Evaluating general solution of Dirac equation

Question

In Ashok Das Lecture on QFT book, pg. 40,  the solution of the Dirac equation for the general motion of a free particle with mass $m$ along an arbitrary direction is given by
$$psi (x)=int d^4p  a(p) delta(p^2-m^2)e^{-ipx}u(p),$$
where $x$, $p$ are the position and momentum 4 vectors.
My question is how does one evaluate such an integral? I know that the dirac delta function $delta(p^2-m^2)$ picks out the value at $p^2=m^2$. However, there can be many different  four vectors $p$ that give $p^2=m^2$. How does one know which $p$ to pick?

Thomas Fritsch · Accepted Answer

With the help of (see Dirac delta function - Composition with a function)
$$delta(x^2-alpha^2)=frac{1}{2|alpha|}left[delta(x+alpha)+delta(x-alpha)right]$$
you can write
$$begin{align}
delta(p^2-m^2)
 &=delta(p_0^2-vec{p}^2-m^2) 
 &=frac{1}{2sqrt{vec{p}^2+m^2}}left[
 deltaleft(p_0+sqrt{vec{p}^2+m^2}right)+deltaleft(p_0-sqrt{vec{p}^2+m^2}
 right)right]
end{align}$$
You can insert this into
$$psi (x)=int d^4p  a(p) delta(p^2-m^2)e^{-ipx}u(p).$$
Then you see, only two values of $p_0$ contribute to the integration over $dp_0$,
and you are left with a sum of two 3-dimensional integrals over $dp_x dp_y dp_z$.

Buzz · Answer

The general solution of the free Dirac equation is not just one plane wave with a well-defined momentum, since that is not the most general state of a single particule.  The general solution is actually a superposition of waves with all possible momenta (and spins*).  So there can be contributions from every $p^{mu}$ that is on the mass shell, $p^{2}=m^{2}$.  There are a continuous infinity of values of $p$ allowed by the $delta$-function. Another way to see this is that that expression contains just a one-dimensional $delta$-function, which reduces the four integrations to just three integrations over the spatial components of $p$.
In practice, the only way to do the integral directly is to convert it into a three-dimensional integral over the spatial momenta and work from there.  That makes that formula not very useful for practical calculations.  What it is useful for is deriving further expressions, in a way that is completely relativistically covariant.  That expression treats space and time on equal footing, so it can be used to derive things like the Feynman propagator, which really is useful for practical computation.
*According to the usual notation, that expression is not actually the most general solution.  It only includes positive-frequency wave components; negative-frequency waves would have to have different spinors, $v(p)$ instead of $u(p)$.

Evaluating general solution of Dirac equation

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