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I would like to demonstrate the following property:

$$(|\psi\rangle\langle\phi|)^{\dagger}=|\phi\rangle\langle\psi|$$

The other properties that concern the adjoint operator I have already been able to demonstrate. I'm missing this one.

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Note $$ \langle \psi_1 | A \psi_2 \rangle^* = \langle A \psi_2 | \psi_1 \rangle = \langle \psi_2 | A^\dagger \psi_1 \rangle $$ Let $A = |\psi \rangle \langle \phi |$. Then, $$ \langle \psi_1 | A \psi_2 \rangle^* = \langle \psi_1 | \psi \rangle^* \langle \phi | \psi_2 \rangle^* = \langle \psi_2| \phi \rangle \langle \psi | \psi_1 \rangle = \langle \psi_2| ( | \phi \rangle \langle \psi | ) | \psi_1 \rangle $$ Thus, $A^\dagger = | \phi \rangle \langle \psi |$.

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You might consider the fact that

$$ \langle \alpha | A \ \beta\rangle = \langle A^\dagger \alpha |\beta\rangle = \overline{\langle\beta|A^\dagger \alpha\rangle}$$

Where the line denotes complex conjugation (as the inner products are complex numbers). This fits a bit more neatly into the bra-ket notation used to define your operator.

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