What is Locality in QM?
John Rhoades
particular, I'd like to see a mathematical definition instead
of, or at least in addition to, a bunch of words. What SPACE is
being referred to? I'd guess it is the 4D manifold we live in, not
the state space, is that right?
/s John Rhoades
john baez
The key thing to keep in mind is that different people mean different
things by this word. There are also, I suspect, plenty of people who
don't know what they mean. This is part of why people are able to argue
forever about whether QM is `local.'
One meaning is important in understanding Bell's inequality. This
is the sense of locality that appears in the term `local realism,' and
by this sense QM is not local. Roughly speaking it amounts to this.
Say a physical system is in a pure state - i.e., we know everything
about it that can be simultaneuously known. Suppose the system is
spread out through space (like they always are). Then we say we have
`local realism' if we can describe the state of the system by first saying
everything that's going on in *this* little patch of space, then in *this*
little patch of space, and so on. I.e., a pure state can be described by
specifying pure state in each region of a partition of space. (Space is
good old 3d space here!)
This fails in QM. In the jargon of algebraic QM, there are pure states
of the tensor product of several algebras of observables that are not
just tensor products of pure states of each algebra.
Another meaning is important in understanding quantum field theory.
This is the sense in which quantum field theory IS local. It goes like
this. The Hamiltonian of a quantum field theory is what describes how things
evolve in time. The Hamiltonian is always an integral over space of a product of
fields and their derivatives *at a single point*. That is, two things only
interact when they're at the same place. Again, the "space" is good old
3d space here.
Closely related to the locality of quantum field theory is its causality.
This says that the state of a system within some region R at time T can
be computed from knowing the state within any region R' at time T' as
long as no paths moving at the speed of light or slower that don't pass through
R' at time T' can pass through R at time T. In other words, signals
propagate no faster than light.
In a sense, traditional quantum field theory is what you get when you try
to do quantum theory in a way that it is local and causal, as well as
invariant under the symmetry group of special relativity.
Paul Budnik uunet!mtnmath!paul
> [...]Closely related to the locality of quantum field theory is its causality.
> This says that the state of a system within some region R at time T can
> be computed from knowing the state within any region R' at time T' as
> long as no paths moving at the speed of light or slower that don't pass through
> R' at time T' can pass through R at time T. In other words, signals
> propagate no faster than light.
I do not see the equivalence between the first part of this statement
and the second part and I do not think the first part is true of
any system in which Bell's inequality is violated. The two would be
equivalent in a classical theory. In quantum mechanics you can influence
the state of a system in a way that is experimentally detectable but
cannot be used to send a signal. This is possible because probability is
claimed to be irreducible in quantum mechanics.
To reproduce the predictions of quantum mechanics about a joint probability
of detection requires that information outside the light cone of at least
one detection influences that detection. If causality as you defined it
were true in QM the configuration space wave function could be mapped into
physical space.
When I speak of information transfer I am not talking about
what is happening physically but only about the mathematics of QM.
This is what I assume you are talking about as well. I do
not want to get in a pointless philosophical debate but we should be
able to agree on what the mathematics says.
Paul Budnik
Dan Seeman
>evolve in time. The Hamiltonian is always an integral over space of a
product of
>fields and their derivatives *at a single point*. That is, two things
only
>interact when they're at the same place. Again, the "space" is good old
>3d space here.
I am confused by this. It seems like you are saying:
"The Hamiltonian is...an *integral* of...[a] *derivative*..."
So are you not left with the original field(s)? How is this a
Hamiltonian?
You mention a product between the derivative and its field, so it seems
this is where I am confused. (Maybe my confusion is this; The solution
to the integral of x is different from that of x^2, x^3 and so on. I
would enjoy reading your response to this question...)
dks.
SCOTT I CHASE
>
>I am confused by this. It seems like you are saying:
>
>"The Hamiltonian is...an *integral* of...[a] *derivative*..."
>
>So are you not left with the original field(s)? How is this a
>Hamiltonian?
The Hamiltonian density is constructed from the *products* of various
fields and/or their derivatives. So you are integrating over products
which may or may not contain derivatives. It all depends on the form
of the interaction. In many cases, you are integrating products of
currents (like the weak V-A interaction) which indeed contain pairs
of derivatives.
-Scott
-------------------- Physics is not a religion. If
Scott I. Chase it were, we'd have a much easier
SIC...@CSA2.LBL.GOV time raising money. -Leon Lederman
Lawrence R. Mead
H = integral {1/2* [del psi(x)]^2 + V(psi(x)) } dx
Note that each factor of psi(x) is evaluated at the same (space-time)
point ,x. It could be that H consisted of ,say, a double integral
where the first factor was evaluated at x, the second at x', or even
more if V(psi) has many factors ( like psi ^4). That would be "nonlocal".
The above is said to be "local".
.
--
Lawrence R. Mead (lrm...@whale.st.usm.edu) | ESCHEW OBFUSCATION !
Associate Professor of Physics
john baez
>In article <35...@galaxy.ucr.edu> john baez, ba...@ucrmath.ucr.edu writes:
>>evolve in time. The Hamiltonian is always an integral over space of a
>product of
>>fields and their derivatives *at a single point*. That is, two things
>only
>>interact when they're at the same place. Again, the "space" is good old
>>3d space here.
>I am confused by this. It seems like you are saying:
>"The Hamiltonian is...an *integral* of...[a] *derivative*..."
>So are you not left with the original field(s)? How is this a
>Hamiltonian?
The words that you replaced with "..." are the ones that make the
difference. A typical Hamiltonian (for a scalar field f(t,x), say)
has terms like
\int (f'(t,x))^2 dx
where ' denotes d/dx, and
\int fdot(t,x)^2 dx
where fdot is df/dt, and
\int f(t,x)^2 dx
None of these terms can be boiled down to the value of the field at a
given point.
I should add that in the example above I'm talking about
a field theory in one space and one time dimension. In general the
integral is over space at a fixed time t, and the f' term is replaced
by the gradient of f dotted with itself, integrated over space.
In general things get a bit more complicated but are still integrals of
products of the field and its derivatives.
Ron Maimon
|>
|> I am confused by this. It seems like you are saying:
|>
|> "The Hamiltonian is...an *integral* of...[a] *derivative*..."
|>
|> So are you not left with the original field(s)? How is this a
|> Hamiltonian?
|>
|> You mention a product between the derivative and its field, so it seems
|> this is where I am confused. (Maybe my confusion is this; The solution
|> to the integral of x is different from that of x^2, x^3 and so on. I
|> would enjoy reading your response to this question...)
|>
the hamiltonian in field theory is an integral of the field functions multiplied
by thier derivatives- e.g. the hamiltonian of a free scalar field is
(integral) .5 [ grad( phi) ^2 + (d phi /dt ) ^2 + m^2 phi ^2 ]
and this cannot be reduced to just the value of the field function at some small
set of points.
This is because not every integral of a derivative is equal to the value of the
original function at the endpoints, for example, in one dimentional calculus
the integral of f'(x)^2 from a to b has no simple expression in terms of f(x)
-Ron Maimon
Hannu Poropudas
starting from the same starting point in curved space-time, that
could have something to do with local concept. When you are trying
to combine two signals starting from two different points in curved
space-time, that could have something to do with global concept.
In this latter case you should probably use quide how to combine
two geodesic flows, which starts from two different points, on the
surface.
Hannu.