# Why Dirac delta function is considered as wave function?

I have read in my textbook that dirac delta function is regarded as wave function. But isn't it violating the condition required for a wave function ie .

It becomes infinite $$δ(x-x_0)$$ at $$x=x_0$$ but wave function should be finite everywhere How can this be explained?? Book- Quantum physics by Hc verma, page 27, reprint 2018 edition.

• A delta function is sometimes considered as a potential well or potential barrier $V(x)$ in the Schrodinger equation. Could that be what you read, rather than $\psi(x)$ being a delta function? – G. Smith Aug 23 '19 at 18:40
• In the book they clearly state that the wavefunction is not realistic; however, it may be useful to conceptually understand how a fully 'localised' particle wavefunction would look like if you collapsed it to a definite position. – Akerai Aug 23 '19 at 19:00
• Who says the wavefunction has to be finite everywhere? – Javier Aug 23 '19 at 19:01
• You are used to delocalized plane waves in coordinate space. Consider such in momentum space and inspect their Fourier transform, representing them in coordinate space. – Cosmas Zachos Aug 23 '19 at 19:20
• @dmckee I understand that. My question was more of a rhetorical one; the point I would have made (if OP replies) is that a finiteness requirement is a bit naive: the value of a PDF at a point doesn't matter, what you want is square integrability. – Javier Aug 23 '19 at 19:49

The delta function here can be thought of a position eigenstate, but as it is not square-integrable, it cannot be an actual wave function. Hence, "unrealistic". An actual, square-integrable wave function can be thought as being made of a superposition of dirac deltas, however: $$\psi(x) = \int \psi(x_0) \delta(x - x_0) dx_0.$$
You can think of this as a superposition of dirac deltas at every possible $$x_0$$ along the real axis, with weight $$\psi(x_0)dx_0$$