What makes a superposition of states a coherent superposition?

[vtahmoorian]

Hi everyone
I am investigating spontaneously generated coherence(SGC), I found that it happens when an excited atomic state decays to one or more closed atomic levels so that atom goes to a coherent superposition of states , Effect of State Superpositions Created by Spontaneous Emission on Laser-Driven Transitions.
J. JAVANAINENEurophys. Lett., 17 (5), pp. 407-412 (1992)

according to this article "spontaneous emission from a single
initial state may give rise to a coherent superposition of two (or more) receiving states"..
Now I have a question,

I am wondering when can we call a superposition of states a coherent one?

 

[rubi]

In QM, you can usually specify a set of physically relevant observables $\mathcal A$. A superposition $\left|\psi\right> = \alpha \left|\psi_1\right> + \beta \left|\psi_2\right>$ is said to be a coherent superposition of $\left|\psi_1\right>$ and $\left|\psi_2\right>$ if there is an $A\in\mathcal A$ such that $\left<\psi_1\right|A\left|\psi_2\right> \neq 0$.

The reason for this definition is that if there is no such $A$, the state can't be physically distinguished from the statistical mixture $\rho = |\alpha|^2 \left|\psi_1\right>\left<\psi_1\right|+|\beta|^2 \left|\psi_2\right>\left<\psi_2\right|$.

 

[vtahmoorian]

Thank you dear Rubi
I understand your first statement , it is related to the coherence condition,which is,having non zero off-diagonal elements of density matrix operator, right?
but can you explain more about your second statement?

 

[rubi]

The density matrix corresponding to the state $\left|\psi\right>$ from my earlier post would be $\rho_\psi = \left|\psi\right>\left<\psi\right|$. It differs from the $\rho$ I wrote earlier in the off-diagonal terms: $\rho_\psi = \rho + \alpha\beta^*\left|\psi_1\right>\left<\psi_2\right| + \alpha^*\beta\left|\psi_2\right>\left<\psi_1\right|$. However, for all physical observables $A\in\mathcal A$, the expectation values are the same: $\mathrm{Tr}(\rho_\psi A) = \mathrm{Tr}(\rho A)$. The off-diagonal terms don't contribute since $\left<\psi_1\right|A\left|\psi_2\right> = 0$, so the pure state $\rho_\psi$ can't be physically distinguished from the mixed state $\rho$. One says that $\left|\psi_1\right>$ and $\left|\psi_2\right>$ lie in different superselection sectors. From the form of $\rho$, you can see that the relative phase between $\left|\psi_1\right>$ and $\left|\psi_2\right>$ cancels out completely, so there can't be any interference.