The solution to the equation

\begin{equation} \int d^3k \; e^{i\mathbf{k}.\mathbf{x}}(k^2-m^2)\phi(\mathbf{k})=0 \end{equation}

(which appears in the Klein-Gordon equation solving) is said to be

\begin{equation} \phi(\mathbf{k})=j(\mathbf{k})\delta(k^2-m^2). \end{equation}

It is easy to see that the arbitrarity of $\textbf{x}$ obligates the overall coefficient of the exponential to be identically zero and that this solution meets this requirement.

My question is: why a $\delta$-function? Whouldn't any Kronecker-type (defined in a continuous set) with some finite coefficient do the trick? Is the arbitrarity of $j$ enough to make the $\delta$ factor to make this the most general solution?

I'm pretty sure my doubt comes from poor formal understanding of distributions, which causes me to not be confortable with just throwing delta functions everywhere (despite this being what life looks like in QFT).


By "Kronecker-type (defined in a continuous set)" I assume you mean a function defined as $$ \kappa(x) = \left\{ \begin{array}{c} 1~{\rm for}~x=0 \\ 0~{\rm elsewhere} \end{array} \right. $$ Such a function is of measure zero. In other words, if the solution is defined in terms of such a function and everything else in the definition is finite, then it would be a zero energy function.

  • $\begingroup$ Thanks! Could you refer me to anything that could make me understand those facts without having to go through a complete measure theory analysis? $\endgroup$ – GaloisFan Dec 16 '17 at 5:19
  • $\begingroup$ @GaloisFan: Would you agree that $\int\kappa(x) dx =0$ for any region of integration? Now insert a function $g(x)$ that is finite at 0 and you would still have $\int g(x) \kappa(x) dx =0$. One can now shift the functions and the result would still remain the same. $\endgroup$ – flippiefanus Dec 16 '17 at 15:45
  • $\begingroup$ Of course, but in our case (forgive if it's too obvious) against what would $\phi$ be integrated whose result shouldn't (physically) be zero? $\endgroup$ – GaloisFan Dec 16 '17 at 21:26
  • $\begingroup$ @GaloisFan: sorry, I don't understand the question. $\endgroup$ – flippiefanus Dec 17 '17 at 14:34
  • $\begingroup$ If the factor which caused my doubt must be a delta because otherwise the integral of $\phi$ times any operator $g$ would vanish, I suppose that $\int g(x)\kappa(x)dx=0$ for all $g$ is not an interesting result. This might be intuitively very easy to agree with, but could you give me one simple example of why that would be the case? $\endgroup$ – GaloisFan Dec 17 '17 at 22:45

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