Mean value of projection operator

Question

In an orthonormal eigenbasis ${ left|psi_1rightrangle , left|psi_2rightrangle}$ of the Hamiltonian operator $hat{H}$ we have the projection operator:
$$ hat{mathbb{P}} = prod_{i neq j} left( frac{hat{H} - lambda_i}{lambda_j - lambda_i} right)$$
and :
$$ hat{H} left|psi_1rightrangle = lambda_1 left|psi_1rightrangle$$
$$ hat{H} left|psi_2rightrangle = lambda_2 left|psi_2rightrangle$$
Dependent upon our choice:
$lambda_i = lambda_1 longrightarrow lambda_j = lambda_2$
$$ hat{mathbb{P}} = left|psi_1rightrangle leftlanglepsi_1right|$$
or $lambda_i = lambda_2 longrightarrow lambda_j = lambda_1$:
$$ hat{mathbb{P}} = left|psi_2rightrangle leftlanglepsi_2right|$$
If we have:
$$ left|psirightrangle = c_1 left|psi_1rightrangle + c_2 left|psi_2rightrangle$$
we just conclude that the mean value of $hat{mathbb{P}}$ is $|c_1|^2$ or $|c_2|^2$ or do we unify it somehow? This is confusing me because of the product, I know we only consider when $i neq j$ but that is dependent on our initial choice for $lambda_i$, right?

SolubleFish · Accepted Answer

In this case, there are $2$ projection operators :
begin{align}
P_1 = |psi_1ranglelanglepsi_1| = frac{H - lambda_2}{lambda_1-lambda_2} 
P_2 = |psi_2ranglelanglepsi_2| = frac{H-lambda_1 }{lambda_2-lambda_1}
end{align}
These to operators satisfy : $P_i P_j = delta_{ij}P_j$, so if the product defining your $mathbb P$ is over both $i$ and $j$, you have $mathbb P = 0$.
The expected value of $P_i$ in the state $|psirangle = c_1 |psi_1 rangle + c_2 |psi_2 rangle$ is $|c_i|^2$.

Mean value of projection operator

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