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Let us assume we have a have a fixed hermitian operator on an Hilbert space $\mathcal{H}$, that written in its eigenbasis is $$ H = \sum_k e_k \vert e_k\rangle \langle e_k \vert $$ and we have a second hermitian operator $V$ $$ V = \sum_j v_j \vert v_j\rangle \langle v_j\vert $$ The two operator do not commute, i.e. $$ [H,V]=0 $$ My question is: is it possible to write the following decomposition $$ V = V_\parallel + V_\perp $$ where $$ [H,V_{\parallel}]=0 $$ and $$ [H,V_{\perp}]\neq 0 $$ with $\text{Tr}[V_{\perp} V_{\parallel}]=0$, i.e. the two operator are orthogonal? Is this last condition on the orthogonality necessary to have the decomposition?

In case we can do that, there is a characterisation in terms of the $\{e_k,\vert e_k\rangle\}$ and $\{v_k,\vert v_k\rangle\}$? In other words, there is a closed formula for $V_\parallel$ in terms of these two pairs of eigenvalues and eigenvector?

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    $\begingroup$ Wouldn't $V_\parallel$ be just the diagonal part (in the $\vert e_k\rangle$ basis) and $V_\perp$ the non-diagonal part of $V$? (I'm assuming no repeated eigenvalues.) $\endgroup$ Dec 8, 2021 at 19:51
  • $\begingroup$ Can you reassure the reader that you have simulated this with 4x4 matrices? $\endgroup$ Dec 8, 2021 at 20:29

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Taking a hint from one of the comments, write $\hat{V}$ in the eigenbasis of $\hat{H}$ as \begin{align} \hat{V} &= \sum_{jk} v_{jk}|e_j\rangle\langle e_k|\,, \end{align} and separate out the diagonal terms, i.e., \begin{align} \hat{V} &= \hat{V}_{\parallel}+\hat{V}_{\perp}=\sum_j v_{jj}|e_j\rangle\langle e_j| + \sum_{j\neq k} v_{jk}|e_j\rangle\langle e_k|\,. \end{align} Clearly, the parallel component commutes with $\hat{H}$, because they share an eigenbasis. Furthermore, since \begin{align} \hat{V}_{\perp}^{\dagger}\hat{H} &= \left( \sum_{j\neq k} v_{jk}^*|e_k\rangle\langle e_j| \right) \left( \sum_n h_{n}|e_n\rangle\langle e_n| \right) = \sum_{j\neq k}\sum_n h_{n}v_{jk}^*|e_k\rangle\langle e_j|e_n\rangle\langle e_n| \\&= \sum_{j\neq k}h_{j}v_{jk}^*|e_k\rangle\langle e_j| \,, \end{align} it is clear that this has no diagonal elements, and so it's trace is zero. Therefore, $\hat{V}_{\perp}$ is orthogonal to $\hat{H}$ under the trace-norm. Similarly, it is orthogonal to $\hat{V}_{\parallel}$.

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  • $\begingroup$ How does it change if H has some degeneracy? Does it simply mean that there is not unique decomposition since there is no unique eigenbasis of H? $\endgroup$ Dec 9, 2021 at 7:20
  • $\begingroup$ Seems that way to me! $\endgroup$
    – march
    Dec 9, 2021 at 7:22
  • $\begingroup$ Also, you tacitly assumed that the operator have the same rank, i.e. that the dimension of the basis of $H$ and of $V$ are the same, since otherwise you would not be able to write $V$ in the eigenbasis of $H$. How does it change if they do not have the same rank? $\endgroup$ Dec 9, 2021 at 8:53
  • $\begingroup$ They're both self-adjoint operators acting on the same space. How could they have different rank? $\endgroup$
    – march
    Dec 9, 2021 at 16:21

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