-1
$\begingroup$

In Quantum Mechanics how can I prove this property?

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

$\endgroup$
0
$\begingroup$

"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

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


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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.