Transcript Optical implementation of the Quantum Box Problem

Optical implementation of the
Quantum Box Problem
AKA: Let's Make a Quantum Deal!
Kevin Resch
Jeff Lundeen
Aephraim Steinberg
Department of Physics,
University of Toronto
Outline
• Motivation: subensembles, postselection,
and "Let's Make a Deal"
• Weak measurements
• The Quantum 3-Box Problem
• An optical implementation
• Can a particle be in 2 places at 1 time?
• Summary
Funding (sources not scared to admit
association with this research):
Motivation
In QM, one can make predictions (probabilities)
about future observables.
Can one "retrodict" anything about past observables?
What can one say about "subensembles" defined
both by preparation & post-selection?
N.B. Questions about postselected subensembles
becoming more and more widespread in
quantum optics, quantum info, etc.
[Cf. Wiseman, PRA 65, 032111 ('02) and quant-ph/0303139]
Pick a box, any box...
A+B+C
A+B–C
What are the odds that the particle
was in a given box?
Weak Measurements
Conditional measurements
(Aharonov, Albert, and Vaidman)
AAV, PRL 60, 1351 ('88)
Prepare a particle in |i> …try to "measure" some observable A…
postselect the particle to be in |f>
i i
Measurement
of A
f f
Does <A> depend more on i or f, or equally on both?
Clever answer: both, as Schrödinger time-reversible.
Conventional answer: i, because of collapse.
Reconciliation: measure A "weakly."
Poor resolution, but little disturbance.
Aw 
f Ai
f i
…. can be quite odd …
A (von Neumann) Quantum
Measurement of A
Initial State of Pointer
Final Pointer Readout
Hint=gApx
System-pointer
coupling
x
x
Well-resolved states
System and pointer become entangled
Decoherence / "collapse"
Large back-action
A Weak Measurement of A
Initial State of Pointer
Final Pointer Readout
Hint=gApx
x
System-pointer
coupling
x
Poor resolution on each shot.
Negligible back-action (system & pointer separable)
Mean pointer shift is given by <A>wk.
The Quantum 3-Box Problem
The 3-box problem
Prepare a particle in a symmetric superposition of
three boxes: A+B+C.
Look to find it in this other superposition:
A+B-C.
Ask: between preparation and detection, what was
the probability that it was in A? B? C?
Aw 
f Ai
f i
PA = < |A><A| >wk = (1/3) / (1/3) = 1
PB = < |B><B| >wk = (1/3) / (1/3) = 1
PC = < |C><C|>wk = (-1/3) / (1/3) = -1.
Questions:
were these postselected particles really all in A and all in B?
can this negative "weak probability" be observed?
[Aharonov & Vaidman, J. Phys. A 24, 2315 ('91)]
Aharonov's N shutters
PRA 67, 42107 ('03)
An Optical Implementation
The implementation –
A 3-path interferometer
Diode Laser
Spatial Filter: 25um PH, a 5cm and a 1” lens
l/2
GP A
BS1, PBS
l/2
MS, fA
GP B
BS2, PBS
GP C
BS4,
50/50
MS, fC l/2
BS3, 50/50
PD
CCD
Camera
Screen
The pointer...
• Use transverse position of each photon as pointer
• Weak measurements can be performed by tilting a
glass optical flat, where effective
Hint  g A A p x
Mode A
q
Flat
gt
cf. Ritchie et al., PRL 68, 1107 ('91).
A negative weak value
1.4
Intensity (arbitrary units)
Perform weak msmt
on rail C.
1.2
A+B–C
(neg. shift!)
Rail C
(pos. shift)
Post-select either A, 1
B, C, or A+B–C.
0.8
Compare "pointer
states" (vertical 0.6
profiles).
0.4
220
Rails A and B (no shift)
200
180
160
140
Pixel Number
120
100
Post-selected state displacement
(Units of RMS Width)
Data for PA, PB, and PC...
2
Rails A
and B
1
0
Rail C
-1
WEAK
STRONG
STRONG
-2
-3
-2
-1
0
1
2
Displacement of Individual Rail
(Units of RMS Width)
3
So can one really "detect" that a particle is
in box A and that it is in box B ????
Measuring joint probabilities
• If PA and PB are both 1, what is PAB?
• For AAV’s approach, one would need an
interaction of the form
Hint  g A A B B p x
OR: STUDY CORRELATIONS OF PA & PB...
- if PA and PB always move together, then
the uncertainty in their difference never changes.
- if PA and PB both move, but never together,
then D(PA - PB) must increase.
Practical Measurement of PAB
Use two pointers (the two transverse directions)
and couple to both A and B; then use their
correlations to draw conclusions about PAB.
Hint  g A A A p x  g B B B p y
We have shown that the real part of PABW
can be extracted from such correlation
measurements:
Re PABW  
2 xy
g Ag Bt
2
- Re(P
*
AW
BBW )
Extracting the joint probability...
anticorrelated
exact calculation particle model
no correlations
(PAB = 1)
And a final note...
The result should have been obvious...
|A><A| |B><B|
= |A><A|B><B|
is identically zero because
A and B are orthogonal.
Even in a weak-measurement sense, a particle
can never be found in two orthogonal states at
the same time.
Summary
• You have won the fabulous vacation!
• We have implemented the quantum box
problem and confirmed the predictions,
including the strange "negative probability."
• New method for joint weak measurements
• Each photon appears to be definitely in each of
two places but never both (cf. Aharonov et al., Phys.
Lett. A 301, 130 (2002) on Hardy's Paradox)
• Much more to explore in the strange magical
land of weak measurements!