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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.

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    $\begingroup$ WP. You've mastered its matrix exponential? $\endgroup$ Sep 15, 2021 at 17:57
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    $\begingroup$ Recall. $\endgroup$ Sep 15, 2021 at 17:59
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    $\begingroup$ "Direct way of simplifying..." $(\sum_m \psi^*_m)(\sum_n \phi_n)$ not direct enough? $\endgroup$ Sep 15, 2021 at 18:18
  • $\begingroup$ Thanks for those comments. The first one pushes the problem to determining $\langle \psi| e^{i \mathbb{J} }|\phi\rangle$, which (to my eyes) does not seem easier. From the second and third, we learn that things are nice if $\mathbb{J}$ is defined in a basis with $|\psi\rangle$ and $|\phi\rangle$ as elements or something similar, which presupposes knowledge of the inner products $\psi_m^*$ (no qualms). Products of sums of CGs don't seem as useful as sums of products of CGs, so there may be no more to do $\endgroup$ Sep 15, 2021 at 18:24
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    $\begingroup$ Note $\mathbb{J}$ is of the form $|\chi\rangle\langle\chi|$. $\endgroup$
    – J.G.
    Sep 15, 2021 at 18:27

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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.

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  • $\begingroup$ I like this perspective. Doesn't help my physical problem but it definitely answers my question! $\endgroup$ Sep 15, 2021 at 18:45
  • $\begingroup$ $\chi$ is not normalized, so these overlaps are not proportional to cos(stuff). $\endgroup$ Jan 26, 2022 at 15:19
  • $\begingroup$ @NorbertSchuch I hope my edit fixes it. $\endgroup$
    – J.G.
    Jan 26, 2022 at 15:24
  • $\begingroup$ Sounds more reasonable. But why do you say these are "internal angles of a triangle"? I'd say these are angles between three (arbitrary!) vectors in C^3, which could sum up to anything <= pi. $\endgroup$ Jan 26, 2022 at 18:42
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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.

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    $\begingroup$ In some basis, yes. It's too bad that the second is basis-dependent. $\endgroup$ Sep 17, 2021 at 16:27

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