"Alfred Einstead" wrote in message
> In quantum theory, of course, the problem is worse since the Haag
> Theorem (= the quantized version of the Leutwyler Theorem) puts severe
> restrictions on the very existence of an interaction picture.
On 4/14/2013 3:11 AM, Jay R. Yablon wrote:
> Fortunately, I found in the end that the Yang-Mills tensors I had discovered
> could be used to get extremely tight parts-per-million predictions of binding
> energies even without having to do the many-body problem. Tensors are
> remarkable things, in that way.
If I can jump in here, there are a few comments I can add at this
point.
Of course, you can't disprove a theorem -- other than by rendering its
preconditions invalid or irrelevant.
In the case of Haag, there is a significant insight to be drawn from
the (apparent) absence of any non-relativistic version of the theorem.
The first thing you should ask when seeing this discrepancy is:
"What's up with this?! I thought we were supposed to have a
Correspondence Limit property going on here. So how can the limit of
'No' be 'Yes'?"
In reality the key word in the above sentence is "apparent". If you
take the idea of a Correspondence Limit seriously, then there should
indeed be a non-relativistic version of Haag's Theorem! Only, its
conclusion will miss the mark and fail to exclude interactions.
So, now, if you start at the non-relativistic limit and *slowly* turn
on the "relativity parameter" alpha = 1/c^2 from 0 on up, you should
be able to keep the pristine situation you found in non-relativistic
theory at least over some range of alpha over an open set surrounding
0. A result should not draw a sharp distinction between 0 and
everything else -- because that would in principle provide a way to
determine the precise value of a continuous parameter ... which is
absolutely verboten.
However, what you (apparently) find is that the very instant alpha
goes from 0 to anything that's non-zero, Haag's conclusion appears to
go from "No" to "Yes" on the question "is there a non-trivial
interaction?"
What this shows, then, is that the root of the problem is that we lost
something when running the correspondence limit in reverse and that
what we call relativity is actually not the complete account of what
relativity should be. In other words, the theory of relativity has not
yet been completely discovered or developed! In particular, the one
thing that's missing is the very thing we need to render Haag
impotent.
So, what's missing? The answer actually is sitting in clear sight, and
has been hiding in clear sight for a very long time. It is this: in
non-relativistic theory, the symmetry group is NOT the Galilean
transformation group. Rather, it is the Bargmann group, which has one
generator in addition to the Galilei group. This discrepancy shows up
when you place the Galilean transformation
x -> x - vt, t -> t, y -> y, z -> z
for coordinates alongside the BARGMANN TRANSFORM(!)
H -> H - v px + v^2/2 m, px -> px - vm, py -> py, pz -> pz, m -> m
for kinetic energy H, momentum p = (px, py, pz) and mass m.
We can, correspondingly, form 2 invariants out of this rather than
just one:
mu = m -- the linear invariant
rho = px^2 + py^2 + pz^2 - 2 m H -- the quadratic invariant.
So see what the second invariant is and what it means, take the case
of a non-relativistic interacting body with momentum p, and energy
potential U:
H = p^2/2m + U.
For rho we then get:
rho = -2mU.
Now, compare this situation to relativity. The kinetic energy H takes
the form:
H = mc^2/root(1 - v^2/c^2) - mc^2.
To put this in a form that more closely compares to the non-
relativistic form, rewrite it as:
H = mv^2/(1 + root(1 - alpha v^2))
with alpha = 1/c^2 being the relativity parameter.
But now look at what happens when you substitute this into the mass
shell invariant of relativity with the total energy E = H + mc^2:
E^2/c^2 - p^2 = m^2 c^2 --> rho = p^2 - 2mH - alpha H^2 = 0.
Look at what happens to the linear invariant:
mu = E/c^2 - alpha H = m.
This is the relativistic version of rho. When taking the non-
relativistic limit, the result is
mu -> m; rho -> 0.
The potential U is gone.
That's what's missing. So, when you run through Haag's theorem or the
classical version (Leutwyler's Theorem), the result is that you end up
drawing the conclusion that U = 0 ... because that's effectively the
assumption you started with!
To fix the problem requires restoring the Correspondence Limit. But
first, this requires recognizing that the Correspondence Limit should
be with the Bargmann group, not the Galilei group. Since Bargmann has
11 generators, then so must the symmetry group in relativity that
leads to Bargmann. In addition, it includes Poincare' as a subgroup.
The Bargmann group is a non-trivial central extension of the Galilei
group that is obtained by modifying the Lie brackets between the
momentum (= translation generator) P and the moment (= boost
generator) K from
[K_i, P_j] = delta_{ij} (alpha H) (= 0 for Galilei)
(where H is the time translation generator), to the form:
[K_i, P_j] = delta_{ij} (alpha H + mu) (= delta_{ij} m for
Bargmann)
Pooincare' only admits "trivial" central extensions. And here, that
"trivial" central charge is just mu itself. Nonetheless, despite its
triviality, the alpha -> 0 limit of the centrally extended group is a
non-trivial central extension. So, to have a valid Correspondence
requires keeping the additional generator "mu" in the Poincare' group.
Consequently, the creed "Poincare' is the symmetry group for
Relativity" is revoked and replaced by the creed "the central
extension of Poincare' is the symmetry group for Relativity".
So, now we come back to Haag. This is where Haag runs afoul. We want a
vacuum state |0> to be boost-invariant. This means that K |0> = 0. We
want it to be homogeneous, which means P |0> = 0. With the Poincare'
group we draw the conclusion that
H |0> = (1/alpha) [K, P] |0> = 0
and we note that the non-relativistic form is immune since alpha was
0.
But with the centrally extended Poincare' group, the only conclusion
we can draw is that
M |0> = 0
where [K, P] = M = mu + alpha H is no longer the energy or time-
translation generator, but the relativistic mass! In other words, we
draw the same conclude in place of Haag (after replacing Poincare' by
its central extension) that the vacuum has no mass.
But the only conclusion to be drawn for the energy H is
H |0> = (M - mu)/alpha |0> = -mu/alpha |0>.
And noting the expression mu = m - alpha U, we simply find:
H |0> = U |0>.
The interaction potential U survives Haag!
The meanings of the generators subtly shift. This is best seen by
writing a coordinate form of the Bargmann transform and then
converting it to relativistic form. We do this by associating a
coordinate u with the mass mu and writing the canonical 1-form:
theta = px dx + py dy + pz dz - H dt + mu du.
It we take the Bargmann transform on (p, H, mu) and require that theta
be an invariant, then we can conclude the corresponding transform for
the coordinate differentials:
theta = (px - v m) dx' + py dy' + pz dz' - (H - v px + v^2/2 m) dt'
+ mu du'
= px dx + py dy + pz dz - H dt + mu du.
From this, we conclude:
dx' = dx - v dt, dy -> dy, dz -> dz, dt -> dt, but also du -> du +
v dx - v^2/2 dt!
The relativistic form of this would be a generalization of Lorentz to
a "Bargmann Lorentz":
dx' = (dx - v dt)/root(1 - alpha v^2), dy' = dy, dz' = dz, dt' =
(dt - alpha v dx)/root(1 - alpha v^2)
du' = du + v dx/root(1 - alpha v^2) - dt/root(1 - alpha v^2) v^2/(1
+ root(1 - alpha v^2))
In the alpha -> 0 limit, you can clearly see the v^2/2 term emerge
from the du transform.
The relativistic version of the canonical 1-form can be written in
either of two ways:
theta = px dx + py dy + pz dz - H dt + mu du
or
theta = px dx + py dy + pz dz - H ds + M du.
In the second form, the (t,u) coordinates combine to give you s = t +
alpha u. This is an invariant and gives us a vestige of the "absolute
time" of non-relativistic theory.
What was originally the time translation generator in relativity E = M
c^2, is here now just the translation generator for the u coordinate,
where s is kept fixed.
So, after Poincare' is centrally extended, the Haag theorem's
assumptions are no longer that the vacuum is isotropic, boost-
invariabnt, spatially homogeneous AND time-translation invariant; but
only that it is the first three of these four AND u-translation
invariant. "Time-translation" of Poincare' becomes "u-translation" of
the centrally extended Poincare' group. So all the theorem allows you
to conclude is that a u-symmetric, isotropic, boost-invariant,
homogeneous vacuum has 0 relativistic mass, 0 momentum, 0 moment and 0
angular momentum ... but not zero energy!
It is out of the extra energy term U that one can safely tuck away all
sorts of nice goodies, like the energy needed for pair production,
interaction potentials a' la non-relativistic theory and so on.