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Thread: Prove A=C_dagger C is Hermitian and positive

  1. #1
    Senior Member oldguynewstudent's Avatar
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    Prove A=C_dagger C is Hermitian and positive

    An operator A is said to be positive if {<\psi|A|\psi>} \ge 0 for all states | \psi> in the space.

    PROVE: If C is an arbitrary operator the products A=C^\dagger C and B=CC^\dagger are Hermitian and positive.

    Since C is an arbitrary operator then {C|\psi>} = {|\phi>} for some |\phi> in the space.

    So <\psi|C^\dagger C|\psi> = <\phi|\phi>

    Then A^\dagger = (C^\dagger C)^\dagger

    Which gives <\psi|(C^\dagger C)^\dagger|\psi> = <\psi|C C^\dagger|\psi> = (|\phi> |(\psi^*) >< (\psi) | <\phi|)^* = <\phi|\phi>

    Can someone verify if I got this correct so far or if I totally botched this up?

    Thanks
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  2. #2
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    Re: Prove A=C_dagger C is Hermitian and positive

    An operator $A$ is Hermitian if $A^{\dagger} = A$.

    It is easy to check that $(C^{\dagger}C)^{\dagger} = C^{\dagger}(C^{\dagger})^{\dagger} = C^{\dagger}C$

    and $(CC^{\dagger})^{\dagger} = (C^{\dagger})^{\dagger}C^{\dagger} = CC^{\dagger}$

    Note that $|\psi\rangle^{\dagger} = \langle \psi|$, so that for $A = C^{\dagger}C$:

    $\langle \psi|A|\psi\rangle = \langle\psi|C^{\dagger}C|\psi\rangle$, so taking $C|\psi\rangle = |\phi\rangle$

    we have $\langle\psi|C^{\dagger}C|\psi\rangle = \langle\phi|\phi\rangle \geq 0$, by the positive-definiteness of the inner product.

    A similar observation holds for $CC^{\dagger}$ taking $|\phi\rangle = C^{\dagger}|\psi\rangle$.

    (Basically we are regarding $\langle x|y\rangle$ as $|x\rangle^{\dagger}|y\rangle$, which is common for complex inner-product spaces).
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  3. #3
    Senior Member oldguynewstudent's Avatar
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    Re: Prove A=C_dagger C is Hermitian and positive

    Thank you so much. I was on the right track but messed up a few details.
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