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"Consider an operator $A = r - a$, where $r$ is an operator and $a$ is a constant. Consider only those state kets $V_i$ in the Hilbert space such that $AV_i = 0$ ($A$ acting on $V_i$). Define a projection operator $P_A$ that projects an arbitrary ket on to the states $V_i$'s, i.e. the subspace of kets annihilated by $A$.

Then $P_AAP_A = 0$. So $P_ArP_A = aP_A$."

Now further in the literature which includes whatever I've quoted above, the following is stated:

Suppose $H = (1/r^2)p_\theta^2 + \lambda(r - a)$ (where $p_\theta$ is an operator). It follows that

$P_AHP_A = (1/a^2)p_\theta^2$.

Could anyone please elaborate how the last implication comes about? Is it that since $rV_i = aV_i$, $(1/r^2)V_i = (1/a^2)V_i$?

Thanks...

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The question is trivial except that the assertion isn't true. $P_A H P_A$ may be rewritten as $p_\theta^2/a^2\cdot P_A$ which differs by $P_A$, an extra factor, from your formula. If the operator only picks the wave functions near $r=a$, you clearly can't eliminate this fact and erase $P_A$, can you? It's not enough that $r$ has already been rewritten as $a$ which is possible if it (or its function) is sandwiched between two $P_A$ operators. One still can't forget $P_A$ because $P_A\neq 1$. – Luboš Motl Jun 21 '12 at 13:53
Yes, thanks! What I wrote in the question is exactly what's written in the paper I quoted. That's why I got confused. I indeed remembered what Ron said and it wasn't compatible with what was written. Thanks for clarifying. – 1989189198 Jun 22 '12 at 11:02

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Lubos's comment aside, the reason that the identity you write works on the subspace of definite eigenvalues is because of the way functions of operators are defined. Given an operator A with eigenvectors $V_i$ and eigenvalues $A_i$, the operator f(A) is defined as the operator with eigenvectors $V_i$ and eigenvalues $f(A_i)$. If f is a power series, this is equivalent to substituting the operator A in the power series, since acting on any eigenvector, A just turns into multiplication by a number.

(but you should fix the question, as Lubos says)

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