1
(
)
2
i Z
Z
\
J
The boxes are transparent for the photon but opaque for the atom. Now let the atom’s Z
−
box
be positioned across the photon’s v path in such a way that the
photon can pass through the
box and interact with the atom inside in a 100% efficiency.
Next let the photon be transmitted by BS
1
:
1
(
)(
)
2
i u
v
i Z
Z
\
Discarding all these cases of the photon’s absorption by the atom (25% of the experiments)
:
1
(
)
2
u Z
i u Z
i v Z
\
Next, reunite the photon by BS
2
:
2
1
(
)
2
BS
v
d
i c
o
2
1
(
)
2
BS
u
d
i c
o
So that:
3
[
(
2
)
]
2
i
c i Z
Z
d Z
\
After the photon reaches one of the detectors, the atom’s Z boxes are joined and a reverse
magnetic field –B is applied to bring it to its final state. Measuring this state's spin yields:
1
1
(
)
( 3
)
4
4
d
i X
X
c
X
i X
\
In 1/16 of the cases, the photon
hits detector D, while the atom is found in a final spin state
of X
-
rather than its initial state X
+
. In every such a case, both particles performed IFM on
one another; they both destroyed each other’s interference. Nevertheless,
the photon has not
been absorbed by the atom, so no interaction seems to have taken place.
Hardy’s analysis stressed a striking aspect of this result: The atom can be regarded as EV’s
“bomb” as long as it is in superposition, and its interaction with the photon can end up with
one out of three consequences:
• The atom absorbs the photon – this is analogous to the explosion in EV’s original device.
• The atom remains superposed – this is analogous to the no-explosion outcome.
• The atom does not absorb the photon but loses its superposition – a third possibility that
does not exist with the classical bomb and amounts to a delicate form of explosion.
Hence, when the last case occurs, it appears that the photon has traversed the u arm of the
MZI, while still affecting the atom on the other arm by forcing it to assume (as measurement
will indeed reveal) a Z
+
spin!
(1)
(2)
(3)
(4)
(5)
(6)
ICNFP 2013
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2.5 EPR Effects between Particles that never interacted in the past
Another elegant experiment by Hardy [21] brings together nearly all the famous quantum
experiments, such as the double-slit, the delayed choice, EPR and IFM – all in one simple
setup (Fig. 2).
Fig. 2. Entangling two atoms that never interact.
Let again a single photon traverse a MZI. Now let two Hardy atoms be prepared as in Sec.
2.3, each atom superposed in two boxes that are transparent for the photon but opaque for the
atom. Then let the two atoms be positioned on the MZI’s two arms such that atom 1’s Z
+
box
lies across the photon’s v path and 2’s Z
−2
box is positioned across the photon’s u path. On
both arms, the photon can pass through the box and interact with the atom inside in 100%
efficiency. Now let the photon be transmitted by BS
1
:
1
1
2
2
3
1
(
)(
)(
)
2
i u
v
i Z
Z
i Z
Z
\
Once the photon was allowed to interact with the atoms, we discard the cases in which
absorption occurred (50%), to get:
1
2
1
2
1
2
1
2
3
1
(
)
2
i u Z
Z
u Z
Z
i v Z
Z
v Z
Z
\
Now, let photon parts u and v pass through BS
2
:
2
1
(
)
2
BS
v
d
i c
o
2
1
(
)
2
BS
u
c
i d
o
(7)
(8)
(9)
EPJ Web of Conferences
00028-p.6
giving
1
2
1
2
1
2
1
2
1
(
2
)
4
d Z
Z
d Z
Z
i c Z
Z
c Z
Z
\
If we If we now post-select only the experiments in which the photon was surely disrupted
on one of its two paths, thereby hitting detector D, we get:
1
2
1
2
1
(
)
4
d
Z
Z
Z
Z
\
Consequently, the atoms, which never met in the past, become entangled in an EPR-like
relation. In other words, they would violate Bell’s inequality [22]. Unlike the ordinary EPR,
where the two particles have interacted earlier or emerged from the same source, here the
only common event in the two atoms’ past is the single photon that has “visited” both of
them.
2.6 EPR Upside-down
Hardy’s abovementioned experiment [21] inspired Elitzur and Dolev propose a simpler
version [23] that constitutes an inverse EPR. Let two coherent photon beams be emitted from
two distant sources as in Fig. 3. Let the sources be of sufficiently low intensity such that, on
average, one photon is emitted during a given time interval. Let the beams be directed
towards an equidistant BS. Again, two detectors are positioned next to the BS:
1
0
u
u
u
p
q
J
I
1
0
v
v
v
p
q
J
I
1
1
1
1
(
)
2
A
i Z
Z
\
2
2
2
1
(
)
2
A
i Z
Z
\
where
1
denotes a photon state (with probability p
2
),
0
denotes a state of no photon (with
probability q
2
), p
≪
1, and p
2
+ q
2
= 1.
Since the two sources’ radiation is of equal wavelength, a static interference pattern will be
manifested by different detection probabilities in each detector. Adjusting the lengths of the
photons’ paths v and u will modify these probabilities, allowing a state where one detector,
D, is always silent due to destructive interference, while all the clicks occur at the other
detector, C, due to constructive interference. Notice that each single photon obeys these
detection probabilities only if both paths u and v, coming from the two distant sources, are
open. We shall also presume that the time during which the two sources remain coherent is
long enough compared to the experiment’s duration, hence we can assume the above phase
relation to be fixed.
Next, let two Hardy atoms be placed on the two possible paths such that atom 1’s Z
+ 1
box
lies across the photon’s u path and 2’s Z
−2
box is positioned across the photon’s v path. After
the photon was allowed to interact with the atoms, we discard the cases in which absorption
occurred (50%), getting
(10)
(11)
(12)
ICNFP 2013
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