# Hermitian Of An Operator

1. Aug 27, 2007

### buraqenigma

How can i show that $$(a_{1} A_{1}+a_{2} A_{2})^{\dagger}=a_{1}^{\ast} A_{1}^{\dagger}+a_{2}^{\ast} A_{2}^{\dagger}$$

notice: $$a_{1},a_{2}\in C$$ and $$A_{1}^{\dagger},A_{2}^{\dagger}$$ are hermitian conjugate of $$A_{1},A_{2}$$ operators

Last edited: Aug 27, 2007
2. Aug 27, 2007

### dextercioby

Use the $\dagger$ to denote the adjoint. Since you're considering the case of bounded operators, just apply the definition of the adjoint operator for a bounded operator.

3. Aug 27, 2007

### buraqenigma

Thanks But ?

I want to know that to show this equations must i use boundered operator.Can i use unbounded (going to +-infinity) operator to show this.

4. Aug 27, 2007

### dextercioby

Let's take for example $A_{1}, A_{2}:\mathcal{H}\longrightarrow \mathcal{H}$ and $\phi,\psi\in\mathcal{H}$. Then the adjoints of the operators exist and are defined everywhere in $\mathcal{H}$ and the same could be said of the sum operator and its adjoint.

$$\langle \phi,left(a_{1}A_{1}+a_{2}A_{2}\right)\psi\rangle$$ is the object that you should start from.

5. Aug 28, 2007

### buraqenigma

I tried something.Is it regular?

$$\left\langle\phi|(a_{1}A_{1}+a_{2}A_{2})^{\dagger}\psi\right\rangle = \left\langle(a_{1}A_{1}+a_{2}A_{2}) \phi|\psi\right\rangle$$ then $$\left\langle\phi|(a_{1}A_{1}+a_{2}A_{2})^{\dagger}\psi\right\rangle =\left\langle(a_{1}A_{1})\phi|\psi\right\rangle + \left\langle(a_{2}A_{2})\phi|\psi\right\rangle$$ so $$\left\langle\phi|(a_{1}A_{1}+a_{2}A_{2})^{\dagger}\psi\right\rangle =a_{1}^\ast \left\langle A_{1}\phi|\psi\right\rangle + a_{2}^\ast \left\langle A_{2}\phi|\psi\right\rangle$$ in the end
$$\left\langle\phi|(a_{1} A_{1}+a_{2}A_{2})^{\dagger}\psi\right\rangle =a_{1}^\ast \left\langle \phi|A_{1}^\dagger \psi\right\rangle + a_{2}^\ast \left\langle \phi|A_{2}^\dagger \psi\right\rangle$$ becauuse of $$a \left\langle\phi| \psi\right\rangle = \left\langle\phi|a \psi\right\rangle$$(a is complex constant) we can write $$\left\langle\phi|(a_{1} A_{1}+a_{2}A_{2})^{\dagger}\psi\right\rangle = \left\langle \phi|a_{1}^\ast A_{1}^\dagger \psi\right\rangle + \left\langle \phi|a_{2}^\ast A_{2}^\dagger \psi\right\rangle$$ in the end $$\left\langle\phi|(a_{1} A_{1}+a_{2}A_{2})^{\dagger}\psi\right\rangle = \left\langle \phi|(a_{1}^\ast A_{1}^\dagger + a_{2}^\ast A_{2}^\dagger) \psi\right\rangle$$. we 've showed that $$(a_{1} A_{1}+a_{2}A_{2})^{\dagger} = a_{1}^\ast A_{1}^\dagger + a_{2}^\ast A_{2}^\dagger$$

Last edited: Aug 28, 2007
6. Aug 28, 2007

### dextercioby

It looks okay. For the first equality that you wrote you used that $A^{\dagger\dagger}=A$ which is okay for bounded operators in Hilbert spaces.

7. Aug 29, 2007

### buraqenigma

Thank you for your help. see you later in another problem.