What makes a superposition of states a coherent superposition?

  • #1
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. JAVANAINEN
Europhys. 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?
 

Answers and Replies

  • #2
rubi
Science Advisor
847
348
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|##.
 
  • #3
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?
 
  • #4
rubi
Science Advisor
847
348
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.
 

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