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Science Forum Index » Physics - Particle Forum » How can a particle be in a superposition of states of differ
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| Tareq |
 Posted: Thu Nov 16, 2006 12:56 pm |
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Guest
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Hello
In classical mechanics, energy, mass, charge,.... are unique properties
of objects. We can't say that one object has two different masses or
energies at the same time, once the reference has been fixed. In
quantum mechanics however, an electron can be in a superposition of
states of different energies. To an observer, the expectation value of
the energy is constant , while for the particle, the energy is not
determined. Is this true? Shouldn't the energy, like charge, be a
unique property of any particle? |
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| Richard Saam |
| Posted: Fri Nov 17, 2006 3:09 pm |
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Guest
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Tareq wrote:
Quote: Hello
In classical mechanics, energy, mass, charge,.... are unique properties
of objects. We can't say that one object has two different masses or
energies at the same time, once the reference has been fixed. In
quantum mechanics however, an electron can be in a superposition of
states of different energies. To an observer, the expectation value of
the energy is constant , while for the particle, the energy is not
determined. Is this true? Shouldn't the energy, like charge, be a
unique property of any particle?
I would say yes and reasoning is as follows:
Define a momentum state 'K' and energy state 'KK'
associated with a particular cell volume
and g and h being functions
of the Lorentz transform sqrt(1-v2/c2) or 'b'
This is a simple extension of the de Broglie hypothesis
http://en.wikipedia.org/wiki/De_Broglie_hypothesis
Further define a resonant system 1, 2, 3, 4 such that
K1*g(b1) + K2*g(b2) <> K3*g(b3) + K4*g(b4)
K1*K1*h(b1) + K2*K2*h(b2) = K3*K3*h(b3) + K4*K4*h(b4)
conservation of energy and momentum (elastic resonant state)
is by definition maintained between state 1,2 and state 3,4.
Under further condition that
g(b1) ~ g(b2) ~ g(b3) ~ g(b4)
h(b1) ~ h(b2) ~ h(b3) ~ h(b4)
and as a consequence
(g(b1) + g(b2)) / (g(b3) + g(b4)) nearly equals one (1)
(h(b1) + h(b2)) / (h(b3) + h(b4)) nearly equals one (1)
Also momentum conditions K1, K2, K3 & K4 are defined such that
K1 ≠ K2 ≠ K3 ≠ K4
so the Lorentz transform conditions g(b) and h(b)
are not consequential factors in a resonant state 1,2,3,4.
In other words the (g(b) and h(b) 's nearly mathematically cancel.
Further logic would dictate
that superluminal state (v > c) is allowable
in the context of
g(i*b1) ~ g(i*b2) ~ g(i*b3) ~ g(i*b4)
h(i*b1) ~ h(i*b2) ~ h(i*b3) ~ h(i*b4)
and
(g(i*b1) + g(i*b2)) / (g(i*b3) + g(i*b4)) nearly equals one (1)
(h(i*b1) + h(i*b2)) / (h(i*b3) + h(i*b4)) nearly equals one (1)
The imaginary (g(i*b) and h(i*b) 's nearly cancel.
This leaves the sub- and super- luminal resonant state solution
in terms of finding satisfactory
momentum and energy states K1, K2, K3, K4 and K1K1, K2K2, K3K3, K4K4.
such that:
K1 + K2 <> K3 + K4
K1*K1 + K2*K2 = K3*K3 + K4*K4
An attempt at doing this is in:
http://arxiv.org/abs/physics/9905007
with listed consequences
of being able to scale over 15 orders of magnitude
from nuclear (superluminal {v > c} condition)
to universe (subluminal {v < c} condition) dimensions
with K associated with a particular space filling lattice and
with energies 'KK' associated
with observed masses (KK/c2)
and observed pressures (KK/volume).
Fundamentally, this approach is a mathematical approximation
which does not negate but suspends the Lorentz transform application
and this approximation is assumed to be applicable in nature
by correlation to standard model and observed universe masses. |
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| Guest |
| Posted: Sat Nov 18, 2006 6:31 pm |
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Quote: determined. Is this true? Shouldn't the energy, like charge, be a
unique property of any particle?
If it is measured, energy or momentum becomes a unique property for the
particle at the expense of losing information about its position which
gets "spread out" over a range of values.
In particle physics, we do deal with charge and mass eigenstates, which
do not necessarily match!
-S |
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| Autymn D. C. |
| Posted: Wed Nov 22, 2006 7:03 am |
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Guest
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souvik1982@gmail.com wrote:
Quote: determined. Is this true? Shouldn't the energy, like charge, be a
unique property of any particle?
If it is measured, energy or momentum becomes a unique property for the
particle at the expense of losing information about its position which
gets "spread out" over a range of values.
I disagree with this. The Planckian condition is satisfied; one does
not lose information or precision as that is the relevant description
of the body.
The superposition means that the body isn't picky, because someone
didn't set it for one state over another. In other words, the observer
doesn't know what the next state is.
-Aut |
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