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