Proving the Hermitian Conjugate Property of Operators

[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

 

[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.

 

[buraqenigma]

Thanks But ?

Thank you for your correction.

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.

 

[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.

 

[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

 

[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.

 

[buraqenigma]

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