I was studying quantum theory and came to know that the unrealistic function representing the position of a particle can be represented in the form of an infinitely peaked Gaussian or simply the Dirac delta function $\delta(x-x_0)$ where $x_0$ is the definite position with zero uncertainty. The Dirac function can also be written as $$\delta(x-x_0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ik(x-x_0)}dk$$ or simply $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ik(x)}dk\tag{1}$$ Now in solving a problem where it was asked to find the momentum distribution of the given position wave function given as $$\psi(x)=Ae^{\frac{i}{\hslash}p_0x}$$ where $A$ is a real positive constant. While solving for the momentum wave function $a(p)$ using Fourier transform I get, $$a(p)=\frac{A}{\sqrt{2\pi\hslash}}\int_{-\infty}^{\infty}e^{ikx}dx$$ where the $k={(p_0-p)}/{\hslash}$ . Now just by using the equation $(1)$, I get, $$a(p)=\sqrt{\frac{2\pi}{\hslash}}A\delta(k)\\\implies a(p)=\sqrt{\frac{2\pi}{\hslash}}A\delta\left(\frac{p_0-p}{\hslash}\right)\tag{2}$$ This was simply my answer, but the book shows a step forward which I am unable to understand how they did this. It shows the final answer as $$a(p)=\sqrt{{2\pi\hslash}}A\delta(p-p_0)\tag{3}$$ which is obviously the expected one as the position wave function $\psi(x)$ given is the basis eigenfunction for momentum, so we can expect to get $a(p)$ as in equation $(3)$ so that we can find the momentum of the particle to be $p_0$ with zero uncertainty. So my question at last is that how to operate such a Dirac function for which the argument is a fraction(as in equation $(2))$?

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    $\begingroup$ $\delta(a x) = \frac{ \delta(x) }{ | a | }$. PS - If this is tripping you up, you need to spend much more time understanding Dirac delta functions before going further into quantum mechanics. Also, you very clearly did not do proper research since its the FIRST equation on the Wiki article for properties of Dirac delta functions - en.wikipedia.org/wiki/Dirac_delta_function#Properties $\endgroup$
    – Prahar
    Jul 10, 2021 at 12:01
  • $\begingroup$ ohh yes I forgot about this totally $\endgroup$
    – Jack
    Jul 10, 2021 at 12:07
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    $\begingroup$ No problem! These things will be second nature in time. $\endgroup$
    – Prahar
    Jul 10, 2021 at 12:14
  • $\begingroup$ @PraharMitra Please post answers as answers, not as comments. $\endgroup$ Jul 10, 2021 at 13:08

1 Answer 1


$\delta(ax) = \frac{1}{|a|} \delta(x)$


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