Download Fourth lecture, 28.10.03 (dispersion cancellation, time measurement

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Transcript
Frequency and time... dispersioncancellation, optical phase, etc.
(AKA: A bunch of loosely connected stuff I decided
I need to cover before I get to some other stuff...)
• Dispersion cancellation in an HOM interferometer
– (more "collapse versus correlations")
– (useful for time measurements)
• What are time measurements?
– (no time operator)
– (indirect measurements)
– (energy-time "uncertainty relation")
• States of an electromagnetic mode
– (number-phase "uncertainty relation")
– (homodyne measurements, et cetera)
– Phase of a single photon...
28 Oct 2003
Entangled photon pairs
(spontaneous parametric down-conversion)
The time-reverse of second-harmonic generation.
A purely quantum process (cf. parametric amplification)
Each energy is uncertain, yet their sum is precisely defined.
(For a continuous-wave pump!)
Each emission time is uncertain, yet they are simultaneous.
Hong-Ou-Mandel interferometer
Remember: if you detect only one photon, the other photon "knows" where
yours came from. Hence there is no interference (each detector sees 1/2 of
the photons, irrespective of any phases or path-length differences).
But: if you detect both photons, there is no way to tell whether both were
reflected or both were transmitted. r2+t2 = (i2 + 12)/2 = 0.
(any lossless symmetric beam splitter has a p/2 phase shift between r and t.)
CAVEAT: there must be no way to tell which occurred.
If the paths aren't aligned right, no interference occurs.
If one photon reaches the beam splitter before the other, no interference occurs.
How long is a photon?
the famous dip
In every experiment to date, the
width of this feature is limited only
by the bandwidth of the photons;
in other words, the photons are as
tightly correlated as they could
possibly be given their own
uncertainty in time (Dt > 1/2Dw).
DELAY TIME (fs)
Instead of an amplitude for each frequency component of each beam,
there is an amplitude for each frequency-correlated pair of photons.
Energy-entangled state.
What's the speed of a photon?
Silly questions about group velocity, phase velocity, "collective" nature
ofthe index of refraction, precursors, et cetera.
More serious questions about how to quantize electromagnetic fields
in a dispersive medium.
Longstanding debate about "superluminal" tunneling.
Problem with propagation
measurements
Quadratic term leads to group-velocity dispersion,
broadening, chirp.
But wait!
So, what can we predict?
Detectors are "infinitely slow" (ns = 106 fs);...
Coinc. probability is an integral over
all times at which D1 and D2 could fire:
UGH! Each E is an integral over its own frequency!
But... if T goes to infinity, life simplifies:
We have only to integrate a†a†aa over all w1, w2.
The physical meaning: calculate the probability for each pair of frequencies which
might reach the two detectors, and then integrate.
Why? No interference between paths leading to different frequencies at the detectors,
because in principle one could go back and measure how much energy had been absorbed.
Note: it took a long time-integral to enforce this. If the detector had been open only for
1 fs, it would be impossible to tell what frequency it had seen.
No, we're not done yet...
The probability of detecting a given frequency pair:
(one can always insert a complete set of states.)
But we only started with 2 photons, so if we annihilate 2, there will be 0 left:
(the square of the two-photon amplitude)
the phase difference is all that concerns us:
Note 1: the GVD terms (quadratic in freq.) cancelled out.
Note 2: the pattern moves at the group velocity.
Note 3: the shape is the Fourier transform of f(w), like the pulse itself.
Nonlocal cancellation of dispersion
(Oh, and by the way...
yeah, single-photon pulses travel at the group velocity.)
The Feynman path picture
REMEMBER: interference
only occurs between two
paths which yield the same
w1 and the same w2.
If in the TT path, a "blue"
photon is detected at D1
(and a "red" at D2), then I
must compare with the phase
of an RR path with the same
outcome...
The two paths are indistinguishable
even though in each case one of the
pulses was broadened.
Perfect interference occurs.
The collapse picture
After D1 fires, it projects the light
in arm 2 into a superposition of two
identical chirped wave packets-these two packets exhibit perfect
interference.
But how do you really measure time?
There is no such thing as a direct time measurement, even classically.
"The position of an electron" has some meaning...
but what is "the time of an electron"?
In QM (as in CM), time is a free parameter, not an operator/observable.
Time is always measured by observing a pointer which is believed to evolve at
a constant rate, and "triggering" its evolution (or at least readout).
Clock
Oscilloscope
x(t):
Time-to-amplitude converter
Vout = V' t
What's the shortest time you can
measure?
For this "clock" system to evolve appreciably in a time t,
it must have an energy uncertainty greater than h/t.
"Energy-time uncertainty principle."
Warning: oddly, the reverse is not true. You can measure energy arbitrarily
well in an arbitrarily short time [Aharonov + Bohm, Phys Rev 122, 1649 (1961)].
Why no rigorous uncertainty principle?
Because time is not an operator.
Energy (the Hamiltonian) is the generator of time-translations.
A canonically conjugate time operator would generate energy-translations.
But energy is bounded (at least from below); no Hermitian T-operator exists.
Can you use photons as a clock?
Of course; Electric field changes very rapidly, on the order of 1015 Hz:
(even microwave -- Cs atomic clock -- yields 9•109 Hz)
But wait... the energy of a photon is completely certain.
Energy eigenstate = stationary state:
In a single-photon state, there "are" oscillations
with a known amplitude... but since their time
origin is completely uncertain, so is E(t).
Compare an eigenstate of the harmonic oscillatorthe maximum X is related to the energy, but the
particle's equally likely to be at +X or -X.
What does this mean?
Remember: two photons meeting at a beam-splitter don't interfere.
|1>
|1>
n1
n2
The cross terms change n1 and n2; unless these
numbers are uncertain, these terms vanish.
Optical phase actually refers (roughly) to the quantum phase difference
between the amplitude to have n photons and the amplitude to have n+1.
Uncertainty relations?
What is phase, except time?
What is photon-number, except energy?
Phase-space picture
complex E = |E| eif e-iwt
= E1 cos wt + E2 sin wt
E2 (think p)
|E|
f
The vacuum isn't empty:
E1 (think x)
minimum phase-space area,
as for X and P of a particle.
E2 (think p)
E2 (think p)
E1 (think x)
A phase-space distribution
for a 1-photon state
E1 (think x)
A "coherent" or "quasiclassical" state.
Electric field of a coherent state
E
t
(nice pictures stolen from Gerd Breitenbach's web page)
How does one measure these things?
With a radio-frequency field, of course, one measures E with an antenna, in real time.
Not possible at optical frequencies - we can only detect power/energy/photons.
Homodyning/heterodyning: interfere a signal with a strong oscillator.
|Es eif + |ELO| |2
|ELO|2 + 2 |ELO|Es cos f + ...
By varying phase of local oscillator, can measure different quadratures
just by measuring intensity.
Note: not so different from RF; your oscilloscope has a local oscillator too.
E
t
E1
E2
n
Back to Bohr & Einstein
(and Young and Taylor,...)
So, does a single photon exhibit
interference, or not?!
|1>
Entangled state
|01> + |10>; no
definite number
in either arm.
Some references
(incomplete for now!)
Dispersion cancellation:
http://home.t-online.de/home/gerd.breitenbach/gallery/
for info & pictures about squeezed-states and their measurement.
Uncountable textbooks & review articles...
see for instance Loudon; Walls & Milburn; Scully & Zubairy.
Glauber for quantum optics formalism
Clauser for early work on classical versus quantum effects
Grangier et al. for experiment on true single-photon interference.
Thought for next week:
atom number is conserved (unlike photon number).
Can it be uncertain?
Should two atom clouds interfere with one another or not?