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Dirac equation and Weyl equation

已有 5134 次阅读 2013-11-21 23:23 |系统分类:科研笔记| and, quot, Equation, amp, Weyl

Quantum mechanics and relativity theory are the two greatest achivements in the history of modern physics starting at the beginning of the 20th century. Each of the two has been so sucessfully examined by experiments and widely applied in various fields. However, in such a harmonious physical world, the two must be consistent (at least not in contradiction) with each other. The efforts of combining quantum mechanics and relativity theory have been vigorous since the early days after their births. Dirac equation for a free relativistic electron is for sure one of them.

In quantum mechanics, the motion of moving particles is described by wave functions $psi(bf{x},rm t) $ that satisfy the Schrodinger equation:
$$ left[- frac{hbar ^2 nabla ^2}{2m}  + V(bf{x}) right] psi(bf{x},rm t) = it{i} rm hbar frac{partial}{partial t} psi(bf{x},rm t) $$
where     <li>$hbar$ is the reduced Plank constant, </li>
          <li>$m$ is the particle mass</li>
         <li>$V(bf{x})$ is the potential the particle feels and for free particles $V(bf{x})=0$</li>
         <li>$nabla ^2 = frac{partial ^2}{partial x^2} + frac{partial ^2}{partial y^2} + frac{partial ^2}{partial z^2}$ is the 2nd-order derivative operator.</li>
The Schrodinger equation was derived from the correspondence between phycical variables and operators
$$ bf{p} rightarrow it{i}hbar nabla , quad E rightarrow it{i}hbar frac{partial}{partial t}$$
by replacing the physical variables with the corresponding operators in the momentum-energy relation
$$ left[- frac{bf{p}}{2m}  + V(bf{x}) right] = E $$
which is valid only for classical particles.
For free relativistic particles, the momentum-energy relation becomes
$$ E^2 = p^2 c^2 + m^2 c^4 $$
where      <li>$E$ is the energy of the particle,</li>
   <li>$p$ is the momentum of the particle,</li>
   <li>$m$ is the rest mass of the particle</li>
   <li>$c$ is the velocity of light.</li>

then the Schrodinger equation for such a relativistic particle is
$$ - hbar ^2 frac{partial ^2}{partial t^2} psi(bf{x},rm t) =left( - frac{hbar ^2 nabla ^2}{2m} c^2 + m^2 c^4 right) psi(bf{x},rm t)$$
$$ left( frac{nabla ^2}{2m} - frac{1}{c^2} frac{partial ^2}{partial t^2} right) psi(bf{x},rm t) = frac{m^2 c^2}{hbar ^2} psi(bf{x},rm t)$$
The above Schrodinger equation is consisten with relativity theory in that all the four components of spacetime vector $(x,y,z,t)$ and their (second order) derivatives symmetrically enter the equation, which is obviously not true for the Schrodinger equations of non-relativistic particles. Howerver, it also produces paradoxes! The particle density defined by
$$ rho = frac{i hbar}{2m} ( psi^* partial _t psi - psi partial _t psi^*)$$
is no longer positive-definite since the initial values for $psi$ and $partial _t psi$ can be freely chosen and the density may thus become negative, which is impropriate.

Increasing the first derivative with respect to time to second order seems not a good choice. Dirac then went the other way around. He tried to make all the derivatives with respect to the four components of spacetime vector be first order. By hard thinking, he found that the wave operator could be square rooted like this:
$$  nabla ^2 - frac{1}{c^2} frac{partial ^2}{partial t^2} = (A partial _x + B partial _y + C partial _z + frac{i}{c} D partial _t) (A partial _x + B partial _y + C partial _z + frac{i}{c} D partial _t) $$
The above equation can hold provided that
$$ AB + BA =0 , ; ldots $$
and $$ A^2=B^2=ldots=1$$.
Then the Schrodinger equation can be rearranged to be
$$ (A partial _x + B partial _y + C partial _z + frac{i}{c} D partial _t) psi(bf{x},rm t)= frac{m c}{hbar} psi(bf{x},rm t)$$
$$ (A,B,C) = i beta alpha _i, D = beta ,$$
we get the common form of Dirac equation
$$ left( beta m c^2 + sum_{k=1}^{3} alpha _k p_k cright) psi(bf{x},rm t) = i hbar frac{partial}{partial t} psi(bf{x},rm t).$$
The relations between the four constants $A,B,C,D$(or $alpha _i, beta$) can be satisfied only when they are all $4times4$ matrice. In consistence with the relativistic invariance of the equation, they were found to be
$$ A=frac{1}{i} left( begin{array}[c|c]
              rm 0   & sigma_x \
            -sigma_x &    0
           end{array} right) , quad B=frac{1}{i} left( begin{array}[c|c]
              rm 0   & sigma_y \
            -sigma_y &    0
           end{array} right) , quad C=frac{1}{i} left( begin{array}[c|c]
              rm 0   & sigma_z \
            -sigma_z &    0
           end{array} right) , quad D=left( begin{array}[c|c]
             rm I_2   &    0 \
               0    &   -I_2
           end{array} right), $$
where $sigma_x,sigma_y,sigma_z$ are Pauli matrice and $I_2=left( begin{array}[c|c] rm 1   & 0 \     0 & 1 end{array} right) .$
The above $4times4$ matrice lead to four-component solutions for the Dirac equation, while the solutions for non-relavistice Shrodinger equation are one-component.

Howerver, when the rest mass of the particle is zero, the Dirac equation will reduce to Weyl equation
$$ left( I_2 frac{1}{c} frac{partial }{partial t} + sigma_x frac{partial }{partial x} + sigma_y frac{partial }{partial y} + sigma_z frac{partial }{partial z} right) psi = 0 $$
which has two-component solutions.




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