An Introduction to the Dirac Delta Function and Applications in Physics
If you have ever studied physics before at an undergraduate level, you have probably come across what is known as the Dirac delta function as it is essential to various fields in physics. However, for those who don’t know about it, or do not understand it really well, I would like to explain how the delta function works and go through some applications in this article (namely applications in electrodynamics and quantum mechanics). You can skip the two application sections if you want as I will be assuming some understanding of slightly more advanced concepts in those but if you are interested, feel free to read them.
Defining the Delta Function
While the delta function has the word ‘function’ in its name, this is actually a misnomer because it is not a function at all and is instead what is known as a ‘generalized function’ or a ‘distribution’. The delta function can be thought of as an infinitesimally high and narrow spike, but has the special property that its area is 1. This is illustrated in the figure above. We can express these properties of the delta function using simple mathematical equations as shown below:
The first equation shows that the delta function is infinitesimally tall and infinitesimally narrow (since it is only nonzero at a single point) while the second equations shows that the area of the delta function is indeed 1. While I have integrated from -∞ to ∞ here, as long as we are covering the center (which is at x = 0) that is all that matters.
Now, we can also shift this delta function horizontally if we wish by changing the x to an x - c as we would do with normal functions, where c is how much we want to shift by to the right. Hence, our shifted delta function becomes δ(x - c) which represents a spike centered at x = c.
The delta function also has a special property that it can ‘pick up’ the value of any ordinary function at some point. To see what I mean, let’s suppose we have some continuous function f(x). The product between f(x) and δ(x - c) must be f(c)δ(x - c) since δ(x - c) is 0 at any other point and hence only the value of f(x) at c matters. We can now integrate this product to get f(c) as shown below:
Where in the last step I used the fact that the area of the delta function is 1. This property of the delta function is something that becomes quite useful in certain situations as we will see later.
Finally, we can also define what is known as the three dimensional delta function:
Where r is the usual position vector. This, as you would expect, defines an infinitesimally tall and narrow spike centered at r = 0. All the properties that I have discussed before in the one dimensional case also follow naturally.
Now that we have defined the delta function in both 1D and 3D, we can begin to see why it is so useful in physics by looking at some applications.
Applications in Electricity and Magnetism
(For this section I will be assuming an understanding of vector calculus)
The Dirac delta function becomes quite important in the field of electricity and magnetism as certain electric fields can behave as delta functions. To see what I mean, consider the vector field, v = (1/r²)r where r refers to the radial unit vector (I would add a hat above if I could but unfortunately Medium does not support LaTeX). These type of vectors fields are very common in electrostatics because Coulomb’s law which is the basis for most electric fields is precisely in this form. Now, if we imagine what this vector field would look like visually, it would look like vectors pointing radially outward, getting weaker by 1/r² as you get further from the origin. The important part here is that we have a vector field that is always point radially outward, and hence we can deduce that it must have some positive divergence.
Well, we can attempt to confirm this by actually calculating the divergence. Using the formula for the divergence in spherical coordinates we can calculate ∇ ⋅ v:
Therefore, if we directly calculate the divergence, we end up getting zero which can’t be true considering how the vector field is always radially pointing outward. However, if we calculate the flux of this vector field through a surface of radius R, we get a nonzero value:
And so, evidently there is some outward flux occurring which contradicts what we just showed with the divergence. In fact, if we use the divergence theorem on this integral we get the following:
But this clearly can’t be true if ∇ ⋅ v is 0. While this might render the divergence theorem incorrect, we can actually resolve this problem quite easily.
This is precisely where the delta function comes into play. If we look at how we calculated the divergence earlier, we can see that this method is actually invalid exactly when r = 0 since we would be dividing by 0. Hence, while at all other r values the divergence is indeed zero, at exactly r = 0 it actually isn’t. This should remind you of the delta function as it has a very similar behavior. In fact, the divergence of v is actually a delta function:
If you plug this back into the equation where I used the divergence theorem, you will get ∫4πδ³(r)dV which is indeed 4π since the integral of the 3D delta function is just 1. Hence using the power of the delta function, we have managed to resolve our problem. Recalling that this vector field, v, that I have defined is quite prevalent in electricity and magnetism, this shows that the delta function can appear as intrinsic properties of electromagnetic fields and are essential in explaining various electromagnetic phenomena.
Applications in Quantum Mechanics
(For this section I will be assuming some understanding of quantum mechanics and linear algebra. This article I wrote before regarding the linear algebra of quantum mechanics may help.)
One last application of the delta function is in quantum mechanics. The delta function actually appears in several areas of quantum mechanics but I will just be looking at two of examples, namely the eigenfunctions of position and the orthonormality for a continuous spectrum of eigenvalues.
To begin, let’s try to determine what the eigenfunctions of position are. If we recall that the corresponding operator for position is just x, we can write the eigenvalue equation as the following:
Where f_λ(x) are the eigenfunctions of the position operator and λ is the corresponding eigenvalue (the textbook I am referencing uses the letter y instead of λ but I think that is a bit confusing since y is often associated with the y-axis). Now, the important thing to notice here is that x is a continuous variable that can take up any value while λ is just a fixed constant. Therefore for any value of x that is not exactly λ, the eigenfunction must be 0 since the coefficients in front are different. However, at exactly x = λ, the eigenfunctions can be a nonzero value since the equation would be satisfied. This behavior resembles the delta functions we have been talking about and hence the eigenfunction can be written as a delta function centered at x = λ:
Where A is the normalization constant which turns out to just be 1 since the area of δ(x - λ) happens to already be 1. This result actually should not be a surprise at all if we recall that the eigenfunction of an operator will have a definite value for whatever the corresponding dynamic variable is. Considering how this is just a probability density function for the position, a delta function is the only possible function that could have a definite position due to it being infinitesimally narrow.
Another way the delta function appears in quantum mechanics is if we consider the orthogonality of the eigenfunctions we just derived. Since the eigenfunctions of an observable operator must be orthogonal to each other, we must end up with some sort of orthogonality. We can confirm this by actually calculating the inner product between two separate position eigenfunctions, δ(x - λ) and δ(x - λ’):
Where in the last line I used ∫ f(x)δ(x - λ)dx = f(λ) by letting f(x) = δ(x - λ’). Notice what the final result tells us. Whenever λ ≠ λ’, the inner product is 0 but exactly when λ = λ’ we get a nonzero value which is precisely the conditions for orthogonality. One way to think of this final result is as a generalization of the Kronecker delta which is how orthogonality is usually defined when dealing with discrete eigenfunctions. However, in our case we have continuous eigenfunctions and so the delta function is required. The textbook I am referencing calls this an ‘ersatz’ orthogonality since it technically isn’t the same as orthogonality as we would usually define it, but I think it still captures the essence quite well. Additionally, this isn’t limited to just position but any other operator with a continuous spectrum of eigenvalues (such as momentum) would yield the same result.
To conclude, I hope you now understand what the delta function really is, and why it is so important in the world of physics. There is still so much more I haven’t talked about and other areas where it could be applied, but this article has gotten a bit too long so I will end it here. Thank you for reading.
References
English. (2023). Graph of a shifted Dirac delta function (1d) | universaldenker.org. Universaldenker.org. https://en.universaldenker.org/illustrations/1101
Griffiths, D. J., & Schroeter, D. F. (2020). Introduction to quantum mechanics. Cambridge University Press.
GRIFFITHS, D. J. (2023). Introduction to electrodynamics. CAMBRIDGE UNIV PRESS.
Shankar, R. (2014). Principles of Quantum Mechanics. Springer.
