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Thread: expectation

  1. #1
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    expectation

    The linear momentum operator of quantum mechanics is expressed as p=ħ∂/i∂x where ħ is Planck’s constant of action divided by 2π, i is the imaginary unit, and /∂x is the partial derivative with respect to one component of 3-space. The Hamiltonian energy operator is expressed as H=-ħ∂/i∂t. It is a negative operator where /∂t is now the partial derivative with respect to time.

    Using these operators, the expectation of both linear momentum and Hamiltonian energy can be expressed as the weighted means sum over all of the plane waves making up the wavefunction Ψ, which is a function both of 3-space and of time: <p>=ſ Ψ(x,t)*p Ψ(x,t) dx / ſ Ψ(x,t)*Ψ(x,t) dx and <H>=ſ Ψ(x,t)*H Ψ(x,t) dx / ſ Ψ(x,t)*Ψ(x,t) dx .
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Re: expectation

    Does it have to be linear momentum? Could it be curved or angular?

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    Re: expectation

    If it's angular then it's already quantized.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Re: expectation

    Does it have to be a straight line. Your formula suggest that it is but I thought in reality there is no such thing as a truly straight line.

    In my class when I use the linear equation I tell my class that accountants assume a straight line but we really agree with the economist that the line is indeed curved but accountants can't handle exponential equations. Besides in the short run it probably is close to linear anyways.

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    Re: expectation

    In the quantum domain, a straight line or path has no definition because of the uncertainty principle between position and linear momentum. Their uncertainties product can never be less than Planck's constant of action.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Re: expectation

    Thats true. That is one for probability theory and one against determinism.

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    Re: expectation

    My hunch is that the TOE must be a deterministic theory without those probabilities. I'm sure Einstein will go along with me on this idea. And this happens at absolute zero.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Re: expectation

    Quote Originally Posted by AntonioLao View Post
    My hunch is that the TOE must be a deterministic theory without those probabilities. I'm sure Einstein will go along with me on this idea. And this happens at absolute zero.
    Sure, but at absolute 0 nothing moves, including probabilities.

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    Smile Re: expectation

    Quote Originally Posted by Profpat View Post
    Sure, but at absolute 0 nothing moves, including probabilities.
    That only applies to the dense physical universe,not to the trancending planes,or even
    the etheric plane of expression,after all the absolute can do whatever it intends to.




    regards michael.
    Humilty,coupled with boldness,surprises truth to
    reveal herself?

  10. #10
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    Re: expectation

    Quote Originally Posted by Profpat
    but at absolute 0 nothing moves
    This can only remain as a subjective conjecture. Mine is that at absolute zero there exist a dynamic equilibrium between all spacetime charges as square of energy.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

 

 
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