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I'm trying to grok more of the adiabatic model. I also really enjoyed O'Donnell's lecture on the Elitzur-Vaidman bomb tester.

The familiar setup involves a test for the presence (or absence) of a bomb. The bomb goes off if a qubit passes through in the $\vert 1\rangle$ state, and does nothing if going through in the $\vert 0\rangle$ state.

Conventionally a qubit is prepared in the $\vert +\rangle$ state to go through the bomb tester, and if it doesn't trigger the bomb, it is measured in the $\{\vert +\rangle,\vert -\rangle\}$ basis. If the bomb is present, there's a 50% it will go off, a 25% chance that it might be inconclusive but a 25% chance that the qubit will be measured as $\vert -\rangle$, proving the presence of the bomb.

In O'Donnell's description of the improved test, a qubit without a bomb present is slowly rotated by $\epsilon$ degrees, from $\vert 0\rangle$ to $\vert 1\rangle$. However if a bomb is present, the qubit will remain in $\vert 0\rangle$ with high probability.

This slow rotation of the qubit feels, to me, a bit like a slowroll of a changing Hamiltonian used in the adiabatic model. But that's where my intuition hits a road-block of not understanding the adiabatic theorem well enough.

Is my intuition close to correct? How would one frame the Elitzur-Vaidman bomb tester in terms of the adiabatic theorem? For example what would be the initial and final Hamiltonians?

Mark Spinelli
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    I'm not familiar with the test, but that does sound like adiabatic evolution, yes. If adding the "bomb" corresponds to creating an energy different between the two states, then it makes sense that slowly "rotating the system" causes the state to remain in the instantaneous ground state, which I guess here would be $|0\rangle$, only when the bomb is there – glS Aug 07 '20 at 22:57
  • Ahh.. are you suggesting that the presence of the bomb "breaks the symmetry" between $\vert 0\rangle$ and $\vert 1\rangle$? The adiabatic theorem and the quantum-Zeno effect seem to be intimately related in ways that I've been grasping at straws over. – Mark Spinelli Aug 08 '20 at 00:23
  • again, I'm not familiar with the way the situation is modelled, but it does seem like that should be the case from your description. A caveat is that, if this is the case, then "$|0\rangle$" would represent the instantaneous ground state, that is, it would represent different states at different times, which admittedly is a bit weird in this notation. I wouldn't say that adiabatic thm and zeno effect are related though. In the former case the state remains in the instantaneous gs without measurements, in the latter it remains in a fixed state due to repeated measurement – glS Aug 08 '20 at 00:28
  • admittedly, now that you mention the zeno effect, that might be another different explanation for the phenomenon. I guess one could regard the fact that the experiment continues as "having measured that the bomb didn't go off", in which case the state would remain in $|0\rangle$ due to repeated measurement, not the adiabatic theorem. Knowing the Hamiltonian describing the system would clarify the issue I'd say. What's the timestamp in the video where they mention this? – glS Aug 08 '20 at 00:29
  • O'Donnell's lecture talks of the "improved" Elitzur-Vaidman tester at around 44:23 or so, although he doesn't mention Hamiltonians at all. – Mark Spinelli Aug 08 '20 at 00:43
  • yes from the video I would say this is a variation of the q.Zeno effect. By measuring the state when the probability of $|0\rangle$ is high, you make it (probably) collapse to $|0\rangle$, and therefore reset the rotation. If you measure frequently enough you end up freezing the state at $|0\rangle$, as it has no time to accumulate sufficient amplitude on $|1\rangle$ to be ever measured there. – glS Aug 08 '20 at 12:45
  • Thanks for all your comments! This “measuring frequently enough” to keep in $\vert 0\rangle$ as in Zeno sounds very analogous to “rotating slowly enough to keep in the ground state”as in the adiabatic theorem. – Mark Spinelli Aug 08 '20 at 14:12
  • Just spit-balling... Maybe we can consider with probability $p$, we live in a world with the bomb, and with probability $1-p$ we live in a world without the bomb. If we put this as the first qubit and pray at the church of the higher Hilbert space somehow, we want to avoid the state $\vert 11\rangle$ (e.g. we want to avoid living in a world with the bomb, and bomb goes off). A ground state of an initial Hamiltonian is $\frac{1}{\sqrt 2}(\vert 10\rangle+\vert 00\rangle)$. A ground state of a final Hamiltonian is $\frac{1}{\sqrt 2}(\vert 10\rangle+\vert 01\rangle)$... – Mark Spinelli Aug 14 '20 at 22:04

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