Find self-adjoint of $P=|a \rangle \langle b |$

Question

Let $P$ be an operator s.a. $P=|a \rangle \langle b |$ and $P|f \rangle = \langle b | f \rangle | a \rangle $.
Find the self-adjoint and the $P^2$ operator.

My attempt:
We know to find the self adjoint operator: $$ \langle g | P | f \rangle = \langle P^{\dagger}g|f \rangle \Leftrightarrow \int_{a}^{b} \langle g^{*} |a \rangle\langle b |f \rangle dx = \int_{a}^{b} P^{\dagger} g^{*} f dx $$ And from here, i can't find anything.

Answer 1

Several things:

• "self-adjoint" is an adjective, that indicates that something is equal to its adjoint. The noun is "adjoint".

• The equalities $P=|a\rangle\langle b|$ and $P|f\rangle=\langle b|f\rangle\,|a\rangle$ (for every $f$) say exactly the same.

• The bra-ket notation is pretty bad to deal with adjoints. The adjoint of $P$ is operator $P^\dagger$ such that $$\langle P^\dagger g,f\rangle=\langle g,Pf\rangle$$ for all $f,g$.

• Now you can calculate (and for the first part, the bra-ket notation is good, tough it can hide understanding): $$ \langle P^\dagger g,f\rangle=\langle g,Pf\rangle=\langle g|a\rangle\langle b|f\rangle. $$ In my view, this is better understood by treating the inner product properly: $$ \langle P^\dagger g,f\rangle=\langle g,Pf\rangle=\langle g,(\langle b|f\rangle)a\rangle=\langle b|f\rangle\,\langle g|a\rangle=\langle\overline{\langle g|a\rangle}\,b|f\rangle =\langle \,\langle a|g\rangle\,b|f\rangle=\langle |b\rangle\langle a|g\big|f\rangle. $$ AS this holds for all $f,g$, $$ P^\dagger=|b\rangle\langle a|. $$

• The whole thing has nothing to do with integrals, and operators of the form $P=|a\rangle\langle b|$ can be defined for $a,b$ in any Hilbert space. It doesn't have to be space of functions, and even then the inner product doesn't have to be defined by an integral.