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Simultaneous recognition of to change and being constant

  1. Jun 9, 2014 #1
    In quantum mechanics, in a stationary eigenstate of Hamiltonian, simultaneous recognition of to change and being constant won’t pose any serious problem, change of state vector, or its future evolution, occurs in an abstract complex numbers space but the real observable facts of energy are and remain constant regardless.

    Is it correct to say that, ontologicaly speaking, the same entity, can be both changing and constant, at the same time?
     
  2. jcsd
  3. Jun 9, 2014 #2
    Yes, according to your definitions. A cute example is f(x) =ex: the change in the function with respect to the values of its domain (aka the derivative) is the same as the function itself.
     
    Last edited: Jun 9, 2014
  4. Jun 9, 2014 #3

    bhobba

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    Energy like any other observable is a Hermitian operator of the form O = ∑ yi |bi><bi|. Its basic axiom is the Born rule which says a yi will result on observation with expected value E(O) = Trace (PO) where P is by definition the state of the system. In general we do not know what yi will result when it's observed, until observed. Its not anything when not observed - although some interpretations have their own take. P is simply a book-keeping device to help in calculating E(O), although again different interpretations have their own take. If its in an eigenstate (ie P is one of the |bi><bi|) you know what it will be if observed, but when not observed the situation is still the same - its not anything. In general of course it will not remain in such an eigenstate so at some later time you cant say what the result of observation will be.

    No. Its a logical absurdity anyway - nothing can be both changing and constant. QM is weird, but not illogical.

    Thanks
    Bill
     
    Last edited: Jun 9, 2014
  5. Jun 9, 2014 #4
    So can we say that when it's in an eigenstate, "that nothing is still changing with respect to time", that is evolving, yet the observed remains constant, which means not changing with respect to time?
     
  6. Jun 9, 2014 #5
    A question as to whether something is changing or is constant (which just means a zero change) depends on it having a definite value at the beginning and the end of the relevant interval. Something in an indefinite state has by definition no definite values; you need to measure your observable to get a definite value, which you are not doing when considering it in its unobserved state. In other words, the question as to whether something in an unobserved, indefinite state is changing or constant is meaningless.
     
  7. Jun 10, 2014 #6
    I don't think that I am referring to an "indefinite" state here. What I am asking is related to the "stationary state". From Wikipedia we see:

    "...a stationary state is not mathematically constant. However, all observable properties of the state are in fact constant."


    The core of the question is if the objects of change, as mentioned above, are two different "entities" or the same thing. I am getting convinced that the mathematical object which is not constant is not indeed "a thing" at all.
     
  8. Jun 10, 2014 #7
    I was merely referring to your reaction to bhobba's remark that "it's not anything when not observed." But as to your question as to whether the objects of change in this case are physical entities, this remains an object of debate; however, the most relevant paper I know concerning this question is http://xxx.lanl.gov/abs/1111.3328
     
  9. Jun 10, 2014 #8

    bhobba

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    What is changing is the state, which is simply an aid in calculating the outcomes of observations. The outcomes of those observations do not exist until observed. They have no value until then.

    My comment was in relation to:

    Energy, being an observable, has no value between observations. The only thing that does is the state - but that is simply an aid to calculating the expected value of observables if you were to observe it. That's from the formalism - different interpretations have a different take.

    You may be getting confused between an observable and a state.

    They are both operators - but have different roles as the Born rule shows.

    Thanks
    Bill
     
    Last edited: Jun 10, 2014
  10. Jun 10, 2014 #9

    bhobba

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    Mathematical objects are defined axiomatically.

    If such is a 'thing' (whatever a 'thing' is) is purely philosophical waffle of zero value as far as science is concerned, and, IMHO, correctly, off topic here.

    In QM a state is simply a mathematical requirement from the definition of observables.

    Its tied up with Gleason's Theorem:
    http://kof.physto.se/cond_mat_page/t...ena-master.pdf [Broken]

    The paper above gives two versions of the theorem.

    The first version is Gleason's original version which is quite difficult but based on resolutions of the identity.

    The second version is much simpler but based on the stronger assumption of POVM's.

    For simplicity I will use the second.

    A POVM is a set of positive operators Ei ∑ Ei =1.

    An observation/measurement with possible outcomes i = 1, 2, 3 ..... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

    Here only by Ei means regardless of what POVM the Ei belongs to the probability is the same. This is the assumption of non contextuality and the rock bottom essence of the probability rule of QM, known as Born's rule.

    You can run through the proof in the link above. Its proof of continuity is a bit harder than it needs to be so I will give a simpler one. If E1 and E2 are positive operators define E2 < E1 as a positive operator E exists E1 = E2 + E. This means f(E2) <= f(E1). Let r1n be an increasing sequence of rational's whose limit is the irrational number c. Let r2n be a decreasing sequence of rational's whose limit is also c. If E is any positive operator r1nE < cE < r2nE. So r1n f(E) <= f(cE) <= r2n f(E). Thus by the pinching theorem f(cE) = cf(E).

    Hence a positive operator P of unit trace exists such that probability Ei = Trace (PEi).

    This is called the Born rule and by definition P is the state of the system.

    Its simply a mathematical requirement that follows from the fundamental axiom I gave.

    You possibly haven't seen it in that form. To put it in a more recognisable form by definition a Von Neumann measurement is described by a resolution of the identity which is a POVM where the Ei are disjoint. Associate yi with each outcome to give O = ∑ yi Ei. O is a Hermitian operator and via the spectral theorem you can recover uniquely the yi and Ei. By definition O is called the observable associated with the measurement. The expected value of O E(O) = ∑ yi probability outcome i = ∑ yi Trace (PEi) = Trace (PO).

    A state of the form |u><u| is called pure. A state that is the convex sum of pure states is called mixed. It can be shown (it's not hard) all states are either mixed or pure. For a pure state E(O) = trace (|u><u|O) = <u|O|u> which is the most common form of the Born rule.

    Just to recap the state is simply something introduced to aid in calculating the probabilities of the outcomes of observation as required by the fact those outcomes are described by a POVM. The only thing that the theory is concerned with is the outcomes of measurements/observations. What's going on between such the formalism says nothing - other than that you have this thing called the state that aids in calculating the probabilities of those outcomes.

    Again I want to emphasize - this is the formalism, interpretations have other things to say.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  11. Jun 10, 2014 #10
    Thanks very helpful.
     
  12. Jun 10, 2014 #11

    What is confusing here is that state vector, even though an aid of calculation, is expressed in terms of time and is evolving into the future. So I did presume that whatever is related to time, should be something, and should therefore exist.
     
  13. Jun 10, 2014 #12
    In your original post you used the word "ontologically" which probably put the question more in the realm of philosophy than physics, supporting bhobba's side remark that the topic may not fit in this forum. In your above text you use the word "exist", which is too vague to respond to: can you give a meaningful definition of physical existence? There seems to be no consensus on this. Max Tegmark has some interesting things to say about this, as does David Deutsch. But, as bhobba remarked, the formalisms of physics do not resolve the question.
     
  14. Jun 10, 2014 #13
    Yes you are right after this explanation based on formalism.
     
  15. Jun 10, 2014 #14

    bhobba

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    Mate this is philosophy territory. What does 'exist' mean? Do probabilities exist like say an electric field does? Do even electric fields exist? The state is like probabilities. When you sort that one out the question may be meaningful. However this is philosophy pure and simple - not science.

    Thanks
    Bill
     
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