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Abstract |
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On the Positivity of Relativistic One-Electron Atoms Hamiltonian of Brown and RavenhallV.I. BurenkovIn the model due to Brown and Ravenhall [\,1\,] a relativistic electron of charge $-e $ and mass $m$ in the field of a nucleus of charge $Ze$ is described with the help of the operator $$B=\Lambda_+\biggl(D_0-\frac{e^2Z}{|x|}\biggr)\Lambda_+$$ acting in the Hilbert space $H=\Lambda_+(L^2({\bf R}^3)\bigotimes {\bf C}^4)$. Here $$D_0=c\alpha\cdot \frac{\hbar}i\nabla+mc^2\beta,$$ where $\alpha=(\alpha_1,\alpha_2,\alpha_3)$, $$ \matrix{ \alpha_j=\left(\matrix {0 & \sigma_j\cr \sigma_j & 0\cr} \right),~~ \sigma_1=\left(\matrix {0 & 1\cr 1 & 0\cr} \right),~~ \sigma_2=\left(\matrix {0 & -i\cr i & 0\cr} \right),~~ \sigma_3=\left(\matrix {1 & 0\cr 0 & -1\cr} \right), } $$ and $$ \matrix{ \beta=\left(\matrix {I & 0\cr 0 & -I\cr} \right),~~ I=\left(\matrix {1 & 0\cr 0 & 1\cr} \right), } $$ is the free Dirac operator, and $$ \Lambda_+=\chi_{(0,\infty)}(D_0) $$ is the projection on the positive spectral subspace of $D_0$ ($c$ is the velocity of light, $\hbar={h\over{2\pi}}$ and $h$ is the Plank's constant).
In [\,2\,] it was proved that on the subspace $\widetilde{\!H}=\Lambda_+(L^2({\bf R}^3,\sqrt{|p|^2+1}dp)\bigotimes {\bf C}^4)$ the operator $B$ is bounded from if and only if $Z\le Z_c=2[({\pi\over2}+{2\over\pi})\alpha]^{-1}$, where $\alpha={e^2\over \hbar c}$ is Sommerfeld's fine structure constant ($Z_c\approx 124,2$). From the point of view of physics, in the framework of the model under discussion, this means that a one-electron atom is stable if and only if $Z\le Z_c$. Moreover in [\,2\,] it was proved that, for $Z\le Z_c,~ B\ge-\alpha Z({\pi\over4}-{1\over\pi})mc^2$ on ${\widetilde{\!H}}.$ This result was improved in [\,3\,], where it was shown that $B\ge (1-{Z\over Z_c})mc^2$ on $\widetilde{\!H}$ for $Z\le Z_c$. In particular, $B\ge0$ for $Z=Z_c$. It was also proved that in this case 0 is not an eigenvalue of $B$. {\bf Theorem.}~~ If $0<Z\le Z_c$, then on ${\widetilde{\!H}}$ $$B>\biggl(1-0,971{Z\over{Z_c}}\biggr)mc^2.$$ From the point of view of analysis the problem reduces to proving the inequality \begin{equation} \int\limits_0^{\infty}\varphi(x)^2\,dx -\frac{\xi}{{\pi^2\over4}+1}\int\limits_0^{\infty} \int\limits_0^{\infty}t(x,y)\varphi(x)\varphi(y)\,dxdy \ge D_{\xi}\int\limits_0^{\infty}\varphi(x)^2\,{dx \over \sqrt{x^2+1}}, \end{equation} where $0<\xi\le1$, and finding the minimal possible value of $D_{\xi}$. Here $$ t(x,y)={1\over2}\Biggl\{ \sqrt{\frac{\sqrt{x^2+1}+1}{x^2+1}} g_0\biggl({x\over y}\biggr) \sqrt{\frac{\sqrt{y^2+1}+1}{y^2+1}} $$ $$ +\sqrt{\frac{\sqrt{x^2+1}-1}{x^2+1}} g_1\biggl({x\over y}\biggr) \sqrt{\frac{\sqrt{y^2+1}-1}{y^2+1}} \Biggr\} $$ with $$ g_0(u)=\ln\biggl|\frac{u+1}{u-1}\biggr|,~~ g_1(u)={1\over2}\biggl(u+{1\over u}\biggr) \ln\biggl|\frac{u+1}{u-1}\biggr|-1,~~u>0. $$ The proof is based on the following statement. {\bf Lemma.}~~ For all functions $h_0,h_1$, which are positive and measurable on $(0,\infty)$, inequality (1) is valid with $D_{\xi}=B_{\xi}(h_0,h_1)$, where $$ B_{\xi}(h_0,h_1) =\inf\limits_{0<x<\infty}\sqrt{x^2+1}\biggl(1-\frac{\xi}{{\pi^2\over2}+2} \biggl( \frac{\sqrt{x^2+1}+1}{x^2+1}\int\limits_0^{\infty}\frac{h_0(y)}{h_0(x)} g_0({y\over x})\,dy $$ $$ + \frac{\sqrt{x^2+1}-1}{x^2+1}\int\limits_0^{\infty}\frac{h_1(y)}{h_1(x)} g_1({y\over x})\,dy\biggr)\biggr). $$ Equality in (1) with $D_{\xi}=B_{\xi}(h_0,h_1)$, for a function $\phi$, which is not equivalent to $0$, can hold if, and only if, for some constants $B_0,B_1,B_2$, $$ \phi(x) =B_0h_0(x)\sqrt{\frac{x^2+1}{\sqrt{x^2+1}+1}} =B_1h_1(x)\sqrt{\frac{x^2+1}{\sqrt{x^2+1}-1}} $$ and $$ \frac{\sqrt{x^2+1}+1}{x^2+1}\int\limits_0^{\infty}\frac{h_0(y)}{h_0(x)} g_0({y\over x})\,dy + \frac{\sqrt{x^2+1}-1}{x^2+1}\int\limits_0^{\infty}\frac{h_1(y)}{h_1(x)} g_1({y\over x})\,dy $$ $$ =\biggl(1-\frac{B_2}{\sqrt{x^2+1}}\biggr)\frac {{\pi^2\over2}+2}{\xi} $$ for almost all $x\in(0,\infty)$. By setting $$h_0(x)={x\over{1+x^2}},~~h_1(x)={1\over x}$$ inequality (1) follows with $$ D_{\xi}=\inf\limits_{0<x<\infty}G_{\xi}(x), $$ where $$ G_{\xi}(x)=\sqrt{x^2+1} \biggl(1-\frac{\xi}{{\pi^2\over4}+1} \biggl({\pi\over2}(\sqrt{x^2+1}+1)\frac{\arctan x}{x} +\frac{(\sqrt{x^2+1}-1)x}{x^2+1}\biggr)\biggr). $$ One can prove that $$ D_1=\lim\limits_{x\to\infty}G_1(x)={\pi\over2}+1-{\pi^2\over4}. $$ Consequently, the estimate $D_{\xi}\ge1-\xi+\xi D_1 >1-0,971\xi$ follows, and, hence, the statement of the Theorem. The author is grateful to C. Tix who sent his manuscript in which the positiveness of the operator $B$ for ${Z\le Z_c}$ was claimed. However a part of his proof was numerical. After an analytical proof of the positiveness of $B$ was given by the author, C. Tix [\,4\,] amended his proof which became completely analytical, though it still contained tedious (analytical) calculations. Moreover his proof gives a better estimate: $B>(1-0.906{Z/{Z_c}})mc^2$. The problem of finding the best possible estimate is still open. \centerline{\sf R e f e r e n c e s} [\,1\,] {\sf G.E. Brown and D.G. Ravenhall.} On the interaction of two electrons. {\it Proc. Royal Soc. London.} {\bf A 208} ({\bf A 1095}), 552 - 559 (1951). [\,2\,] {\sf W.D. Evans, P. Perry and H. Siedentop.} The spectrum of relativistic one-electron atoms according to Bethe and Salpeter. {\it Commun. Math. Phys.} {\bf 178}, 733 - 746 (1996). [\,3\,] {\sf V.I. Burenkov, W.D. Evans.} On the evaluation of the norm of an integral operator associated with the stability of one-electron atoms. {\it Math. Phys. Prepr. Archive mp arc} {\bf 97 - 247} (1997); {\it Proc. Royal Soc. Edinburg}, {\bf 128A}, 993-1005 (1998). [\,4\,] {\sf C. Tix}. Strict positivity of a relativistic Hamiltonian due to Brown and Ravenhall. {\it Bull. London Math. Soc.} {\bf 30}, 283-290 (1998). |
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12/04/99 |