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Physical observables, locality, and a preferred basis

  1. Apr 25, 2013 #1
    Quantum mechanics says that physical observables are self-adjoint operators. Is this correspondence a bijection, ie can we realize any such operator as a physical observable? There are obvious practical concerns with physically realizing certain contrived operators. But are there any theoretical limitations (eg locality) to measuring in an arbitrary basis?

    Suppose we place an electron in a box with two compartments, left and right, and insert a barrier between the compartments, so that the electron is in a superposition of being in either. When we open the box and measure the electron's position by simply looking, we find the electron either in the L state or in the R state. This is because the "simply looking" measurement process corresponds to the "position" operator, which has eigenstates L and R, and these eigenstates are the possible results of the measurement.

    Of course, "a superposition of left and right" is one of many ways to describe the electron's state before measurement. We can choose another basis consisting of the eigenstates of another operator. I can easily write an abstract operator with (L+R)/√2 and (L-R)/√2 as eigenstates, but can I observe the electron in one of these states?

    With some effort, yes. We can position a lens at the opening of each compartment and direct the two signals (light from the electron) to a system (such as a one-way mirror) that combines the signals. We can do this such that, due to constructive interference, (L+R)√2 yields a signal while, due to destructive interference, (L-R)/√2 does not yield a signal. We have measured in the new eigenbasis.

    That's great. But now consider Schrodinger's Cat. Can I design an experiment (however contrived) to measure in the (Alive±Dead)/√2 basis?

    Why, in the theoretical sense, is this measurement extremely difficult (if not impossible)?

    My guess is that it has something to do with the locality of physical interactions. Somehow (?), locality gives preference to the position basis (eg L and R). Since it is impossible to fundamentally overcome this preference, we must instead convert (via contrived experiments) arbitrary states into position states (eg (L+R)/√2 becomes "YES, I observe a signal through the lens").

    Am I on the right track with locality and the preferred basis?
     
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  3. Apr 26, 2013 #2

    Simon Bridge

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    Not quite - only that the eigenvalues of self-adjoint operators correspond to physically observable properties. But it is usually considered to go both ways - yep.

    The method of looking is quite important.

    But what you are describing is preparing two systems in a classically uncertain state ... this would be a "hidden variables" system. Within each "side" the electron may have any combination of the energy eigenstates of the system it finds itself in... inside it's side of the box, it may have a range of positions.

    How you do the detection is important. You seem to be describing a situation where there are two electron detectors - each covering half the combined box. Each 100% accurate.

    What I am trying to say is that the setup is unlikely to be careful enough to serve in the thought-experiments you are trying to use.

    eg. Lets say the electron is prepared in the ground state of one of two infinite square wells with a 50:50 classical probability. When the barrier is removed, the energy state of the electron becomes uncertain - the result of a subsequent measurement will depend on the time evolution of initial position wave-function under the new situation.

    What to expect of a measurement of position will depend on the evolution from either of the initial states using classical statistics to decide between them.
     
  4. Apr 26, 2013 #3
    I think you're correct that I didn't set up my experiment carefully enough. Perhaps I could have gone with some form of the double slit experiment; maybe there would also be problems here. I haven't given the physical details much thought.

    For the sake of argument, let's suppose that, somehow, I have a system in a pure state ψ that is a quantum superposition of eigenstates of some nice observable (maybe position or energy?). (To clarify, I don't mean to involve mixed states or classical uncertainty.)

    Is it always possible (through use of optics or whatever) to make a measurement in a basis that includes ψ? Or does nature have a tendency to see certain ψ as superpositions of a preferred eigenbasis? If it does, why?
     
  5. Apr 26, 2013 #4

    Simon Bridge

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    Yeh - I figured you want to go for something a bit different for your example - rather than have us go into depth with one that wont serve.
    If soething is in a superposition of eigenstates, then it is in a mixed state. You cannot have it both ways.

    It is always possible to expand a wavefunction in terms of eigenstates of any operator.
    The choice of basis is determined by the measurement process.
    In general, we use whatever expansion makes the math easy.

    There is a lot of discussion of this around 2-state systems, but to follow it properly you need to get used to the math. What I am concerned about is that you appear to be trying to get to some quite tricky QM concepts from pop-accounts of Schodinger's cat. I could be mistaken.

    To advance:
    Study the ideal "2-state system" (search terms in quotes), the "Stern-Gerlach experiment" (quantum measurement), and the "neutral kaon oscillations".
     
  6. Apr 26, 2013 #5

    DrClaude

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    What definition of mixed states are you using? The one I know is that for a pure state
    $$
    \mathrm{Tr}\{\rho^2\} = 1
    $$
    otherwise it is a mixed state. That definition of a pure state does not preclude the system to be in a superposition of eigenstates. Moreover, excluding superpositions would mean that you could only classify a state as pure/mixed for a given basis.
     
  7. Apr 26, 2013 #6

    Simon Bridge

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    Any mixture of quantum states ... but there is a stricter use eg.
    http://en.wikipedia.org/wiki/Mixed_state_(physics)#Mixed_states

    Pure state can be represented as a single vector.
    Mixed states must be represented by a density matrix.

    So - returning the post #3 in this context, OP wants to consider only states that can be represented as a single ket. This would allow a superposition of kets. But that's pretty much how I continued in post #4 (I hope).

    I'd like to hear OPs response before going into details.

    Perhaps you have a different take on the original question to share?
     
  8. Apr 26, 2013 #7

    DrClaude

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    I just wanted to make sure things were clear. And it doesn't look like the rest of your answer is affected at all.

    Well, maybe I should just point out that I don't see what locality has to do with it. You can imagine that the cat would look exactly the same if it were asleep or dead, and therefore that the cat's wave function wouldn't even collapse when opening the box, but only after measuring the expectation value of the operator "heartbeat" (which commutes with the operator "living" :tongue2:).

    But I certainly agree with you that it makes more sense to consider actual physical systems for with we can write operators instead of guessing what the difference would be in measuring ##|\text{alive}\rangle + |\text{dead}\rangle## versus ##|\text{alive}\rangle - |\text{dead}\rangle##.
     
  9. Apr 26, 2013 #8

    Simon Bridge

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    Heh!
    Well the main point with the cat was that the wavefunction was not for the cat exactly, but for an actual quantum system for which we can write operators ... and some contrivance connects a measurement with the life of the cat.

    Penrose suggested that the cat was part of the detector - the cat itself being complex enough to be self-measuring. The issue then being about when the wave-function "collapsed". i.e. rig the cat with a heart-rate monitor which prints out a message with the current state of the cat - but on one of those secure envelopes you get your payslip in... so the wavefunction collapses when someone reads the message?

    My grad student picks it up and brings it to me, but I'm in a lecture so it gets left in my in-box... a stray breeze blows it to the floor and when I got to my office I accidentally kick it under the desk where it languishes for months until the cleaner finds it ... the cleaner opens it and reads it but does not under stand it ... throws it away but mentions what it said on physics-forums and someone reads the post and tells them what it means and thus collapses the wavefunction.

    But maybe it collapse long ago when the box started to smell?

    Perhaps the wavefunction collapsed - then set off a chain of events which kills the cat...
    You can drive yourself nuts with this sort of thing.

    BTW: I did see a lecture at one point where locality was argued to be important to this.
    Notice that there is a time lag for the information to get to me?
     
  10. Apr 26, 2013 #9

    kith

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    That's not correct. Superpostions are as pure as eigenstates. The property of being an eigenstate depends on the basis, which is given by the observable in the case of measurement. Every state is an eigenstate as well as a superposition - just wrt a different basis.

    The OP's question is actually a really good one. How do we build a measurement apparatus given a mathematical observable? Do general principles determine what classes of mathematical observables can be measured easily or even at all? Why does it seem easy to measure in the {|dead>, |alive>} basis but not in the {|dead>+|alive>, |dead>-|alive>} basis?

    I think the answer is given by decoherence and einselection. In order to perform a measurement, the final eigenstates have to be stable against perturbations from the environment. The nature of the interactions between the system of interest and the environment determines which basis is "preferred". IIRC, it can be shown that the Coulomb interaction leads to the position basis.
     
  11. Apr 26, 2013 #10
    My apologies about the confusion, everyone. When I said "mixed state", I meant what DrClaude and kith have described: a single ket. I acknowledge that a single ket can be expanded as a superposition of eigenkets. But in doing so, classical probability is not invoked.

    kith puts my original question most eloquently:

    I read the Wikipedia article for einselection, and it's exactly what I wanted. I'd be interested to hear further thoughts about the role of the Coulomb interaction in selecting pointer states. But now that I have new words to google, I may just do this.
     
  12. Apr 27, 2013 #11

    vanhees71

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    One should really clarify this issue before discussing further! Quantum theory is confusing only if not discussed properly!

    A quantum system is described by a Hilbert space and an algebra of observables. The state of the quantum system is described by a positive semidefinite self-adjoint operator [itex]\hat{R}[/itex] of trace 1. The system's properties are determined completely if this self-adjoint operator is a projection operator. Then it takes the form [itex]\hat{R}=|\psi \rangle \langle \psi|[/itex] with a normalized Hilbert-space vector [itex]|\psi \rangle[/itex]. This is called a pure state. If this is not the case the state is called mixed.

    A system can be prepared in a pure state by determining the values of a complete set of compatible observables. Then it's state is given by the common eigenvector of the corresponding operators representing these observables.
     
  13. Feb 17, 2014 #12
  14. Feb 17, 2014 #13

    bhobba

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    Quite right.

    That's why the OP needs to see a proper axiomatic treatment of QM.

    For that see the first three chapters of Ballentine - Quantum Mechanics - A Modern Development.

    Thanks
    Bill
     
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