Solutions to Dirac equation


are usually obtained by acting from the left with complex conjugate Dirac operator


and realizing that this is a wave equation. I wonder if the additional multiplication by $(i\gamma^\mu\partial_\mu+m)$ above just restricts the full set of solutions to plane waves only, and so whether the original first order equation might have other types of solutions as well? Is it proven that plane waves are the complete set of solutions to the Dirac equation?

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    $\begingroup$ Anything which satisfies the Dirac equation also satisfies the Klein-Gordon equation, so this 'additional multiplication' is, if anything, going to give us spurious solutions rather than cause us to lose solutions. Indeed, we can take our plane wave solutions to the KG equation and substitute them back into the Dirac equation, and we will find an additional constraint on the components of $\psi$. $\endgroup$
    – gj255
    Nov 7 '16 at 22:22
  1. We first check that $$ \psi(x)=\sum_s\int\mathrm d\vec k\ u_s(\vec k)c_{\vec k}\mathrm e^{-ikx}+v_s(\vec k)d_{\vec k}^\dagger\mathrm e^{ikx}\tag{1} $$ is a solution of the Dirac equation. To do so, substitute this expression into $(-i\not\partial+m)\psi(x)$, and use $(\not p-m)u=(\not p+m)v=0$.

  2. Second, we realise that the relation above can be inverted: $$ \begin{aligned} c^{s\dagger}_{\vec k}&=\int\mathrm d\vec x\ \mathrm e^{ikx}\psi^\dagger(x)u_s(\vec k)\\ d^{s\dagger}_{\vec k}&=\int\mathrm d\vec x\ \mathrm e^{ikx}v^\dagger(\vec k)\psi(x) \end{aligned}\tag{2} $$

  3. Finally, noting that the Dirac equation is linear, we see that $(1)$ is indeed the most general solution, as it solves any initial value problem (as per $(2)$).


Srednicki's book, http://web.physics.ucsb.edu/~mark/qft.html. For $(1)$ see page 239, and for $(2)$ see page 263.

  • $\begingroup$ For some reason I was thinking that being a four component equation, it roughly corresponds to a fourth order single component equation and would require four initial conditions to specify a solution. with the $c$ and $d$ we have two conditions... but I guess I was wrong. $\endgroup$
    – Kagaratsch
    Nov 7 '16 at 22:40
  • $\begingroup$ @Kagaratsch it does require four initial conditions: the four components of $\psi(0,\vec x)$. You need these four functions of $\vec x$ to specify $\psi(x)$ for all $t$! $\endgroup$ Nov 7 '16 at 22:41
  • $\begingroup$ Oh, but don't we only have two constraints in your equation $(2)$? Is that sufficient to fix all four components in $\psi$ in the most general case? $\endgroup$
    – Kagaratsch
    Nov 7 '16 at 22:43
  • $\begingroup$ @Kagaratsch recall that $u^\dagger \psi=u_1\psi^1+u_2\psi^2+u_3\psi^3+u_4\psi^4$, so that the inner products $u^\dagger\psi$ and $\psi^\dagger v$ contain the four functions $\psi_1,\psi_2,\psi_3,\psi_4$. You need to know these four functions to know the value of $c,d$. Once you calculate $c,d$ with $(2)$, you can plug their value back to $(1)$ to find $\psi$ for all $t$. But to use $(2)$ you need to known the initial value of the four components of $\psi$, thus four initial conditions. I hope it's clear now; otherwise tell me and I'll try to explain it again $\endgroup$ Nov 7 '16 at 22:45
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    $\begingroup$ @Kagaratsch well I was thinking right now how to answer to your question from yesterday, but I guess it's clear already :-) glad to hear that $\endgroup$ Nov 8 '16 at 21:27

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