# Derivation of the Klein-Gordon equation [closed]

My question arises from studying section 8.1.3 of Sakurai. I am confused on the way Klein–Gordon equation is created. In Sakurai, the book said that it was derived by taking another time derivative to the Schrodinger equation to obtain a covariant derivative form (as explained in equation 8.5 and 8.6 of Sakurai). This derivation looks to me as stating implicitly that Schrödinger equation works in the relativistic case. However, we knew that the Schrödinger equation is not compatible with special relativity. Even though Klein-Gordon did explain some experiments (as stated in 8.1.4), I don't understand how this derivation is legal, and I am wondering whether the strong limitation of Klein–Gordon equation is related to this ill-defined presumption.

• To people with the same question, there is an alternative source that may be useful: Chapter 1 and 2 of "Relativistic Quantum Mechanics" by Greiner. I found it in this post: link Commented Jun 6, 2023 at 2:23
• While Greiner's series are wonderful and he definitely knows his stuff, the whole conception of RQM is indefensible. One cannot have RQM without being able to create and destroy particles (this is also true of statistical thermodynamics) and so only in QFT may we have a tolerable union of SR and QM. Commented Jun 6, 2023 at 16:44

No, the issue with the Klein–Gordon equation does not come from this derivation, and the Klein–Gordon equation actually predates the Schrödinger equation (to the best of my knowledge, Schrödinger got it first, but discarded it due to the issues it predicts).

The difficulty with the KGE has to do with it being relativistic. When you bring QM and special relativity together, you get to the possibility of creating and annihilating particles, but the KGE itself can't predict this. If you understand the wavefunction as we usually do in regular QM, it describes the probability associated with a single particle, and you end up with difficulties. The solution is to perform something called second quantization: we start treating the wavefunction itself as an operator that can create and annihilate particles. This is the basis of quantum field theory, which does employ the Klein–Gordon equation and obtains correct results. The Higgs boson, for example, is described by the Klein–Gordon equation.

Notice also that you can't really derive the Klein–Gordon equation, or the Schrödinger equation: you can only guess it and then test it experimentally. Sakurai's argument is indented to illuminate why one would expect the KGE to solve some issues with Schrödinger's equation, but it shouldn't be seen as a proof of the KGE. The proof of the pudding is in the eating, and the proof of a physical law is in an experiment. Schrödinger guessed his equation by drawing in inspiration from other bits of physics, but at the end of the day that is at most a clever guess. What's truly remarkable is that the equation actually is useful to describe the world.

Bonus: but how do QM and SR predict the creation and annihilation of particles?

I'll give a sketch argument, that is purposed to give intuition, but is also not quite a proof.

Suppose you put a particle of mass $$m$$ in a small box of size $$L$$, as we often do in QM. Now, since we know the particle is inside this box, Heisenberg's uncertainty principle tell us that $$L \Delta p \geq \frac{\hbar}{2},$$ and hence the particle's momentum must be such that $$\Delta p \geq \frac{\hbar}{2 L}.$$

Now, if we make $$L$$ small enough, it is reasonable to assume the particle is in the ultrarelativistic limit, and hence it has $$E \approx pc$$. We then find that the particle has $$\Delta E \geq \frac{\hbar c}{2 L}.$$

Notice that the uncertainty in $$E$$ grows as $$L$$ diminishes. If we make the box small enough, we can get $$\Delta \geq 2 mc^2$$ when $$L \leq \frac{\hbar}{4 mc} \sim \lambda_{\text{Compton}}.$$ Hence, when the box is smaller than the particle's Compton wavelength $$\lambda_{\text{Compton}} = \frac{h}{mc}$$, the energy uncertainty is about enough to create a particle-antiparticle pair, at which point it is no longer reasonable to assume we are dealing with a fixed number of particles. Hence, when dealing with relativistic quantum mechanics, we expect to need to deal with creation and annihilation of particles, and the Klein–Gordon equation is not capable of doing this without quantum field theory.

• Thank you for your picture on particle and anti-particles. I do have a further question on this topic. Sakurai also introduced the idea of anti-particles by turning the KGE into two SE like equations. Is this the way we look for anti particles for spinless particles? If possible, could you please give a bit more intuition on why this does not work for particles with spins? Thank you so much! Commented Jun 5, 2023 at 1:24
• @H-JJeffWang I'm not really familiar with this viewpoint. In field theory, while the KGE is there and it is important, the solution techniques are vastly different and we hardly ever deal directly with solutions to the KGE. I suggest posting this as a new question =) Commented Jun 5, 2023 at 3:06
• @H-JJeffWang I skimmed over Sakurai's discussion and I'm not really familiar with this point of view (it does resemble a lot of what I've seen of relativistic quantum mechanics with the Dirac equation, but I'm not used to thinking like that in field theory). Nevertheless, there are analogous discussions for spin-1/2 fields, for example. I haven't checked, but I find it likely that Sakurai will do that in the section about the Dirac equation Commented Jun 5, 2023 at 3:11
• Got it, thank you so much! Commented Jun 5, 2023 at 3:14

Firstly, in general the Klein-Gordon equation can't be derived from non-relativistic QM. If it could, it would not be relativistic, as you stated. It is never possible to derive a "bigger" theory from a "smaller" one, at least not without some further assumptions (for example it is possible to derive the laws of relativistic dynamics from classical mechanics if you further assume that special relativity is correct and especially the Lorentz transformations work the way they do).
Secondly, all this derivation assumes is "$$E = i \hbar \partial_t$$", it doesn't require Schrödinger's equation. For Schrödinger's equation, $$\hat H$$ has a certain form which is not assumed in Sakurai's derivation of the Klein-Gordon equation. In fact, he later uses $$H^2 = p^2 + m^2$$ which is not true in non-relativistic dynamics.
So, in other words, he says "$$E^2 | \psi \rangle = - \partial_t^2 | \psi \rangle \Rightarrow (\square + m^2) | \psi \rangle = 0$$, since $$E^2 = p^2 + m^2 = - \nabla^2 + m^2$$.

• Thank you so much! Can I understand the motivation of this "derivation" as we made a wild guess that we can replace the Hamiltonian in the RHS of Schrodinger equation with barely the "energy operator" to fit it into SR? Commented Jun 5, 2023 at 1:05
• No problem. Yes, you can understand the "derivation" like that. But also in non-relativistic QM the Hamilton operator normally is the "energy operator" (If the Lagrangian doesn't depend on the velocities, so in most practical cases without a magnetic field). Commented Jun 5, 2023 at 12:12
• The wilder guess is that we took the square. The (at first glance) easier ansatz would be $i \partial_t | \psi \rangle = \sqrt{- \nabla^2 + m^2} | \psi \rangle$. But that equation has a lot of problems, for example it is not relativistically covariant (since the square root is defined by it's Taylor series and therefore the RHS contains second, third, forth etc. derivatives in x while the LHS only contains a first derivative in t). All those problems vanish if you "take the square" of that equation, however other problems arise, namely the probability density is no longer positive definite. Commented Jun 5, 2023 at 12:12