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I have to prove that $(|\Psi\rangle \langle \Phi|)^\dagger= |\Phi\rangle \langle \Psi|$

Hint: Prove that $\langle f|(|\Psi\rangle \langle \Phi|)^\dagger|g\rangle = ... = \langle f||\Phi\rangle \langle \Psi||g\rangle$

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  • $\begingroup$ show your attempt please, otherwise this is like "do my homework" $\endgroup$ – Physor Sep 28 '20 at 22:33
  • $\begingroup$ @Physor I know that $\langle f|(|\Psi\rangle \langle \Phi|)^\dagger|g\rangle = \langle (|\Psi\rangle \langle \Phi|)f||g\rangle$ or either $\langle f|(|\Psi\rangle \langle \Phi|)^\dagger|g\rangle = [\langle g|(|\Psi\rangle \langle \Phi|)|f\rangle]*$ I just don't know how to continue $\endgroup$ – schrodingal Sep 28 '20 at 22:56
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    $\begingroup$ I added another notation for clarity $\endgroup$ – Physor Sep 29 '20 at 0:09
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Let $A = |\Psi\rangle \langle \Phi|$, then $$ \langle f|A^\dagger|g\rangle = \langle g|A|f\rangle^* = \big(\langle g|\Psi\rangle \langle \Phi|f\rangle\big)^* = \langle g|\Psi\rangle^* \langle \Phi|f\rangle^* = \langle\Psi|g\rangle\langle f|\Phi\rangle = \langle f|\Phi\rangle \langle\Psi|g\rangle, $$ so $A^\dagger = |\Phi\rangle \langle\Psi|$.

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The notation obsecures the meaning but I'll try to cope with it by writing the definition of the adjoint of $A$ as $$ \langle f|A|g\rangle = \overline{\langle g|A^\dagger|f\rangle}\\ $$ The given operator is in a tensor (dyad) form. It is to be understood as scaling the lefthand ket by the inner product of the right bra with the argument, (on which the operator is applied) $$ \langle f||\Phi\rangle \langle \Psi||g\rangle = \overline{\langle \Phi||f\rangle \langle g||\Psi\rangle} = \overline{ \langle g||\Psi\rangle \langle \Phi||f\rangle} = \overline{ \langle g|(|\Phi\rangle \langle\Psi |)^\dagger|f\rangle} $$

In a more clear notation and for better explanation we do it without the Dirac notation but the meaning of the operator is as above $$ A = |\Phi\rangle \langle \Psi| = \Psi\langle\Phi,\cdot\rangle $$ this acts on $x$ like $$ Ax = \Psi\langle\Phi,x\rangle $$ The Definition of the adjoint of $A$ is $$ \langle f, Ag \rangle = \langle A^\dagger f, g \rangle $$ Proving the statement $$ \langle f, Ag \rangle = \langle f, \Psi\langle\Phi,g\rangle \rangle = \langle f, \Psi \rangle \langle\Phi,g\rangle = \langle \overline{\langle f, \Psi \rangle} \Phi,g\rangle = \langle \langle \Psi, f \rangle \Phi,g\rangle = \langle\Phi \langle \Psi, f \rangle ,g\rangle = \langle A^\dagger f ,g\rangle $$ Where $$ A^\dagger = \Phi \langle \Psi, \cdot \rangle = |\Psi\rangle \langle \Phi| $$

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