\begin{align} 0 &= i \gamma^\mu \partial_\mu \psi(x) - m \psi(x) \\ &= \int \frac{d^4 k}{(2\pi)^4}e^{-i k x}\left( \gamma^\mu k_m \tilde{\psi}(k) - m \tilde{\psi}(k) \right) \\ &= \left( \gamma^\mu k_\mu - m \right)\tilde{\psi}(k), \qquad \text{i.e.} \quad (\cancel{k} - m)\tilde{\psi}(k)=0 \end{align}

How does it get from line 2 to line 3?

  • $\begingroup$ I removed the picture and put in real TeX instead. Please do this in future questions so that the links can't rot later on. Please note that I did not know how to do the slash through the $k$, but did a simple Google search to find out (and got an answer for the TeX Stack Exchange!). $\endgroup$ – DanielSank Apr 18 '15 at 3:27
  • $\begingroup$ cheers for the help. i am very bad at TeX coding $\endgroup$ – Clauston Joe Apr 18 '15 at 12:25

Line (2) states that the Fourier transform of a function is zero. Then that function is also zero, at least almost everywhere.

  • $\begingroup$ +1 for using "almost everywhere" (and, heck, it's Friday night). $\endgroup$ – Alfred Centauri Apr 18 '15 at 1:48
  • $\begingroup$ is there a page or website which i can refer to for this theorem? $\endgroup$ – Clauston Joe Apr 18 '15 at 12:26
  • $\begingroup$ It follows from that Fourier transform, considered as a linear transformation between function spaces, is invertible. To make the statement fully rigorous you need to think about which function spaces. This is covered in any functional analysis book that covers the Fourier transform. $\endgroup$ – Robin Ekman Apr 18 '15 at 14:13

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