Finding the adjoint of a linear operator using Dirac notation.

Trying to answer this question and I am fairly new to Dirac Notation:

Let $|\psi\rangle$ and $|\phi\rangle$ be two states in a Hilbert Space and consider the linear operator $A:=|\psi\rangle\langle\phi|$. Show that the adjoint is given by $A^\dagger=|\phi\rangle\langle\psi|$.

Here's my go at it.

Using the definition for the adjoint as $$\langle A^{\dagger}x|y\rangle=\langle x| Ay\rangle$$ Using the right side we achieve $\langle x|\psi\rangle\langle \phi|y\rangle$. Now $$\langle x|\psi\rangle\langle \phi|y\rangle=\langle A^{\dagger}x|y\rangle$$ Inputting the given expression for $A^\dagger$ I get $$\langle x|\psi\rangle\langle \phi|y\rangle=\langle|\phi\rangle\langle\psi|x\rangle|y\rangle$$

I am unsure of what happens at $\langle|\phi\rangle$ and can't see how these two equate. If anyone could point me in the right direction I would greatly appreciate it.

• Notice that the dot product is sesquilinear. We then have $\langle A^{\dagger}x|y\rangle=\langle y| A^{\dagger}x\rangle ^*=\langle y| \phi\rangle ^*\langle\psi|x\rangle^*=\langle x| \psi \rangle\langle \phi | y\rangle$, as required. – vnd Nov 9 '15 at 20:17

Let $H$ be the Hilbert space and $\langle \cdot, \cdot \rangle$ the inner product. Fix $\psi, \varphi \in H$. Define the operator $A: H \to H$ by $$Ax = \psi \langle \varphi, x \rangle$$ and the operator $A^\dagger: H \to H$ by $$A^\dagger x = \varphi \langle \psi, x \rangle\,.$$ Now $$\langle x, Ay \rangle = \langle x, \psi \rangle \langle \varphi, y \rangle$$ and $$\langle A^\dagger x, y \rangle = \langle \psi, x \rangle^* \langle \varphi, y \rangle = \langle x, \psi \rangle \langle \varphi, y \rangle\,,$$ where I assumed that the inner product is linear in the second variable and conjugate-linear in the first variable. Thus $A^\dagger$ is the Hermitian adjoint of $A$.
Edit: The problem with your computation is that you misinterpreted $\langle A^\dagger x|$. Because $| A^\dagger x \rangle = | \phi \rangle \langle \psi | x \rangle$ we get $$\langle A^\dagger x| = \langle x | \psi \rangle \langle \phi |\,.$$
I would use that $\langle x|y\rangle=\langle y|x\rangle^*.$ Hence $$\langle A^{\dagger}x|y\rangle=\langle y|A^\dagger x\rangle^*=(\langle y|\phi\rangle\langle \psi|x\rangle)^*=\langle y|\phi\rangle^*\langle \psi|x\rangle^*=\langle \phi|y\rangle\langle x|\psi\rangle=\langle x|Ay\rangle.$$
If $B=|\phi\rangle\langle\psi|,$ then $$\langle Bx|y\rangle=\langle y|B x\rangle^*=(\langle y|\phi\rangle\langle \psi|x\rangle)^*=\langle y|\phi\rangle^*\langle \psi|x\rangle^*=\langle \phi|y\rangle\langle x|\psi\rangle.$$ Hence we can expand Dirac notation to $\langle Bx|y\rangle,$ where $B=|\phi\rangle\langle\psi|,$ in such a way that $$\langle Bx|y\rangle=\langle \phi|y\rangle\langle x|\psi\rangle.$$ And since it follows from properties it is the only reasable way to write it.