Some bombs are good and some are duds; a good bomb has a sensor that sets it off if a single photon hits it.

I first heard of this problem from a friend who was taking an
introductory class at Harvard.  In their first section, their teaching
fellow, Dave Morin, posed this problem to them and told them that by
the end of the semester they should be able to solve it.  The solution
that follows is written by Dave Morin.

-- Chang Chan <cschan@scws26.harvard.edu>

[ Editor: the collected copy of the posting is damaged: the diagrams
and several passages are missing, and the gaps are marked [...] below.
The text is otherwise as posted. ]

[...]s bomb-testing problem was originally posed and solved by A.
Elitzur and L. Vaidman.  Their reference, along with others, is listed
at the end of this solution.  Any errors, ambiguities, etc., presented
here are of course mine; it is strongly recommended that one takes a
look at the references below.

The upshot of the solution is this: we will show that it is possible to
find out what would have happened if a photon had hit a bomb, even if
it really didn't.

First, forget about the bombs, and consider the following setup of a
photon beam-splitter.  The splitter is set up such that the two beams
of light travel along the sides of a square:

[...]

If a [...]an arrange the relative phases of the two paths such that one
of the detectors, say D1, will definitely register a photon, and the
other, D2, definitely will not.  We'll just accept here that such
interference can take place, and not go through the justification (see
the Elitzur-Vaidman paper or Penrose's book, listed below).  Let us
assume that we have arranged the setup such that only D1 will click
when a photon is sent in.

Now let's bring a bomb into the setup[...]

[...]his happens with probability 1/2.
      A (very violent) measurement has taken place, and one can
      say that the photon ended up at the lower right-hand corner
      of the square and set the bomb off there.  This turn of events
      doesn't help us much; let's just rebuild our apparatus and wait
      for the second possibility:
  ii) The bomb does not explode.  This happens with probability 1/2.
      In this case, a (very non-violent) measurement has taken place
      (the lack of an explosion is indeed a measurement; we have
      gained information that the photon did not hit the bomb's
      sensor), and we may conclude that the photon reaches the final
      beam-splitter via ONLY the upper "top-side" route of the square;
      the [...]

[...] modification, it can be shown that we can cut the waste down to
1/2 of the good bombs.  With other similar modifications, it turns out
that one can't do any better than this 1/2.  However, P. Kwiat, et al.,
have demonstrated an ingenious method by which essentially none of the
good bombs are wasted.

References:
The original papers:

A. Elitzur and L. Vaidman, FOUNDATIONS OF PHYSICS, Vol. 23, p. 987
(1993).

P. Kwiat, H. Weinfurter, T. Herzog, A. Zeilinger, and M. Kasevich,
  PHYSICAL REVIEW LETTERS, Vol. 74, p. 4763 (1995).

Very readable accounts dealing with these two papers can be found,
respectively, in

Roger Penrose, "Shadows of the Mind" (Oxford University Press, 1994).

C. Bennett, NATURE, Vol. 371, p. 479 (Oct. 6, 1994).

-- Dave Morin

[ Editor: a complete copy of Dave Morin's solution, reposted by Chang
Chan on 3 October 1995 ("Sorry, my previous posting does not contain
the solution in its entirety.  Here it is again.  Written by Dave
Morin.").  It fills the gaps in the damaged copy above; the text is
given in full so that it reads continuously. ]

This bomb-testing problem was originally posed and solved by A. Elitzur
and L. Vaidman.  Their reference, along with others, is listed at the
end of this solution.  Any errors, ambiguities, etc., presented here
are of course mine; it is strongly recommended that one takes a look at
the references below.

The upshot of the solution is this: we will show that it is possible to
find out what would have happened if a photon had hit a bomb, even if
it really didn't.

First, forget about the bombs, and consider the following setup of a
photon beam-splitter.  The splitter is set up such that the two beams
of light travel along the sides of a square: If a beam of light coming
from the left hits a 50-50 beam-splitter located at the lower left-hand
corner of the square, then half the light is reflected along the left
side and half is transmitted along the bottom side of the square.  Full
mirrors are placed at the upper left-hand and lower right-hand corners
of the square, and the beams are then reflected along the top and right
sides of the square.  At the upper right-hand corner of the square
there is another 50-50 beam-splitter.  This splitter is situated such
that the light coming along the top and right sides of the square ends
up either at a detector, D1, just above this splitter, or at a
detector, D2, just to the right of this splitter.

Consider the case where we send one photon into this apparatus.  It is
well known that we can arrange the relative phases of the two paths
such that one of the detectors, say D1, will definitely register a
photon, and the other, D2, definitely will not.  We'll just accept here
that such interference can take place, and not go through the
justification (see the Elitzur-Vaidman paper or Penrose's book, listed
below).  Let us assume that we have arranged the setup such that only
D1 will click when a photon is sent in.

Now let's bring a bomb into the setup.  We will put a mirror on the
bomb's sensor and place the bomb such that this mirror is the "lower
right-hand mirror" in the above setup.  We will then send a photon into
the apparatus.  This photon has a $1/\sqrt{2}$ amplitude of being
reflected along the left side of the square, and a $1/\sqrt{2}$
amplitude (times some definite relative phase) of being transmitted
along the bottom side of the square.  What is the result of this
experiment?  There are two cases to consider:
1) The bomb is a dud.  In this case, the mirror on the sensor is a
   fixed mirror.  The situation is therefore exactly the same as our
   original setup, so detector D1 will click 100 percent of the time.
2) The bomb is a good bomb.  In this case there are two sub-cases:
   i) The bomb explodes.  This happens with probability 1/2.
      A (very violent) measurement has taken place, and one can
      say that the photon ended up at the lower right-hand corner
      of the square and set the bomb off there.  This turn of events
      doesn't help us much; let's just rebuild our apparatus and wait
      for the second possibility:
  ii) The bomb does not explode.  This happens with probability 1/2.
      In this case, a (very non-violent) measurement has taken place
      (the lack of an explosion is indeed a measurement; we have
      gained information that the photon did not hit the bomb's
      sensor), and we may conclude that the photon reaches the final
      beam-splitter via ONLY the upper "top-side" route of the square;
      the photon has zero amplitude to be on the right side of the
      square if the (good) bomb does not explode.
        But if the photon arrives at the second beam-splitter via only
      one route, then no interference takes place, so detectors D1 and
      D2 will each click 50 percent of the time.  Therefore, if D2
      clicks, we have found ourselves a good bomb (because if it was a
      dud, then only D1 could have clicked), and the problem is solved.

Comment: If the bomb does not go off and D1 clicks, then we can't tell
if the bomb is a good one or a dud.  In this event, we can just keep
sending photons through until either the bomb explodes or D2 clicks.
With this strategy, we end up wasting 2/3 of the good bombs and
obtaining 1/3 of them.  With a slight modification, it can be shown
that we can cut the waste down to 1/2 of the good bombs.  With other
similar modifications, it turns out that one can't do any better than
this 1/2.  However, P. Kwiat, et al., have demonstrated an ingenious
method by which essentially none of the good bombs are wasted.

[ Editor: the references are as given above. ]

-- Dave Morin