# Quantifying coherence with density matrices

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.

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

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.
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.