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Vector spaces

  1. Jul 30, 2005 #1
    what is the reason behind choosing the linear vector spaces in representing the state of a system? why is it convenient ? and why do we actually need a linearity ???
  2. jcsd
  3. Jul 30, 2005 #2
    At the beginning there were the natural numbers (surely to make money), thereafter someone introduced the addition (surely to win more money). From the addition one deduced the multiplication for efficiency (3+3+...3 takes more time to calculate than 100*3: the productivity). The linearity and the business were born :biggrin:

  4. Jul 30, 2005 #3
    The Schrödinger equation is a linear differential equation, so the sum of the solutions is still a solution. Thefore, quantum states that are solutions of the SE form a vector space (actually Hilbert space).
  5. Jul 30, 2005 #4


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    We exist in a multi-dimensional world/universe, and vector analysis provides a convenient tool for dealing with n-dimensional models/state spaces.
  6. Jul 30, 2005 #5
    Good Lord how I love this forum! So manyh good questions and answers here.

    The best answer to your question, imho, is this - We chose math to describe Nature. Nature chose to be linear. Why? Nobody really knows. Its simply rooted in the axioms of quantum mechanics.

  7. Jul 30, 2005 #6
    It's true.
    It's true
    If you give me the permission, I think the nature also shows non linear phenomenon and further more chaotic phenomenon. I think it is not the nature but the human laziness or a human principle of the lowest coast that explain the historic introduction of linear ways of thinking.
    Once more time I will try, just for fun, the contradiction. If you consider that the principle of uncertainty (Heisenberg) also is rooted in the axioms of quantum mechanics and if you consider any trinome of a variable (a.x² + b.x + c = 0), you know that the latter owns 0, 1 or 2 solutions depending on the delta. The 2 solutions [delta non zero; x = (-b/2a) + or - ...] are in someway centred on the double one [delta = 0; x = (-b/2a)] and if the coefficient c is not constant in the nature, then the 2 different solutions can be less or more distant from the double one; there is a kind of amplitude, of imprecision in the solutions... Don't you think that it is a mathematical argument to introduce at least bilinearity in the quantum mechanics?
    As said, just :rolleyes: for fun. Blackforest.
  8. Jul 30, 2005 #7


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    This argument assumes the solutions of the quadratic must be real. Quantum Mechanics solutions are all complex numbers (until the state vector is acted on by a Hermitian operator with real eigenvalues). So the quadratic always has two solutions, perhaps a repeated one counted twice. Fundamental Theorem of Algebra: Over the complex field every polynomial of degree n has n roots, counting repeated ones as many times as they are repeated.
  9. Jul 30, 2005 #8


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    I disagree with that. WE choose linear models because they are far easier than non-linear models. In fact, one can argue that "modern" physics, quantum physics and relativity consist in replacing linear models with more accurate but harder, slightly non-linear, models.
  10. Jul 30, 2005 #9
    I'm not certain whether I agree/disagree with that ... yet. Please give an example of how you'd choose the axioms on QM such that they are stated in terms of a non-linear model.

  11. Jul 30, 2005 #10
    I was speaking only of QM.
    If you mean that they can be derived from the axioms then I agree.
    "the delta"? What do you mean by this? A measure of the associated parabola perhaps?
    Can you give me an example of what this parabola represents?

    Sometimes you have to restrict the mathemetical solutions to those which describe the physical phenomena. Therefore rather than there being some sort of "bilinearity" that you suspect it may be that nature does not allow something you assumed to be true before you got to that point. But I'm not all that clear on what you're talking about with this parabola thing above.

    If you're refering to linearity then that equation you gave is irrelevant since it is the operator x which is supposed to be linear and not any equation such as the one you gave.

    Last edited: Jul 30, 2005
  12. Jul 30, 2005 #11


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    His "delta" is the discriminant [tex]\sqrt{b^2 - 4ab}[/tex], often denoted [tex]\Delta[/tex].
  13. Jul 31, 2005 #12
    Yes, but Sorry sir : [tex]\sqrt{b^2 - 4ac}[/tex]
  14. Jul 31, 2005 #13
    Oh I was only trying to explain how one could, perhaps, introduce something else than linearity, f.e here bilinearity, in our way of thinking. "x" was only any variable without any condition, not necessary an operator. It was effectively just a kind of "parable". The contra-argumentation of self adjoint is a bad point for me, except if one could find some real situations, as you suggest yourself for the reality of the things, where the n solutions are automatically centred on one of them ... I repeat: it was a try just for fun and to show that one can perhaps develop other ways of thinking. Best regards
  15. Jul 31, 2005 #14
    ... Said with other words, I was trying to connect the dispersion of the (n) solutions with the uncertainty concerning the value of the variable x (which is a central question in QM). For the bilinearity (x is solution of a trinome as given above) the idea appears quite more clearly than for the case where x is solution of a polynome of degree n.
  16. Jul 31, 2005 #15
    You were discussing the HUP right? The terms which appear in that expression must have an associated operator to even be physically meaningful though. If you're speaking about a physical observable then it is an axiom that the operator corresponding to all physical observables are Hermetian operators. These operators are linear. There are non-linear equations in all fields of physics that I know of. But the one that you're speaking of has no meaning to me as far as introducing non-linearity into QM since that quite literally means to me that you're introducing non-linear operators. Otherwise your comments have no meaning to me and you'll have to "dumb it down" for me. :tongue:
    Again, you're speaking of the "reality of things." But this can only mean that you're refering to something which has an operator. All physically observable quantities in QM must have a corresponding Hermetian operator - That's an axiom.

  17. Jul 31, 2005 #16
    The reasons why I was introducing:
    a) this discussion
    b) the equation of the parabola
    are easy to understand.

    a) to develop the initial question of preeto283;
    b) because
    f(t) = 1/2. acceleration. (time)² + speed. time + initial position (at t = 0)
    is a meaningfull non linear equation in physics depending on the time.

    And now I am the pupil and if you agree I ask: Is there an operator associated to the time in QM? Which one? How does it work?

    Concerning the Heisenberg's principle, you are totally right, it is not really an axiom because it can be mathematically demonstrated starting with considerations depending on the dispersion of variables (another delta of course).
  18. Jul 31, 2005 #17
    May be the ideal description of the nature must be non-linear but the non-linear equation is hard to solve. Because the states in space-time with random variables or random parameters with hidden indexes is translated to the abstract linear space of wave-functions?
  19. Jul 31, 2005 #18
    Firstly, let's remember that QM is a model. So we choose how it works, and we choose linearity because it works reasonably well enough and makes our lives much easier.

    But if we look at the fields that make up nature, they couple to each other in highly non-linear ways, just looking at the Lagrangians for interactions betweens fields will tell you that.

    Furthermore, even the classical gravitational field is non-linear. Einstein's field equations are non-linear; again just looking at the equations (and understanding what the terms are) will tell you the same thing.

    So nature, as far as we can tell is non-linear. QM is not nature -- it is a convenient mathematical model, invented by humans, that describes nature remarkably well.

    But there is also added confusion here. The original poster is speaking of the linearity of the space of states. Well what is a state? What we define it to be. So of course we can say that the states form a linear complex vector space if we want to (equipped with an inner product). But fundamental equations in QM can still be non-linear - we can easily have non-linear terms in the Schrodinger equation, for example, depending on how we choose to describe the energy of the system.
  20. Jul 31, 2005 #19


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