From: <[email protected] <mailto:[email protected]>>
On Tuesday, April 24, 2018 at 12:14:06 AM UTC, [email protected]
<mailto:[email protected]> wrote:
On Monday, April 23, 2018 at 11:54:15 PM UTC, Bruce wrote:
From: <[email protected]>
On Monday, April 23, 2018 at 7:38:30 PM UTC,
[email protected] wrote:
On Monday, April 23, 2018 at 1:20:05 PM UTC,
[email protected] wrote:
On Monday, April 23, 2018 at 8:58:53 AM UTC, Bruce
wrote:
From: <[email protected]>
On Monday, April 23, 2018 at 5:53:59 AM UTC,
Bruce wrote:
From: <[email protected]>
Let's agree that electrons A and B form a
singlet entangled system. Let's further
agree that they are non separable. What do
you do with the fact that when their spins
are measured, they ARE in different spatial
locations, not even space separated in Bell
experiments. How do we deal with this FACT? AG
What do you want me to do with the fact? I
learn to live with facts that I can't do
anything about. The fact that the system is
non-local is a fact that you just have to
come to terms with.
Bruce
*ISTM that when you have a theory that seems
correct and in some sense is well tested, but
there are facts which contradict it, in this
case a key fact right in front of your nose
which contradicts it -- the fact that we see as
plain as daylight that the subsystems as
spatially separated -- invariably the theory
must be wrong. AG*
I wish you luck with your project to prove
quantum mechanics wrong.
Bruce
*Right now I have a more modest goal. Starting from
the postulates of QM, how do you justify writing the
wf of the singlet state as a superposition of tensor
product states? TIA AG *
*What it's not. It's not the SWE. It's not Born's Rule.
It's not the operator correspondence with observables. AG *
*I suppose it could be traced to the superposition principle;
that the state vector of the singlet state is a linear
combination of the states which are members of the
corresponding Hilbert space of the system. But why are these
states tensor product states? AG*
Why try worrying these things out for yourself? The easiest
thing is to go and look up a text book.
Bruce
*Recall when I asked whether entanglement necessarily implies non
locality. You replied "not necessarily" and gave the classical
example of elastic scattering of billiard balls where the momentum
of its constituents and the whole system is known exactly. No
uncertainty. In the wf for the singlet system you assume a
definite net spin angular momentum, zero. How can you treat the
singlet system quantum mechanically and at the same time assume
you know its spin momentum exactly? Do you think this question
could be answered in a text book? How could I even pose it to an
inert, non responsive medium? AG *
*I just took a quick look at chapter 15, section 4 of Merzbacher,
Quantum Mechanics (Third Edition). The tensor equation can't be
copied. It appears in the blank lines below. Immediately you can see
the problem with this kind of treatment. It doesn't explain WHY, from
First Principles, the tensor product can be used to describe the
composite system. It's virtually impossible to find an explanation
from First Principles. AG*
4. Quantum Dynamics in Direct Product Spaces and Multiparticle
Systems. Often the state vector space of a system can be regarded as
the direct, outer, or tensor product of vector spaces for simpler
subsystems. The direct product space is formed from two independent
unrelated vector spaces that are respectively spanned by the basis
vectors /A;) and I B;) by constructing the basis vectors
Although the symbol @ is the accepted mathematical notation for the
direct product of state vectors, it is usually dispensed with in the
physics literature, and we adopt this practice when it is unlikely to
lead to misunderstandings. If n1 and n2 are the dimensions of the two
factor spaces, the product space has dimension nl X n2. This idea is
easily extended to the construction of direct product spaces from
three or more simple spaces.
Quite right. And what else are you going to use for many-particle
systems that have independent Hilbert spaces -- you multiply them
together, of course.
Bruce
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