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In 2-fold degenerate perturbation theory we can find appropiate "unperturbate" wavefunctions by looking for simultaneous eigenvectors (with different eigenvalues) of and H° and another Hermitian operator A that conmutes with H° and H'.

Suppose we have the eingenvalues of H° are ##E_n = a(-n)^2 ##, with eivenvectors ##\vert n \rangle ## and ##\vert -n \rangle ## ( where a is some complex number). Since the squared minus sign is what is causing the degeneracy would it be valid to choose ##A = \sqrt{H°}## so that its eigenvalues would simply be ##E_n = a(n) ## and ##E_n = a(-n) ##?

Thank you for your time.

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# Conmutative Hermitian operator in degenerate perturbation theory

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