Solution of the Dirac equation for a free particle

The Dirac Hamiltonian takes the form

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

where

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

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

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

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 $t$:

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

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

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

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

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

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:

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

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

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

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:

$$u_{{\bf p}} = \left(\matrix{\phi_{{\bf p}} \cr \cr \chi_{{\bf p}}}\right)$$

Therefore, writing the equation in matrix form, we find

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

or

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

which yields two equations

$$\left(E-mc^2\right)\phi_{{\bf p}} - c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\chi_{{\bf p}} = 0$$

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

From the second equation:

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

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

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

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

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

However,

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

Hence, we have the condition

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

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

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

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

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

If this is the case, then

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

However,

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

so that

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

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

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

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

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

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

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

so that

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

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

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

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

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

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

$$E_{{\bf p}}\approx mc^2 + {{\bf p}^2 \over 2m}$$

so that

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

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

$$\chi_{{\bf p}}\ll \phi_{{\bf p}}$$

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

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

the eigenfunctions reduce to

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

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.