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When calculating the emitted intensity of an excited $\Lambda$-type or $V$-type atom with a quantized field (instead of a semiclassical approach with only quantized energy levels) one finds that there is a beat modulation on top of the emitted intensity for $V$-type but not for $\Lambda$-type atoms. The calculations are quite clear, however when it comes to the intuitive explanation of the result I'm not sure if I understand it correctly. The wikipedia page on quantum beats says

This difference is caused by quantum mechanical uncertainty. A $V$-type atom decays to state ${\displaystyle |c\rangle }$ via the emission with ${\displaystyle \omega_{a}}$ and ${\displaystyle \omega_{b}}$. Since both transitions decayed to the same state, one cannot determine along which path each decayed, similar to Young's double-slit experiment. However, ${\displaystyle \Lambda }$-type atoms decay to two different states. Therefore, in this case we can recognize the path, even if it decays via two emissions as does $V$-type. Simply, we already know the path of the emission and decay.

I heard a similar argument about the Fano resonances that arise in the photo electron spectrum of helium, when there is a resonant level (see e.g. ref. 1). In this case the intuitive explanation was the uncertainty whether the ionized electron was directly ionized by absorbing two photons or first excited to the resonant level by one photon and then ionized by another one.

Question: In the case of $V$-type atoms, what is meant by the "path" along which the atom decays and where, according to this explanation, does the interference come into play? Is it that, when observing only the final state, we cannot know which decay has happened for $V$-types?

v-typelambda-type


  1. Fano resonances observed in helium nanodroplets, Laforge et.al., Phys. Rev. A 93, 050502 (2016), arXiv:1510.05473.
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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.

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