Siddharth Bhat
I’ve been studying Quantum Mechanics from Shankar’s Principles of Quantum Mechanics recently, and came across the derivative of the Dirac delta function that had me stumped.
$ \delta'(x - x') = \frac{d}{dx} \delta(x - x') = -\frac{d}{dx'} \delta(x - x') $
I understood neither what the formula represented, how the two sides are equal.
Thankfully, some Wikipedia and Reddit (specifically /r/math and /u/danielsmw) helped me find the answer. I’m writing this for myself, and so that someone else might find this useful.
Terminology
I will call $\frac{d}{dx} \delta(x - x')$ as the first form,
and $-\frac{d}{dx'} \delta(x - x')$ as the second form
Breaking this down into two parts:
- show what the derivative computes
- show that both forms are equal
1. Computing the derivative of the Dirac Delta
Since the Dirac Delta function can only be sensibly manipulated in an integral, let’s stick the given form into an integral.
$$
\delta'(x - x') = \frac{d}{dx} \delta(x - x') \\
\int_{-\infty}^{\infty} \delta' (x - x') f(x') dx' \\
= \int_{-\infty}^{\infty} \frac{d}{dx} \delta(x - x') f(x') dx' \\
$$
Writing out the derivative explicitly by taking the limit,
$$
= \int_{-\infty}^{\infty} \lim{h \to 0} \; \frac{\delta(x - x' + h) - \delta(x - x')}{h} f(x') dx' \\
= \lim{h \to 0} \; \frac{ \int_{-\infty}^{\infty} \delta((x + h) - x') f(x') dx' - \int_{-\infty}^{\infty} \delta(x - x') f(x') dx'}{h} \\
= \lim{h \to 0} \; \frac{f(x + h) - f(x)}{h} \\
= f'(x)
$$
Writing only the first and last steps,
$$
\int_{-\infty}^{\infty} \delta' (x - x') f(x') dx' = f'(x)
$$
This shows us what the derivative of Dirac delta does. On being multiplied with a function, it “picks” the derivative of the function at one point.
2. Equivalence to the second form
We derived the “meaning” of the derivative. Now, it’s time to show that the second form is equivalent to the first form.
Take the second form of the delta function as the derivative,
$$
\delta'(x - x') = - \frac{d}{dx'} \delta(x - x') \\
\int_{-\infty}^{\infty} \delta' (x - x') f(x') dx' \\
= \int_{-\infty}^{\infty} - \frac{d}{dx'} \delta(x - x') f(x') dx' \\
$$
Just like the first time, open up the derivative with the limit definition
$$
= \int_{-\infty}^{\infty} \lim{h \to 0} \; - (\frac{\delta(x - (x' + h)) - \delta(x - x')}{h}) f(x') dx' \\
= \lim{h \to 0} \; \frac{ \int_{-\infty}^{\infty} \delta((x - h) - x') f(x') dx' - \int_{-\infty}^{\infty} \delta(x - x') f(x') dx'}{h} \\
= \lim{h \to 0} \; - \frac{f(x - h) - f(x)}{h} \\
= \lim{h \to 0} \; \frac{f(x) - f(x - h)}{h} \\
= f'(x)
$$
Conclusion
That shows that the derivate of the Delta Function has two equivalent forms, both of which simply “pick out” the derivative of the function it’s operating on.
$$
\delta'(x - x') = \frac{d}{dx} \delta(x - x') = -\frac{d}{dx'} \delta(x - x')
$$
Writing it with a function to operate on (this is the version I prefer):
First form:
$$
\int_{-\infty}^{\infty} \delta' (x - x') f(x') dx' = \\
\int_{-\infty}^{\infty} \frac{d}{dx}\delta(x - x') f(x') dx' = \\
f'(x)
$$
Second form:
$$
\int_{-\infty}^{\infty} \delta' (x - x') f(x') dx' = \\
\int_{-\infty}^{\infty} -\frac{d}{dx'}\delta(x - x') f(x') dx' = \\
f'(x)
$$
A note on notation
In a violent disregard for mathematical purity, one can choose to abuse notation and think of the above transformation as -
$$
\delta'(x - x') = \delta(x - x') \frac{d}{dx}
$$
We can write it that way, since one can choose to think that the delta function transforms
$$
\int_{-\infty}^{\infty} \delta'(x - x')f(x') dx' \to \\
\int_{-\infty}^{\infty} \delta(x - x')\frac{d}{dx}f(x')dx' = \\
\int_{-\infty}^{\infty} \delta(x - x') f'(x') = \\
f'(x)
$$
The original forms and the rewritten one are equivalent, although the original is “purer” than the other. Which one to use in is up to you :)
So, to wrap it up:
$$
\delta'(x - x') = \frac{d}{dx} \delta(x - x') = -\frac{d}{dx'} \delta(x - x') = \delta(x - x') \frac{d}{dx}
$$