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The bloch vectors are given by

$$u=\frac{1}{2}(\rho_{ge}+\rho_{eg})$$ $$v=\frac{1}{2i}(\rho_{eg}-\rho_{ge})$$ $$w=\frac{1}{2}(\rho_{gg}-\rho_{ee})$$

Where $\rho$ is the density matrix for a two level system. The first two elements, $u$ and $v$, are proportional to the in phase and quadrature components of the atomic dipole moment.

Why is this and what does this mean physically?

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  • $\begingroup$ This is extremely unclear and very hard to parse. It will be much easier to get answers if you provide a good deal more context for the question and details about what you're asking. $\endgroup$ – Emilio Pisanty May 27 '18 at 9:33
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It is assumed that $e$ and $g$ in the subscripts represent "excited state" and "ground state". In that case, one can equate $$\hat{\rho}_{eg} = |e\rangle \langle g| = a^{\dagger}|g\rangle \langle g| $$ and $$\hat{\rho}_{ge} = |g\rangle \langle e| = a|e\rangle \langle e| . $$

In a crude sense (where one neglects the projection operators $|g\rangle \langle g|$ and $|e\rangle \langle e|$), one can represent $u$ and $v$ as $$u = \frac{1}{2}\left(\hat{\rho}_{ge}+\hat{\rho}_{eg}\right) \sim \frac{1}{2}\left(a+a^{\dagger}\right) = \hat{q}$$ and $$v = \frac{i}{2}\left(\hat{\rho}_{ge}-\hat{\rho}_{eg}\right) \sim \frac{i}{2}\left(a-a^{\dagger}\right) = \hat{p} , $$ assuming appropriate definitions of the quadrature operators $\hat{q}$ and $\hat{p}$. Therefore, $u$ and $v$ can be associated with the quadrature operators (in a crude sense).

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