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I know people try to solve Dirac equation in a box. Some claim it cannot be done. Some claim that they had found the solution, I have seen three and they are all different and bizarre. But my main issue is what would make the particle behave differently (in the same box). How useful is it, aside from the physicist curiosity.

To improve the question I am adding some clarification.

In this paper Alhaidari solves the problem, but he states in the paper that “In fact, the subtleties are so exasperating to the extent that Coulter and Adler ruled out this problem altogether from relativistic physics: “This rules out any consideration of an infinite square well in the relativistic theory”[4].

Unfortunately, I have no access to this reference

Ref [4] B. L. Coulter and C. G. Adler, Am. J. Phys. 39, 305 (1971)

But also there are attempts in this paper.

And I quote from this paper (page 2).

“A particular solution may be obtained by considering the Dirac equation with a Lorentz scalar potential [7]; here the rest mass can be thought of as an x-dependent mass. This permits us to solve the infinite square well problem as if it is were a particle with a changing mass that becomes infinite out of the box, so avoiding the Klein paradox [8].”

So my question is why all these discrepancies.

QSA
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1 Answers1

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By several reasons explained in textbooks, the Dirac equation is not a valid wavefunction equation. You can solve it and find solutions, but those solutions cannot be interpreted as wavefunctions for a particle [1].

I have checked the three articles linked by you and I do not find any discussion of this. For instance, if $\psi(x)$ is a solution to the Dirac equation then $|\psi(x)|^2$ is not the probability density of finding the particle at $x$ because $x$ in Dirac theory is not observable [2]. Moreover, their treatment is far from being completely relativistic. They are working in a pseudo-relativistic approach as in the Coulomb-Dirac approach.

[1] This is the reason why the solutions to the Dirac equation are re-interpreted as operators in QFT.

[2] This is the reason why $x$ is downgraded from operator status to parameter in QFT.

juanrga
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