Constructive Quantum Mechanics (Relativistic)

[pirillo]

Some have suggested, (I'd say most) that the Hilbert space for a
single relativistic
massive particle is the set of L^2 functions on the mass hyperboloid,
with the measure
on the mass hyperboloid thats given by the embeding in Minkowsky
space. Now once this is done I consider 4 functions on the Hyperboloid
(Coordinate functions w.r.t. a given Lorenz observer) p_1, p_2, p_3,
p_0 (actually p_0) is related to the other three. Multiplication
by each p_i gives us our momentum operators. Now for the position
operators, we make
the usual Newton-Wigner construct. Now I'm wondering, for the
coordinate conjugate to
p_0 (time). How do i define the "position operator" --time really!

Also the x_0, ..., x_3 approach requires [p_o, x_0] = -1, now does
having -1 ,
instead of 1, in these commutation relations fix any of the objections
that
people have against [H, T] = 1 (in the construct I outlined avove,
H=p_0 certainly has "below bounded spectrum") . I almost doubt it cuz
[T, H] = -1 , so do the people who say
[p_o, x_0] = -1 overlook this?

[Oh No]

Thus spake pirillo <ultraman2002@hotmail.com>
>Some have suggested, (I'd say most) that the Hilbert space for a
>single relativistic
>massive particle is the set of L^2 functions on the mass hyperboloid,
>with the measure
>on the mass hyperboloid thats given by the embeding in Minkowsky
>space. Now once this is done I consider 4 functions on the Hyperboloid
>(Coordinate functions w.r.t. a given Lorenz observer) p_1, p_2, p_3,
>p_0 (actually p_0) is related to the other three. Multiplication
>by each p_i gives us our momentum operators. Now for the position
>operators, we make
>the usual Newton-Wigner construct. Now I'm wondering, for the
>coordinate conjugate to
>p_0 (time). How do i define the "position operator" --time really!

You don't. Time is a parameter, even in relativistic quantum mechanics.
>
>Also the x_0, ..., x_3 approach requires [p_o, x_0] = -1,


In a recent thread, "Query about commutation of Scalar rest mass with
Position", here and on s.p.f. Jay Yablon found an amusing paradox which
showed that one cannot use this commutation relation. See responses on
both groups.

Regards

--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)

http://www.teleconnection.info/rqg/MainIndex

[Rock Brentwood]

On Jun 11, 11:22 am, pirillo <ultraman2...@hotmail.com> wrote:
> Also the x_0, ..., x_3  approach requires [p_o, x_0] = -1, now does
> having -1 ,
> instead of 1, in these commutation relations fix any of the objections
> that
> people have against [H, T] = 1

This is the wrong way of looking at things. Coordinate-momentum
conjugacy is *not* the pattern here that generalizes, but something
quite different is the pattern that generalizes. In fact, coordinate-
momentum conjugacy is a red herring. Moreover, the generalization is
not just non-relativistic -> relativistic; but tardyon -> luxon &
other sectors (in *both* non-relativistic and relativistic cases).
It's a 2-fold generalization.

Defining the parameter z = (1/c)^2 for the relativistic case, z = 0
for the non-relativistic case, z = -1 for the 4-D Euclidean case, the
following sectors turn out to emerge:
* Tardyon, z < 0
* Tardyon, z = 0 (here, a 3-fold Heisenberg algebra emerges)
* Tardyon, z > 0 (here, also, a 3-fold Heisenberg algebra emerges)
* "Synchron", z = 0 (the infinite velocity non-relativistic sector)
* Luxon, z > 0 (similar in some respects to the Synchron sector)
* Tachyon, z > 0 (similar in other respects to the Synchron sector)

For the synchron sector, there *is* an energy-time uncertainty
relation; but only 2 sets for coordinate-momenta.

So, let's start at the beginning.

A "particle", at least in the parlance of foundational physics, is not
a corpuscular entity that travels on worldlines (though it may come
out to that). Its initial conception is as a *system* of some
otherwise unspecified nature whose characteristic property is that its
state space is an irreducible sector that is completely spanned by the
underlying spacetime symmetry group.

To classify such systems means to determine what the invariants of the
symmetry group are and then to go through the cases, delimiting the
possibilities based on each set of values the invariants can take on.
(And in certain special cases, where the invariants take on certain
values, further invariants emerge, but I won't get into that here).

For the Galilei group (extended to include the mass m, as a "central
charge"), there are 3 invariants:
P^2 - 2mT, |mJ + PxK|^2, m
Hence, for instance, the Schroedinger equation for the free mass is
actually seen as a quadratic equation not linear:
(P^2 - 2mT) psi = 0.

The same applies to the Poincare' and Euclidean group, where m is
replaced by the "relativistic mass" M. One then has 3 invariants
P^2 - 2MT + z T^2, |MJ + PxK|^2 - z (P.J)^2, M - zT
where J represents angular momentum, K represents moment, P momentum,
and T kinetic energy.

For non-zero z, "relativistic mass" M can be eliminated in favor of
the "total" energy E = M/z and the kinetic energy can be eliminated in
favor of E, by using the 3rd invariant. So there are then just 10
generators (3 components of P, 3 of J, 3 of K, and E) left and 2
invariants.

The key distinction (tardyon vs. luxon/synchron vs. tachyon) is
captured by the (derived) invariant)
M^2 - z P^2 = (M - zT)^2 - z (P^2 - 2MT + z T^2).
If positive, one has tardyons; if 0, one has synchrons/luxons, if
negative one has tachyons.

For each set of values of the invariants, the symmetry group reduces
to a subgroup that preserves those values. The subgroup for the
tardyons in the Galilei case is: (Heisenberg on 3 generator pairs) x
SO(3) x E(1); the E(1) being for internal energy, SO(3) for spin and
Heisenberg being for position and momentum.

For the non-relativistic case, where M = m, and m is not 0, one can
then proceed to DEFINE the position (up to a 1-parameter's worth of
ambiguity) by
r = (K - Ps)/m.
The parameter s is otherwise indeterminate.

The *spin* then emerges from the decomposition of the angular momentum
J into its orbital and intrinsic part:
S = J - r x P.
It's independent of s and can be written as S = (mJ + PxK)/m.

The velocity is obtained by applying Hamilton's equation to the
invariant (P^2 - 2MT + zT^2), adopting T as the Hamiltonian:
0 = d/dP (P^2 - 2MT + z T^2) = 2P - 2(dM/dP) T - 2M (dT/dP) + 2z dT/
dP T
= 2P - 2(dM/dP) T - 2Mv + 2zvT
and to the 3rd invariant
0 = d/dP (M - zT) = dM/dP - z dT/dP --> dM/dP = zv.
Hence, substituting into the other equation, you get:
P - (zv) T - Mv + zv T = 0 --> P = Mv.

The first invariant for the non-relativistic case reduces to
P^2 - 2MT = 2m (mv^2/2 - T)
thus yielding the decompotion of the kinetic energy T = 1/2 mv^2 +
internal energy.

Thus, one has position-momentum, spin (S) and internal energy for the
non-relativistic tardyon.

There is no 4th set of commutators. But the reason there is no 4th set
*has nothing to do with this being non-relativistic*.

It has to do with the mass-energy invariant, P^2 - 2MT. As an ultimate
result, the "4th coordinate" t is just the parameter, s, itself. In
effect, the commutator [H, t] = [H, s] = 0.

The way you can tell that this has nothing to do with this being the
non-relativistic case, is that the *same thing* happens for the
relativistic Tardyon case. One has a similar decomposition of the
symmetry group, though the position operator is more difficult to
define.

Another way that one can tell that the conception you raised is
actually a red herring is that in the non-relativistic case, there
*is* an energy-time uncertainty relation! But it's for the synchrons.

The synchrons has m = 0, but P non-zero. (Thus, since the velocity is
v = P/m, it follows that v is infinite). This undercuts the definition
of a position operator. Instead, defining the x-axis as the direction
of P, one can only define the position r up to a multiple, s, of the
unit vector in the x-direction. The extra "parameter" s becomes
associated with x, instead of t.

As a result there are only 2 sets of commutators for spatial
coordinates vs. momenta; because [p_x, x] = [p_x, s] = 0. The 3rd
commutator is for energy and time. The time *operator* is just t = K.P/
P^2. In place of spin, one has a kind of "helicity" operator W = PxK.
From W and t, one can reconstruct K. The Poisson bracket {t,T} = 1,
here, instead of 0.

For this sector, the symmetry group reduces to one that consists of:
* The Heisenberg group on one pair of generators, (t,T)
* The Abelian group spanned by P, W, subject to the constraint P.W = 0
* The group SO(3), defined with the natural action (rotation) on P and
W.

With a little more work, you can reduce this to a combination of E(2)
for "helicity", Heisenberg on 2-pairs of coordinate-momentum
generators; Heisenberg on (t,T), and E(1) for the "internal momentum"
in the x-direction. The x coordinate is just the parameter s.

[pirillo]

Charles,

Hi, you say

> You don't. Time is a parameter, even in relativistic quantum mechanics.

Yet I've seen in many books (especially on strings) when they describe
the relativstic particle
as a warmup, that they have x_0, to x_3 say, i.e., t,x,y,z operators,
and the states are wavefunctions
of txyz (the eigenvalues of the aforemantioned operators ) (in
covariant quantization--- this is after they've
eliminated tau by the constraint) and they claim that the commutation
relations
are [x_m, p_n] = i eta_{mn} (for n, m=0123 eta00 being -1.) So I
was wondering, how can we
make an explicit construction of these operators, the Hilbert space
etc?

[Arnold Neumaier]

pirillo wrote:
>
> I've seen in many books (especially on strings) when they describe
> the relativstic particle
> as a warmup, that they have x_0, to x_3 say, i.e., t,x,y,z operators,
> and the states are wavefunctions
> of txyz (the eigenvalues of the aforemantioned operators ) (in
> covariant quantization--- this is after they've
> eliminated tau by the constraint) and they claim that the commutation
> relations
> are [x_m, p_n] = i eta_{mn} (for n, m=0123 eta00 being -1.)

Assuming these conditions implies that p_0 is an unbounded operator,
so that it cannot describe the causal time evolution. Thus while p_0
is an operator in these spaces, it is not the Hamiltonian operator
responsible for the physics.


> So I was wondering, how can we
> make an explicit construction of these operators, the Hilbert space
> etc?

Most of current quantum field theory (i.e., everything with exception
of 2D and 3D constructive field theory - which doesn't even cover QED)
does not have a well-defined Hilbert space at all, in which a
time operator would be defined.

Well-defined are only the asymptotic Hilbert spaces of in and out
states for scattering experiments. These are Fock spaces of
free particles, and hence defined on a mass shell.
There is a basic result called Haag's theorem which states that
these asymptotic Fock spaces cannot carry a nontrivial local dynamics,
as would be required for a field theory.

The full dynamics can be defined only indirectly, via CTP (closed
time path) integration, and subject to all interpretation problems
of the renormalization procedures.


Constructing for a relativistic field theory a physical
Hamiltonian which is bounded below is really difficult, and has
been achieved only in less than 4D theories.

The construction is usually based on a preferred time coordinate
which is needed in all cases I am familiar with;
- in the Foldy-Wouthuysen transformation (for the Dirac equation,
where p_0 also fails to have the right properties),
- in the Newton-Wigner construction (for single particles in
an arbitrary massive irreducible representation of the
Poincare group) and
- in the Osterwalder-Schrader reconstruction theorem (for
Lorentz-invariant field theories from Euclidean field theories).

While the Hilbert space and the Hamiltonian depend on the choice of
the time coordinate, the physics is independent of it since all these
Hilbert spaces are isomorphic via isomorphisms that maps the
Hamiltonians into each other.


See also the entries
S2g. Particle positions and the position operator
S2h. Localization and position operators
S6c. Functional integrals, Wightman functions, and rigorous QFT
S6d. Is there a rigorous interacting QFT in 4 dimensions?
S6e. Constructive field theory
S6h. Hilbert space and Hamiltonian in relativistic quantum field
theory
S9c. What about relativistic QFT at finite times?
in my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt


Arnold Neumaier

[Igor Khavkine]

On Jun 12, 9:52 am, pirillo <ultraman2...@hotmail.com> wrote:
> Charles,
>
> Hi, you say
>
> > You don't. Time is a parameter, even in relativistic quantum mechanics.
>
> Yet I've seen in many books (especially on strings) when they describe
> the relativstic particle
> as a warmup, that they have x_0, to x_3 say, i.e., t,x,y,z operators,
> and the states are wavefunctions
> of txyz (the eigenvalues of the aforemantioned operators ) (in
> covariant quantization--- this is after they've
> eliminated tau by the constraint) and they claim that the commutation
> relations
> are [x_m, p_n] = i eta_{mn} (for n, m=0123 eta00 being -1.) So I
> was wondering, how can we
> make an explicit construction of these operators, the Hilbert space
> etc?

This is an application of constrained system quantization to the
relativistic particle. The idea is to represent x_m and p_n operators
(m,n=0,1,2,3) on a Hilbert space such that [x_m,p_n] = i eta_{m,n} is
satisfied, along with a representation of the Poincare group (Lorentz
group + translations) acting on the same Hilbert space, such that x_m
and p_n also transform in the appropriate way. The next step is to
construct the operator C = (eta^{m,n} p_m p_n + m^2) and isolate its 0-
eigenspace (which can be rather technical). The 0-eigenspace of C
together with all the operators that commute with C will constitute
the Hilbert space of a relativistic particle together with its algebra
of observables. If you do this carefully, this construction is
equivalent to taking L^2 functions on the future hyperboloid, as you
mentioned in your original post.

Note, the p_n operators commute with C, while the x_m do not. If you
go through the above exercise, you should think about what variations
of x_m will commute with C and become observables.

For a good overview of constrained systems and their quantization,
check out these books:

Rovelli, _Quantum Gravity_ (first half of the book)
Henneaux & Teitelboim, _Quantization of Gauge Systems_

Hope this helps.

Igor

[pirillo]

So, Igor,

You mean to tell me that after all the dust settles, In this x_i
(i=0123) I'll have left three
operators , call'em q_i , that satisfy the usual relativistic QM
commutator relations with
the four p_i , and I'll just have to introduce an external parameter t
(pulled out of where?)
by which I'll time evolve observables e.g., the q_i so I'll be back
to the usual relativistic QM
picture.

Again, why would I be necesitated to use the external parameter t in
this setup?

[Igor Khavkine]

On Jun 29, 11:52 pm, pirillo <ultraman2...@hotmail.com> wrote:
> So, Igor,
>
> You mean to tell me that after all the dust settles, In this x_i
> (i=0123) I'll have left three
> operators , call'em q_i , that satisfy the usual relativistic QM
> commutator relations with
> the four p_i , and I'll just have to introduce an external parameter t
> (pulled out of where?)
> by which I'll time evolve observables e.g., the q_i so I'll be back
> to the usual relativistic QM
> picture.
>
> Again, why would I be necesitated to use the external parameter t in
> this setup?

You are not. In fact, you don't have a single preferred time
parameter, there's a lightconeful of them. You'll have a Hilbert space
with a representation of the Poincare group on it. As such, you'll
have a generator for time-like translations for each vector in the
light cone. Pick any observable O(q,p), and you get a 1-parameter
family O(q(t),p(t)) = U(t)* O(q,p) U(t), where U(t) is the unitary
operator corresponding to a time-like translation. That's how the
parameter t appears.

Of course, you can generate other 1-parameter families. For example.
O(q(s),p(s)) = U(s)* O(q,p) U(s), where U(s) is a translation by s in
some space-like direction. You can do the same with rotations.
However, the usefulness of making these families look like "time
evolution" is questionable.

Any particular canonically conjugate set q_i and p_i is associated to
a specific space-like hyperplane. That's why it would seem that there
is a preferred direction time-like translation perpendicular to that
plane. However, you can apply a canonical transformation to get q'_i
and p'_i that are associated to a different hyperplane.

Hope this helps.

Igor

[pirillo]

Igor,

Originally I thought (Now I'm beginning to think I was wrong) that for
the x_i(tau) approach
(i.e., [x_i, p_j] = eta_{ij} ) with i=0123, that while the x_i are
different than the x_i usually
used for relativistic QM, that the p_i of this approach were the
same (albeit extensions) of the p_i
used in the standard approach for relativistic QM.
Lets call the x_i of the usual approach q_i, and the p_i of the usual
approach k_i . We'll reserve x_i, p_i
for the [x_i, p_j] = eta_{ij} approach.

Now, I'm begining to think that while for i=123 the the p_i are
extensions of the k_i,
the p_o is very, very different from k_0. The idea, roughly is that
q_i = x_i + 1/2 k_i/k_o^2 (i=123)
and assuming k_i = p_i for all i=0123, then wed get that q_i = x_i +
1/2 p_i/p_o^2, but then
[q_i, p_j] = [ x_i + 1/2 p_i/p_o^2, p_j] = [x_i, p_j] which gives
the wrong commutation relations
for the [q_i, p_o] case, if p_o is to be k_o. For the other indices
its correct to have k_j = p_j.

Another rough way to view this, is that while a fourier transform
sends p_i to k_i for i=123
It doesnt send k_o = sqrt{k^2 + m^2} to p_o .The fourier transform
sends the operators x_i to
differentiation which is what we use when constructing the Newton-
Wigners q_i .

But again my question is is it true that k_i =p_i (When we push
things over approprartely (say fourier))
while k_o is not the push of p_o .
_

[pirillo]

I guess an answer to my question is that it depends how you
explicitly define your, x, p , and your fourier transform. If you
define the fourier transform by

f(x, t) = \int e^{i(x, p)} e^{- sqrt{p^2 + m^2}} g(p) (1/sqrt{p^2
+m^2}) dp

and we define in (x-t) space the operators: x_i by multiplication and
p_i = -i eta_ii partial_i

Here, p-space is seen as the standard representation for relativistic
QM.
Then the usual p_i in p-space push to these p_i in (x, t)-space while
the correct-x in p-space push to some complicated looking expression
of the x, p of (x-t) space.

Now with a different "fourier transform" then the p in the p-space
representation may not push so nicely.