- #1
rbnphlp
- 54
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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
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