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If we write the Klein-Gordon equation in this form \begin{equation*} c^2 \hbar^2 \nabla^2 \Psi = \hbar^2 \ddot{\Psi} + 2i\hbar (U - mc^2) \dot{\Psi} + U (2mc^2 - U) \Psi \end{equation*} we have a pleasant sense of continuity from the non-relativistic to the relativistic treatment of quantum particle (we use the Schrödinger formalism, and to get the NR solutions we only have to put $c\to\infty$). The (not Lorentz invariant) equation has to be handled with care because the manipulations I used in order to obtain it, included squaring conservation of energy, so we can get spurious solutions too. But I think that for zero-spin particles it works, because I found it in page 42 of Wachter's Relativistic Quantum Mechanics (written slightly differently).

If we suppose that $|\Psi|^2$ is stationary (i.e. the solution has the form $\Psi (\mathbf{r},t) = \psi (\mathbf r) e^{Ct} $ with $C$ purely imaginary) the equation takes the time-independent form: \begin{equation*} -c^2 \hbar^2 \nabla^2 \Psi = [U^2 - 2(E+mc^2)U + E^2+2Emc^2] \Psi \end{equation*} (if you are interested in the proofs search sr.pdf in my Home Page, I don't transcribe here because, more than a question, this should become an article)

My question:

Suppose using this equation with a finite monodimensional hole: \begin{equation*} U(x) = \left\{ \begin{array}{ll} -V_0 & \quad \textrm{if }-a<x<a\\ 0 & \quad \textrm{if }|x|>a \end{array} \right. \end{equation*} In the internal region $\Psi$ is sinusoidal (with the not restrictive condition $E>-V_0$), but in the external region we get \begin{equation*} \Psi'' = k^2 \Psi ;\qquad k = \frac{\sqrt{-E (E + 2 mc^2)}}{c \hbar} \end{equation*} If $-2mc^2 <E <0$, $k \in \mathbb{R}^+$, otherwise $k$ is purely imaginary, the wave function is sinusoidal and the normalization is impossible. Not surprising that for $E>0$ we don't have stationary states with that finite hole, but:

  • what about the case $E<-2mc^2$? What does it mean?

The only reasonable interpretation I found, is that in this case the particle is totally confined into the hole. - Is this wrong?

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  • $\begingroup$ Your Klein-Gordon equation, in presence of potential is not correct, it is (($\hat E - \hat U)^2 - \hat P^2 - m^2) \psi=0$ $\endgroup$ – Trimok Aug 8 '14 at 11:20
  • $\begingroup$ My equation is Wachter's one with V=U-mc^2, but except this case, I really never found K-G equation written with potential, neither mine nor your. Can you say to me more about your K-G equation? What is the P? Where I can find this way of writing K-G equation? $\endgroup$ – Fausto Vezzaro Aug 8 '14 at 12:26
  • $\begingroup$ $\hat P^i$ is the momentum operator $(-i\hbar \dfrac{\partial}{\partial{x^i}})$, and $\hat P^2 = \sum\limits_i \hat P^i \hat P^i$ is the squared norm operator, that is $ (-\hbar^2 \nabla^2)$. For the Klein-Gordon equation with potential, see for instance equations $(45), (46)$ page $7$ in this paper. In the paper, $U=e\phi$ is the electromagnetic potential energy. $\endgroup$ – Trimok Aug 8 '14 at 13:23
  • $\begingroup$ I fear we're using two different formalism, and I don't know your (with 4-vector). Now I'm leaving: I'll try to reflect upon what you wrote. $\endgroup$ – Fausto Vezzaro Aug 8 '14 at 14:20
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    $\begingroup$ If in my stationary equation you replace $U$ with $U+mc^2$, and set $c=1$, you obtain your equation. Despite two different choice in unit and in setting zero $U$, we wrote the same thing. But the choice of $U=0$ shouldn't play any role. If we add arbitrary $\xi$ to $U$ in the above finite hole, with my equation we find normalization condition $\xi - 2mc^2 <E<\xi$, while using your we find $\xi - 2mc^2 <E - mc^2<\xi$. But in your equation $E$ is the total relativistic energy so our normalization condition are the same: if the hole is sufficiently deep (and large) the confinement seems possible. $\endgroup$ – Fausto Vezzaro Aug 11 '14 at 7:13
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If we suppose that $\Psi$ is confined we must have $\Psi=0$ in the external region, so in the border of internal region we must have $\Psi=0$ and $\frac{d\Psi}{dx}=0$. But with our flat $U$, stationary K-G equation say that $\Psi$ is sinusoidal or exponential, so this requisites cannot be satisfied sumultaneously. Conclusion: the confinement of $\Psi$ is impossible (simply, stationary states with $E<-2mc^2$ are impossibile).

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