The reason is simple, but is by some silly coincidence hidden quite well in most books.

If the product of a spinor and its 'adjoint' (taken at the same space-time point x) is to be a Lorentz-invariant (scalar) quantity, the 'adjoint' can't be simply the Hermitean conjugate (of a column-matrix) but has to be the product of this Hermitean conjugate and gamma_0. A chapter on 'bilinear covariants' should have this information.

I am looking for a more elaborate answer -- since I am reviewing in more depth this chapter, see my paper: http://www.hindawi.com/journals/ahep/2013/391741/ -- The spinorial part of the boost is evidenty non-unitary and there must be some physics to this (parallel to boost) elongation. Lorentz invariance does not leave me totally cold, but neither says too much. Gamma^0 happens to be the spinorial part of space inversion, its action has physics to it.

In your paper you conclude that real eigenvalues enforce a matrix to be Hermitean. This is, of course, only correct if the eigenvectors are known to be orthogonal (with respect to the scalar product of matrix calculus). Also in other places your paper seems to misrepresent the relation between scalar product and the adjoint.

you are right, the paper was hurried and there are a few "left-overs" (in the introduction, then the part showing that real eigen values _may_ lead to the conclusion shows only approximately why things are the way they are. I am putting out a new paper that shows much simpler (without the "a_i and M_ij unjustifyable asymmetries") why \gamma^0 is the adjunctator, but I want to add more to it about \gamma^0, physics-wise). In any case, returning to \gamma^0 - the invariant "thing" is a scalar operator times \gamma^0 is Lorentz invariant, or a vector/tensor operator times \gamma^0 behaves (under expectation value) as Lorentz vector/tensor. Take "1" - what the relation means is that the expectation value of the positive spinor parity part minus the negative spinor parity part is Lorentz invariant (i.e. < psi | gamma^0 | psi > ). That is a "thing" hard to understand physics wise as "just what quantity it represents". The BS about mass-eigen-states and flavor-eigen-states is just hoopla-woopla.

Spinors are no quantum mechanical states (only spinor-valued functions of momentum or of space-time can be states, see the irreducible unitary representations of the Poincarè group). Who speaks of 'expectation values of spinors' as a self-evident operation, even one 'understood physics wise' raises the suspicion to speak about only partially understood things.

I was looking for some ideas, not textbook algebra. It is evident what we are talking about: what is \gamma^0 messing the scalar product for. I exemplified what the average of "\gamma^0" means, perhaps in cumbersome words - for this is Lorentz invariant, but devoid of physical meaning. The average of "1" does have physical meaning, but transforms as part of the "\gamma^0 \gamma^\mu" pseudo 4-vector. Like so (example) "\gamma^0 p^\mu" transforms also as a pseudo 4-vector. While "p^\mu" has physical meaning, it does not transform as a 4-vector, but "\gamma^0 p^\mu" - with what (?!) physical meaning - does. So the question is "just WHAT" is that \gamma^0 in there physics wise ? Spare me textbook algebra.

> ???? Isn't it the 4-vector par excellence (in textbook algebra, admittedly)?

S\dagger U^\dagger p^\mu U S

=

S U^{-1} p^\mu US

=

\lambda^\mu_\nu p^\nu S^2

due to the S^2, it does not.

the quantity that does is: \gamma^0 p^\mu

the transformation of operators and wave-vectors is in Hilbert space - in physical understanding we need the averages - hence the problem: U^\dagger U = 1, but S^\dagger S is not unitary (that is the problem, and the question - what does the \gamma^0 that "kludges" the problem actually do - physically).

You seem to assume that p^\mu is a matrix in spinor space. Strange! It acts (as differentiation) on an argument x that you don't take into account. See Bjorken/Drell I, 4th equation past (2.11) (I don't have the Engish edition, so I better give no page numbers). That S is not unitary does matter by no means here.

If you nevertheless have something valid (in a context unfamiliar to me) in mind please tell me (preferably by formulas in which no symbols are missing). As far as I understand your calculations they are simply beginner's misconceptions. This is an observation which I don't understand taking into account your profil. Perhaps you can convince me of the contrary.

Sir, I make no assumptions about you - and since this is a (sort of) scientific forum - it could be worth considering reciprocity.

Second - I live in my own world. The exterior interests me only to the degree that it can provide some contribution / idea. Hence I am not in the business of convincing people - maybe someone like Boltzmann could do that for you.

P^\mu is an operator on a space. The qualifying space upon which it acts is a tensor product of the traditional Hilbert space (x) spinors. Its action is trivial (1) on spinors, but that does not mean they do not exist. The Lorentz transformation has 2 parts, one traditional and one on spinors (U and S). The S is not unitary and it is known to create a problem. Said problem has been "kludged" with \gamma^0 as a "new scalar product" (which is not a proper scalar product anyway - and devoid of physical meaning also).

Any quantity makes physical sense as: < psi | A | psi > / < psi | psi >, not as

< \gamma^0 psi | A | psi > / < \gamma^0 psi | psi >.

Any study of the above average shows the problem that S makes. (Which, by the way, becomes a projection operator for infinite boost, losing part of the wavefunction - evidently the scalar product < psi | phi > cannot be Lorentz invariant under such action, hence we need not have a Lorentz invariant < \gamma^0 psi | phi > that is so much sought - for that would be outright wrong.)

If we are seeing this problem, I would be interested in some idea regarding the physical meaning - and what does adding \gamma^0 mean, physically. (That would be all.)

>

This is the case only if one expects Lorentz invariance for averages that are not true scalar products of true physical states. That the scalar products which have to be Lorentz invariant (even Poincaré invariant) on physical grounds are actually invariant is shown in all relevant text books; a particularly nice argument is Schweber: An Introduction to Relativistic Quantum Field Theory p. 81, (97), (98), (102).

To make the essential point explicit: The invariant scalar product does not only involve sumation over spinor indexes but also integration (over position or momentum). Invariance depends crucially on the transformation properties of the volume element involved. The 'scalar product' of spinors is not allwed to be invariant under boosts since it has to compensate the changes coming from the volume element (e.g. Lorentz contractions).

My answer to your question thus is: The physical scalar product is free of \gamma_0 (see Schweber (97)). There are, however, suggestive reformulations (Schweber (98)) in which the gamma's appear. Gamma_0, however, as a square: just as a result of having it, first, "klutged" to the adjoint spinor and, second, having selected the 0-component of a 4-vector.

A physical meaning of \psi^\dagger \gamma_0 occurs in the construction of Lagrangian densities for Lorentz invariant theories. Here the integration does not extend over space-like hyperplanes but over domains of Minkowski space and the volume element of integration is dx^0 dx^1 dx^2 dx^3. The mass term m \psi^\adjoint \psi is acceptable in a Lorentz invariant theory iff

\psi^\adjoint \psi dx^0 dx^1 dx^2 dx^3

is Lorentz invariant. As we know, the simplest solution for this is

\psi^\adjoint = \psi^\dagger \gamma_0.

Once we have a scalar product, it must perform for any

observer.

IF we have < \gamma^0 psi | ... | psi >, admitedly this

performs a weird Lorentz invariance, however, for a non-

boosted observer S = S_rot (unitary) the extra \gamma^0

in there is evidently in offside !

There can be a rotated-boosted observer ... even more

complicated.

We cannot have "piece-wise" physics: for boosts we place

the \gamma^0, for rotations we take it out, for rot-boosts

we half-place-it, half-take-it-out.

That would be monkey business !

1) Space volume element is Lorentz invariant:

dt'^dx'^dy'^dz' = det(\Lambda) dt^dx^dy^dz

det(\Lambda) = 1

dt' = gamma dt

dx'^dy'^dz' = dx^dy^dz/gamma

2) I do not understand \psi^\adjoint and \psi^\dagger.

There is only one Bargmann-Pauli adjunctator, it is

proven unique, there are not possible two definitions.

The "new"-(not-a)-scalar-product has none of the prop.'s

needed for a scalar product. That is desastrous for the

norm, which is fundamental for the physics.

This has gone "by the waist side" for almost a century

(of "kludging-the-kludge").

3) I looked up Schweber - p. 81 is the only one relevant.

There he restricts integration to a space-like sheet.

I have objections (physically) to this, but math-wise

it does not change anything.

I prefer to be able to integrate also in time. One cannot

define 4-current with a 3D \rho, but only with a 4D one.

1. My '\dagger' is the same as in your contributions and '\adjoint' was intended to denote a star operation (as in star algebras) the definition of which is open at the moment of introduction and that will be determined by an explicitely stated requirement (here to act as a contribution to a Lagrangian density of a Lorentz invariant therory). In German one calls such tentative introductions 'Ansatz' and the word is sometimes also used in English.

2. I cited no other page in Schweber than p. 81. The other numbers refer to numbered equations. Inspecting just these equations helps to understand my argument. It shows that your statement > is not the case for the way in which Lorentz invariance of the Dirac equation is understood in the textbooks. The point is that the sub-manifold of Minkowsi space over which one integrates depends on the observer. This is just what is to be expected from a physical point of view.

3. I did not advocate different scalar products for rotations and boosts. Only the statement that the non-invariance of the 'scalar product' of spinors has to be compensated by a variation of the volume element is restricted to non-rotations.

4. > Schweber, Blorken/Drell, and Itzikson/Zuber show that the Poincaré invariant scalar product for the Dirac theory is done by integrating the 0-component of a 4-current over a spacelike hyper-plane or, using the whole 4-vector when integration is preferred to run over a general hyper-surface.

5. Of course, the 4-dimensional volume element is Lorentz invariant. My argument was that as a result of just this the expression \bar{\psi}(x) \psi(x) dx⁰ dx¹dx² dx³ is an contribution to a Lorentz-invariant action integral if the expression \bar{\psi}(x) \psi(x) is Lorentz invariant.

1 - i do not understand

2 - i prefer psi to describe the system, and use | x > < x | = \rho if I want to extract point info. Or | L > < L | if i want density of left-helicity, etc. I do not like to go outside < psi | A | psi > / < psi | psi > with restrictions on domains, etc.

3 - yes, the solution should be somewhere in this. I am not desperate for some invariant quantity (Lagrangian) for that is an artifact of other conditions that determine the physics. (Berry phase - http://arxiv.org/abs/quant-ph/9912045 )

4 - j^0, agreed. Note: j^0 as 4D density.

5 - ok

Great, we seem to have found a mode of communication that works for us.

1. You wrote > You still not understand? No problem since the content is essentially that of 5.

2. Of course, you may. I argued always in the framework of Schweber p. 81where \psi is the wave function of a Dirac particle.

3.,4.,5 no comments

ok, we have something to think over Gluehwein at Christmas :)

- Wein

- Nelkenpfeffer

- Gewürznelken-Baum

- Zimt

I like this conclusion and wish you good thoughts over Glühwein!

@ Mihai Dima

1st:

is a mathematical object: µ is an element of C^4,

is the scalar product on that Hilbert space, invariant under U(4).

: A is one of these 15 gamma matrices and is well defined and real, if A is hermitian. if A is similar to diag(1,1,-1,-1) then A defines an indefinite scalar product, then invariant under SU(2,2) which contains the Lorentzgroup (in reducible spinor representation).

2nd:

A² = 1, 1 is unit 4,4 matrix. Then = transforms as the zero component of this well known 4-current gamma mu This current is supposed to be conserved which is tru only for a subgroup of SU(2,2). This subgroup is SO(3,2) in spinor representation, which contains SO(3,1), the Lorentz group.

3rd: the physical interpretation is straightforward:

we have two "scalar products":

a standard "mathematical", the ordinary Hilbertspace scalar product, which is the zeroth component of a (conserved) current,

and an indefinte "physical" scalar product, which is the - invariant - charge of the particle.

4th: the "Dirac" equation singles out a "physical" Hilbert space, whose basis is not necessarily identical to that of the "mathematical" Hilbert space, where its elements are classified by the eigenvalues of three commuting hermitian gamma matrices, usually sign of energy, spin and magnetic moment for the particle "at rest".

note: SU(2,2) is a rank 3 Lie group.

1) on said space I need a scalar product,

none of the requirements (sesquilinearity

and positive definiteness) require, or

refer to, invariance under U(4).

Pauli observed that the adjoint operators

(obtained w/ help of sesquilinear/+def.

inner-prod.) form another repres. of the

same space, hence related via BXB^-1, where

B = "adjunctator". Question then arose ...

"who is B ?" (Bargmann-Pauli adj.) and is

it unique ?

Pauli offered a repres. dependent answer

why B = gamma^0 - which excluded anti-symm

B (however any e^{i\phi} B satisfies same).

More recent works, Arminjon/Reifler have

begun exploring repres.-indep. the Dirac

eq. (as it should have been from the very

start).

I have shown in:

http://www.hindawi.com/journals/ahep/2013/391741

(with above discussed limitations) why is

B = gamma^0 (mainly to satisfy positive

definiteness).

However one aspect came across, that on the

positive-helicity half it is possible to

extract a (sort-of) square root of the

adjunctator, and hence ANY adjunctator

is possible - this being related to Dirac

currents - implies possible definition of

any number of such currents on said half.

Therefore, I have purposely swayed away from

invariance discussions, as these ignore the

more basic foundation of the space.

2) what Lorentz transforms the Dirac eq. is:

S \gamma^\mu S^{-1} = \lambda^\mu_\nu \gamma^\nu

what transforms the scalar product is:

S \gamma^0 \gamma^\mu S = \lambda^\mu_\nu \gamma^\nu

the physical average does NOT Lorentz transform:

< psi | \gamma^0 \gamma^\mu | psi > / < psi | psi >

... but ! It transforms (unsurprisingly) as a 4-velocity !

And in fact I can construct a number of quantities that

ALL transform as 4-velocities ...

SO(3,2) - what does this help with ?

3) we have 1 scalar product and 1 not-a-scalar-product, which

happens to be zero for non-zero input states, try (L|L) or

(R|R) !... (in antithesis = = 1, as they should be !)

and this is NON-TRIVIAL to explain physics wise, for a scalar

product has a very specific physical meaning wrt overlaps and

averages.

first of all, Dirac theory is plain linear algebra on C^4 space.

(an almost trivial affair, sorry!)

even the hydrogen may be formulated algebraically in terms of so(4)xso(2,1) algebra.

I defined above the necessary mathematical objects. after this, Dirac theory is completely geometrizied, and geometry may be considered as theory of certain invariant objects.

A unitary C^4 space allows numerous "scalar products", these are certain endomorphisms into C, and these "scalar products" remain invariant under well defined transformations, all subgroups of GL(4,C).

we consider you should be aware, this is an U(4) invariant object.

we know what kind of observable this is in DIRAC theory. (zero comp. of a current).

but Dirac theory is not "all" theory.

SU(4) is standard on ISO symmetry, say a nucleonic pair. (or 4 quarks)

if you consider the intersection of SU(4) and SU(2,2) you have a nice

physico-geometrical iso-Lorentz system.

++++++++++++++++

this physico-geometrical picture is simple, but not well known to the huge majority of the scientific community.

????????????

how was your Glühwein?

1. - Sir, I find no papers of yours on this (or on any subject for that matter ... ).

2. - there is no representation independent definition of transposition, hence C does not exist.

3. - we are aware of the benefits of group theory, WHERE NEEDED. For offering lectures on the subject, please refer to the Math Dept.

4. - I am interested in pertinent ideas on the subject of the question stated, not on what the "huge majority of the scientific community" knows or not-knows.

PS - have I forgot something ?... oh yes, we are not dealing with unitary transformations.

@ MD/CERN:

1. what I publish and what not is my private affair

2. you're wrong: geometrical invariants are independent of representation, bases, coordinates and all that. it is the best possible choice you can do. this C you may even find in standard (physical) textbooks, but a bit mysterious presented.

note: these terms SO(n) or SU(n) or Sp(2n) fix a set of properties of the generators of the Lie algebras, transposition belongs to all of them, coordinate free.

in an elaborate language: geometrical objects can be defined by their INNER properties, like norm, distance, incidence ...

you don't really have to know, how these famous 2x2 Pauli spin matrices look like.

3. we are not dealing with group theory, but with geometrizied quantum theory.

For mathematicians, theoretical or mathematical physics is almost trivial, if not wrong or not well defined (strings). as a matter of fact, in these posts this funny gamma 0, nobody really understands it.

4. pertinent ideas: you should worry about covariant quantum mechanics, otherwise you run into serious trouble f.i. with your CERN neutrinos, sent to Ereditato, faster than light, or they may stop oszillating after becoming massless, or if massive, you conserve parity in weak decay.

have a nice day.

AS./TUB

Slightly revised answer given on different thread by Dima here on Researchgate https://www.researchgate.net/post/Can_anyone_help_with_Dirac_matrices_choice:

Short answer: The (indefinite) Hermitian form \bar \psi \phi is the invariant Hermitian form, unique up to real factor, defined by the abstract data of a Clifford algebra and an irreducible spinor representation. Here invariant means that the vector space of which the Clifford algebra is build, acts Hermitian.

The \gamma^0 in

\bar\psi \phi = \psi^\dagger \gamma^0 \phi

comes from choosing an orthonormal basis and (skew) Hermitian \gamma matrices (i.e. a Hermitian gamma^0 for the one timelike and skew Hermitian matrices \gamma^i for the space like basis vectors) This then defines an explicit spinor representation and the textbook form above of the invariant Hermitian form.

Long answer

For a real even dimensional vector space V with a non degenerate quadratic form g, consider the abstract Clifford algebra Cl(V,g). It has a unique [1] irreducible spinor representation \Delta. It carries a unique (up to real scalar) _but- _not_ _necessarily_ _positive_ _definite_ Hermitian form h such that the action of a vector in V is Hermitian (see below for a construction). This is the form that physicists write as

\bar\psi\phi = h(\psi, \phi).

If we already have _some_ Hermitian inner product ( , ) on \Delta then we can represent h as

h(\psi, \phi) = (\psi, H\phi)

for some ( , ) hermitian linear map H: \Delta \to \Delta which is determined by

(v\cdot)^* H = H (v\cdot)

where ^* is the hermitian conjugate with respect to ( , ). This also shows why h is unique up to a scalar: if h' is a different invariant form, then the H (with respect to h') commutes with all elements of V, hence commutes with the whole Clifford algebra, hence is a (necessarily real) scalar.

Now how come H is always gamma^0 in physics textbooks?

First physicists choose an orthonormal basis e^a with g(e^a, e^a) = 1 for a = 1 .. p (the time like basis vectors) and g(e^a, e^a) = -1 for the remaining q = n - p (the space like basis vectors). Then they _choose_ a matrix representation on C^{2^n/2} to get "gamma matrices"

\gamma^a = \gamma(e^a)

Having chosen spinors concrete complex column vectors we have available the normal inner product ( , ) = ( )^\dagger ( ), but because the spinor basis defined by the columns is arbitrary (we can think of them as coming from some abstract spinor representation \Delta and then choosing a completely arbitrary basis \psi_1, \psi_2, ...\psi_{2^{n/2}} for \Delta and unlike, the (functorial) tensor representations, the choice of a basis for V does not in any way induce a choice of basis for \Delta ) , it has has, nothing to do with the invariant hermitian form h except that, as above, h can be written as

h(\phi, \psi) = phi^\dagger H \psi

for some Hermitian matrix H which is determined by

\gamma^a^\dagger H = H\gamma^a for i = 1, ... , n (*)

Now we can always choose a matrix representation such that the \gamma^a are Hermitian for the time like and skew Hermitian matrices for the space like basis vectors ((in the usual sense that \gamma^a^\dagger = \pm \gamma^a). Then if p = 1, we see by inspection that H = \gamma^0 is a solution to equation (*). Thus, the \gamma^0 in the textbook form of the invariant form actually comes from choosing gamma matrices that are (skew) Hermitian and having just one time dimension.

So why can we choose the gamma matrices Hermitian or skew Hermitian depending on the sign of g(e^a, e^a)? This is standard but looks suspiciously like what we are trying to prove so lets prove it here. By going to the complexification we can get a new complex "genuinely" orthonormal basis f^1 \cdots f^n with g(f^a, f^b) = \delta^{ab} of V\tensor \C, by judiciously multiplying the basis e^1, ... e^n with factors of i = \sqrt{-1}. Clearly, a matrix representation of Cl(V,g)\tensor \C = Cl(V\tensor \C, g\tensor \C) such that the \gamma(f^a) are Hermitian is the same thing as a representation of Cl(V, g) such that the \gamma(e^a) = \gamma^a are (skew) Hermitian. But since (f^a)^2 = 1 we have (f^a)^{-1} = f^a, so finding a representation with \gamma(f^a) Hermitian is the same thing as finding a representation such that the \gamma(f^a) are unitary. But such a representation is easy to find: just consider the Clifford group G generated by the f^a inside CL(V\tensor \C, g\tensor \C). By the commutation relations

G = \{\pm 1, f^a, f^af^b (a< b), f^af^bf^c (a < b< c), ....\}

is finite and so its representations can be chosen to be unitary.  

short answer - in a representation independent view there are no skews. The only object there is the scalar product (metric - if you wish). This is chosen diagonal, though if can be arbitrary, in particular "elongated" such that all systems of reference give the same measure of norm (or overlap) between states. There is no universal norm, just like there is no Euclidian metric. There is a Minkowsky (non-definite) (non)-norm to which in spin space corresponds \gamma^0.

The overlap is important, hence we want to be able to relate between systems of reference - and for that we need to know the relative velocity. In particular we want to know if states are orthogonal in the system's center of mass (which in relativity is a fairly complex concept).

The Lorentz invariant objects: \gamma^0 * (something = scalar, vector, tensor) - in the proper metric (or without \gamma^0 in the (non)-metric) - are also important, but they do not refer to measurements.

further answer: we have gamma^0 for nothing.

it is plain wrong.

likeway with the minkowski (non)-metric is used with Lorentz contractions.

as announced there is a mathematically sound explanation:

http://dx.doi.org/10.1007/s10773-015-2739-3