Hermitian operators

  • Thread starter rbnphlp
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
470
58
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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