# 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? Commented Aug 23, 2019 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. Commented Aug 23, 2019 at 19:00
• Who says the wavefunction has to be finite everywhere? Commented Aug 23, 2019 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. Commented Aug 23, 2019 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. Commented Aug 23, 2019 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 $$\psi(x_0)dx_0$$