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I'm not familiar with physics jargons compare to those who study physics professionally, but to remark my knowledge regarding this question, the Klein-Gordon equation is a relativisitc equation for particles (or matter waves) derived from the relativistic energy equation (here the units are natural) $$E^2=p^2+m^2$$ by looking $E$ and $p$ as operators acting on complex functions (which are in a Hilbert space) as follows $$\hat E\doteq i\frac{\partial}{\partial t}\qquad\text{and}\qquad \hat p\doteq -i\nabla$$ thus we get $$\left(\square+m^2\right)\psi=0.$$ My question is, I don't know how this equation can be applied to tackle real world problems, such as the wave functions of electrons in a hydrogen atom, with the absence of potential term (thus I can only think of the free particles case). To compare this with the Schrödinger equation, what you need to do is identify the coulomb potential field and put it in the equation then solve, so you can say the solutions are pretty much potential dependant, which I think is quite straightforward to understand, at least the idea of what-we-are-doing itself.

Remark. (I somewhat found an answer to this question, but as the question has been closed, I leave the answer here.) The Klein-Gordon equation is for free particles/fields. One needs to add interaction terms to introduce interaction with other fields. As an example, for the hydrogen atom, the photon field can be introduced to do the physics (per quantum electrodynamics, quantum field theory).

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closed as too broad by Qmechanic Jul 18 '18 at 4:57

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Klein-Gordon equation can only describe spinless particles. If you use Klein-Gordon equation to solve the Hydrogen atom, you will get incorrect results because you are not taking into account the spin of the electrons. $\endgroup$ – Andrei Geanta Jul 18 '18 at 4:55
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    $\begingroup$ @AndreiGeanta Does it give us completely incorrect results or upto some specific order? I mean, the spin of electrons is also ignored in the Schrödinger equation, isn't it? But I really have no idea how KG can be subject to deal with a hydrogen atom. $\endgroup$ – user575201 Jul 18 '18 at 5:01
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    $\begingroup$ The Klein-Gordon equation is the simplest relativistic equation (spin zero), and is used in almost all areas of physics. Please be more specific. $\endgroup$ – Qmechanic Jul 18 '18 at 11:52
  • $\begingroup$ @Qmechanic I am asking how, I mentioned wave functions in a hydrogen atom as an example. Nevertheless I edited the question to be more specific to make it clearer for some, apparently. $\endgroup$ – user575201 Jul 18 '18 at 14:01
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    $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$ – Qmechanic Jul 18 '18 at 14:04