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Hermitian Of An Operator

  1. Aug 27, 2007 #1
    How can i show that [tex](a_{1} A_{1}+a_{2} A_{2})^{\dagger}=a_{1}^{\ast} A_{1}^{\dagger}+a_{2}^{\ast} A_{2}^{\dagger}[/tex]

    notice: [tex]a_{1},a_{2}\in C[/tex] and [tex]A_{1}^{\dagger},A_{2}^{\dagger}[/tex] are hermitian conjugate of [tex]A_{1},A_{2} [/tex] operators
    Last edited: Aug 27, 2007
  2. jcsd
  3. Aug 27, 2007 #2


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    Use the [itex]\dagger [/itex] 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.
  4. Aug 27, 2007 #3
    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.
  5. Aug 27, 2007 #4


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

    [tex]\langle \phi,left(a_{1}A_{1}+a_{2}A_{2}\right)\psi\rangle [/tex] is the object that you should start from.
  6. Aug 28, 2007 #5
    I tried something.Is it regular?

    [tex]\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[/tex] then [tex]\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 [/tex] so [tex]\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 [/tex] in the end
    [tex]\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 [/tex] becauuse of [tex] a \left\langle\phi| \psi\right\rangle = \left\langle\phi|a \psi\right\rangle [/tex](a is complex constant) we can write [tex]\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 [/tex] in the end [tex]\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 [/tex]. we 've showed that [tex](a_{1} A_{1}+a_{2}A_{2})^{\dagger} = a_{1}^\ast A_{1}^\dagger + a_{2}^\ast A_{2}^\dagger [/tex]
    Last edited: Aug 28, 2007
  7. Aug 28, 2007 #6


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    It looks okay. For the first equality that you wrote you used that [itex] A^{\dagger\dagger}=A [/itex] which is okay for bounded operators in Hilbert spaces.
  8. Aug 29, 2007 #7
    Thank you for your help. see you later in another problem.
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