2.7: Solution of the Dirac equation for a free particle
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The Dirac Hamiltonian takes the form
where
Using \({\bf P}= (\hbar/i)\nabla\), in the coordinate basis, the Dirac equation for a free particle reads
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 \(t\):
which is called a four-component Dirac spinor. In order to generate an eigenvalue problem, we look for a solution of the form
which, when substituted into the Dirac equation gives the eigenvalue equation
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:
where \(u_
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Since the matrix on the left is expressible in terms of 2\(\times\)2 blocks, we look for \(u_
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Therefore, writing the equation in matrix form, we find
or
which yields two equations
\(\displaystyle \left(E-mc^2\right)\phi_ (click for details) - c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\chi_ Callstack:
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(click for details) \) Callstack:
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| \(\textstyle =\) | \(\displaystyle 0\) | |
\(\displaystyle - c\stackrel{\rightarrow}{\sigma}\cdot{\bf p}\phi_ (click for details) + \left(E+mc^2\right)\chi_ Callstack:
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(click for details) \) Callstack:
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| \(\textstyle =\) | \(\displaystyle 0\) |
From the second equation:
Note, one could also solve the first for \(\phi_
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Using the first of these, then a single equation for \(\phi_
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However,
Hence, we have the condition
Since \(\phi_
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We see that the eigenvalues can be positive or negative. A plot of the energy levels is shown below:
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We will show that for \(E>0\), an appropriate solution is to take
If this is the case, then
However,
so that
so that the full solution \(u_
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Note that when \({\bf p}=0\), the third and fourth components of \(u_
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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
so that
By the same reasoning, the solution for \(u_
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so that in the limit \({\bf p}=0\), and \(E_0=-mc^2\),
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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In the non-relativistic limit, for \(E>0\), we have
so that
since \(mc^2 \gg {\bf p}^2/2m\), it follows that
Neglecting it, and recalling that for \(E>0\),
the eigenfunctions reduce to
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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