2.7: Solution of the Dirac equation for a free particle

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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_

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\) 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_

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\) 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

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

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

which yields two equations

\(\displaystyle \left(E-mc^2\right)\phi_

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- c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\chi_

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$\textstyle =$

$\displaystyle 0$

\(\displaystyle - c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\phi_

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+ \left(E+mc^2\right)\chi_

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$\textstyle =$

$\displaystyle 0$

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_

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\) 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_

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\) 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_

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\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:

\(E_

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>mc^2\) (turquoise) and for \(E_

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<-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_

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\) 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_

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\) 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_

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\) 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_

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\) and \(\chi_

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\) of \(u_

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\) 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_

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\) is called the minor component and the upper component \(\phi_

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\) is called the major component.

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