Quantum measurements and quantum erasers

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Quantum measurementsand quantum erasers
(AKA no more dull than the last lecture?)

• von Neumann measurements
• (entanglement and decoherence)
• The Quantum Eraser
• Equivalence of collapse and correlation pictures
• EPR correlations
• An application of the collapse picture
• EPR correlations for nonlocal dispersion
  cancellation
• (AKA something to leave over for next week
  again...)
• Slides, and some other useful links, are still
  being posted at
• http//www.physics.utoronto.ca/steinberg/QMP.html

21 Oct 2003
2
Recap decoherence arises from throwing away
information
Taking the trace over the environment retains
only terms diagonal in the environment variables
i.e., no cross-terms (coherences) remain if
they refer to different states of the environment.
(If there is any way even in principle to
tell which of two paths was followed, then no
interference may occur.)
?? ?? ?? ??
?s when env is ?
?s when env is ?
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So, how does a system become "entangled" with a
measuring device?

• First, recall Bohr we must treat measurement
  classically
• Wigner why must we?
• von Neumannthere are two processes in QM
  Unitary and Reduction. He shows how all the
  effects of measurement we've described so far
  may be explained without any reduction, or
  macroscopic devices.
• Of course, this gets us a diagonal density
  matrix classical probabilities without
  coherence but still can't tell us how
  those probabilities turn into one occurrence
  or another.

To measure some observable A, let a "meter"
interact with it, so the bigger A is, the more
the pointer on the meter moves. P is the
generator of translations, so this just means
we allow the system and meter to interact
according to Hint ? A P.
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An aside (more intuitive?)
Suppose instead of looking at the position of our
pointer, we used its velocity to take a reading.
In other words, let the particle exert a force on
the pointer, and have the force be proportional
to A then the pointer's final velocity will be
proportional to A too.
F g A U(x) g A X Hint g A X
This works with any pair of conjugate
variables. In the standard case, Hint g A Px ,
we can see
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A von Neumann measurement
Initial State of System
Initial State of Pointer
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A von Neumann measurement
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Entangled (nonseparable) states
Consider the state resulting from this
interaction with a pointer P
If the different states Pi are orthogonal, no
such product could yield terms like 1P1 and 2P2
without yielding 1P2 etc. The canonical example
is the EPR spin state ??? - ???.
IN OTHER WORDS if you ask a question just about
the system on its own, there exists no quantum
state vector which can fully describe it.
Effectively, we have a mixed state, and need the
density matrix obtained by tracing over the
pointer.
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A von Neumann measurement
Effective state of system (if pointer ignored)



OR
OR
OR
A
Unless the pointer is somehow included in the
interferometer, interference will never again be
observed between these different peaks we may as
well suppose a collapse has really occurred, and
one peak or another has been selected at random.
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Back-Action
In other words, the measurement does not simply
cause the pointer position to evolve, while
leaving the system alone.
The interaction entangles the two, and as we
have seen, this entanglement is the source of
decoherence.
It is often also described as "back-action" of
the measuring device on the measured system.
Unless Px, the momentum of the pointer, is
perfectly well-defined, then the
interaction Hamiltonian Hint g A Px looks like
an uncertain (noisy) potential for the particle.
A high-resolution measurement needs a
well-defined pointer position X. This implies (by
Heisenberg) that Px is not well-defined. The more
accurate the measurement, the greater the
back-action. Measuring A perturbs the variable
conjugate to A "randomly" (unless, that is, you
pay attention to entanglement).
(For future thought note that my entanglement
argument needed to assume that the pointer
states were orthogonal.)
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Summary so far...
We have no idea whether or not "collapse" really
occurs. Any time two systems interact and we
discard information about one of them, this can
be thought of as a measurement, whether or not
either is macroscopic, whether or not there is
collapse. The von Neumann interaction shows how
the two systems become entangled, and how this
may look like random noise from the point of view
of the subsystem. The "reduced density matrix"
of an entangled subsystem appears mixed, because
the discarded parts of the system carry
away information. This is the origin of
decoherence of the measured subsystem.
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Quantum Eraser(Scully, Englert, Walther)
Suppose we perform a which-path measurement using
a microscopic pointer, z.B., a single photon
deposited into a cavity. Is this really
irreversible, as Bohr would have
all measurements? Is it sufficient to destroy
interference? Can the information be erased,
restoring interference?
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Some mathematics...
But what if we select (project) out, not A, and
not B, but an equal superposition?
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A microscopic measurement
The "i" photons provide which-path information,
and destroy the interference. Can this
information be "erased"?
If it's no longer possible to tell whether the
photon came from s1 or s2, then interference is
restored!
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But it is still possible...
In fact, this should have been obvious. If
combining the i photons at a beam-splitter could
restore fringes on the right, nothing would
prevent me from combining them a year after you
looked at your detectors. Could I change whether
or not you had seen fringes ?! UNITARY EVOLUTION
CANNOT DESTROY INFORMATION! ORTHOGONAL STATES
REMAIN ORTHOGONAL FOR ALL TIME. Obviously,
nothing you do to the idlers can affect the
signals.
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Sorry, that was another lie.
Nothing unitary I do to the idlers affects the
signals. Measurement is not unitary in other
words, if I only keep some events and throw out
others, perhaps I can restore your interference.
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Don't overlook the symmetry...
Detectors 1 and 2 are equally likely to fire,
regardless of the phase setting.
When the "i1-i2" detector fires, this may tell me
that detector 1 will fire instead of detector 2.
Of course, have the time, the "i1i2" detector
fires, telling me that detector 2 will fire
instead of detector 1.
...or is it that half the time, detector 1 fires,
collapsing the "i" photon into "i1-i2"...
...and that half the time,
detector 2 fires, collapsing the "i" into
"i1i2"...?
Which is the system and which is the measuring
apparatus?
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Making it look more complicated...
Ou, Wang, Zou, Mandel, Phys Rev A 41, 566
(1990).
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Plus ça change...
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What if you combine the idlers sothey've got
nowhere else to go?
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A polarisation-based quantum eraser...
t2 r2 1/2 - 1/2 0 no coincidence counts.
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The polarisation quantum eraser
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Interference going away...
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And coming back again!
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How complicated you have to makeit sound if you
want to get it published
"Calculations are for those who don't trust their
intuition."
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Simple collapse picture

• Suppose I detect a photon at
• here. This collapses my photon
• into H cos q V sin q.
• This means an amplitude of cos q
• that the other photon was V, and
• of sin q that it was H.
• Being careful with reflection phase
• shifts, this collapses the other output
• port into V cos q - H sin q, which of
• course is just (q p/2).

M1
SOURCE
signal
V
BS
idler
H
M2
Here I'm left with a photon 900 away from
whatever I detected. Now I just have linear
optics to think about. Of course I get
sinusoidal variation as I rotate this polarizer.
"...and experiment is for those who don't trust
their calculations."
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Polarisation-dependence of rateat centre of
H-O-M dip...
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But did I need to invoke collapse?(and if so,
which photon did the work?)
This is the Bell-Inequality experiment done by
ShihAlley and OuMandel.
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Hong-Ou-Mandel Interferenceas a Bell-state
filter (Viennese delicacy)
r2t2 0 total destructive interf. (if photons
indistinguishable). If the photons begin in a
symmetric state, no coincidences. Exchange
effect cf. behaviour of fermions in analogous
setup! The only antisymmetric state is the
singlet state HVgt VHgt, in which each photon
is unpolarized but the two are orthogonal.
Nothing else gets transmitted. This
interferometer is a "Bell-state filter," used for
quantum teleportation and other applications.
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Some references
Quantum measurement theory the quantum
eraser some early QE experiments.
Bell-inequality tests dispersion
cancellation newer QEs (atom interferometry de
layed choice).
Y.H. Kim et al., Phys. Rev. Lett. 84, 1
(2000) T. Pfau et al., Phys. Rev. Lett. 73, 1223
(1994)