Hermitian Conjugate of an Operator

First let us define the Hermitian Conjugate of an operator $H$ to be $H^\dagger$. The meaning of this conjugate is given in the following equation.

$$\bgroup\color{black} \langle\psi\vert H\vert\psi\rangle=\int\... ...psi\rangle \equiv \langle H^\dagger\psi\vert\psi\rangle \egroup$$

That is, $H^\dagger$ must operate on the conjugate of $\psi$ and give the same result for the integral as when $H$ operates on $\psi$.

The definition of the Hermitian Conjugate of an operator can be simply written in Bra-Ket notation.

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

Starting from this definition, we can prove some simple things. Taking the complex conjugate

$$\bgroup\color{black} \langle\psi \vert A^\dagger \phi\rangle=\langle A\psi\vert\phi\rangle \egroup$$

Now taking the Hermitian conjugate of $A^\dagger$.

$$\bgroup\color{black} \langle\left(A^\dagger\right)^\dagger \psi \vert\phi\rangle=\langle A\psi\vert\phi\rangle \egroup$$

$$\bgroup\color{black} \left(A^\dagger\right)^\dagger=A \egroup$$

If we take the Hermitian conjugate twice, we get back to the same operator.

Its easy to show that

$$\bgroup\color{black} \left(\lambda A\right)^\dagger=\lambda^* A^\dagger \egroup$$

and

$$\bgroup\color{black} \left(A+B\right)^\dagger=A^\dagger + B^\dagger \egroup$$

just from the properties of the dot product.

We can also show that

$$\bgroup\color{black} \left(AB\right)^\dagger=B^\dagger A^\dagger .\egroup$$

$$\bgroup\color{black} \langle \phi\vert AB\psi\rangle=\langle ... ...\rangle=\langle B^\dagger A^\dagger\phi\vert\psi\rangle \egroup$$

* Example: Find the Hermitian conjugate of the operator $a+ib$.*

* Example: Find the Hermitian conjugate of the operator ${\partial \over \partial x}$.*