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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.

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    $\begingroup$ I'm definitely not an expert in this subject, but one obvious difficulty with any 'relativistic particle-in-a-box' problem is that in QFT the number of particles is not fixed. The smaller the box, the higher the ground-state energy and the more probability that quantum fluctuations can create another Dirac fermion. Once the size of the box becomes of the order of the Compton wavelength of the particle then the single-particle approximation completely breaks down. David Tong gives a lovely exposition of this point in his first lecture here $\endgroup$
    – Mark Mitchison
    Nov 14, 2012 at 11:53
  • $\begingroup$ Heuristically potential gradients imply pair creation. In this case you have an infinite gradient. If you have a Dirac field in a potential, pairs would spontaneously form out of nowhere. This violate the applicability of the wave equation which is a single particle equation. $\endgroup$
    – Prathyush
    Nov 15, 2012 at 0:41
  • $\begingroup$ @Prathyush , you right, but there are quite few physicists like I have listed that think otherwise. They seem to say That there is a solution as long as you don't go below certain distance or potential, not sure really. $\endgroup$
    – QSA
    Nov 15, 2012 at 1:00
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    $\begingroup$ The edits help, but what discrepancies are you talking about? I think the question could still use some clarification. $\endgroup$
    – David Z
    Nov 15, 2012 at 4:38
  • $\begingroup$ @DavidZaslavsky, Well, as you can see. Adler claims that there is no solution, Alhaidari finds a particular solution and Vidal Alonso finds a solution where mass is proportional to the width of the box, and there are others. So it is not clear to me what is going on. I usually try to ask questions related to fundamental issues that seem to be either glossed over or not well explained in the textbooks and literature. $\endgroup$
    – QSA
    Nov 15, 2012 at 15:08

1 Answer 1

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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.

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  • $\begingroup$ thanks for going through the papers. What you are saying is of course textbook stuff. But my question( I reformulate) is why are they going about it i.e. how useful is it. and what is wrong with what is been proposed. Something along these lines. $\endgroup$
    – QSA
    Nov 15, 2012 at 23:19
  • $\begingroup$ I cannot know "why are they going about it", somehow as I cannot know why some few people still believes that Earth is flat. I already said in my answer what is wrong with their work, e.g. their solutions to the Dirac equation have no physical meaning and I avoided objectionable statements about particles with changing masses that become infinite out of the box. What is then the utility of their work? Well, it is useless for me. $\endgroup$
    – juanrga
    Nov 16, 2012 at 18:15
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    $\begingroup$ @QSA: Actually, it's all the excuse you need in Mathematics. If an equation is challenging, bears some elegance or some kind of mystery, finding its solution is interesting. It doesn't matter to the mathematician that it doesn't work with physics of our universe. While it's useless to a physicist, it does push the frontier of Mathematics a little, and besides is it entirely wrong or just inaccurate? Sometimes an order of approximation is all an engineer needs too, mr. Physicist :D $\endgroup$
    – SF.
    Jan 14, 2013 at 23:26
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    $\begingroup$ @CiroSantilli郝海东冠状病六四事件法轮功 Mandl & Shaw mention in their textbook about QFT that $x$ is not an operator in QFT (unlike in QM). And Landau and Lifshitz explain why position is not an observable in their book about QED (Volume 4 of their course of theoretical physics) and the "only observable quantities are the properties (momenta, polarizations) of free particles". $\endgroup$
    – juanrga
    Aug 31, 2020 at 8:35
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    $\begingroup$ Without position observables, there is no ordinary wavefunctions because the "physical significance of the wave function $\Psi(x)$ is that the square of its modulus gives the probability of finding a particular value of the electron coordinate as the result of a measurement made at a given instant. The concept of such a probability clearly requires that the coordinate can in principle be measured with any specified accuracy and rapidity, since otherwise this concept would be purposeless and devoid of physical significance." $\endgroup$
    – juanrga
    Aug 31, 2020 at 8:36

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