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.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Define the context. Also notation is just notation. $\endgroup$– MauricioSep 12, 2021 at 18:52
-
$\begingroup$ It's Gram Schmidt procedure. $\endgroup$– mumSep 12, 2021 at 19:06
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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.
-
$\begingroup$ What is $\langle f| $ in $|A\rangle = \sum_f a_f|f\rangle\langle f|$? $\endgroup$– mumSep 12, 2021 at 19:13
-
$\begingroup$ @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 $\langle e\vert$? $\endgroup$ Sep 12, 2021 at 20:09