TITLE: Derivation of $\langle \psi|\psi_n\rangle\langle \psi_n|\psi\rangle = |\langle \psi_n|\psi\rangle|^2$
QUESTION [0 upvotes]: I'm trying to find a derivation for the relationship between $$\int|\Psi(\vec{r},t)|^2\ d^3r = 1 = \int \Psi^*(\vec{r},t)\ \Psi(\vec{r},t) \ d^3r.$$
Is this considered a postulate or is there an intuitive way to explain this?
The reason is the fact that the Bra-ket notation depends on the fact that $\langle \psi|\psi_n\rangle\langle \psi_n|\psi\rangle = |\langle \psi_n|\psi\rangle|^2$, which I can't get to work out by drawing the projections like below, where blue is $|\psi\rangle$ and red is $|\psi_n\rangle$.

As one can see in the drawing, the projections do not have the same length, therefor, how can one conclude with $\langle \psi|\psi_n\rangle\langle \psi_n|\psi\rangle = |\langle \psi_n|\psi\rangle|^2$?
I have assumed that $\langle \psi_n|\psi\rangle$ means the projection of $\psi$ on $\psi_n$ as stated in the textbook.

REPLY [3 votes]: This is just standard multiplication of complex numbers. If we have a complex number $z$, then $z^*z=|z|^2$ 
$$z=a+bi$$
$$z^*=a-bi$$
$$z^*z=(a-bi)(a+bi)=a^2+b^2=|z|^2$$
In our case, $z=\langle\psi_n|\psi\rangle$, and $z^*=\langle\psi|\psi_n\rangle$, so the result you want is true. It doesn't matter what the projections actually are.
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