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I Particles from a thermal source

  1. Feb 22, 2016 #1

    A. Neumaier

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    In the following, I want to consider both photons in a sharply focussed, monochromatic beam of light (''type P'') and electrons in an electron beam (''type E'') on the same footing. In the following, X is either P or E. If we only concentrate on the internal degrees of freedom, both kinds of particles can be described as a 2-level system, with a 2-dimensional Hilbert space.

    Consider a stationary thermal source that emits particles of type X in a sharply focussed, monochromatic beam. Due to the thermal nature of the source, one has to use statistical mechanics to model its statistical behavior. Thus a ##2\times 2## density matrix ##\rho## gives a complete account of the statistics of the ensemble of particles emitted by the source. The density matrix is Hermitian and positive semidefinite of trace 1, such that ##\langle A\rangle=\mbox{tr} \rho A## gives the mean response of a particle received in a detector measuring ##A##. In our particular case, the density matrix is given by ##\rho=\pmatrix{0.5 & 0\cr 0 &0.5}## since the thermal setting implies that no transversal direction is preferred. We say that the beam is unpolarized. (For the electron, see spin polarization.)

    By passing the beam through various equipment (filters) we may prepare modified states. I am considering only perfect filters that project to a 1-dimensional subspace in direction of a normalized state vector ##\phi##, which is adjustable by the experimenter. Such a filter can be realized for photons (type X=P) by a polarizer, and for electrons (X=E) by deflecting the beam by a magnet and passing the two resulting daughter beams through a slit that absorbs the electrons in one of the daughter beams.

    If we denote by ##P=\phi\phi^*## the associated orthogonal projector, the effect on the filtering is that particles pass the filter with a probability of ##\langle P\rangle=\mbox{tr} \rho P =\phi^*\rho\phi##, and the density matrix of the ensemble of particles that passed the beam is ##P=\phi\phi^*##. This corresponds to a pure state, and implies that the modified beam is completely polarized, with polarization (if X=P) resp. spin direction (if X=E) determined by the filter setting.

    The above is the shut-up-and-calculate description of the experiment. It only refers to macroscopic objects and to properties that can be directly checked by statistical measurements. The latter consist of counting sufficiently many detection events of a particle counter placed after the filter, using enough filter vectors ##\phi## to be able to determine the density matrix of an arbitrary beam.

    I'd like to know what some interpretations of quantum mechanics assert about states and observables of individual particles and about which properties they are silent,
    • (a) direcly after emission,
    • (b) directly after having passed the filter,
    • (c) directly after having been measured.
    In particular, I invite the currently most outspoken proponents on PF of three interpretations (minimal ensemble, @vanhees71; Copenhagen, @atyy; Bohmian, @Demystifier) to state their views about individual particles regarding the above experimental setting.
     
    Last edited: Feb 23, 2016
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  3. Feb 22, 2016 #2

    vanhees71

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    Well, I've the most simple task, because you've already answered it for me in your question! I only strongly disagree in principle with the claim that you can ignore the space-time part of the quantum fields, but for this matter-of-principle discussion it's enough.

    So you say, you have a thermal source, which means you are in thermal equilibrium. That implies that the statistical operator to use is the one of maximum von Neumann entropy (because that's the definition of "thermal source" in this theory context). The Hilbert space is then simply the unitary space in 2 dimensions. That means in both cases your density matrix is indeed ##\hat{\rho}=1/2 \mathbb{1}_2## (to write it in my private notation, which I find pretty clear).

    Now let ##|a \rangle## be the normalized eigenbasis of the polarization/spin in the measured direction. Then
    $$P(a)=\langle a|\hat{\rho}|a \rangle=\mathrm{Tr} (\hat{\rho} |a \rangle \langle a|)$$
    is the probability to measure the polarization/spin to take the value ##a##.

    There's no other content in the formalism of quantum theory. That's all you can say about the system given the state it is prepared in. The claim that it is in this state can only be proven by measuring a complete set of compatible observables of a (sufficiently large) ensemble.
     
  4. Feb 22, 2016 #3

    A. Neumaier

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    I find it confusing. Could you please explain? Ah, from the latex I see that you mean half the identity matrix of size 2. But MathJax converts it to something incomprehensible...
     
    Last edited: Feb 22, 2016
  5. Feb 22, 2016 #4

    vanhees71

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    I don't understand. It's just a convenient notation. It's ##\mathbb{1}_2## is the identity operator in 2D complex vector space.
     
  6. Feb 22, 2016 #5

    A. Neumaier

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    Next time you'd do some extra formatting. ##\frac12 \cdot{\bf 1}_2## reads much better than ##1/2\mathbb{1}_2##.
     
  7. Feb 22, 2016 #6

    vanhees71

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    I don't know, why black-board fonts don't work. Sorry for the confusion.
     
  8. Feb 22, 2016 #7

    A. Neumaier

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    So you no longer want to assert anything about the state of the individual particles in the ensemble, as in another recent thread (can't find the precise post anymore)?
     
  9. Feb 22, 2016 #8

    vanhees71

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    Of course, still the physical definition of state must be operative. The formalized way to express this is to say that a state is defined by a preparation procedure (or more precisely an equivalence class of preparation procedures). An operation precedure of course must refer to a single system, and you have to demand that you can repeat this procedure arbitrarily often in the same way to prepare as large an ensemble of independently (i.e., uncorrelatedly) prepared systems.

    "Preparation" can be very simple. E.g., to define the equilibrium state of the various ensembles (microcanonical, canonical, grand canonical) you simply have to take a (closed, to a heat bath coupled, and to a heat-particle bath coupled) system and wait long enough until the system becomes stationary in the appropriate macroscopic observables. Other preparations can be quite complicated and subtle, e.g., to prepare a single-photon state you need, e.g., a laser and parametric-down-conversion material like certain birefringent crystals.

    What I was referring to is the pure meaning of the various notions of the formalism. Of course you have to fill the words "the system is prepared in a state described by the Stat. Op. ##\hat{\rho}##" and "measuring an observable ##O##, represented by a self-adjoint operator ##\hat{O}##" with physical meaning. As in classical physics, I leave this to the experimentalists. As a theorist I can just take it for granted that a clever enough experimentalist can define a preparation procedure for the state and a measuring device to measure the observable(s) in question.

    The meaning of "state" is purely probabilistic and given by the Born rule, which implies FAPP that the state refers to ensembles only since probabilities can be determined only on a large ensemble. Thus the claim that the system is prepared in a state described by ##\hat{\rho}## must verify at least the probabilities for a complete set of compatible observables. A detailed discussion about how to completely determine a state through measurements is given in Ballentines textbook.
     
  10. Feb 22, 2016 #9

    A. Neumaier

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    In the present case, the preparation procedure is completely describe by saying that the source is stationary and thermal, so that the state is ##\frac12 \cdot{\bf 1}_2##. This can be checked independent of the particular source [you can buy a black box source with the properties claimed, and check them in this way] by measuring (in random, sufficiently long time intervals) the detection rate with the filter in four arbitrary random (or appropriately chosen) directions [so that their projectors form a complete system of Hermitian operators] and verifying that within the statistical error, the detection rates agree. No reference to the single system is necessary in this preparation and calibration procedure.
     
    Last edited: Feb 22, 2016
  11. Feb 22, 2016 #10

    Demystifier

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    Can you be more specific why are you interested in the Bohmian interpretation in this particular case? Do you suspect that there might be some particular problem with the Bohmian interpretation in this case?
     
  12. Feb 22, 2016 #11

    A. Neumaier

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    The motivation for starting this thread came from reading an old paper by U. Fano, Description of states in quantum mechanics by density matrix and operator techniques, Reviews of Modern Physics 29 (1957), 74.
    (''thereby'' refers to that the density matrix is well handled by operator techniques.)

    The particular setting described is just the simplest nontrivial case where density matrices appear. (Fano also considers a number of more complex examples, among them - in 1957, long before Bell! - coincidence polarization measurements on thermal entangled photon pairs from positron annihilation.) That the setting is not described by a pure state gives all interpretations an extra twist with respect to the relation between the single system and the ensemble. I think it is an excellent example for comparing interpretations without yet having to deal with more sophisticated instances of quantum weirdness.
    Possibly yes since position seems to be irrelevant in the above setting - it is not even represented in the state, which factors with respect to all variables except for polarization/spin. But in the Bohmian interpretation, position plays a distinguished role. Thus I wonder to which extent the interpretation is different for electrons (where a position operator exists) and for photons (where it doesn't exist). Also I haven't studied the Bohmian take on mixtures, so I'd learn something new. (In this thread, I don't want to be agressively critical, and will only ask questions that help clarifying the claims.)
     
  13. Feb 23, 2016 #12

    vanhees71

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    I disagree. You have to assume that a specific preparation of your black box leads to a well determined state. In this case it's an equilibrium state and you can verify this as you said on an ensemble. If you don't assume this, the state definition doesn't make physics sense in any interpretation!

    A nice example are ultrarelativistic heavy-ion collisions. The claim is that the particle abundancies follow that of a (grand-)canonical ensemble. This is verified by measuring particle abundancies in a lot of single collisions and average over them. Then you can check the particle ratios by fitting to a grand canonical ensemble leading to a temperature, baryo-chemical potential, and volume of the fireball at chemical freeze-out. This works amazingly well, and thus one concludes that the fireball chemically freeze out close to the pseudocritical temperature for the confinement-deconfinement cross over transition as calculated from lattice QCD. At the LHC the baryo-chemical potenial is very small.
     
  14. Feb 23, 2016 #13

    A. Neumaier

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    Suppose that the instruction sheet asserts that, after switching on the source, setting the intensity to a given level, and waiting for 5 minutes, it will produce thermal particles at this rate. The manufacturer may have needed to make assumptions to model his source design to ensure that it works as described.

    But users don't need any assumptions: They can check by proceeding according to the instructions, and test the resulting output beam according to the recipe given in my post #9. If it works as predicted the users can be confident that the source produces a stream of thermal particles as claimed; if not they can complain in the shop where they bought the source. Or they correct the description given in the instruction sheet by noting the state reconstructed from the measurements, and its deviations from stationarity, thus calibrating the source behavior. No reference to any property of the particles is needed to verify any claim about the state of a particular preparation of a sufficiently stationary source, only references to control settings and to detection events. Indeed, this is the reason why the minimal interpretation works!

    This also holds for your more complex example, and for anything done in the labs.
     
    Last edited: Feb 23, 2016
  15. Feb 23, 2016 #14

    Demystifier

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    The reason why Bohmian interpretation works for measurement of any observable is the fact that any measurement eventually reduces to an observation of a position. For example, in the Stern-Gerlach "measurement of spin", what you really observe is the position of the detector which "clicks".

    There is no much difference in practice, as long as the macroscopic detector is made of atoms. You don't observe the position of the photon, you observe the position of some macroscopic pointer made of atoms.

    But still, in a Bohmian interpretation you may want to calculate the photon trajectories, even if you don't observe them. Unfortunately, photons are relativistic and there is no unique relativistic Bohmian mechanics. In the simplest version one introduces a preferred Lorentz frame, which, among other things, allows you to define a Lorentz non-invariant photon position-operator. This means that the interpretation will depend on the choice of the preferred frame, but it can be shown that measurable predictions do not depend on it.

    Bohmian mechanics always takes a view that the full closed system is in a pure state, even if an open subsystem is in a mixed state.

    Good! :smile:
     
  16. Feb 23, 2016 #15

    A. Neumaier

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    Ok. So what are the properties that Bohmian mechanics claims in my setting about the state or the polarization/spin of the single particles (rather than the full closed system), at the points (a), (b), (c) of post #1?
     
  17. Feb 23, 2016 #16

    Demystifier

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    It claims the following:
    - At each point, the particle has a well defined position and velocity. Yet, in the absence of full information about the full closed system, this position and velocity cannot be precisely calculated.
    - The particle does not have polarization/spin at any point. Polarization/spin is a property of the wave function, not of the particle.
     
  18. Feb 23, 2016 #17

    vanhees71

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    Sure, but the point is that you claim that the state is not associated to a single system but only to the ensemble, but that's a contractio in adjecto, because if the state is not associated to the single systems that make up an ensemble, it can also not be assiciated to this ensemble as a whole, because by definition the preparation of the single systems making up the ensemble must be independent.
     
  19. Feb 23, 2016 #18

    A. Neumaier

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    Well, I actually associate the state to the beam; talking about particles and ensembles in addition to the beam was just a concession to tradition. To phrase it without this concession:

    We prepare the stationary light beam or electron beam (i.e., whatever leaves the source under discussion), observe sufficiently many macroscopic detection events in dependence on the filter setting, calculate from it the appropriate statistics, compare it with the shut-up-and-calculate predictions, and find agreement with each other within the statistical uncertainty. Thus all physics is preserved!

    This is much more minimal than what you like to call the minimal interpretation! Neither the ensemble nor the particles, nor any independence assumption plays any role in the experimental setting or its analysis, except to color the intuitive picture. By Ockham's razor, one can eliminate them completely from the language, without impairing the shut-up-and-calculate part and the comparison with the detection statistics!
     
  20. Feb 24, 2016 #19

    vanhees71

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    If you take the electron beam as a whole, you are right. Then you take the entire ensemble as one preparation. I don't see any difference between your and my interpretation then. It's just semantics.

    The independence assumption is crucial. There is a difference between an ensemble of equally prepared single systems, which is by definition prepared such that there are no correlations between the single preparations, and preparing one big many-body system, which may contain correlations. The latter represents a different state than the single particle state you like to investigate with your preparation of an ensemble for this single-particle state. This is part of any proper definition of statistics for empirically checking probabilistic assertions (within the frequentist interpretation of probabilities).

    This is the whole point of some of the loopholes in Bell experiments. E.g., sometimes the experimenters want to demonstrate the correlations in entangled far-distant entities and exclude the possibility that any classical (perhaps non-local) information can cause the corresponding correlations in randomly choosing what to measure (e.g., the direction of a polarization measurement on a photon) very close before detection, so that there's no signal between the measurements at the far distant places possible (provided that relatistic causality holds true). This, however, is not so simple, because you cannot exclude the possibility that the "random-number generators" (which can be quantum systems themselves) used to make the "random choice" of the measured observable are somehow correlated (although it's very unlikely).
     
  21. Feb 24, 2016 #20

    A. Neumaier

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    Of course, since these are described by different states. Thus given the complete state, the degree of independence is encoded in it and needs no additional specification. Thus it seems we have reached complete agreement?
     
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