Question 7.4: Find an operator that satisfies the requirements of the prec...

The perturbation  H^{'} has less symmetry than H^{0}. H^{0} had continuous rotational symmetry, but H= H^{0}+H^{'} is only invariant under rotations by integer multiples of π. For A, take the operator  R(\pi) that rotates a function counterclockwise by an angle π. Acting on our stats \psi_{a} and \psi _{b} operator R(\pi) that rotates a function counterclockwise by an angle π. Acting on our stats \psi_{a} and \psi_{b} we have

R(\pi )\psi ^{0}_{a}(x,y)=\psi ^{0}_{a}(-x,-y)=-\psi ^{0}_{a}(x,y)

R(\pi )\psi ^{0}_{b}(x,y)=\psi ^{0}_{b}(-x,-y)=-\psi ^{0}_{b}(x,y)

That’s no good; we need an operator with distinct eigenvalues. How about the operator that interchanges x and y? This is a reflection about a 45° diagonal of the well. Call this operator D. D commutes with both H^{0} and H^{'}, since they are unchanged when you switch x and y. Now,

D\psi ^{0}_{a}(x,y)=\psi ^{0}_{a}(y,x)=\psi ^{0}_{b}(x,y)
D\psi ^{0}_{b}(y,x)=\psi ^{0}_{b}(x,y)=\psi ^{0}_{a}(x,y)

So our degenerate eigenstates are not eigenstates of D. But we can construct linear combinations that
are:

\psi ^{0}_{\pm}=\pm \psi ^{0}_{a}+\psi ^{0}_{b}      (7.39)

Then

D(\pm \psi ^{0}_{a}+\psi ^{0}_{b})=\pm D\psi ^{0}_{a}+D\psi ^{0}_{b}=\pm\psi ^{0}_{b}+\psi ^{0}_{a}=\pm(\pm \psi ^{0}_{a}+\psi ^{0}_{b})

These are “good” states, since they are eigenstates of an operator D with distinct eigenvalues (\pm 1) , and D commutes with both H^{0} and  H^{'}.