Property of the adjoint operator in the array element

Question

In Quantum Mechanics how can I prove this property?

$$<\psi|A^{\dagger} |\phi>=<\phi|A|\psi>^{*}$$

Answer 1

"Sandwiching" an operator between a bra and a ket is standard notation, but I'm going to use somewhat clearer notation and then introduce the "sandwiching" at the end to make this more clear.

The inner product between a state $\psi$ and a second state $\phi$ is written like this:

$$\langle \psi | \phi\rangle$$ If I operate on $\phi$ with the operator $A$ before I take the inner product, then I would write it like this:

$$\langle \psi | A\phi\rangle$$

On the other hand, if I operate on $\psi$ with the operator $B$ before I take the inner product, then I would write that like this:

$$\langle B \psi | \phi\rangle$$

Lastly, because of the properties of the inner product, taking the complex conjugate is equivalent to flipping the order of the states, so

$$ \langle \psi | \phi \rangle^* = \langle \phi | \psi \rangle$$

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Now that that's been taken care of, the Hermitian conjugate of an operator $A$, which we denote by $A^\dagger$, is defined as follows: for any two states $\phi$ and $\psi$,

$$\langle\phi | A \psi\rangle = \langle A^\dagger \phi | \psi \rangle$$

Therefore, $$\langle \phi | A | \psi \rangle^* \equiv \langle \phi | A\psi\rangle^* = \langle A^\dagger \phi | \psi \rangle^* = \langle \psi | A^\dagger \phi \rangle \equiv \langle \psi | A^\dagger | \phi \rangle$$

where, at the beginning and the end, I've used the "sandwich" notation. Such notation is quite common and the reasons why it's questionable are fairly technical, so I wouldn't worry about them for now.