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I'm aware that the coherence of a statistical ensemble can be measured by looking at the off-diagonal elements of its density matrix, which is computed as the weighted sum of the density matrices of all the possible quantum states that it could be in. However, I recently thought about what would happen if we summed a statistical ensemble of density matrices, each with significant magnitudes of its off-diagonal terms, and the off-diagonal terms interfered destructively to produce zero coherence. Is this a problem with using ensemble density matrices to determine how coherent a system is? Or if there is destructive interference in the way mentioned above, does that really mean that the system is incoherent?

For example, say that a system has a $50 \%$ chance of being in the pure state $\frac1{\sqrt2}\left(|1\rangle+|2\rangle\right)$, with density matrix $\begin{pmatrix}\frac12&\frac12\\\frac12&\frac12\end{pmatrix}$, and a $50\%$ chance of being in the pure state $\frac1{\sqrt2}\left(|1\rangle-|2\rangle\right)$, with density matrix $\begin{pmatrix}\frac12&-\frac12\\-\frac12&\frac12\end{pmatrix}$. Both of these two options exhibit nonzero coherence, and yet the weighted sum of their density matrices is $\begin{pmatrix}\frac12&0\\0&\frac12\end{pmatrix}$, which implies zero coherence.

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    $\begingroup$ Can you add some equations? I think in this way it will be a lot easier for others to follow your thoughts. $\endgroup$ Feb 19, 2023 at 14:43
  • $\begingroup$ @TobiasFünke Just did! $\endgroup$ Feb 19, 2023 at 15:21

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Are you familiar with von Neumann entropy? It is a standard way of capturing coherence. It is given by: $$ S = -\text{Tr }(\rho\ln\rho) $$ Intuitively, a pure state is maximally coherent and the normalised identity is maximally incoherent. This is captured by the entropy as it is minimal, $S=0$, on pure states and maximal on the normalised identity, $S=\ln d$ ($d$ the dimension of the Hilbert space).

More generally, $S$ is concave, so this translates how convex combinations of density states increase decoherence. In particular, a convex combination of pure state will typically have a positive entropy even if individually it is null. This captures the phenomenon you observed with off diagonal entries.

Hope this helps.

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The non-diagonal elements for the density matrices that you write for $(|1\rangle\pm|2\rangle)/\sqrt{2}$ are simply the result of a basis transformation. If we worked in the same basis in which your last density matrix is written (i.e., $(|1\rangle\pm|2\rangle)/\sqrt{2}$), they would have forms $$ \begin{pmatrix}1&0\\0&0\end{pmatrix}\text{ and } \begin{pmatrix}0&0\\0&1\end{pmatrix} $$ Both correspond to a pure state, and this does not change when we transform them to basis $|1\rangle,|2\rangle$.

When you assemble a system that has 50% probability to be in one of these states, you make an implicit assumption that the coherence is zero. Indeed, there are other possibilities - e.g., state $|1\rangle$ corresponds to 50% probability to be in either of $(|1\rangle\pm|2\rangle)/\sqrt{2}$, and so does state $|1\rangle$. However, if we assume that the density matrix is either in $|+\rangle=(|1\rangle+|2\rangle)/\sqrt{2}$ or in $|-\rangle=(|1\rangle-|2\rangle)/\sqrt{2}$, then our density matrix is already specified as lacking coherence $$ \rho=0.5|+\rangle\langle +|+0.5|-\rangle\langle -| = 0.5\begin{pmatrix}1&0\\0&1\end{pmatrix}, $$ and no basis transformation can cure that. That the sum in different bases is the same is the result of the equal probabilities that you chose - this makes density matrix proportional to the identity matrix, which does not change under basis transformations.

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