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Hermitian operators

  1. Oct 10, 2012 #1
    If A and B are hermitia operators , then prove (A+B)^n is also hermitian.

    Justw ondering if this would suffice ?

    ∫ ψ^*(A+B) ∅ dt= ∫((A+B) ψ)^* ∅ dt assuming (A+B) is hermitian

    I can do that again
    ∫ ψ^*(A+B) ∅ dt= ∫((A+B) ψ)^* ∅ dt

    multiply them together
    ∫((A+B) ψ)^(2*) ∅^2 dt

    and we contine to multiplying till n aand theyre stil hermitian .

    what do you guys think ? should I do soemthing else
     
  2. jcsd
  3. Oct 10, 2012 #2
    I don't think your multiplication of integrals has been done correctrly as you should use a new variable of integration for every integral you multiply. However, you can firstly see that if an operatore C is hermitian then C^n is hermitian. In fact, if you consider a scalar product [itex](\cdot,\cdot)[/itex]:

    $$(x,Cy)=(Cx,y)\Rightarrow (x,C^2y)=(Cx,Cy)=(C^2x,y)$$

    and you can repeat the same thing n times.
    So if A and B are hermitian then:

    $$(x,(A+B)y)=(x,Ay)+(x,By)=(Ax,y)+(Bx,y)=((A+B)x,y)$$

    If you call A+B=C then you have your proof.
     
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