## The Elitzur-Vaidman bomb test

So I’ve been quite busy which I why I haven’t yet responded to the comments on my previous post, but now it’s Saturday!  Realizing that not everyone knows the Elitzur-Vaidman bomb test, I’ll give a full description to further discussion along!

Imagine a simplified Mach-Zender interferometer in which a beam of light passes through a beamsplitter (usually some kind of crystal) becoming two beams of light.  The two beams bounce off a couple of mirrors and meet again at a second beamsplitter, each splitting again with some from each merging.  Thus, in the diagram below, the yellow beams are each a mixture of the green and red beams (sort of).

If you set up the beamsplitter’s properly, you can create interference such that detector D0 always registers photons and D1 never does.  In other words, the probability that D0 detects a photon is 1 and the probability that D1 does is 0.  This is because we have set it up such that we have constructive interference leading to D0 and destructive interference leading to D1.

Suppose now that we stick our hand in the path of the green beam such that the photons are absorbed by our hand.  This destroys any interference effects.  Any single photon entering the device at the beginning, then, has a 50-50 chance of hitting our hand.  If it doesn’t, it goes through the red beam and then has an equal chance of hitting D0 or D1.  So the probabilities are

• Photon reaches D0 = 1/4
• Photon reaches D1 = 1/4
• Photon reaches hand = 1/2

Notice that by blocking the green beam with a hand, we have actually increased the probability that the photon reaches D1.

Now suppose there is a factory that produces bombs triggered by a single photon of light.  Because of manufacturing defects, though, some come off the assembly line without working triggers.  Photons passing through these wouldn’t do anything and the bombs would be labeled as duds.  How do the factory managers, however, tell if some of the bombs are in working order without triggering them?  They want at least some working bombs when they’re done otherwise the Army won’t be too pleased.

They do it using the above Mach-Zender interferometer that is preset to always produce constructive interference at D0 and destructive at D1.  Instead of placing a hand in the green beam, a bomb (actually its trigger) is placed there.  The trigger acts like a photon detector and the results will be

Bomb is a dud

• Photon reaches D0 = 1
• Photon reaches D1 = 0
• Bomb explodes = 0

Bomb is working

• Photon reaches D0 = 1/4
• Photon reaches D1 = 1/4
• Bomb explodes = 1/2

Now put a bomb fresh off the assembly line into the device (and back away – far, far away).  If the bomb explodes, it was in working order but has now been wasted (the cost of quality control under such conditions is losing a few bombs).  If the photon is detected by D0, the test is inconclusive and can be run again (if, after many repeated trials, the photon keeps showing up at D0, the bomb is probably a dud).  But if the photon ever reaches D1 – which is should do 25% of the time – then the managers know that the unexploded bomb is in working order even though the bomb never detects the passage of the photon!

Presumably someone out there has done a calculation to figure out the maximum percentage of usable bombs one can hope to get out of this by repeating the D0 tests.  I don’t know what the answer is but it’s not quite as simple as it looks since it is not clear exactly what percentage of total photons is detected at D0 (remember, it’s only 1/4 if the bombs are all working!).

### 23 Responses to “The Elitzur-Vaidman bomb test”

1. The diagram is a nice addition … it helps us realize that even this simple problem is built upon a wealth of subtle engineering details.

For example, let’s simplify it to use, not photons, but individual neutrons (so we don’t have to worry about second quantization, Bose statistics, or special relativity, but instead we can just use the one-particle Schroedinger equation throughout). And let’s further simplify it to be 1D+time (but with a loop topology).

Then we *still* have to specify the dynamical details of a single neutron emitter, and the starting state of that emitter (OK … a potential box of variable depth).

And also, we have to specify the microscopic dynamics of the various detectors, some of which are absorbing detectors, others of which detect-and-reflect (OK, they’ll be Lindblad processes).

And then, after all this work … we find that the detector doesn’t work as we expect … the dispersive dynamics of the Schroedinger equation goof-up the simple picture we had in-mind (OK, so to eliminate the dispersion, we have to go back to massless photons, hence second quantization, Bose statistics … and we end-up reinventing most of cavity QED).

It’s been my experience that quantum mysteries … the spukhafte Fernwirkungen of the 1930s … are particularly prone to arise whenever attempts are made to leap over the technical details associated to (continuous) measurement processes.

The overall point is that quantum Hamiltonian dynamics is tough … but not any tougher than classical Hamiltonian dynamics. What’s novel about quantum physics is not the Hamiltonian dynamics, but rather the Lindbladian dynamics associated to measurement processes … and this is exactly the dynamics that typical spukhafte Fernwirkungen examples gloss-over.

We can think about it is this way: was there ever a mysterious thought-experiment in quantum mechanics that didn’t involve a measurement process? No? … then we’d better do a thorough job of teaching Lindbladian dynamics!

2. The usual discussions about measurement “hits” don’t effectively deal AFAICT with these Renninger-style null measurements: If the object is not interacted with, then the output somewhere else is different than it would be (like this Elitzur-Vaidman interaction-free bomb detection, etc.) The WF does not “interact” with the target object, yet is reallocated on account of the target’s presence. Some writers say this makes it hard to integrate into many worlds ideas (if the alternative is not an actual “result” then how is it separated), etc.

3. It was Sherlock Holmes who said: “It is a capital mistake to theorize before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts.”

Transposing this theme into quantum physics: “It is a capital mistake to philosophize before one has a solid mathematical framework. Insensibly one begins to twist mathematics to suit philosophy, instead of philosophy to suit mathematics.”

Here the point is that our present mathematical framework for describing quantum dynamics coexists very uneasily with our understanding of general relativity, causality, and separability … it is likely that substantial changes in this mathematical framework are in the offing. The risk for philosophers is that these shifts in our mathematical description of quantum dynamics (not to mention, experiments!) may in coming decades render irrelevant large chunks of the philosophical literature.

Still, we all have to do what we can … mathematicians, scientists, and philosophers stumbling forward together … and be ready for surprises! 🙂

4. Neil Bates Says:

I think the implications of this experiment are wrongly framed. I don’t think the E-V setup would test to see which bombs are good or duds. The test only shows that something is “in” the path of the split photon. But if we get a hit in normally-dark channel B, then the fuse is not “hit” by the photon whether it is active or not. The photon, as already stated, did not “interact” and the fuse wouldn’t go off anyway! The whole point is that only if a photon is actually absorbed or reflected by the fuse then it will go off. So we cannot distinguish duds versus lives unless the photon really does impact – in which case either the bomb blows, or nothing happens (it is a dud.)

5. IMHO, among the trickiest notions in this problem is the notion of a “photon path”. So let’s explore the microscopic description of a photon path.

In the context of (lossless) cavity QED, the notion of a path is not natural because photons reflect back-and-forth in the cavity many times.

We can recover the naturality of “photon path” by introducing an absorbing detector in the cavity… but in cavity QED it is nontrivial to specify the design of a detector that is perfectly “black” at all photon frequencies. Similarly, by time-reversing a single photon detector, we obtain a single-photon emitter … this too is nontrivial to specify microscopically.

When we equip perfect cavities with perfect detectors and perfect emitters, we find that the cavity modes are divergently renormalized … and this provides a microscopically natural explanation of why the presence of a detector anywhere in the cavity, is detectable everywhere in the cavity.

6. Neil Bates Says:

So John, do you think I’m right that the E-V setup doesn’t really test for light-sensitivity after all? A body in the path prevents interference because the other leg of the WF can’t get to BS2 for interference, but that by no means would tell if the object responds to photons in a particular way or not. As we imagine, the WF “sniffs out” there being a barrier – and if it is not absorbed there, then it might as well have never “come near” (or whatever the hell it does ….)

Note also that the setup doesn’t depend on perfect absorption. A white or mirror object will block the path and just have a chance of sending a photon back to BS1 (where it won’t go back up towards BS2 anyway. Since the reflected photon then faces the diagonal the other way, it goes down the other way.

Actually the most fascinating outcome IMHO is if we put a gray filter in one path. Then, even though the filter didn’t absorb the photon anyway, an attenuated amplitude reaches BS2 (AFAIK) and we get an attenuated interference pattern. I wonder, if we could prove the photon wasn’t absorbed by the filter (like, no energy change) if that would change the result. If not it seems absurd to have affected the outcome, but if it did then we could change the outcome at a distance.

7. Neil, in cavity-language the idea seems simpler than in path-language. Optical modes are renormalized by a detector if-and-only-if the detector is sensitive to that mode. And to create and launch “photons” along a “path” requires the sophisticated design, and the coherent manipulation, of a *lot* of these renormalized modes.

Maybe that is why there are no books on quantum optics (that I know of) that are short *and* useful for detector design.

8. I have always have this question when reading the EV-bomb problem: Is the bomb a classical or quantum object? Manufacturers make classical bombs, and describing it using a quantum state is akin to schroedinger’s cat issue…
But if it is a quantum object, then detecting it by not triggering it seems no big deal conceptually, but just a matter of finding a clever method to do it, which in this case is just the EV-bomb method.

9. quantummoxie Says:

Well, the bomb itself is classical, but the big question is the trigger. The whole apparatus is supposed to be classical, but if the trigger is detecting single photons then it must have some quantum properties, I would think.

10. John S, I think it’s good to talk about cavity modes etc. to get away from rigid Schroedinger thinking. I may be missing something, but I still don’t see how the setup would show that a fuse was “live” but wouldn’t set it off. Again, the reallocation of the WF by the presence of the fuse is not “an interaction” and IMHO wouldn’t set off a live fuse. All we’re doing here, despite the title given by the proponents, is to show that “something is there” – just “there”, not anything about its reactivity etc. I do note that if the fuse was reflective, then we have a chance of getting light going back out. But then a reflection would just set the bomb off.

As best I can tell, the photon either:
1. Really does interact and therefore a live bomb goes off, and a dud doesn’t.
2. Does not interact. Hence the photon being reallocated into a channel it couldn’t have reached via clear paths, shows nothing about the state of the fuse. A dead or live fuse would have the same effect – neither would let the photon go thru it (or did I misunderstand) and either would allow the photon to appear in channel D1. (Daniel, see the point?)

If the wording of the problem is clarified, to say that a dud is transparent (which would, given proper phase relations, maintain the interference pattern) then the thought experiment can be viable.

11. (OK, it seems that the experiment is conceived properly since we see here:
“Photons passing through these [duds] wouldn’t do anything …” So the duds are transparent, but they have to have to maintain phase. Not all descriptions of the EVBT made this clear, pardon the metaphor. But now we understand, the optical effect on the paths is critical and it’s not a matter of directly showing that the active fuse could have worked. It’s testing the difference between the effect of clear v. opaque insertions in the path. It is fascinating we can find this out without always passing an active photon through that insertion.)

12. quantummoxie Says:

Neil,

I think this really gets at the heart of the problem – where does the universe make that transition from quantum to classical? (Of course, Bob Griffiths thinks the universe is entirely quantum, but that’s another argument…).

• what ?

• quantummoxie Says:

Meaning: at what level of magnification or in what processes (and where within those processes)?

13. Neil,
I think it is incorrect to say the photon either interacts or not. It is both. Since it’s in a superposition of upper and lower path.
This is the point of the problem as I see it: we cannot talk abt the path of photon. It is on both and the bomb (the quantum part of it) changes the phase between the path.
and I think the issue has nothing whatsoever to do about the intricacies of cavity and things like that. We may as well change the example into one where we are to determine whether there’s a (real) bomb inside a box or the box is empty. Or we can change it into one where we shoot electron and not photon, and we are to determine whether there’s a positron at the lower arm.

14. Well Daniel you have a point. However, normally when we say that “interaction” with a photon would set something off, we mean some measurable alteration of the target. That means, an absorption or a reflection able to detect the momentum ping (which brings up questions in its own right.) So just having the bomb inside the MZ doesn’t show that a photon “interacted” with the fuse, by mere virture of the fuse just being there during a photon “shot.” If the WF ends up taking the other path and causes some hits in D1, that implies to me that it just didn’t “hit” the fuse. That’s why I said it was so important to claim that the dud fuse was transparent, as implied by the wording (albeit maybe the wording doesn’t make that “perfectly clear” itself!) In any case, we don’t really know what that “great smoky dragon” is doing in that MZ, do we?

As for reflection, that reminds me of a paradox I set up. Let a BS split a photon wave. Add to the MZ tow parallel mirror tracks, one on each side. Let then the two channels involve repeated multiple reflections against the mirror walls surrounding the MZ as above (like zig-zag, running around the outer parts.) So classically, a wave is pushing outward against each side of the enhanced MZ. So, is it in outward tension and presumably inner compression too? There is enough momentum. Force would equal 2h*nu*(1/c)dn/dt for n reflections/second, in principle, or not? (The “two” from reversal of direction.) Or, “no effect” until we actually detect the photon one place or another?

PS: Sorry I wasted some space-time wrangling over the framing of the bomb testing issue before we got the “no effect” issue ironed out (did we?) But I can see it’s still debatable overall.

15. I find it very helpful to conceive pretty much any quantum system as a lattice of spins-j particles. An oscillator then is just a large-j spin … the optical mode of a cavity is itself a spin-oscillator … the two-state atoms in a laser cavity are j=1/2 spins … it is perfectly feasible to extend this picture all the way to lattice gauge theories.

This picture is very convenient for numerical work … pretty much any quantum dynamical system can be conceived rather naturally in terms of Bloch dynamics-and-relaxation.

However, this system works rather badly for a physical system that typically is one of the first that textbooks treat: the hydrogen atom.

The reason is simply that the hydrogen atom has a large group of symmetries that no spin lattice respects … and it is these symmetries that allow ODE techniques to work so effectively back in the 1920s. Indeed, the entire hydrogen-atom system can be solved using group-theoretic methods.

In the real world though, we often want to tackle dynamical systems that have little or no symmetry … beginning as early as … uh … helium?

In which case, numerical/perturbative methods become essential. IMHO students might as well be solidly grounded in this practical reality, right from the start.

16. […] Elitzur is a quantum physics rock star, and to be honest, I understood only his jokes (most of them); if the professor ever runs out of […]

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18. yes but also according to quantum theories, i think the Hawthorne effect or something like that, then even the act of observing this experiment will cause the results to change. Photons will be brought into this system to just view it, also causing destructive interference on the anti quark being produced (universe splitting probability theory), so this would null the negative beam of light (assuming all possibilities happen at the same time)(finite divided by infinite) which reflected through the crystal causing the bomb to explode half the time. However if the bomb had a big enough explosion to kill the viewer, each time the universe splits into all possible probabilities, the probability in which the universe allowed the viewer to be killed would exactly 0(not impossible). Because although this is theoretically inevitable, the viewer would not be aware of his/her death in a parallel universe, he/her would only be aware of the universe in which he/she lived. When the universe spits the viewer will only be aware of the one he/she lives, therefore the bomb would never explode, revealing a probability .50 its a dud and .50 it should theoretically work.

19. Sunil Pai Says:

1/3 of live bombs will be detected…1/4*(1+1/4+…) = 1/4 * 4/3 = 1/3, where the 1/4 represents the ambiguous case for some live bomb.

20. It’s like a double slit experiment with a polarizer system behind the left slit to see if a photon went through the left slit. Send one photon through the double-slit. If it lands on the screen at a point behind the right slit, then you know it didn’t go through the left slit. You have no chance that repeated experiments would ever build up an interference pattern when you map all the points along the screen where photons hit. The wave-like behaviour due to superposition is destroyed by the presence of the polarizer – even when the photon does not take the left-slit path. So you can say that the polarizer system collapsed the wavefunction (destroyed the superposition). But think about what that means when you only send one photon through the double slit. If it makes it to the screen, it lands behind the right slit. The wavefunction collapsed due to an interaction with the polarizer system even though the photon never went through the polarizer system. Collapse can happen when a “null observation” is effectively made by the polarizer system. Same deal with the E-V bomb tester. A working trigger can collapse the wave function even when the trigger doesn’t absorb the photon; if you see that the photon lands at detector D1 (when it would go to D0 if the bomb were absent) then you know the photon did not take the bomb’s path. The trigger (or the bomb, if you prefer) has made a null observation, forcing wavefunction collapse (at the trigger) and telling you which way the photon went (along the empty path). I might be wrong, but this is the only way this experiment makes sense to me.

• quantummoxie Says:

I think you explained it well. The null observation is just as real as the others in this case.