# Is the Dirac $\delta$-function necessarily symmetric?

The Dirac $$\delta$$-function is defined as a distribution that satisfies these constraints:

$$\delta (x-x') = 0 \quad\text{if}\quad x \neq x' \quad\quad\text{and}\quad\quad \delta (x-x') = \infty \quad\text{if}\quad x = x'$$

$$\int_{-\infty} ^{+\infty} \delta(x-x')\, dx = 1$$

Some authors also put another constrain that that Dirac $$\delta$$-function is symmetric, i.e., $$\delta(x)=\delta(-x)$$

Now my question is, do we need to separately impose the constraint that the Dirac $$\delta$$-function is symmetric or it automatically comes from other constrains?

Well, to illustrate my query clearly, I am going to define a function like that: $$ξ(t)=\lim_{\Delta\rightarrow0^+} \frac{\frac{1}{3}{\rm rect}\left(\frac{2x}{\Delta}+\frac{1}{2}\right)+\frac{2}{3}{\rm rect}\left(\frac{2x}{\Delta}-\frac{1}{2}\right)}{\Delta}$$ where $${\rm rect}(x)$$ is defined as: $${\rm rect}(x)= 1 \quad\text{if}\quad |x| < \frac{1}{2} \quad\quad\text{and}\quad\quad {\rm rect}(x)= 0 \quad\text{elsewhere}.$$ $$ξ(t)$$ is certainly not symmetric, but it does satisfy the following conditions, $$ξ(t)= 0 \quad\text{if}\quad t \neq 0 \quad\quad\text{and}\quad\quad ξ(t)= \infty \quad\text{if}\quad t = 0$$ $$\int_{-\infty} ^{+\infty} ξ(t)\,dt = 1$$

Now, my question is, can we define $$ξ(t)$$ as Dirac Delta function or not?

• I suggest you have a look at generalized functions, (theory of distributions), where Dirac delta is defined rigorously. Jan 11, 2021 at 14:05
• Thanks for your suggestion. Jan 11, 2021 at 14:27
• Generically, a distribution with support in a single point does not need to be symmetric, A example is < T | f > = f ' (0). Surely a good answer should go along classification of such distributions. Jan 12, 2021 at 1:12
• Could you edit your function, to show the dependence on $t$? The typo is confusing.. Were you trying to define the distribution $1/3 f(0^-) + 2/3 f(0^+)$ ? Jan 12, 2021 at 15:34
• Yes, I was trying to define the distribution as $\frac{1}{3}f(0^{-})+\frac{2}{3}f(0^{+})$ Jan 13, 2021 at 4:21

"Delta function" is not a function, but a distribution. Distribution is a prescription for how to assign number to a test function. This distribution may but does not have to have function values in the ordinary sense. In case of delta distribution, it does not have function values.

So statement like

$$\delta(x) = \delta(-x) \quad\text{for all }x \tag{*}$$ meaning "value of $$\delta$$ at $$x$$ equals value of $$\delta$$ at $$-x$$" is meaningless/invalid.

But statement $$\int dx~ \delta(x) f(x) = \int dx~\delta(-x) f(x) \quad \text{for all functions }f \tag{**}$$ may be valid.

You can easily verify that the function of $$\Delta$$ and $$x$$ ( the expression after the limit sign in definition of $$\xi$$) does not satisfy either of these two statements (in the role of $$\delta$$). So it is not "symmetric".

The delta distribution can hypothetically satisfy only the second statement. Does it do so?

We can evaluate both sides of the equality. The left-hand side has value, by definition of $$\delta(x)$$, $$f(0)$$.

We can transform the right-hand side integral into $$\int dx~\delta(-x) f(x) = \int dy~\delta(y) f(-y)$$ By definition of $$\delta(y)$$,value of this integral is $$f(0)$$, the same as the left-hand side. So (**) is satisfied.

The equation $$\delta(x) = \delta(-x)$$ is thus consequence of the definition of $$\delta(x)$$, it is not independent assumption.

Your function $$\xi$$ may actually obey the second statement too (and thus be symmetric in that sense), even though the $$\Delta$$-dependent expression after the limit sign does not. This is similar for other approximations of delta distribution; the approximation may not have properties of $$\delta$$ (such as symmetry), but the limit does.

• Treating it as a function is good for trolling mathematicians because it's continuous across an asymptote. Jan 11, 2021 at 23:37

The symbol $$\delta(x\!-\!y)\tag{A}$$ with two arguments $$x,y\in\mathbb{R}$$ is an informal kernel notation for the Dirac delta distribution $$u~\in~ D^{\prime}(\mathbb{R}^2)\tag{B}$$ defined as

$$u[f]~:=\int_{\mathbb{R}}\!\mathrm{d}z~f(z,z)\tag{C}$$

for testfunctions $$f~\in~ D(\mathbb{R}^2).\tag{D}$$ It follows that the Dirac delta defined as above is symmetric $$\delta(x\!-\!y)~=~\delta(y\!-\!x), \tag{E}$$ cf. OP's title question.

• I knew the Dirac distribution as defined on a domain of real functions of one real variable. Why a domain of functions $f$ depending on two variables? Jan 11, 2021 at 14:46
• Because two variables are closer to most physics applications. The symmetric condition $\delta(x)=\delta(-x)$ can of course also be formulated & proven in one dimension in a similar fashion. Jan 11, 2021 at 15:26

Delta function is a distribution, defined on a set of functions. Mathematicians usually express this by using the bra-ket notation, where delta function is the bra $$<\delta|$$ and $$<\delta| f> = \int \delta(x) f(x) dx = f(0)$$

Were you speaking of the set of continuous functions, I believe you would not need the symmetry requirement. But this is not usually the case. In quantum mechanics, we use the set of square integrable functions; this is a mild requirement, that allows for discontinuities.

Now, if you are considering functions that can be discontinuous at zero then you need to define explicitly what to do, the symmetric delta distribution should be

$$<\delta | f > = \frac{f(0^+)+f(0^-)}2$$

and you could have another different "delta functions" that work the same in continuous functions but work differently in the case of discontinuity.

BONUS: in one dimensional quantum mechanics, you have a whole set of "delta - like potential barriers" defined by the multiple ways to connect $$\Psi'(0^+),\Psi(0^+)$$ to $$\Psi'(0^-),\Psi(0^-)$$. Nomenclature is a nightmare here, due to errors in textbooks. Each "delta" or "barrier supported in a single point" can be seen as a rule to join the intervals $$(-\infty, 0)$$ and $$(0, \infty)$$.