- #1
dyn
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Hi.
I have just looked at a question concerning a free particle on a circle with ψ(0) = ψ(L). The question asks to find a self-adjoint operator that commutes with H but not p.
Because H commutes with p , i assumed there was no such operator.
The answer given , was the parity operator. It acts as PψN = ψ-N.The answer then states it can be seen to be self-adjoint by writing it as a matrix in this basis.
I do not know how to write the parity operator as a matrix in this basis. For starters is the basis not infinite-dimensional ?
Any help would be appreciated. Thanks
I have just looked at a question concerning a free particle on a circle with ψ(0) = ψ(L). The question asks to find a self-adjoint operator that commutes with H but not p.
Because H commutes with p , i assumed there was no such operator.
The answer given , was the parity operator. It acts as PψN = ψ-N.The answer then states it can be seen to be self-adjoint by writing it as a matrix in this basis.
I do not know how to write the parity operator as a matrix in this basis. For starters is the basis not infinite-dimensional ?
Any help would be appreciated. Thanks