Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

"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 $rV_i = aV_i$, $(1/r^2)V_i = (1/a^2)V_i$?


share|cite|improve this question
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
up vote 1 down vote accepted

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)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.