From: George Jones
Subject: Re: Lorentz invariace of inner products
Date: 30 Oct 2001 06:01:57 GMT
Newsgroups: sci.physics.research
Summary: Adapting quantum mechanics constructions to relativistic case
Iuval clejan wrote:
> John Baez wrote:
>
> > In article <3BD1B63B.9C9326E9@mindspring.com>,
> > Iuval clejan wrote:
> >
> > >The thing that bothers me is
> > >that under a passive boost (change of reference frames) this spinor
> > >does not transform in a unitary manner, for example it does not
> > >preserve its norm (actually this applies to all finite dimensional
> > >irreps of SO(3,1), not just spin 1/2 spinors) [...]
> >
> > This is a fact of life, there's no point getting bothered by it. :-)
>
> It may not be bothersome to a mathematician, but to a physicist this is a
> major problem (until it is resolved). But not worrying about it does not
> make it go away.The problem is not that the finite dimensional reps are
> non unitary.
Yes.
> The problem is that the finite dimensional reps represent
> particles and the fact that they are non unitary mean observers in
> different frames disagree on measurements for which no disagreement
> should occur (i.e. they should be invariant).
No, there is no problem. Let me elaborate a little on what John Baez
wrote. First, though, consider the inner product for standard
non-relativistic quantum mechanics. If f and g are wavefunctions, then
their inner product is
/
= |f*(x) g(x) dx.
/
Note that f and g are elements of an (subset of) infinite-dimensional
function space F given by
F = { f | f : R^3 -> C },
where R^3 represents all of space, and all the elements of F satisfy
the Schroedinger equation. So, wave functions are complex-valued
functions of space. Note that the inner product of wave functions is
defined on elements of F, not on elements of C, which is the range of
the functions in F.
Now let's move to the relativistic case. The wave functions are still
going to be functions that have common domains and ranges, and the
inner product is still going to be defined on wave function space,
and not on the common range of the functions in wave function space.
Wave function space is still going to be infinite-dimensional. It
would be strange if in going from the non-relativistic to the
relativistic we go from an infinite-dimensional wave function space
to a finite-dimensional wave-function space. What changes? The domain
and range of the wave functions. The domain for all the wavefunction
changes to all of spacetime (R^4 in a particular inertial coordinate
system), and the common range is (Dirac-)spinor space D, which is
4-dimensional. Thus, wave function space is now
G = { f | f : R^4 -> D },
where all the elements G satisfy the Dirac equation.
So, wave functions are spinor-valued functions of spacetime. Note that
while D is finite-dimensional, G isn't!. Again, the inner product
needs to be defined on wave function space, not on the common range of
the wavefunctions.(Sorry for repeating myself.)
I think part of the confusion comes from the following. Consider the
action of a rotation on non-relativistic wave functions f and g. Under
a rotation R, f -> f' and g -> g' defined by
f'(x) = f( R^(-1) x), g'(x) = g( R^(-1) x),
where x is a vector in R^3. Clearly, = for every f and g,
so the rotation R has a unitary action on wave function space F.
Now consider the action of a boost on relativistic wave functions f and
g. Under a boost B, f -> f' and g -> g' defined by
f'(x) = L f( B^(-1) x), g'(x) = L g( B^(-1) x),
where L is the transformation on spinor-space D that corresponds to the
boost B. At the the risk of being pedantic, x and x' = B^(-1) x are
4-vectors, x' = B^(-1) x , f(x') is a spinor, f'(x) = L f(x') is a
spinor, and f and f' are wavefuntions. Unfortunately, some books
abbreviate the above by leaving out the arguments, i.e., they write
f' = L f. This gives the impression that wave functions are spinors
(elements of D), but they're not, they're spinor-valued functions
(elements of G).
In the non-relativistic case, leaving out the arguments results in
f' = f, which is non-sensical, but actually just tries to say that a
rotation leaves elements of C (complex scalars) unchanged. This doesn't
happen in the relativistic case because elements of D (spinors) are not
left unchanged by the action of the Lorentz group.
For the relativistic wave functions above, an appropriately defined
inner product on the infinite-dimensional wave function space G
results in = for every f and g, so the boost B has a
unitary action on wave function space G.
Actually this allows the Dirac equation to, in some sense, be
derived. (If you're going to split an infinitive, you might as well
really split it.) Start with a Hilbert space formulation of quantum
theory. Require a unitary action of the Lorentz group so that
transition probabilities are preserved. Require that the action
correspond to spin 1/2 and be irreducible since electrons are
elementary. The Dirac equation is a projection equation that makes
this happen. I first learned about this from a book on general
relativity!
Regards,
George
==============================================================================
From: George Jones
Subject: Re: Lorentz invariace of inner products
Date: Fri, 2 Nov 2001 21:58:59 +0000 (UTC)
Newsgroups: sci.physics.research
Mike Mowbray wrote:
> Pardon my density... what exactly is the "appropriately defined
> inner product" that you mention in the above paragraph? (I presume
> it's not the same as the simple inner product mentioned near the
> start of your post?)
First, let me correct something I said in the last paragraph of my
previous post in this thread. When I said "unitary action of the Lorentz
group," I meant the (cover of the) of the Poincare group, which is the
the semi-direct product of the Lorentz group with spacetime
translations. Now on to the inner product.
Things are easier in momentum space. The inner product is defined in a
manner similar to the non-relativistic case, but relativity throws in a
few slight subtleties. In the former, the inner product integration is
over all of space, while in the latter the integration is not over all
of (spacetime) momentum space. This is because an electron has a
Lorentz-invariant rest mass m. In its rest frame the electron has
4-momentum (m,0,0,0). A Lorentz transformation takes this to 4-momentum
(p^0 , p^1 , p^2, p^3) subject to the constraint
(p^0)^2 - [(p^1)^2 + (p^2)^2 + (p^3)^2] = m^2,
i.e., the 4-momentum of the electron must stay on the mass hyperboloid.
Also, since in the elctron's rest frame, p^0 = m > 0, and only
orthochronous Lorentz transformations are allowed, p^0 > 0 always, i.e.,
the electron's 4-momentum always stays on the same branch of the
hyperboloid. These 2 constraints are enforced by taking the integration
measure to be not dp^0 dp^1 dp^2 dp^3, but the Lorentz-invariant measure
delta( p^mu p_mu - m^2) theta(p^0) dp^0 dp^1 dp^2 dp^3
where theta is the standard Heaviside step function.
To see what happens in any integration over p^0, let
h(p^0) = (p^0)^2 - [(p^1)^2 + (p^2)^2 + (p^3)^2 + m^2]
Contributions from the delta function occur at the zeros of h, and
h(w) = 0 gives
w = +- sqrt[(P^1)^2 + (P^2)^2 + (P^3)^2 + m^2].
Theta picks out the positive root, and (d/dp^0)h(p^0) = 2 p^0, so after
a p^0 integration,
1/(2w) dp^1 dp^2 dp^3
is left.
P^0 > 0 also puts constraints on the spinor wave function. Let
f(p^0, p^1, p^2, p^3) be the spinor-valued wave function on momentum
space for a free electron. The momentum space Dirac equation is
gamma^mu p_mu f(p) = m f(p), which in the electron's rest frame becomes
gamma^0 f(m,0,0,0) = f(m,0,0,0), since then p = (m,0,0,0). Take the Weyl
representation, so (in 2x2 blocks)
_ _
gamm^0 = | 0 1 | ,
|_1 0_|
and let the wave function Dirac spinor be f(m,0,0,0) =(a,b,c,d)^t (^t
for transpose). The Dirac equation then gives c = a and d = b, and
consequently the space of all possible f(m,0,0,0) is a 2-dimensional
subspace of spinor space D. 4-component Dirac spinors are needed to
implement parity transformations, and the Dirac equation serves to
project out unnecessary degrees of freedom.
To find f(p) for a general p define
_ _
f(p) = | L 0 | f(m,0,0,0) ,
|_0 (L^-1)^+_|
where a unique boost B takes (m,0,0,0) to p, and L is the SL(2,C) version
of B.
The inner product on spinor wave function space G = { f | f : R^4 -> D }
is defined to be
/
= |f(p)^+ gamma^0 g(p) d mu ,
/
where, as above, the integration measure is given by
d mu delta( p^mu p_mu - m^2) theta(p^0) dp^0 dp^1 dp^2 dp^3 .
Now define the action of an arbitrary Lorentz transformation B on an
arbitrary spinor wave funtion by
_ _
f'(p) = | L 0 | f(B^(-1) p) ,
|_0 (L^-1)^+_|
where L is an SL(2,C) matrix and B is the Lorentz derived from L.
_ _
f'(p)^+ gamma^0 = f(B^(-1) p)^+| L^+ 0 | gamma^0
|_ 0 L^(-1)_|
_ _
= f(B^(-1) p)^+ gamma^0 | L^(-1) 0 |
|_ 0 L^+_|
Therefore,
/
= | f(B^(-1) p)^+ gamma^0 g(B^(-1) p) d mu
/
/
= |f(p)^+ gamma^0 g(p) d mu
/
= ,
since d mu is Lorentz-invariant. Taking Fourier transforms gives the
configuration space inner product
/
= |f(x)^+ g(x) dx^1 dx^2 dx^3 ,
/
where f(x) and g(x) are both evaluated at any fixed x^0.
In momentum space, spacetime translations only introduce phase factors,
so the inner product is Poincare-invariant as well as Lorentz-invariant.
Again, I'm really talking about covers.
Regards,
George
==============================================================================
From: ark
Subject: Re: Lorentz invariance of inner products
Date: Sat, 3 Nov 2001 03:17:52 GMT
Newsgroups: sci.physics.research
On 2 Nov 2001 07:33:16 GMT, Iuval clejan
wrote:
>ohn Baez wrote:
>
>>
>>
>> We all agree that it's good to have a norm, or more precisely an inner
>> product, to interpret a vector space as a space of "state vectors"
>> in quantum theory. However, the spin-1/2 representation does not
>> have a Lorentz-invariant norm. What this means is that in the context
>> of special relativity, it is quite artificial to pick any particular
>> norm on this space - whatever you pick, a Lorentz boost will change it!
>
>>
>>
>> And this is consistent with the fact that we should not attempt to
>> use this space as a space of state vectors in the context of special
>> relativity + quantum theory.
>
>So what space should we use? The infinite dimensional unitary reps of the
>Lorentz group have all possible spin values, so they are not what we want when
>we talk about spin 1/2 particles.
Let's concentrated on these things that can be handled mathematically,
that is free fields. A free relativisttic quantum field has a particle
number operator which splits the Fock space into n-particle subspaces.
For n=0 we have vacuum which carries a trivial representation of
SL(2,C). For n=1 we have usually a direct sum of few irreducible
representations of the same spin. (why "few"? because we have
particles and antitiparticles, and possibly some other "internal
symmetries"). Apart of possible subtleties the n=1 (one-particle)
space is isomorphic to the space of solutions of the appropriate
wave equation. The subtleties involve the fact that solutions
of classical wave quation split into positive and negative energies.
Also, the classical wave equation can be real, while in quantum field
theory we need complex Hilbert space.
> The infinite dimensional (time and light-like)
>unitary reps of the Poincare group are probably OK though (I am looking at my
>group theory book), but since the Lorentz group is just the Poincare group
>without translations, these should be equivalent, so I am still confused.
>
>> >There is absolutely no relation between the double cones in Minkowski
>> >and quantum logic except a mathematical one. There is no physics there
>> >at all.
And why so? Here is an example, When we look for a relativistic
position operator, we need into account causal structure of the
Minkowski space. The natural regions for localization are double
cones (diamonds). We want states to be orthogonal when localized in
spacelike diamonds. These diamonds form a non-Boolean lattice.
We want to represent this non-Boolean lattice, together with the
symmetry group. The natural lattice to represent a non-Boolean
lattice is the lattice of subspaces of a Hilbert space. This way
you can rediscover Dirac equation.
>> This relation probably *is* irrelevant to your question, so I don't
>> know why ark mentioned it. However, it's *not* irrelevant to physics!
I mention it, becuase there are finite dimensional representation of
SO(3,1), but with indefinite metric. Thus they can not have a standard
quantum mechanical probabilistic interpretation. However more and more
people are looking into a possibility of using signed measures, thus
"negative probabilities" (Feynman wrote a paper on this subject,
recently Noyes wrote a paper too). This is NOT a standard theory
to be told kids, but this is a valid research avenue.
>> The point is this: Minkowski spacetime and the space of 2x2 hermitian
>> matrices are equivalent representations of the Lorentz group. The
>> causal ordering in Minkowski spacetime corresponds to the usual
>> ordering on the space of 2x2 hermitian matrices, where A < B if B - A
>> has positive eigenvalues. And if we restrict attention to
>> projection operators, which correspond to "propositions" in quantum
>> logic, this ordering corresponds to "implication": we obtain a
>> complete orthomodular lattice of propositions.
I am not talking about representing points. I am talking about
generalization of Mackey imprimitivity theorem, when spectral
measure is replaced by the lattice of diamond sets in the Minkowski
space, with its natural orthogonal complement operation.
There is no such a theory in fact. But there are partial results.
ark