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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?

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  • $\begingroup$ Did you try to write a 3-dimensional integral over $\vec p$ with $p^0=\pm\sqrt{\vec p.\vec p+m^2}$? $\endgroup$
    – Christophe
    Commented Dec 18, 2020 at 7:33
  • $\begingroup$ With that I'd have $\delta({p^0}^2-\vec{p}^2-m^2)=\delta(\vec{p}^2+m'^2-\vec{p}^2-m^2)=\delta(m'^2-m^2)$. Now the same problems arise. Many different $p$ can give $m'=m$. $\endgroup$
    – TaeNyFan
    Commented Dec 18, 2020 at 8:28
  • $\begingroup$ Unfortunately I can't access the source, what's $a(p)$? $\endgroup$
    – Nihar Karve
    Commented Dec 18, 2020 at 8:43
  • $\begingroup$ @NiharKarve It is just a function of the 4 vector p. It is there to normalise the wave function. $\endgroup$
    – TaeNyFan
    Commented Dec 18, 2020 at 8:44
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    $\begingroup$ Usually, one performs the integration over $p^0$ first, writing $\delta((p^0)^2 - \vec p^2 - m^2)$. The delta is not simple but depends non-linearly on $p^0$ and one has to use the defintion of the Dirac delta in this case (see wikipedia). Since a quadratic equation has two roots, it will give a sum of two deltas, except only positive $p^0$ are permitted (this should also be assumed, e.g. by a Heaviside distribution $\theta(p^0)$). $\endgroup$
    – Cream
    Commented Dec 18, 2020 at 8:45

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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 $$\begin{align} \delta(p^2-m^2) &=\delta(p_0^2-\vec{p}^2-m^2) \\ &=\frac{1}{2\sqrt{\vec{p}^2+m^2}}\left[ \delta\left(p_0+\sqrt{\vec{p}^2+m^2}\right)+\delta\left(p_0-\sqrt{\vec{p}^2+m^2} \right)\right] \end{align}$$

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 $dp_x\ dp_y\ dp_z$.

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  • $\begingroup$ For example, $\psi(x)=\int d^4 p \ a(p) \delta(p^0+ \sqrt{\vec{p}^2+m^2}) = \int d^3 \vec{p}\ a(-\sqrt{\vec{p}^2+m^2}, \vec{p})$, right? $\endgroup$
    – TaeNyFan
    Commented Dec 18, 2020 at 9:23
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    $\begingroup$ @TaeNyFan yes, something like that. $\endgroup$
    – Thomas Fritsch
    Commented Dec 18, 2020 at 9:27
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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 $u(p)$.

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