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Quantum measurements and quantum erasers

Feb 25, 2016

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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...) - PowerPoint PPT Presentation

  • Quantum measurementsand quantum erasersvon Neumann measurements (entanglement and decoherence)The Quantum EraserEquivalence of collapse and correlation picturesEPR correlationsAn application of the collapse pictureEPR 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.html21 Oct 2003(AKA: no more dull than the last lecture?)

  • Recap: decoherence arises from throwing away informationTaking the trace over the environment retains only terms diagonalin the environment variables i.e., no cross-terms (coherences) remainif they refer to different states of the environment.(If there is any way even in principle to tell which of twopaths was followed, then no interference may occur.)s when env is s when env is

  • So, how does a system become "entangled" with a measuring device? First, recall: Bohr we must treat measurement classically Wigner why must we? von Neumann:there 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" interactwith it, so the bigger A is, the more the pointer on themeter moves.P is the generator of translations, so this just means weallow the system and meter to interact according toHint A P.

  • 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'sfinal velocity will be proportional to A too.F = g AU(x) = g A XHint = g A XThis works with any pair of conjugate variables.In the standard case, Hint = g A Px , we can see

  • A von Neumann measurementInitial State of PointerInitial State of System

  • A von Neumann measurement

  • Entangled (nonseparable) statesIf the different states Pi are orthogonal, no such product couldyield terms like 1P1 and 2P2 without yielding 1P2 etc. Thecanonical example is the EPR spin state | - |.IN OTHER WORDS: if you ask a question just about the systemon its own, there exists no quantum state vector which can fullydescribe it. Effectively, we have a mixed state, and need the density matrixobtained by tracing over the pointer.Consider the state resulting from this interaction with a pointer P:

  • A von Neumann measurementAEffective state of system(if pointer ignored)ORORORUnless the pointer is somehow includedin the interferometer, interference willnever again be observed between thesedifferent peaks; we may as well supposea collapse has really occurred, and one peakor another has been selected at random.+++

  • Back-ActionIn other words, the measurement does not simply cause thepointer position to evolve, while leaving the system alone.The interaction entangles the two, and as we have seen, thisentanglement is the source of decoherence.It is often also described as "back-action" of the measuringdevice on the measured system. Unless Px, the momentumof the pointer, is perfectly well-defined, then the interactionHamiltonian 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.)

  • Summary so far...We have no idea whether or not "collapse" really occurs.

    Any time two systems interact and we discard information aboutone of them, this can be thought of as a measurement, whetheror not either is macroscopic, & whether or not there is collapse.

    The von Neumann interaction shows how the two systems becomeentangled, and how this may look like random noise from the pointof view of the subsystem.

    The "reduced density matrix" of an entangled subsystem appearsmixed, because the discarded parts of the system carry awayinformation. This is the origin of decoherence of the measuredsubsystem.

  • Quantum Eraser(Scully, Englert, Walther)Suppose we perform a which-path measurement using amicroscopic pointer, z.B., a single photon deposited intoa cavity. Is this really irreversible, as Bohr would have allmeasurements? Is it sufficient to destroy interference? Canthe information be erased, restoring interference?

  • Some mathematics...But what if we select (project) out, not A, and not B, but an equal superposition?

  • A microscopic measurementIf it's no longer possible to tell whether the photon came from s1 or s2, then interference is restored!The "i" photons provide which-path information, and destroy the interference.Can this information be "erased"?

  • But it is still possible...In fact, this should have been obvious.If combining the i photons at a beam-splitter could restore fringeson the right, nothing would prevent me from combining them a yearafter you looked at your detectors. Could I change whether or not youhad 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.

  • 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 andthrow out others, perhaps I can restore your interference.

  • 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 fireinstead of detector 2.Of course, have the time, the "i1+i2" detector fires, telling me that detector 2will 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 "i1+i2"...?Which is the system and which is the measuring apparatus?

  • Making it look more complicated...Ou, Wang, Zou, & Mandel,Phys Rev A 41, 566 (1990).

  • Plus a change...

  • What if you combine the idlers sothey've got nowhere else to go?

  • A polarisation-based quantum eraser...t2 +r2 = 1/2 - 1/2 = 0;no coincidence counts.

  • The polarisation quantum eraser

  • Interference going away...

  • And coming back again!

  • How complicated you have to makeit sound if you want to get it published"Calculations are for those who don't trust their intuition."

  • Simple collapse pictureM1M2BSSOURCEsignalidlerVHSuppose I detect a photon at here. This collapses my photoninto H cos q + V sin q.This means an amplitude of cos q that the other photon was V, andof sin q that it was H. Being careful with reflection phaseshifts, this collapses the other outputport into V cos q - H sin q, which ofcourse is just (q + p/2).Here I'm left with a photon 900 away from whateverI 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."

  • Polarisation-dependence of rateat centre of H-O-M dip...

  • But did I need to invoke collapse?(and if so, which photon did the work?)This is the Bell-Inequality experiment done by Shih&Alley and Ou&Mandel.

  • Hong-Ou-Mandel Interferenceas a Bell-state filter (Viennese delicacy)r2+t2 = 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 |HV> |VH>, in which each photon is unpolarized but the two are orthogonal. Nothing else gets transmitted.

    This interferometer is a "Bell-state filter," usedfor quantum teleportation and other applications.

  • Y.H. Kim et al., Phys. Rev. Lett. 84, 1 (2000)

    T. Pfau et al., Phys. Rev. Lett. 73, 1223 (1994)Some references Bell-inequality tests;dispersion cancellation;newer QEs (atom interferometry;delayed choice).Quantum measurement theory;the quantum eraser;some early QE experiments.