Dirac's Delta Function and its Most Important Properties
Table of contents
The delta function is a useful mathematical object that finds application in many areas of theoretical physics. Starting in electrodynamics in the description of electric point charges as a unit charge density concentrated in a single point, up to quantum field theory in the description of quantum fields as operators.
Why do I need the delta function?
Consider a onedimensional electric charge density \(\rho(x)\) smeared out on a line. This can be as well a mass density or any other density function. We look at the charge density here as an example.
To calculate how large the total charge \(Q\) on this line is, we have to integrate the charge density \(\rho(x)\). Here we choose the integration limits \( a \) and \(b\) so that the total line charge \(Q\) is included:
What if we do not have a smeared charge density, but a single point charge \(Q\) sitting at the origin \(x=0\)?
The total charge density is then concentrated in a single point and zero everywhere else. How can we describe such a charge density mathematically?
The charge density must satisfy two properties if it is to describe a single point charge:

Charge density \(\rho(x)\) must be zero at any location \(x\) except where the point charge is located, that is, at \(x=0\):
$$ \begin{align} \rho(x) ~=~ 0, ~~~ x \neq 0 \end{align} $$ 
The line integral over the charge density must give us the value \(Q\) if the point charge is within the integration limits \(a\) and \(a\):
$$ \begin{align} \int_a^b \rho(x) \, \text{d}x ~=~ Q \end{align} $$On the other hand, if the charge is not somewhere between the points \(x=a\) and \(x=b\), then the integral should yield zero.
If we normalize the charge to the value \(Q=1\) and consider these two properties, then we write down the density with a Greek delta \(\delta\) and call it Dirac's delta function:
\int_a^b \delta(x) \, \text{d}x ~=~ 1, ~~~ a < 0 < b \end{align} $$
Even if the name may suggest, the delta function is mathematically not a function, but another mathematical object, which can be understood for example as a socalled Dirac's delta distribution or as a Dirac measure. Here, we are not interested in a mathematically clean definition. We just want to know how to work with the delta function.
The delta function is illustrated with an arrow located at the position of the point charge. The height of the arrow is usually chosen to represent the value of the integral, in this case 1.
Delta function at the origin
Let us now consider an integral of the delta function together with a test function \(f(x)\) (by "test function" is meant that it can have any "good" properties one can imagine. For example it is infinitely often differentiable):
Such an integral is very easy to calculate, because according to Eq. 4
you know that the delta function is zero everywhere except at the point \(x=0\). Thus, the product is also zero everywhere, except at the point \(x=0\). Only the function value \(f(0)\) remains. Since \(f(0)\) no longer depends on \(x\), we can place it in front of the integral:
The integral over the delta function is 1, according to Eq. 4
, if \(x=0\) lies between \(a\) and \(b\) (otherwise the integral is zero). So we know what the delta function does in the integral 5
when multiplied by a function \(f(x)\). It picks the value of the function at the position \(x=0\):
Shifted delta function
We can, of course, move the charge to another location on the \(x\)axis, for example to the positive location \( x = x_0 \). Then the delta function must be zero everywhere except at the new position. We change the argument of \(\delta(x)\) to \(\delta(xx_0)\). "Why \( x_0\)? Because we have moved the delta function to the positive direction. Also here the integral over the delta function is equal to 1 if \( x_0 \) lies between the integration limits \(a\) and \(b\). After all, we have only shifted the delta function to \(x_0\), so the result of the integral is the same as in the case of \(\delta(x)\):
Now what if the shifted delta function in the integral is multiplied by another function \(f(x)\)? \(\delta(xx_0)\) is zero everywhere except at the point \(x_0\). The shifted delta in the integral must pick as the function value \(f(x_0)\) at the position where the delta function is located.
You can prove this by substituting \(xx_0\) with \(y\): \( y = x  x_0 \).
 Then \(x\) becomes \(y+x_0\) in the argument of the function.
 The \(x  x_0\) in the argument of the delta function becomes \(y\).
 The derivative \( \frac{ \text{d}y }{ \text{d}x } \) is 1. Thus \(\text{d}x\) changes to \(\text{d}y\).
 The integration limits become \(ax_0\) and \(bx_0\).
As we know, \(\delta(y)\) picks the value of the function at the point \(y=0\). Therefore the integral gives the value \( f(0 + x_0) = f(x_0)\):
Note that we do not want to consider the case where the charge lies exactly on the boundary. Either we include a charge in the integration range or not at all. Nevertheless, it should be said that this exotic case can also be considered.
Delta function is symetric (even)
Next, let's look at what happens when we have \(x\) in the delta function:
Let's make a substitution \( y = x\):
 Thus \(f(x)\) becomes \(f(y)\).
 \(\delta(x)\) becomes \(\delta(y)\).
 The derivative of \( \frac{ \text{d} y}{ \text{d} x } \) is 1. Thus we substitute the \(\text{d} x\) with \(  \text{d} y \).
 The lower limit of integration becomes \(a\) because of the substitution. The upper limit becomes \(b\).
So the signs of \(a\) and \(b\) are swapped. If \(a\) was negative, \(a\) now becomes positive and if \(b\) was positive before, \(b\) is now negative. To make the integration start again from negative to positive, we reverse the integration boundaries and thus the sign of the integral. In this way the two minus signs cancel each other out.
We know what \(\delta(y)\) does on the the right hand side in the integral. It picks the value of \(f\) at the point \(y=0\). So the integral gives the value \(f(0) = f(0)\):
So it makes no difference whether we use \(\delta(x)\), as in Eq. 7
, or \(\delta(x)\), as in Eq. 15
, in the integral. The result in both cases is \(f(0)\):
Scaled argument of the delta function
Now what happens if we scale the \(x\) in the delta by a factor \(k\)?
If \(k\) is negative, then we can write it as a negative magnitude \( k = k\). Since the delta function is symmetric, we can omit the minus sign in front of the magnitude: \(\delta(k\,x) = \delta(k\,x)\). So we can also write the integral as follows:
Let's make again a substitution \( y = k \, x \):
 Thus \(f(x)\) becomes \(f\left(\frac{1}{k}\, y\right)\).
 \(\delta(k\,x)\) becomes \(\delta(y)\).
 The derivative of \( \frac{\text{d}y}{\text{d}x} \) is \(k\). Thus we substitute \(\text{d}x\) with \( \frac{1}{k}\,\text{d}y \).
 The lower integration limit becomes \( k \, a \) and the upper one becomes \( k \, b \).
With this we get:
Here you must note that \(k\) cannot be zero, so that you do not divide by zero.
Comparison with the Kronecker Delta
The properties of the Dirac delta \(\delta(xx_0)\) may remind you a bit of the definition of Kronecker delta \(\delta_{km}\), if we rename the letters: \(x:=k\) and \(x_0:=m\). Recall what Kronecker delta, does in a product with a vector component \(v_k\): \(v_k \, \delta_{km}\). It picks the \(m\)th vector component.
If you compare 25
with 11
, you see that a discrete summation (sum sign) in the case of the Kronecker delta is replaced by a continuous summation (integral sign) in the case of the delta function:
\int f(k) \, \delta(k  m) \, \text{d}k ~&=~ f(m) \end{align} $$
While with the Kronecker delta we can pick a vector component out of finitely many vector components, with the delta function we can pick a function value out of infinitely many function values.
 The Kronecker delta comes into play, when we mess around with vectors \(\boldsymbol{v}\) and their finitely many vector components \(v_k\).
 The delta function comes into play when we mess around with functions \(f\) and their indefinitely many function values \(f(k)\).
Threedimensional delta function
So far, we have considered only a onedimensional delta function. The charges or other point objects are usually located in a threedimensional space. If the unit point charge is located at the origin \((x,y,z) = (0,0,0)\), then we can describe its charge density with the product of three delta functions:
The line integral in Eq. 4
becomes a volume integral with this product of three delta functions and yields 1 when we integrate over a volume \(\mathcal{V}\) that includes the origin:
In order not to have to write three deltas, we combine them to one delta with a 3 on top and in the argument we write the position vector \(\boldsymbol{r}\) and for the displacement we write for example the vector \(\boldsymbol{r}_0\):
If the 3d delta function appears in the integral in a product with a scalar 3d function \(f(x,y,z)\), then the delta function works analogously as in the onedimensional case. It picks the value of the function at the position \(f(0,0,0)\):
Now you should have a solid knowledge of the Dirac delta. Next, you should work through a few exercises involving the delta function.
Exercises with Solutions
Use this formula eBook if you have problems with physics problems.Exercise: Simplify Integrals with the Delta Function
Calculate the following integrals containing a delta function \(\delta(x)\):
 $$\int^5_1 \left( 2x^2  x + 1 \right) \, \delta(x3) \, \text{d}x$$
 $$\int^{3}_0 x^3 \, \delta(x+2) \, \text{d}x$$
 $$\int^4_0 \cos(x) \, \delta(x\pi) \, \text{d}x$$
 $$\int \ln(x+3) \, \delta(x+1) \, \text{d}x$$
 $$\int^2_{3} \left( 6x + 2 \right) \, \delta(3x) \, \text{d}x$$
 $$\int^b_{\infty} 3 \, \delta(xa) \, \text{d}x$$
Solution to the exercise #1.1
We want to calculate the following integral: $$ \int^5_1 \left( 2x^2  x + 1 \right) \, \delta(x3) \, \text{d}x $$ Here \(f(x) = 2x^2  x + 1 \). And the position of the delta function is at \(x=3\).
First, we ask ourselves if the position of \(\delta(x3)\) lies in the integration interval [1, 5]. It does. Therefore, the integral is not necessarily zero and we must evaluate the function \(f(x)\) at the position \(x=3\) where the delta function is found: \begin{align} \int^5_1 \left( 2x^2  x + 1 \right) \, \delta(x3) \, \text{d}x ~&=~ 2\cdot 3^2  3 + 1 \\ ~&=~ 16 \end{align}
Solution to the exercise #1.2
We want to calculate the following integral: $$ \int^{3}_0 x^3 \, \delta(x+2) \, \text{d}x $$ Here \(f(x) = x^3 \). And the delta function is shifted to the negative and is at \(x=2\).
First, we ask ourselves if the position of \(\delta(x+2)\) lies in the integration interval [0, 3]. It does not. The integral is zero: \begin{align} \int^{3}_0 x^3 \, \delta(x+2) \, \text{d}x ~=~ 0 \end{align}
Solution to the exercise #1.3
We want to calculate the following integral: $$ \int^4_0 \cos(x) \, \delta(x\pi) \, \text{d}x $$ Here \(f(x) = \cos(x) \). And the position of the delta function is at \(x=\pi\).
First, we ask ourselves if the position of \(\delta(x\pi)\) lies in the integration interval [0, 4]. It does. Therefore, the integral is not necessarily zero and we must evaluate the function \(f(x)\) at the position \(x=\pi\) where the delta function is found: \begin{align} \int^4_0 \cos(x) \, \delta(x\pi) \, \text{d}x ~&=~ \cos(\pi) \\ ~&=~ 1 \end{align}
Solution to the exercise #1.4
We want to calculate the following integral: $$ \int \ln(x+3) \, \delta(x+1) \, \text{d}x $$ Here \(f(x) = \ln(x+3) \). And the position of the delta function is at \(x=1\).
First, we ask ourselves if the position of \(\delta(x+1)\) lies in the integration interval \([\infty, \infty]\). It does. Therefore, the integral is not necessarily zero and we must evaluate the function \(f(x)\) at the position \(x=1\) where the delta function can be found: \begin{align} \int \ln(x+3) \, \delta(x+1) \, \text{d}x ~&=~ \ln(1 + 3) \\ ~&=~ 0.693... \end{align}
Solution to the exercise #1.5
We want to calculate the following integral: $$ \int^2_{3} \left( 6x + 2 \right) \, \delta(3x) \, \text{d}x $$ Here \(f(x) = \left( 6x + 2 \right) \). And the position of the delta function is at \(x=0\). The scaling factor is \( k = 3 \).
First we question whether the position of \(\delta(3x)\) lies in the integration interval \([3, 2]\). It does. Therefore, the integral is not necessarily zero and we need to evaluate the function \(f(x)\) at the position \(x=0\) and multiply it by the factor \(1/k\): \begin{align} \int^2_{3} \left( 6x + 2 \right) \, \delta(3x) \, \text{d}x ~&=~ \frac{1}{3}\, \left( 6\cdot 0 + 2 \right) \\ ~&=~ \frac{2}{3} \end{align}
Solution to the exercise #1.6
We want to calculate the following integral: $$ \int^b_{\infty} 3 \, \delta(xa) \, \text{d}x $$ Here \(f(x) = 3 \) is a constant function. And the position of the delta function is at \(x=a\).
First we question whether the position of \(\delta(xa)\) is in the integration interval \([\infty, b]\). This depends on whether \(a\) is greater or smaller than \(b\).
 If \(a\) is GREATER than \(b\), then the delta function is outside the integration limits and the integral is zero in this case.
 If \(a\) is SMALLER than \(b\), then the delta function is within the integration limits and the integral is not necessarily zero in this case. Here we have to evaluate the function \(f(x)\) at the point \(x=a\): $$ f(a) = 3 $$
The overall result is thus: $$ \int^b_{\infty} 3 \, \delta(xa) \, \text{d}x \begin{cases} 3, &\mbox{} a < b \\ 0, &\mbox{} a > b \end{cases} $$