Phy5646/Group3RelativisticProb

Free Relativistic Particle

For the Dirac Equation, find plane wave solutions for a free particle.

$i\hbar {\frac {\partial }{\partial t}}\psi =-i\hbar c{\vec {\alpha }}\cdot \nabla \psi +\mathrm {B} mc^{2}\psi$

$\alpha _{k}=\left({\begin{array}{cc}0&\sigma _{k}\\\sigma _{x}&o\end{array}}\right);\beta =\left({\begin{array}{cc}I&0\\0&-I\end{array}}\right)$

So, we seek solutions of the form

$\psi ({\vec {r}},t)=Aue^{\frac {i({\vec {p}}\cdot {\vec {r}}-Et)}{\hbar }}$

which is an eigenfunction of both the position and momentum operators. Note that $A$ is a constant, and $u$ is a four-component spinor, independent of the position of the particle.

Putting this general $\psi$ into the Dirac Equation gives a matrix equation:

$(c{\vec {\alpha }}\cdot {\vec {p}}+\mathrm {B} mc^{2})u=Eu$

Now, it becomes convenient to write the four-spinor u using two two-component spinors:

Failed to parse (unknown function "\begin{array}"): {\displaystyle u = \left(\begin{array}c u_a \\ u_b \end{array} \right); u_a = \left(\begin{array}c u_1 \\ u_2 \end{array} \right); u_b = \left(\begin{array}c u_3 \\ u_4 \end{array} \right) }

Our equation is now written:

$\left({\begin{array}{cc}mc^{2}I&c{\vec {\sigma }}\cdot {\vec {p}}\\c{\vec {\sigma }}\cdot {\vec {p}}&-mc^{2}I\end{array}}\right)\left({\begin{array}{c}u_{a}\\u_{b}\end{array}}\right)=E\left({\begin{array}{c}u_{a}\\u_{b}\end{array}}\right)$

Which gives that $u_{a}$ and $u_{b}$ are related by:

$u_{a}={\frac {c{\vec {\sigma }}\cdot {\vec {p}}}{E-mc^{2}}}u_{b};u_{b}={\frac {c{\vec {\sigma }}\cdot {\vec {p}}}{E+mc^{2}}}u_{a}$

Substituting these equations into each other gives:

$c^{2}({\vec {\sigma }}\cdot {\vec {p}})^{2}u_{a}=({\vec {p}}\cdot {\vec {p}})c^{2}u_{a}=(E^{2}-m^{2}c^{4})u_{a}$ and $c^{2}({\vec {\sigma }}\cdot {\vec {p}})^{2}u_{b}=({\vec {p}}\cdot {\vec {p}})c^{2}u_{b}=(E^{2}-m^{2}c^{4})u_{b}$

And, solving for the eigenenergies yields (from the two roots):

$E_{+}=(m^{2}c^{4}+{\vec {p}}\cdot {\vec {p}}c^{2})^{\frac {1}{2}}$ $E_{-}=-(m^{2}c^{4}+{\vec {p}}\cdot {\vec {p}}c^{2})^{\frac {1}{2}}$

Where each one occurs twice, once for each component of the two-component spinors.

Now, solving the two-component matrix equations for $u_{a}$ and $u_{b}$ gives for the four components of the spinor:

$u_{1}=N\left({\begin{array}{c}\alpha \\{\frac {c{\vec {\sigma }}\cdot {\vec {p}}}{E_{+}+mc^{2}}}\alpha \end{array}}\right);u_{2}=N\left({\begin{array}{c}\beta \\{\frac {c{\vec {\sigma }}\cdot {\vec {p}}}{E_{+}+mc^{2}}}\mathrm {B} \end{array}}\right)$

$u_{3}=N\left({\begin{array}{c}-{\frac {c{\vec {\sigma }}\cdot {\vec {p}}}{-E_{-}+mc^{2}}}\alpha \\\alpha \end{array}}\right);u_{4}=N\left({\begin{array}{c}-{\frac {c{\vec {\sigma }}\cdot {\vec {p}}}{-E_{-}+mc^{2}}}\mathrm {B} \\\mathrm {B} \end{array}}\right)$

Which are clearly normalized by:

$N=\left({\frac {E_{+}+mc^{2}}{2E_{+}}}\right)^{2}$.