Inner product of two continuous states

Question

I am revisiting this topic since I've neglected it the first few times I studied it.

A state of a particle can be represented as:

$$|\psi\rangle = \int dx \psi(x)|x\rangle$$

and naturally, its hermitian conjugate is: $$\langle \psi | = \int dx'\langle x'|\psi^*(x')$$

So shouldn't their inner product be:

$$\langle\psi|\psi\rangle = \iint dxdx' \langle x'|x \rangle \psi^*(x')\psi(x)~?$$

Answer 1

It looks like you answered yourself but i think it would be interesting to show you a different way to get there.

First we note that $\langle x|\psi \rangle=\int\psi(x') \langle x|x' \rangle dx'=\int\psi(x) \delta(x-x') dx'=\psi(x) $

We write the identity operator $I=\int|x \rangle \langle x|dx$.

We can quickly check that this is really the identity operator $$I|\psi \rangle =\iint\psi(x) |x'\rangle \langle x'|x \rangle dx'dx=\iint\psi(x) |x'\rangle \delta(x'-x) dx'dx=\int\psi(x') | x' \rangle dx'=|\psi\rangle$$

Now we can do some trickery using the identity operator $$\langle \psi|\psi \rangle = \langle \psi|I|\psi \rangle =\int \langle \psi|x\rangle \langle x|\psi\rangle dx=\int\psi^*(x) \psi(x) dx$$

Answer 2

Okay so I think I got it:

We can simplify:

$$\langle \psi | \psi \rangle = \iint dx dx' \langle x' | x \rangle \psi^*(x')\psi(x) =\iint dx dx' \delta(x-x') \psi^*(x')\psi(x)$$ $$ =\int dx \psi(x) \int dx'\delta(x-x')\psi^*(x')=\int dx \psi(x) \psi^*(x) = \int \psi^*(x) \psi(x) dx$$

And hence the known formula.