# Projection using Inner product

why is projection of A along e is
$$\langle{e}|{A}\rangle |{e}\rangle$$and not $$\langle{A}|{e}\rangle |{e}\rangle$$ ?
e is a unit vector.

• Define the context. Also notation is just notation. Sep 12, 2021 at 18:52
• It's Gram Schmidt procedure.
– mum
Sep 12, 2021 at 19:06

In this rather sloppy language, what you (implicitly) assume is that you can write $$|A\rangle = \sum_f a_f|f\rangle\langle f|$$, with the set of basisvectors $$|f\rangle$$ forming a complete set, i.e. $$\langle e|f\rangle = \delta_{ef}$$.
Then $$\langle e|A\rangle|e\rangle = \langle e|\sum_f a_f |f\rangle\langle f|e\rangle = \sum_f a_f \langle e|f\rangle\langle f|e\rangle = \sum_f a_f \delta_{ef}\langle f|e\rangle = a_e$$, which intuitively makes sense.
• What is $\langle f|$ in $|A\rangle = \sum_f a_f|f\rangle\langle f|$?
• @mum there’s a problem here… if you don’t know what is $\langle f\vert$ then what is the meaning of your $\langle e\vert A\rangle$, which contains $\langle e\vert$? Sep 12, 2021 at 20:09