I'm trying to better understand the off-diagonal terms of the density matrix – an often brought up question on this site I realise. Specifically my confusion at the moment concerns the interpretation of the real and imaginary components of these off-diagonal terms. The diagonal terms represent the populations, and the off-diagonal terms the coherences in the system. I'm reluctant to say the off-diagonal terms represent the probability of occupancy of the coherent superposition states because their values are complex. But they clearly represent something to that effect. Does anyone have some better intuition on this?

# [Physics] Density matrices, off-diagonal terms, coherences and correlations

coherencedensity-operatorsuperposition

## Best Answer

So long story short, those terms are basis-relative and cannot be given a truly deep philosophical meaning: being Hermitian, the density matrix is diagonal in

someorthogonal basis; you just aren't looking in the right one.With that said we can certainly look a little more in-depth at a state of the form $$\rho = \begin{bmatrix}p&q\\q^* & r\end{bmatrix}$$at which point we begin to see that $p = \operatorname{Tr} \big(\rho ~ |0\rangle\langle 0| \big)$ and $r = \operatorname{Tr} \big(\rho ~ |1\rangle\langle 1| \big)$ are precisely the values of these indicator observables which tell us how much this state $\rho$ occupies the $|1\rangle$ or $|0\rangle$ states on average.

If one were instead to use the state $|+\rangle = \sqrt{\frac12} |0\rangle + \sqrt{\frac12} |1\rangle$ for the indicator one would find that the occupation was instead $\frac12 + \operatorname{Re}(q),$ while the occupation for $|-\rangle = \sqrt{\frac12} |0\rangle - \sqrt{\frac12} |1\rangle$ is going to be just $\frac12 - \operatorname{Re}(q).$ If one instead uses $|0\rangle \pm i|1\rangle$ as one's orthogonal sets this becomes $\frac12 \pm \operatorname{Im}(q)$ as well.

In that sense we can take $q=x + i y$ and read two probabilities off of it: $\frac12 \pm x$ and $\frac12 \pm y.$ They tell us those probabilities for the $x$ and $y$ axes of the Bloch sphere whereas $p, r$ tell us about the $z$-axis. Just to unify these in notation you could write this as $$\rho = \frac12~I + \delta_x~\sigma_x + \delta_y~\sigma_y + \delta_z~\sigma_z,$$ with six probabilities $\frac12 \pm \delta_\bullet.$

Any off-diagonal term can be viewed this way indirectly; one can restrict one's view of the system to just those two states specified by the row and column, and one has $\frac12 (p + r) \pm \operatorname{Re} q$ for the residence on the $x$-quadrature of that two-level system.