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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...)
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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...) Slides, and some other useful links, are still being posted at: http://www.physics.utoronto.ca/~steinberg/QMP.html Quantum measurementsand quantum erasers (AKA: no more dull than the last lecture?) 21 Oct 2003
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
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" 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.
The pointer position evolves at a rate proportional to <A>. 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
Final state of both Hint=gApx A A System-pointer coupling x x A von Neumann measurement Initial State of System Initial State of Pointer
Initial State of System Final state of both (entangled) Hint=gApx System-pointer coupling Initial State of Pointer x x A A A von Neumann measurement
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.
Initial State of System Hint=gApx System-pointer coupling Initial State of Pointer x A 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.
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.)
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.
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?
A superposition state: Probabilities: Interference terms Now consider a larger Hilbert space, including a Measuring Apparatus: New probabilities: NO INTERFERENCE! INTERFERENCE RETURNS! Some mathematics... But what if we select (project) out, not A, and not B, but an equal superposition?
M1 i2 SOURCE det. 1 s1 BS s2 det. 2 i1 M2 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!
(i1+i2) + (i1- i2) = i1 M1 i2 SOURCE i1+i2 s1 BS s2 i1- i2 i1 M2 (i1+i2) - (i1- i2) = i2 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.
i2 Trigger on "i1+i2" events – no longer any way to tell whether they were i1 or i2, no matter what! M1 SOURCE det. 1 s1 i1 BS s2 det. 2 M2 "i1+i2" "i1- i2" Together 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.
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 "i1+i2" 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 "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).
Zou, Wang, Mandel, PRL 67, 318 (1991). What if you combine the idlers sothey've got nowhere else to go?
First, a familiar picture: the Hong-Ou-Mandel interferometer A polarisation-based quantum eraser... t2 +r2 = 1/2 - 1/2 = 0; no coincidence counts.
Half-wave plate H H V H Polarizers (why 2?) H V tt distinguishable; no interference. V H rr The polarisation quantum eraser
How complicated you have to makeit sound if you want to get it published "Calculations are for those who don't trust their intuition."
HWP 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."
M1 SOURCE HWP 1 signal BS idler 2 M2 In coincidence, only see |HV> - |VH> .... that famous EPR-entangled state. Of course we see nonlocal correlations between the polarisations. These joint-detection probabilities can be calculated directly, without collapse; add the amplitudes from HV and VH: P(q1,q2) = |cos(q1)sin(q2)-sin(q1)cos(q2)|2 =sin2(q1 - q2). Vs Hi (V2 + i V1) (H1 + i H2) = 1H 2V - 1V 2H + i [1H 1V + 2H 2V] 250 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) r t + t r 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," used for quantum teleportation and other applications.
Some references Quantum measurement theory; the quantum eraser; some early QE experiments. Bell-inequality tests; dispersion cancellation; newer QEs (atom interferometry; delayed choice). Y.H. Kim et al., Phys. Rev. Lett. 84, 1 (2000) T. Pfau et al., Phys. Rev. Lett. 73, 1223 (1994)
