What happens when you replace an identity matrix with a matrix full of ones? In physics, we often use resolutions of identity
$$\sum_n |n\rangle\langle n|=\mathbb{I}$$ to simplify expressions. Sometimes, the "full matrix" (for lack of a better term) $$\sum_{m,n}|m\rangle\langle n|\equiv\mathbb{J}$$ appears instead. This has properties like
$$\mathbb{J}^N=\mathbb{J}\,\mathrm{Tr}[\mathbb{J}]^{n-1}$$ instead of the usual $\mathbb{I}^N=\mathbb{I}$. Can we say anything conclusive about the relationship between
$$\langle \psi|\mathbb{J}|\phi\rangle\qquad \mathrm{and}\qquad \langle \psi|\mathbb{I}|\phi\rangle=\langle\psi|\phi\rangle,$$ or is there no direct way of simplifying $\langle \psi|\mathbb{J}|\phi\rangle$?
This question came up in simplifying a sum of Clebsch-Gordan coefficients
\begin{align}
\sum_J \langle j_1,k_1;j_2,k_2|J,k_1+k_2\rangle\langle J,k_1^\prime+k_2^\prime|j_1^\prime,k_1^\prime;j_2^\prime,k_2^\prime\rangle&=\sum_{J,l,l^\prime} \langle j_1,k_1;j_2,k_2|J,l\rangle\langle J,l^\prime|j_1^\prime,k_1^\prime;j_2^\prime,k_2^\prime\rangle\\
&= \langle j_1,k_1;j_2,k_2|\mathbb{J}|j_1^\prime,k_1^\prime;j_2^\prime,k_2^\prime\rangle.
\end{align} (My scenario has $j_1^\prime=j_1$ and $j_2^\prime=j_2$ so the sum over $J$ is unique.) It would be nice if this constrained the possible relationships between the $k$s. The obvious problem is that $\mathbb{J}$ is basis-dependent, so I doubt any more simplications can arise.
 A: Never forget to think about statistics when considering quantum mechanics. Your question is related to the three correlations between pairs of three variables. Famously, "is correlated with" isn't as transitive as we'd like to think.
Let $|\chi\rangle:=\sum_m|m\rangle$. Write $\sim$ between complex numbers of the same modulus. We can choose three angles so$$\begin{align}\langle\psi|\chi\rangle&\sim\sqrt{\langle\psi|\psi\rangle\langle\chi|\chi\rangle}\cos\theta_{\psi\chi},\\\langle\chi|\phi\rangle&\sim\sqrt{\langle\phi|\phi\rangle\langle\chi|\chi\rangle}\cos\theta_{\chi\phi},\\\langle\psi|\phi\rangle&\sim\sqrt{\langle\psi|\psi\rangle\langle\phi|\phi\rangle}\cos\theta_{\psi\phi}.\end{align}$$In particular,$$\begin{align}\langle\psi|\mathbb{J}|\phi\rangle&\sim\langle\chi|\chi\rangle\sqrt{\langle\psi|\psi\rangle\langle\phi|\phi\rangle}\mathrm{Tr}(\mathbb{J})\cos\theta_{\psi\chi}\cos\theta_{\chi\phi},\\|\cos\theta_{\psi\phi}-\cos\theta_{\psi\chi}\cos\theta_{\chi\phi}|&\le|\sin\theta_{\psi\chi}\sin\theta_{\chi\phi}|.\end{align}$$These don't provide much in the way of constraints (but they're the best we can do), because the angles in question could be the internal angles of any triangle.
A: Perhaps, I do not quite understand the question, but the properties of the two matrix are rather obvious:
$$\mathbf{a}^T\mathbb{I}\mathbf{b}=\sum_{i,j}a_i\delta_{i,j}b_j=\sum_i a_i b_i=\mathbf{a}\cdot\mathbf{b}$$
$$\mathbf{a}^T\mathbb{J}\mathbf{b}=\sum_{i,j}a_ib_j=\sum_i a_i \sum_jb_j$$
This latter is sometimes useful for concize notation.
