Maylis said:
If the inner product is defined for ##f(x)## and ##g(x)##
$$ \langle f|g \rangle = \int_{-\infty}^{\infty} f^{*}(x)g(x) \hspace {0.03 in} dx,$$ then what is wrong with
$$ \langle \psi |\psi \rangle = \int_{-\infty}^{\infty} \psi^{*}(x) \psi(x) \hspace {0.03 in} dx?$$
Nothing, but that's not what you wrote before.
It may help to connect the new notation to what you've learned before. Suppose you have a vector ##\vec{a}##. In the new notation, you'd write ##\lvert a \rangle##. To keep things simple and familiar, let's say it lives in three-dimensional space. Now you decide how you're going to orient the x, y, and z axes. In linear algebra speak, you're choosing a basis. Once you've done this, it makes sense to talk about the x-component of ##\vec{a}## and so on, and you can now write
$$\vec{a} = a_1 \hat{e}_1 + a_2 \hat{e}_2 + a_3 \hat{e}_3 = \sum_{i=1}^3 a_i \hat{e}_i$$ where the ##\hat{e}_i##'s correspond to the unit vectors (basis vectors) pointing along the various axes. In the new notation, you'd write
$$\lvert a \rangle = a_1 \lvert 1 \rangle + a_2 \lvert 2 \rangle + a_3 \lvert 3 \rangle = \sum_{i=1}^3 a_i \lvert i \rangle.$$ To find ##a_i##, you'd calculate ##a_i = \hat{e}_i \cdot \vec{a}##. In Dirac notation, you have ##a_i = \langle i \vert a \rangle##. So finally we can write
\begin{align*}
\vec{a} &= \sum_{i=1}^3 \hat{e}_i (\hat{e}_i \cdot \vec{a}) \\
\lvert a \rangle &= \sum_{i=1}^3 \lvert i \rangle\langle i \vert a\rangle
\end{align*}
Now let's go back to quantum mechanics… You have the state vector ##\lvert \psi \rangle##. The set of vectors {##\lvert x \rangle##} form a basis; they correspond to {##\lvert i \rangle##} in the previous example where ##x## acts as a label as ##i## did earlier. ##x## is continuous, so we replace the summation by an integral
$$\lvert \psi \rangle = \int \lvert x \rangle \langle x \vert \psi \rangle\,dx.$$ Just as each ##a_i = \langle i \vert a \rangle## is a number, ##\langle x \vert \psi \rangle## is a number. It's equal to the wave function ##\psi## evaluated at ##x##. That is, ##\psi(x) = \langle x \vert \psi \rangle##. You need to distinguish between the state vector ##\lvert \psi \rangle##, the wave function ##\psi##, and ##\psi(x)##, the value of the wave function maps ##x## to.
Now consider the dot product of two vectors ##\vec{a}## and ##\vec{b}##. Once we choose a basis, we can write
\begin{align*}
\vec{a}\cdot\vec{b} &= \sum_{i=1}^3 a_i b_i = \sum_{i=1}^3 (\vec{a}\cdot\hat{e}_i)(\hat{e}_i \cdot \vec{b}) \\
\langle a \vert b \rangle &= \sum_{i=1}^3 a_i b_i = \sum_{i=1}^3 \langle a \vert i \rangle \langle i \vert b \rangle
\end{align*} In the continuous (and complex-valued) case, we have
$$\langle f \vert g \rangle = \int f^*(x)g(x)\,dx = \int \langle f \vert x \rangle \langle x \vert g \rangle.$$ In what you wrote back in post 6, you didn't put ##\psi(x)## in the integral; you used ##\lvert \psi \rangle##. That's why you got a non-sensical expression. It was analogous to saying
$$\vec{a}\cdot\vec{a} = \sum_{i=1}^3 \vec{a}\cdot\vec{a} = \sum_{i=1}^3 (a_1^2+a_2^2+a_3^2) = 3(a_1^2+a_2^2+a_3^2).$$ In the very first step, instead of correctly using ##a_i## and regular scalar multiplication, I inserted ##\vec{a}## and used vector multiplication, which leads to an erroneous result.
