Let's clarify the problem a bit. For the V type system, suppose the system starts in state $|\psi \rangle = \alpha |a\rangle + \beta |b\rangle$. When the state $|a\rangle$ decays to $|c\rangle$, it emits a photon in mode $a$: ie., the photon mode $a$ starts in the vacuum state $|0\rangle_a$ and gets a single photon: $a_a^\dagger|0\rangle_a = |1\rangle_a$. Similarly, when state $|b\rangle$ decays to $|c\rangle$, it emits into photon mode $b$: $|0\rangle_b \to a_b^\dagger|0\rangle_b = |1\rangle_b$.

Now, let's walk through the decay: we start with system $$|\psi\rangle \otimes |0\rangle_a |0\rangle_b = (\alpha |a\rangle + \beta |b\rangle) \otimes |0\rangle_a |0\rangle_b$$ That is, the atom is in state $|\psi\rangle$, and both photon fields are vacuum. However, when the atom decays, this changes:
$$
\to \alpha |c\rangle \otimes |1\rangle_a |0\rangle_b + \beta|c\rangle \otimes |0\rangle_a |1\rangle_b
$$
$$
= |c\rangle \otimes \left( \alpha|1\rangle_a |0\rangle_b + \beta|0\rangle_a |1\rangle_b\right)
$$
The key here is that the atom is now completely disentangled with the photon field, which has one photon in a superposition of two states (in fact we can think of this as transferring the atomic superposition state onto the photon field!)

Let's contrast this with the $\Lambda$ system. Here we start in state
$$|a\rangle \otimes |0\rangle_c |0\rangle_b$$
Now the atom decays:
$$
\to \gamma |c\rangle \otimes |1\rangle_c |0\rangle_b + \beta|b\rangle \otimes |0\rangle_c |1\rangle_b.
$$
Here the atom is entangled with the emitted photons! Indeed, this is one variation on how people can intentionally entangle atoms with photons. But here's the catch: if you try to measure the photon field, you collapse the atomic state into either state $|c\rangle$ or $|b\rangle$, which also collapses the measured photon to be firmly in one of the two modes (either photon mode $c$ or mode $b$).

The fundamental distinction between these two systems is that in the V system, the atom is not entangled with the photon field after decay, whereas for the $\Lambda$ system the atom *is* entangled with the photon field.

This post imported from StackExchange Physics at 2016-12-17 15:04 (UTC), posted by SE-user Harry Levine