent to this that in relativistic energy-momentum relation one formally replaces p, E and so on by their corresponding quantum operators, and once she writes the "wave function" (which is really a field) behind these operators, she will get the massive free Klein-Gordon equation? – user129968 Oct 17 '16 at 21:44
• Quasi-invariant means that, after a transformation, you get a measure which is only equivalent to the starting one, but not exactly the same measure. The above argument can be seen as the mathematically precise way of supporting the argument that energy and momentum can be replaced by the corresponding differential operators in the relation $p^2-m^2=0$. – Phoenix87 Oct 17 '16 at 21:48
• @Phoenix87 Fine; in addition, just one more note. I read that the Klein-Gordon (and other similar equations such as Dirac equations and so on) is subject to an additional quantization as well, because it is not a complete quantum mechanical equation (?). Would you please explain this? – user129968 Oct 17 '16 at 22:00
• @Phoenix87 Yes, the question was about the second quantization. I see.. So by the above argument, in fact, we precisely obtain a single-particle massive field equation (which quantum mechanically can describe a single particle)? – user129968 Oct 17 '16 at 22:40
• In second quantisation $\psi(x)$ is a system of infinitely, but countably many quantum harmonic oscillators and therefore it is interpreted as a field rather than a wave-function – Phoenix87 Oct 18 '16 at 21:01

Strictly speaking, the Klein-Gordon equation is not a relativistic version of the Schrödinger equation. The stationary Klein-Gordon equation is obtained by replacing the relativistic pulse with the momentum operator $p_{relativistic} \mapsto -i\hbar\triangledown$ in the expression for the coupling of the energy and momentum of the STR. The Klein-Gordon equation has many drawbacks. For example, the erroneous value of the critical charge of the nucleus is Z = 68. If we act in another way and in the expression for the connection of the energy and momentum of the SRT, we replace the usual pulse by the momentum operator $p_{Nonrelativistic} \mapsto -i\hbar\triangledown$, then we obtain a completely different equation M2. Equation M2 does not have these drawbacks. More details on the derivation of the M2 equation can be found in the publication: http://vixra.org/abs/1609.0086