On 4/29/2025 6:58 AM, Alan Grayson wrote:
On Tuesday, April 29, 2025 at 4:46:58 AM UTC-6 John Clark wrote:
On Mon, Apr 28, 2025 at 9:21 PM Alan Grayson <[email protected]>
wrote:
/> Presumably, if Maxwell's equations are solved in empty
space, in the absence of charged particles, the E and B fields
will be identically zero./
*Yes.*
/> If the same problem is solved in Quantum EM theory, won't
we also get the same classical result,/
*No. If the electric, magnetic, gravitational, or any other type
of field was precisely zero then you would know exactly what the
energy density was in any arbitrarily short amount of time, zero.
And quantum mechanics says that is impossible. So the amount of
energy must be greater than zero for short amounts of time, the
shorter the time the greater the energy. This is why there are
virtual particles, they can't be detected directly but they
produce effects that can be detected, the simplest and most
dramatic is the Casimir Effect.*
/> but then won't the HUP fail to be satisfied?/
*I don't know about that butHUP fails to satisfy me because IHA. *
Never heard of the HUP, the Heisenberg Uncertainty Principle? How can
it be applied in quantum EM field theory other than on an ad-hoc
basis. That is, can the HUP be derived from the princples of the
theory, or is it just ad hoc? As for the Casimir Effect, it can be
predicted classically, so it proves nothing about virtual particles. AG
If you make some rather abstract assumptions about the physics, then you
can derive the HUP. You
assume that the state of a physical system is described by a normalized
vector ψ in a complex
valued vector space of high dimension. This vector is a function of the
physical variables, q , of the
system, ψ(q). So the values of physical variables are given by
projections of this state vector onto
the axes of the vector space. Then we consider unitary transformations,
ψ′=Uψ of the state vector.
Unitary means the norm isn’t changed, so it’s a rotation, U=exp( θi ).
In two dimensions θ is just an
angle, but in this higher dimension space it needs to be an operator,
i.e. a matrix. Let it be an
infinitesimal transformation θ=−ϵG. Then the infinitesimal unitary
transformation is given
by U=1− ϵi G where G is called the generator of the transformation.
ψ′=ψ−iϵGψ
Then we compare this to the form of a change of basis, q′μ=qμ−ϵμ
ψ′(q‘)=ψ(q−ϵ)=ψ(q)−ϵ∂ψ(q)/∂q
It is seen that the generator is
G=-i∂/∂q
If q is position then G is the generator of position, i.e. momentum.
Then to make this look more familiar we define p as:
p≡ℏG=-iℏ ∂/∂q
Note that ℏ is just an arbitrary constant we introduce because we want
to measure p in units other
than the inverse of q . Then it follows that
[p,q]ψ=-iℏψ.
So Heisenberg’s uncertainty relation comes out of the mathematics AFTER
you’ve put some
assumptions about how physical systems are represented in a high
dimensional vector space by a
normalized vector that just rotates. The HUP is just the statement that
the infinitesimal rotation
corresponding to a change in a variable q fails to commute with the
conjugate variable p , we call
“momentum” along q , which is not surprising since rotations don’t
commute in general.
Brent
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