### Physics

Author: S. Dolev, A.C. Elitzur

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\begin{document}

\title{NON-SEQUENTIAL BEHAVIOR OF THE WAVE
FUNCTION}

\author{Shahar Dolev}

\email{ shahar_dolev@email.com}

\author{Avshalom C. Elitzur}

\email{avshalom.elitzur@weizmann.ac.il}

\affiliation{Unit of Interdisciplinary Studies,

\begin{abstract}

An experiment is presented in which the alleged
progression of a photon's wave function is ``measured'' by a row of superposed
atoms. The photon's wave function affects only one out of the atoms, regardless
of its position within the row. It also turns out that, out of $n$ atoms, each
one has a probability which is higher than the classical probability $1/n$ to
be the single affected one. These results indicate that the wave function
manifests not only non-local but also non-sequential characteristics.

\end{abstract}

\pacs{ 03.65.Ta, 03.65.Ud, 03.65.Xp, 03.67.-a}

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section {Introduction}

\begin{figure}

\begin{center}

\includegraphics[scale=0.5]{fig1.eps}

\label{fig:1}

\vspace{0.3cm}

\caption{A photon in a Mach-Zehnder Interferometer
interacting with a superposed atom.}

\end{center}

\end{figure}

When a single photon goes through a Mach-Zehnder
Interferometer, its behavior indicates that it has somehow traversed both arms.
However, when its position is measured during this passage, it turns out to
have traversed only one arm. This is one of the notable manifestations of the
measurement problem, for which several competing interpretations have been
proposed. These can be crudely divided into two groups: ``collapse'' (\eg

Both groups, however, seem to share one assumption.
The photon -- whether in the form of wave-plus-particle or of a wave function
evenly spread over all available positions -- is believed to proceed from the
source to the detector sequentially through space-time. Hence, if a few objects
are placed along its path, the photon is expected to interact with them one
after another, according to the order of their positions.

In this Letter we show an experiment in which
space-time sequentiality does not hold.

\section {Mutual IFM}

Interaction-Free Measurement (IFM) \cite{Elitzur93}
highlights the way two interferometer arms, or even a myriad of them
\cite{Elitzur01}, are ``felt'' by a single particle. Its essence lies in an
exchange of roles: the quantum object, rather than being the subject of
measurement, becomes the measuring apparatus itself, whereas the macroscopic
detector (or super-sensitive bomb in the original version) is the object to be
measured.

In their paper
\cite{Elitzur93}, Elitzur and Vaidman (EV) mentioned the possibility of
an IFM in which both objects, the measuring and the measured, are single
particles, in which case even more intriguing effects can appear. This
proposition was taken up in a seminal paper by Hardy \cite{Hardy92a,Hardy93}.
He considered an EV device (Fig.\ 1) where a single photon traverses a
Mach-Zehnder Interferometer (MZI) and interacts with an atom in the following
way: A spin $^1\!/_2$ atom is prepared in a spin state $\ket{X+}$ (that is,
$\sigma_x=+1$), and split by a non-uniform magnetic field $M$ into its two Z
components. The box is then carefully split into two halves, each containing
either the $\ket{Z+}$ or the $\ket{Z-}$ part, while preserving their
superposition state:

\beq

\Psi=\ket{\gamma} \cdot
\frac{1}{\sqrt{2}}(\ket{Z+}+\ket{Z-}).

\eeq

Now let the photon be split by $BS_1$:

\beq

\Psi=\frac{1}{2} (i\ket{u}+\ket{v}) \cdot
(\ket{Z+}+\ket{Z-}).

\label{eq:hardy_uv}

\eeq

The boxes are transparent for the photon but opaque
for the atom. The atom's $Z+$ box is 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 100\% efficiency. Discarding all cases of the photon's scattering by
the atom (25\%) removes the term \hbox{$\ket{v}\ket{Z+}$}, leaving:

\bea

\Psi &=&\frac{1}{2} (i\ket{u}\ket{Z+} +
i\ket{u}\ket{Z-} + \ket{v}\ket{Z-}) \nncr

&& \quad + \ket{scattering}.

\eea

Next, let us reunite the photon by $BS_2$:

\bea

&&\ket{v}\stackrel{BS_2}{\longrightarrow}\frac{1}{\sqrt{2}}
(\ket{d}+i\ket{c})\\

&&\ket{u}\stackrel{BS_2}{\longrightarrow}\frac{1}{\sqrt{2}}
(\ket{c}+i\ket{d}),

\eea

so that

\beq

\Psi=\frac{1}{\sqrt{2}^3}\cdot [i\ket{c}
(\ket{Z+}+2\ket{Z-})-\ket{d}\ket{Z+}].

\eeq

Once the photon reaches one of the detectors, the
atom's $Z$ boxes are joined and a reverse magnetic field $-M$ is applied to
bring the atom to its final state $\ket{F}$. Measuring $F$'s $\sigma_x$ gives:

\bea

\Psi&=&\frac{1}{4} \ket{c}\cdot
(3\ket{X+}-\ket{X-})\nncr

&&\quad-\frac{1}{4} \ket{d}\cdot
(\ket{X+}+\ket{X-}).

\label{eq:hardy_cd}

\eea

Here, it can happen that the photon hits detector
$D$, while the atom is found in a final spin state of $\ket{X-}$ rather than
its initial state $\ket{X+}$. In such a case, both particles performed IFM on
one another, destroying each other's interference. Nevertheless, the photon has
not been scattered, so no interaction between the photon and the atom seems to
have taken place.

Hardy's analysis revealed the striking consequence
of this result: The atom can be regarded as EV's ``bomb'' as long as it is in a
superposition, whereas a measurement that forces it to assume a definite Z spin
(to ``collapse'') amounts to ``detonating'' it. However, the photon's hitting
detector $D$ indicates that it has been disturbed too. And yet, in the absence
of scattering, no interaction seems to have occurred between it and the atom.
That seems to indicate that the photon has traversed the $u$ arm of the MZI
while ``detonating'' the atom on the other arm, forcing it to assume (as
measurement indeed confirms) a definite $Z+$ spin!

Hardy argued that this case supports the guide-wave
interpretation of QM. His reasoning was that the photon’s corpuscle plus
half-wave took the $u$ arm of the MZI while its other, empty half-wave took the
$v$ arm and broke the atom's superposition. However,

All the above analyses, however, seem to assume
space-time sequentiality. To show how this assumption can become strained, let
us reconsider Hardy's experiment with a slight yet crucial addition. Let a
macroscopic object be placed after the atom on the $v$ arm of the photon MZI
(``B'' on Fig.\ 1). Here Eq.\ (\ref{eq:hardy_uv}) becomes:

\beq

\Psi=\frac{i}{2} \ket{u} \cdot (\ket{Z+}+\ket{Z-}),

\eeq

and consequently Eq.\ (\ref{eq:hardy_cd}) changes
into

\beq

\Psi=\frac{1}{2} (i\ket{c}-\ket{d}) \cdot \ket{X+}.

\eeq

The atom has retained its $\ket{X+}$ state,
indicating that the peculiar effect Hardy pointed out can appear only if the
two halves of the photon’s wave function are allowed to reunite. In other
words, the alleged ``empty guide wave’’ or ``collapsing wave function’’ will
not exert their effect unless path $v$ is allowed, {\it later}, to reach
$BS_2$. Here, ordinary temporal notions are defied, and this defiance will
become more prominent in what follows.

We shall next point out a more peculiar effect of
the wave function for which all the above interpretations, due to their
sequentiality assumption, seem to be insufficient.

\section {IFM with one photon and several atoms}

\label{sec:several}

\begin{figure}

\begin{center}

\includegraphics[scale=0.5]{fig2.eps}

\label{fig:2}

\vspace{0.3cm}

\caption{One photon MZI with several interacting
atoms.}

\end{center}

\end{figure}

Consider the setup given in Fig.\ 2. Here too, one
photon traverses the MZI, but now it interacts with three superposed atoms
rather than one. Formally:

\beq

\Psi=\ket{\gamma}\ket{X_1+}\ket{X_2+}\ket{X_3+}.

\eeq

After the photon's passage through $BS_1$ and the
atoms' splitting according to $\sigma_z$:

\bea

\Psi&=&\frac{1}{4} (i\ket{u}+\ket{v})\cdot
(\ket{Z_1+}+\ket{Z_1-})\nncr

&&\quad\cdot(\ket{Z_2+}+\ket{Z_2-})\cdot
(\ket{Z_3+}+\ket{Z_3-}).

\eea

Let us denote:

\bea

\psi^\equiv&=& \ket{Z_1-} \ket{Z_2-}
\ket{Z_3-}, \\

\phi&=&(\ket{Z_1+}+\ket{Z_1-})\cdot(\ket{Z_2+}+\ket{Z_2-})\nncr

&&\quad\cdot (\ket{Z_3+}+\ket{Z_3-}).

\eea

As in the previous experiment, we discard all the
cases (44\%) in which scattering occurs:

\beq

\Psi = \frac{1}{4} (i\ket{u} \cdot \phi + \ket{v}
\cdot \psi^\equiv) + \ket{scattering}.

\eeq

Note that the $\phi$ part -- where each atom has a
50/50 probability to be found in the $Z+$ or the $Z-$ box -- is coupled to the
photon traversing the upper route $u$, while $\psi^\equiv$ -- where all three
atoms are found in the $Z-$ state -- is attached to the photon traversing the
lower route $v$.

Now let us pass the photon through $BS_2$ and
select only these cases in which it has lost its interference, hitting detector
$D$:

\beq

\Psi = \frac{1}{4\sqrt{2}} \cdot \ket{d} \cdot
(\psi^\equiv - \phi).

\label{eq:atomsatz}

\eeq

Measuring the 3 atoms' spins now will yield, with a
uniform probability, all possible results, except for the case where all the
atoms are found in their $\ket{Z-}$ boxes, which will never occur due to a
destructive interference between $\psi^\equiv$ and $\phi$.

Reuniting the atoms' $Z$ boxes and measuring their $\sigma_x$
will yield all possible combinations of $\ket{X+}$ and $\ket{X-}$ in uniform
probability, except the case of all three atoms measuring $X+$ which has a
higher probability. This is not surprising, as these atoms are supposed to have
interacted either with the guide wave, or with the real particle itself (see
\cite{Hardy92a}) or with the uncollapsed wave function
\cite{Clifton92,Pagonis92}. [the point you wanted to add is already here???]

Let us, however, return to the stage before uniting
the $Z$ boxes (as per Eq.\ (\ref{eq:atomsatz})). We know that at least one atom
must be in the $\ket{Z+}$ box to account for the loss of the photon's
interference. Let us, then, measure atom 2's spin, and proceed only if it is
found to be $\ket{Z_2+}$ (57\% of the cases):

\bea

\Psi&=&\frac{1}{4\sqrt{2}} \cdot \ket{d}
\cdot (\ket{Z_1+} + \ket{Z_1-} ) \nncr

&& \quad\quad \cdot \ket{Z_2+} \cdot
(\ket{Z_3+} + \ket{Z_3-} ).

\label{eq:a}

\eea

Now unite the Z boxes of atoms 1 and 3 and apply
the reverse magnetic field $-M$:

\beq

\Psi=\frac{1}{2\sqrt{2}} \cdot \ket{d} \cdot
\ket{X_1+} \cdot \ket{Z_2+} \cdot \ket{X_3+}.

\label{eq:b}

\eeq

Contrary to classical intuition, these atoms will
{\it always} exhibit their original spin undisturbed, just as if no photon has
ever interacted with them.

In other words, only one atom is affected by the
photon in the way pointed out by Hardy, but that atom does not have to be the
first one, nor the last; it can be any one out of the atoms. The other atoms,
whose half wave functions intersected the MZI arm before or after that
particular atom, remain unaffected.

We can prove, however, that although atoms 1 and 3
seem to be totally unaffected by the photon, {\it something} must have passed
through them. As in the previous section, let a macroscopic object be placed
further along the $v$ route, after the three atoms (object ``B'' on Fig.\ 2). The
above results will never show up. Here, all the atoms will give either $Z-$
(when the photon hits the obstacle), or $X+$ (when it does not). Hence,
something must have passed through all three atoms, yet it has left the first
and last unaffected.

Moreover, that ``something'' that seems to have
passed through all the atoms must have done that at the precise moment. Let us
place the atoms within sealed boxes, with apertures at the $v$ path, which open
only for the minute interval during which the photon's wave function is
supposed to pass through them. The slightest failure in the timing of any
aperture's opening will ruin the predicted result.

The next result will deal the final blow on any
realistic account in which a particular atom is affected by the photon at the
moment of their interaction. We noted above that if we pick one atom, measure
its position and find it to reside in the $Z+$ box, then that measurement will
disentangle the two other atoms and their spins will reveal no trace of
interaction with the photon. One might think that there is, prior to
measurements of the atoms, one particular atom that ``has been'' affected, and
that the experimenter only has to be lucky to pick up that ``right'' atom that
yields $\ket{Z+}$. Not so: rather than the normal 33\% probability to find the
``right'' atom, expected when there are 3 atoms, the probability is 56\%!

In other words, every atom chosen by the
experimenter, regardless of its position within the row, has the 57\%
probability to be ``the only atom that has been affected by the photon.'' And
once this atom gives this result, the other atoms will become nonlocally
disentangled \footnote{ No superluminal communication is entailed. In all cases
in which the particular atom is not detected in the intersecting path, the
probability for one of the other atoms to reside in that path increases to 1.
The overall result is Lorentz-invariant. Still, the correlation is Bell-like.}.

Note that the above analysis does not depend on the
number of atoms or the index of the tested atom. For $n$ atoms, the probability
for any atom to be ``the right atom'' is $P={2^{(n-1)} \over 2^n-1}$ instead of
the expected $P=1/n$, approaching $1/2$ as $n$ increases.

\section{Conclusions}

From a formalist’s point of view, there seems to be
no surprise in the fact that the photon's $v$ branch disappeared, leaving only
part of the $\phi$ state, where the unmeasured atoms remain undisturbed. This
is because the photon’s $v$ branch was coupled to the term $\psi^\equiv$ (all
three atoms are in the $Z-$ state) which contains $Z_2-$. By post-selecting
$Z_2+$ we eliminated this term altogether, leaving the others atoms superposed.

However, it is the attempt to reconstruct a
comprehensible scenario from these correlations that gives a highly
counterintuitive picture. For, if it is the measurement of the second atom that
have cancelled the photon’s $v$ term, then, for the photon to reach that atom,
it must have first pass through the first atom, and, later, through the third
as well. If one tries to visualize this result obliged by the formalism, then,
a single photon's wave function seems to ``skip'' a few atoms that it
encounters, then disturb the $m^{th}$ atom, and then again leave all next atoms
undisturbed. Ordinary concepts of motion, which sometimes remain implicit
within prevailing interpretations, are inadequate to explain this behavior. The
most prudent description of this result is that a wave function, when
interacting with a row of other wave functions one after another, does not seem
to comply with ordinary notion of causality, space and time \footnote{This is,
in fact, one out of a family of peculiar effects yielded by experiments of this
kind, when a quantum mechanical object interacts with a macroscopic measuring
apparatus not directly but through another, intermediate quantum mechanical
object. For an ``inverse EPR’’ effect yielded by this method see \cite{Elitzur02}}.

\acknowledgments{We thank Yakir Aharonov, David
Tannor, Terry Rudolph and an anonymous referee for very helpful comments. It is
a pleasure to thank Anton Zeilinger and all participants of the Quantum
Measurement Conference at the Schr\"odinger Institute in

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\end{document}