I am trying find $$\iiint_{\mathbb{R}^3} d^{3}q ~\delta^{3}(\vec{q})\frac{(\vec{p}\cdot\vec{q})^2}{q^{2}},$$ where $\vec{p}$ is some fixed vector.

The answer should be $\frac{p^2}{3}$. Below is my attempt, which seems to lead to the wrong answer $\frac{p^2}{2}$.

Attempt: Let's align $q_{z}$ with $\vec{p}$, so we measure $\theta$ wrt $\vec{p}$. Since there is no $\phi$ dependence so I can write $$\delta^{3}(\vec{q})=\frac{\delta(q)\delta(\theta)}{2\pi q^{2}\sin(\theta)}.$$

Therefore I have

$$p^{2}\int_{0}^{\infty} dq \delta(q)\hspace{1mm}\int_{-\pi}^{\pi}d\theta\hspace{1mm} \delta(\theta)\cos^2\theta .$$

I understand $$\int_{0}^{\infty}\delta(q)dq = \frac{1}{2},$$ if we treat $\delta(q)$ as a limiting case of a symmetric Gaussian distribution. While the $\theta$ integral is $1$. So my answer to my question is $\frac{p^2}{2}$. Which is different from the correct answer $\frac{p^2}{3}$.

So my questions are:

  1. What went wrong in my derivation?

  2. How do you derive and justify the answer $\frac{p^2}{3}$ from first principles?

  • 1
    $\begingroup$ Cross posted: math.stackexchange.com/q/1030142 $\endgroup$
    – Kyle Kanos
    Commented Nov 23, 2014 at 3:44
  • $\begingroup$ @Kyle Kanos: Since this triple integral is usually not defined in mathematics, this might be a case where a physicist's perspective is called for, cf. meta.physics.stackexchange.com/q/5713/2451 $\endgroup$
    – Qmechanic
    Commented Nov 23, 2014 at 12:44
  • 1
    $\begingroup$ Why complicate this with cylindrical coordinates? just use Cartesian coordinates, and let $\delta^{3}({\vec p}) = \delta(p_{x})\delta(p_{y})\delta(p_{z})$ $\endgroup$ Commented Nov 23, 2014 at 14:49
  • $\begingroup$ By definition, delta distribution $\delta^{(3)}(\vec q)$ returns the value of the integrated function at $\vec{q}=\vec{0}$. Unless $\vec{p}=\vec{0}$, the function $|\vec{p}\cdot\vec{q}|^2/q^2$ has no value at $\vec{0}$. Hence the delta cannot operate. Check how this integral prescription was obtained. There is probably an error somewhere. $\endgroup$ Commented Nov 23, 2014 at 14:57
  • $\begingroup$ @Jerry : I tried that already, doesn't simplify anything. $\endgroup$
    – HuShu
    Commented Nov 23, 2014 at 15:23

2 Answers 2



  1. In mathematics, a distribution is usually only defined wrt. smooth testfunctions. However the function ${\bf q}\mapsto({\bf q}\cdot{\bf p})^2/q^2$ is not continuous at the origin ${\bf q}={\bf 0}$. Nevertheless, we can e.g. try to evaluate the triple integral using the following representation of the 3D Dirac delta distribution $$\tag{1} \delta^3({\bf q})~=~ \lim_{\varepsilon\to 0^+} \frac{1}{4\pi} \frac{3\varepsilon}{(q^2+\varepsilon)^{\frac{5}{2}}}, \qquad q~:=~|{\bf q}|,$$ where it is implicitly understood that the limit $\lim_{\varepsilon\to 0^+}$ should be taken after the triple integration.

  2. For given $\varepsilon>0$, the integrand is integrable on $\mathbb{R}^3$. And it is bounded at the origin ${\bf q}={\bf 0}$, so we can use spherical coordinates. As OP mentions, in spherical coordinates with ${\bf p}$ along the $z$-axis, we have $$\tag{2}\frac{({\bf q}\cdot{\bf p})^2}{q^2}~=~p^2\cos^2\theta.$$

  3. Substitute $q\to \sqrt{\varepsilon}q$ in the triple integral. The $\varepsilon$-dependence disappears. Perform the triple integral.

  • $\begingroup$ Thanks! But I would like to ask why is your definition of Dirac delta independent of angles, since I have a $\cos^2\theta$ in the integrand? $\endgroup$
    – HuShu
    Commented Nov 23, 2014 at 15:17

$$δ^3(q⃗ )=\frac{δ(q)δ(\theta)}{2\pi q^2\sin(\theta)}$$ is wrong. The delta function is spherically symmetric, and thus has no θ dependence. Simply use: $$d^3(q⃗ )=\frac{δ(q)}{2\pi q^2}$$ instead. Use the Jacobian when you switch coordinate systems (from Cartesian to spherical) ($r^2 \sin(\theta)$), and you should get the result.

  • $\begingroup$ Thanks! However, I am confused about the $\cos^2\theta$ term, I thought that made it not spherically symmetric. What do you mean by the dirac delta being spherically symmetric, isn't a dirac delta by definition always spherically symmetric? $\endgroup$
    – HuShu
    Commented Nov 23, 2014 at 3:40
  • $\begingroup$ It is by definition spherically symmetric. The cos($theta$) term comes from the dot product multiplying the delta function, not from the delta function itself... $\endgroup$ Commented Nov 23, 2014 at 17:17
  • $\begingroup$ I agree, but isn't the most general form of $\delta (\vec{r})=\frac{\delta(r)\delta(\theta)\delta(\phi)}{r^2\sin\theta}$ in spherical coordinates? Isn't the $\theta$ from the dot product related to this $\theta$? $\endgroup$
    – HuShu
    Commented Nov 23, 2014 at 17:36

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