Understanding the Equivalence of Dirac Delta Functions in Quantum Mechanics

[Mike2]

Dirac developed his delta function in the context of QM. But there are various functions under the integral that give the delta function. My question is does one Dirac delta function equal any other? Are all ways of getting the Dirac delta function equivalent? Thanks.

 

[Hurkyl]

In the theory of tempered distributions (which is one way to formalize the Dirac delta function), the identity of a distribution is entirely determined by the results you get by convolving it with Schwartz functions.

Therefore, if you compute a distribution g that satisfies the relation²
\int_{-\infty}^{+\infty} g(x) f(x) \, dx = f(0)
for every Schwartz function f, then g is equal to the Dirac delta distribution¹. And equality is transitive: if you compute another distribution h that also satisfies that relation, then g = h.


1: I will use this name, since the Dirac delta isn't a function
2: Note that this integral is a generalization of what you learned in calculus. In particular, it's not defined as a limit of Riemann sums.

 

[Meir Achuz]

DDFs can be defined in terms of a restricted class of functions f in Herky's integral. If the f's are different, the DDFs could be dlfferent.