introduction to tunneling times and to weak measurements

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Introduction to tunneling times and to weak measurements How does one actually measure time ? (recall: there is no operator for time) How long does it take a particle to tunnel through a forbidden region? Classically: time diverges as energy approaches barrier height. "Semi"classically: kinetic energy negative in tunneling regime; velocity imaginary? Wave mechanics: this imaginary momentum indicates an evanescent (rather than propagating) wave. No phase is accumulated... vanishing group delay? Odd predictions first made in the 1930s and 1950s (MacColl, Wigner, Eisenbud), but largely ignored until 1980s, with tunneling devices. This was the motivation for us to apply Hong-Ou-Mandel interference to time-measurements: to measure the single-photon tunneling time. 18 Nov 2003

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Introduction to tunneling times and to weak measurements. How does one actually measure time ? (recall: there is no operator for time) How long does it take a particle to tunnel through a forbidden region? Classically: time diverges as energy approaches barrier height. - PowerPoint PPT Presentation

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Introduction to tunneling timesand to weak measurements

• How does one actually measure time ?(recall: there is no operator for time)

• How long does it take a particle to tunnel through a forbidden region?

• Classically: time diverges as energy approaches barrier height.• "Semi"classically: kinetic energy negative in tunneling regime; velocity imaginary?• Wave mechanics: this imaginary momentum indicates an evanescent (rather than propagating) wave. No phase is accumulated... vanishing group delay?• Odd predictions first made in the 1930s and 1950s (MacColl, Wigner, Eisenbud), but largely ignored until 1980s, with tunneling devices.• This was the motivation for us to apply Hong-Ou-Mandel interference to time-measurements: to measure the single-photon tunneling time.

• How does one discuss subensembles in quantum mechanics?• Weak measurement• How can the spin of a spin-1/2 particle be found to be 100?• How can a particle be in two places at once?• Where is a particle when it's in the forbidden region?

18 Nov 2003

How Long Does Tunneling Take?

We frequently calculate the tunneling rate, e.g., in a two-well system.But how long is actually spent in the forbidden region?Classically, time diverges as E approaches V0; the "semiclassical" time

(whatever it means) behaves the same way...Since the 1930s, group-velocity calculations yielded strange results:

evanescent waves pick up no phase, so no delay is accumulatedinside the barrier?

1980s: Büttiker & Landauer and others propose many other times.

What's the speed of a photon?Can tunneling really be nearly instantaneous? Group-delay prediction saturates to a finite value as barrier thickness grows. For thick enough barriers, it would then be superluminal ( < d/c).

Recall that the Hong-Ou-Mandel interferometer can be used to compare arrival times of single-photon wavepacket peaks.

We used one to check the delay time for a photon tunneling through a barrier.

tunnel barrier

The results

How can this be?

QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.

n1n2 .......

Very little light is transmittedthrough a tunnel barrier (aquarter-wave-stack dielectricmirror, in our experiment).

But how that's all classical waves...how fast did a given photon travel?

Interaction Times

• Büttiker and Landauer: "no law guarrantees that a peak turns into a peak."

• Ask instead how long the particle interacted with something in the barrier region

• (More relevant to condensed-matter systems anyway)

Larmor Clock (Baz', Rybachenko, and later Büttiker)

B

e- e-

Tx

+

z

-z

But in fact: =

x+

z

-z

Which is "the" tunneling time?

Ty? Tz? Tx2 = Ty

2 + Tz2 ?

Disturbing feature... Ty is still nearly insensitive to d, and often < d/c.

Büttiker therefore preferred Tx... which also turns out < d/c, but rarely!

z

x

y

Ty

zTz

Too many tunneling times!

Questions which seem unambiguous classically may have multiple answers in QM – in other words, different measurements which allyield "the time" classically need not yield the same thing in the quantum regime.In particular: in addition to affecting a pointer, the particle itself may be affected by it.

Various "times":

group delay"dwell time"Büttiker-Landauer time

(critical frequency of oscillating barrier)Larmor times (three different ones!)

et cetera...

Okay -- so let's consider specific measurements.

What is this measurement?A few things to note:

• This -˚B interaction is a von Neumann measurement of B (which in turn stands in for whether or not the particle is in the region of interest)

• Since Bz couples to z , the pointer is the conjugate variable (precession of the spin about z) –– Note that this measurement is thus just another interference effect, as the precession angle is the phase difference accumulated between and .

• We want to know the outcome of this von Neumann measurement only for those cases where the particle is transmitted.

• "Being transmitted" doesn't commute with "being under the barrier"; is it valid to even ask such post-selected questions? If so, how can you do so without first collapsing the particle to be under the barrier?

• Note: this Larmor precession could not determine for certain whether or not the particle had been in the field, or for how long; only on a large ensemble can the precession angle be measured to better accuracy than 180o .

Predicting the past ?

Standard recipe of quantum mechanics:1. Prepare a state |i> (by measuring a particle to be in that state; see 4)2. Let Schrödinger do his magic: |i> |f>=U(t) |i>, deterministically3. Upon a measurement, |f> some result |n> , randomly4. Forget |i>, and return to step 2, starting with |n> as new state.

Aharonov’s objection (as I read it):No one has ever seen any evidence for step 3 as a real process;

we don’t even know how to define a measurement.Step 2 is time-reversible, like classical mechanics.Why must I describe the particle, between two measurements (1 & 4)

based on the result of the first, propagated forward,rather than on that of the latter, propagated backward?

Conditional measurements(Aharonov, Albert, and Vaidman)

Prepare a particle in |i> …try to "measure" some observable A…postselect the particle to be in |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.

ii ffMeasurement of A

Reconciliation: measure A "weakly."Poor resolution, but little disturbance.

AAV, PRL 60, 1351 ('88)

"weak values"

A (von Neumann) Quantum Measurement of A

Well-resolved statesSystem and pointer become entangled

Decoherence / "collapse"Large back-action

Initial State of Pointer

x x

Hint=gApx

System-pointercoupling

Final Pointer Readout

A Weak Measurement of A

Poor resolution on each shot. Negligible back-action (system & pointer separable)

Hint=gApx

System-pointercoupling

x

Initial State of Pointer

x

Final Pointer Readout

Strong:

Weak:

if

iAfA =w

Bayesian Approach to Weak Values

Note: this is the same result you get from actuallyperforming the QM calculation (see A&V).

Ritchie, Story, & Hulet 1991

Very rare events maybe very strange as well.

Weak measurement & tunneling times

Conditional probability distributions

A problem...These expressions can be complex.

Much like early tunneling-time expressions derived viaFeynman path integrals, et cetera.

A solution...

Conditional P(x) for tunneling

QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.

What does this mean practically?

Predicting the past...

A+B

What are the odds that the particlewas in a given box (e.g., box B)?

B+C

A+B

It had to be in B, with 100% certainty.

Consider some redefinitions...

In QM, there's no difference between a box and any other state (e.g., a superposition of boxes).

What if A is really X + Y and C is really X - Y?

Then we conclude that if you prepare in (X + Y) + B and postselect in (X - Y) + B, you know the particle was in B.

But this is the same as preparing (B + Y) + X and postselecting (B - Y) + X, which means you also know the particle was in X.

If P(B) = 1 and P(X) = 1, where was the particle really?

The 3-box problemPrepare 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?

PA = < |A><A| >wk = (1/3) / (1/3) = 1PB = < |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?

if

iAfA =w

[Aharonov & Vaidman, J. Phys. A 24, 2315 ('91)]

Remember that test charge...

e-

e-

e-

e-

Aharonov's N shutters

PRA 67, 42107 ('03)

Some references Tunneling times et cetera:

Hauge and Støvneng, Rev. Mod. Phys. 61, 917 (1989)Büttiker and Landauer, PRL 49, 1739 (1982)Büttiker, Phys. Rev. B 27, 6178 (1983)Steinberg, Kwiat, & Chiao, PRL 71, 708 (1993)Steinberg, PRL 74, 2405 (1995)

Weak measurements:Aharonov & Vaidman, PRA 41, 11 (1991)Aharonov, Albert, & Vaidman, PRL 60, 1351 (1988)Ritchie, Story, & Hulet, PRL 66, 1107 (1991)Wiseman, PRA 65, 032111Brunner et al., quant-ph/0306108Resch and Steinberg, quant-ph/0310113

The 3-box problem:Aharonov et al J Phys A 24, 2315 ('91);

PRA 67, 42107 ('03)Resch, Lundeen, & Steinberg, quant-ph/0310091