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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?

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  • $\begingroup$ The real and imaginary part of the coherences are not a great measure. To see why, look at a superposition of two basis states on bases that differ by a complex phase in one of the two components. $\endgroup$ – Emilio Pisanty Nov 3 '17 at 23:24
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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 some orthogonal 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.

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  • $\begingroup$ Nice answer -- I'm not sure I agree that basis dependence robs a mathematical object of deep philosophical significance though. A choice of basis ultimately corresponds to a choice of configuration for how we will try to observe a system, with different representations in different bases corresponding to various probabilities we may observe upon many trials. This is pretty philosophical in and of itself, no? I agree that coordinate-independent tensor forms or whatever coordinate-free mathematics one likes present some fun philosophical questions, but coordinates aren't all that bad! $\endgroup$ – Alec Rhea Nov 4 '17 at 6:38
  • $\begingroup$ So in the state-basis ($|1\rangle,|0\rangle$) the real part of the off-diagonal element contributes to the occupancy of the $|+\rangle,|-\rangle$ coherences then if I understand correctly. Presumably then the imaginary component contains information regarding the phase relationship between the two states . $\endgroup$ – quasi_physicist Nov 7 '17 at 0:36

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