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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.