# A commuting decomposition with respect to a fixed operator

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?

• 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.) Dec 8, 2021 at 19:51
• Can you reassure the reader that you have simulated this with 4x4 matrices? Dec 8, 2021 at 20:29

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 $$\hat{V}_{\parallel}$$.
• 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? Dec 9, 2021 at 8:53