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: Introduction to total angular : lecture_7 : lecture_7

Solution of the Dirac equation for a free particle

The Dirac Hamiltonian takes the form

\begin{displaymath}
H = c\stackrel{\rightarrow}{\alpha}\cdot{\bf P}+ \beta mc^2
\end{displaymath}

where

\begin{displaymath}
\stackrel{\rightarrow}{\alpha}= \left(\matrix{ & 0 & & \stac...
...rix{& {\rm I} & & 0 & \cr & & & \cr & 0 & & -{\rm I} &}\right)
\end{displaymath}

Using ${\bf P}= (\hbar/i)\nabla$, in the coordinate basis, the Dirac equation for a free particle reads

\begin{displaymath}
\left[-i\hbar c\stackrel{\rightarrow}{\alpha}\cdot\nabla + \...
...i({\bf r},) = i\hbar{\partial \over
\partial t}\Psi({\bf r},t)
\end{displaymath}

Since the operator on the left side is a 4$\times$4 matrix, the wave function $\Psi({\bf r},t)$ is actually a four-component vector of functions of ${\bf r}$ and $t$:

\begin{displaymath}
\Psi({\bf r},t) = \left(\matrix{ \Psi_1({\bf r},t) \cr \Psi_2({\bf r},t) \cr \Psi_3({\bf r},t) \cr
\Psi_4({\bf r},t)}\right)
\end{displaymath}

which is called a four-component Dirac spinor. In order to generate an eigenvalue problem, we look for a solution of the form

\begin{displaymath}
\Psi({\bf r},t) = \psi({\bf r})e^{-iEt/\hbar}
\end{displaymath}

which, when substituted into the Dirac equation gives the eigenvalue equation

\begin{displaymath}
\left[-i\hbar c\stackrel{\rightarrow}{\alpha}\cdot\nabla + \beta mc^2\right]\psi({\bf r}) = E\psi({\bf r})
\end{displaymath}

Note that, since $H$ is only a function of ${\bf P}$, then $[{\bf P},H]=0$ so that the eigenvalues ${\bf p}$ of ${\bf P}$ can be used to characterize the states. In particular, we look for free-particle (plane-wave) solutions of the form:

\begin{displaymath}
\psi_{{\bf p}}({\bf r}) = u_{{\bf p}}e^{i{\bf p}\cdot{\bf r}/\hbar}
\end{displaymath}

where $u_{{\bf p}}$ is a four-component vector which satisfies

\begin{displaymath}
\left[c\stackrel{\rightarrow}{\alpha}\cdot{\bf p}+ \beta mc^2\right]u_{{\bf p}} = Eu_{{\bf p}}
\end{displaymath}

Since the matrix on the left is expressible in terms of 2$\times$2 blocks, we look for $u_{{\bf p}}$ in the form of a vector composed of two two-component vectors:

\begin{displaymath}
u_{{\bf p}} = \left(\matrix{\phi_{{\bf p}} \cr \cr \chi_{{\bf p}}}\right)
\end{displaymath}

Therefore, writing the equation in matrix form, we find

\begin{displaymath}
\left(\matrix{mc^2 & & c\stackrel{\rightarrow}{\sigma}\cdot{...
...t)
\left(\matrix{\phi_{{\bf p}} \cr \cr \chi_{{\bf p}}}\right)
\end{displaymath}

or

\begin{displaymath}
\left(\matrix{E-mc^2 & & -c\stackrel{\rightarrow}{\sigma}\cd...
...left(\matrix{\phi_{{\bf p}} \cr \cr \chi_{{\bf p}}}\right)
= 0
\end{displaymath}

which yields two equations
$\displaystyle \left(E-mc^2\right)\phi_{{\bf p}} - c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\chi_{{\bf p}}$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle - c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\phi_{{\bf p}} + \left(E+mc^2\right)\chi_{{\bf p}}$ $\textstyle =$ $\displaystyle 0$  

From the second equation:

\begin{displaymath}
\chi_{{\bf p}} = {c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\over E+mc^2}\phi_{{\bf p}}
\end{displaymath}

Note, one could also solve the first for $\phi_{{\bf p}}$ and obtain

\begin{displaymath}
\phi_{{\bf p}} = {c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\over E-mc^2}\chi_{{\bf p}}
\end{displaymath}

Using the first of these, then a single equation for $\phi_{{\bf p}}$ can be obtained

\begin{displaymath}
\left(E-mc^2\right)\left(E+mc^2\right)\phi_{{\bf p}} -
c^2\...
...krel{\rightarrow}{\sigma}\cdot{\bf p}\right)^2\phi_{{\bf p}}=0
\end{displaymath}

However,

\begin{displaymath}
\left(\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\right)^2 = ...
...stackrel{\rightarrow}{\sigma}\cdot({\bf p}\times{\bf p}) = p^2
\end{displaymath}

Hence, we have the condition

\begin{displaymath}
\left[E^2 - \left((mc^2)+c^2p^2\right)\right]\phi_{{\bf p}}=0
\end{displaymath}

Since $\phi_{{\bf p}}\neq 0$, the equation is only satisfied if the quantity in the brackets vanishes, which yields the eigenvalues

\begin{displaymath}
E = E_{{\bf p}} = \pm\sqrt{p^2 c^2 + m^2 c^4}
\end{displaymath}

We see that the eigenvalues can be positive or negative. A plot of the energy levels is shown below:




図 1:


There is a continuum for $E_{{\bf p}}>mc^2$ (turquoise) and for $E_{{\bf p}}<-mc^2$ (periwinkle). There is also a gap between $-mc^2$ and $mc^2$.

We will show that for $E>0$, an appropriate solution is to take

\begin{displaymath}
\phi_{{\bf p}} = {\left(\matrix{1 \cr 0}\right)}\;\;\;\;\;\;...
...;\;{\rm or}\;\;\;\;\;\;\;\;\;\;{\left(\matrix{0 \cr 1}\right)}
\end{displaymath}

If this is the case, then

\begin{displaymath}
\chi_{{\bf p}} = {c\stackrel{\rightarrow}{\sigma}\cdot{\bf p...
...t{\bf p}\over E_{{\bf p}}+mc^2}{\left(\matrix{0 \cr 1}\right)}
\end{displaymath}

However,

\begin{displaymath}
\stackrel{\rightarrow}{\sigma}\cdot{\bf p}= \left(\matrix{p_z & p_x-ip_y \cr \cr p_x+ip_y & -p_z}\right)
\end{displaymath}

so that

\begin{displaymath}
\chi_{{\bf p}} = \left(\matrix{cp_z/(E_{{\bf p}}+mc^2) \cr \...
.../(E_{{\bf p}}+mc^2) \cr \cr
-cp_z/(E_{{\bf p}}+mc^2)}\right)
\end{displaymath}

so that the full solution $u_{{\bf p}}$ is

\begin{displaymath}
u_{{\bf p}} = \left(\matrix{1 \cr \cr 0 \cr \cr cp_z/(E_{{\b...
.../(E_{{\bf p}}+mc^2) \cr \cr
-cp_z/(E_{{\bf p}}+mc^2)}\right)
\end{displaymath}

Note that when ${\bf p}=0$, the third and fourth components of $u_{{\bf p}}$ vanish. In this case, energy is just $E_0 = mc^2$ and the full time-dependent wave function becomes

\begin{displaymath}
\Psi(t) \longrightarrow
\left(\matrix{1 \cr \cr 0 \cr \cr 0...
...atrix{0 \cr \cr 1 \cr \cr 0 \cr \cr 0}\right)e^{-imc^2t/\hbar}
\end{displaymath}

which are both forward propagating solutions. These correspond to particle solutions, in particular, a spin-1/2 particle propagating forward in time with an energy equal to the rest mass energy.

When $E<0$, we take

\begin{displaymath}
\chi_{{\bf p}} = {\left(\matrix{1 \cr 0}\right)}\;\;\;\;\;\;...
...;\;{\rm or}\;\;\;\;\;\;\;\;\;\;{\left(\matrix{0 \cr 1}\right)}
\end{displaymath}

so that

\begin{displaymath}
\phi_{{\bf p}} = {c\stackrel{\rightarrow}{\sigma}\cdot{\bf p...
...cdot{\bf p}} \over \vert E_{{\bf p}}\vert+mc^2} \chi_{{\bf p}}
\end{displaymath}

By the same reasoning, the solution for $u_{{\bf p}}$ is

\begin{displaymath}
u_{{\bf p}} = \left(\matrix{-cp_z/(\vert E_{{\bf p}}\vert+mc...
...2) \cr \cr
cp_z/(E_{{\bf p}}+mc^2)\cr\cr 0 \cr \cr 1}\right)
\end{displaymath}

so that in the limit ${\bf p}=0$, and $E_0=-mc^2$,

\begin{displaymath}
\Psi(t) \longrightarrow
\left(\matrix{0 \cr \cr 0 \cr \cr 1...
...matrix{0 \cr \cr 0 \cr \cr 0 \cr \cr 1}\right)e^{imc^2t/\hbar}
\end{displaymath}

which describes particles moving backward in times. Thus, the interpretation is that the negative energy solutions correspond to anti-particles, the the components, $\phi_{{\bf p}}$ and $\chi_{{\bf p}}$ of $u_{{\bf p}}$ correspond to the particle and anti-particle components, respectively. Thus, the Dirac equation no only describes spin but it also includes particle and the corresponding anti-particle solutions!

In the non-relativistic limit, for $E>0$, we have

\begin{displaymath}
E_{{\bf p}}\approx mc^2 + {{\bf p}^2 \over 2m}
\end{displaymath}

so that

\begin{displaymath}
\chi_{{\bf p}} = {c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\over 2mc^2 + {\bf p}^2/2m}\phi_{{\bf p}}
\end{displaymath}

since $mc^2 \gg {\bf p}^2/2m$, it follows that

\begin{displaymath}
\chi_{{\bf p}}\ll \phi_{{\bf p}}
\end{displaymath}

Neglecting it, and recalling that for $E>0$,

\begin{displaymath}
\phi_{{\bf p}} = {\left(\matrix{1 \cr 0}\right)}\;\;\;\;\;\;\;\;\;\;{\rm or}\;\;\;\;\;\;\;\;\;{\left(\matrix{0 \cr 1}\right)}
\end{displaymath}

the eigenfunctions reduce to

\begin{displaymath}
\psi_{{\bf p}}({\bf r}) =
\left(\matrix{1 \cr \cr 0 \cr \cr...
...\cr 0 \cr \cr 0 \cr \cr}\right)
e^{i{\bf p}\cdot{\bf r}/\hbar}
\end{displaymath}

The lower component has become reduntant, and the eigenfunctions just correspond to those of a free particle with an attached spin eigenfunction ${\left(\matrix{1 \cr 0}\right)}$ or ${\left(\matrix{0 \cr 1}\right)}$ for $m_s=\hbar/2$ or $-\hbar/2$, respectively. For $E>0$, the lower component, $\chi_{{\bf p}}$ is called the minor component and the upper component $\phi_{{\bf p}}$ is called the major component.


next up previous
: Introduction to total angular : lecture_7 : lecture_7
Mark Tuckerman 平成17年2月18日