Dirac Delta Function in Complex Plane

February 19, 2023  508 words 3 mins read  Join the Discussion

In this post, I’ll be deriving the formal integral expression for the Dirac delta function in a complex plane, $\delta^{(2)}(\alpha - \beta)$ where $\alpha$ is a complex variable, and $\beta$ is a complex number. I came across it for the first time while I was taking Quantum Optics in my Master’s studies. Then, I wondered how to derive it. So, the post is the result of this quest which I derived on the same day. Until now, I forgot to share it on this blog. I hope you will find it useful. Let’s get started!

In order to derive it, we need three ingredients:

  1. Dirac delta function $\delta^{(2)}(x, y)$ is independent of the coordinate basis in $(x, y)$-space. This allows us to do the seperation of variables. i.e. $$ \begin{aligned} \delta^{(2)}(x, y) = \delta(x) ~\delta(y). \end{aligned} $$ Note that I put $2$ inside the bracket (i.e. _$^{(2)}$) to warn you that it’s not a square of one dimensional Dirac delta function, but to represent that it’s in the two dimensions. We do not denote one dimensional Dirac delta function as $\delta^{(1)}(x - a)$, but we simply say $\delta(x - a)$.
  2. The Dirac delta function is an even distribution. i.e. $$\delta(-x) = \delta(x).$$
  3. Integral definition of the one dimensional Dirac delta function, $$\delta(x - a) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i p(x - a)} dp.$$

In the complex plane, we can write it as

$$ \begin{aligned} \delta^{(2)}(\alpha - \beta) &\equiv \delta^{(2)}(\Re(\alpha) - \Re(\beta), \Im(\alpha) - \Im(\beta))\\ &= \text{Using 1., we get} \\ &\hspace{0.5cm} \delta(\Re(\alpha) - \Re(\beta))~\delta(\Im(\alpha) - \Im(\beta))\\ &= \delta(\Re(\alpha - \beta)) ~ \delta(\Im(\alpha - \beta)) \\ &= \text{Using 2., we get} \\ &\hspace{0.5cm} \delta(-\Re(\alpha - \beta)) ~ \delta(\Im(\alpha - \beta)) \\ &= \text{Using 3., we get} \\ &\hspace{0.5cm} \left( \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-i p \Re(\alpha - \beta)} d p\right) \left( \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i q \Im(\alpha - \beta)} dq\right) \\\\ &= \text{Since, } \Re(z) = \frac{z + z^{*}}{2} \text{and } \Im(z) = \frac{z - z^{*}}{2i}.\\ &\hspace{0.4cm}\text{So, } \Re(\alpha - \beta) = \frac{(\alpha - \beta) + (\alpha - \beta)^{*}}{2} \\ &\hspace{0.4cm}\text{and } \Im(\alpha - \beta) = \frac{(\alpha - \beta) - (\alpha - \beta)^{*}}{2i}.\\ &\hspace{0.4cm}\text{We can write} \\ &\hspace{0.5cm} \frac{1}{4\pi^2} \int_{-\infty}^{\infty}e^{-ip \frac{(\alpha - \beta) + (\alpha - \beta)^{*}}{2}} ~e^{\cancel{i}q \frac{(\alpha - \beta) - (\alpha - \beta)^{*}}{2\cancel{i}}} dp~dq\\ &= \frac{1}{\pi^2} \int_{-\infty}^{\infty} e^{\left( \frac{q}{2} - i \frac{p}{2} \right)(\alpha - \beta)} e^{-\left( \frac{q}{2} + i \frac{p}{2} \right)(\alpha - \beta)^{*}} d\left(\frac{p}{2} \right) d\left(\frac{q}{2} \right)\\ &= \text{Define a complex variable } u := \frac{q}{2} + i \frac{p}{2} \text{. We get} \\ &\hspace{0.5cm} \frac{1}{\pi^2} \int_{-\infty}^{\infty} e^{u^{*} (\alpha - \beta)} ~ e^{-u (\alpha - \beta)^{*}} d(\Im(u))~d(\Re(u))\\ &= \text{Define } d(\Im(u))~d(\Re(u)) := d^{2}u. \\ &\hspace{0.5cm} \text{Using } (\alpha - \beta)^{*} = \alpha^{*} - \beta^{*} \text{. We get} \\ &\hspace*{-2.2cm}\therefore \boxed{\delta^{(2)}(\alpha - \beta) = \frac{1}{\pi^2} \int e^{u^{*} (\alpha - \beta) - u (\alpha^{*} - \beta^{*})} d^{2}u}. \end{aligned} $$

Heuristically, we can think of the above expression as a two-dimensional probability distribution. Thus, the important property is that the integral of $ \delta^{(2)}(\alpha - \beta)$ over the entire complex plane is equal to $1$. i.e. $$ \boxed{\int_{\mathbb{C}} \delta^{(2)}(\alpha - \beta) ~d^{2}\alpha = 1}. $$

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  • Damodar Rajbhandari
    Written by Damodar Rajbhandari, a PhD candidate in Theoretical Physics at the Jagiellonian University, Poland, working in the Causal Dynamical Triangulations group at the Institute of Theoretical Physics.
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