E² - p² = m²: from E = 0 --> p = m = 0, and s = 0. next, ψ = 0, a vector of H (= Hilbert space)

I am not understanding.  DE describes a Free Electron without any Electric and Magnetic Potential.  But it says L = J  + S =0.  So S = -J, on average. 

So the free electron still has S = plus or minus 1/2.  Since Elecrton is not in a bound orbit, J = 0.  So the only way we can do this that the electron S is going from + to - 1/2 and hence on average S = 0.

(ap + βm)ψ = Eψ, square this and E = 0: --> p² + m² = 0.

or solve it for E = 0 --> (ap + βm)ψ = 0. now det ( (ap + βm)) = 0 only for p = m = 0.

if p = m = 0:---> ψ = 0, and, trivially s = 0.

resume: in your question you start with E = 0!, this s trivial.

HD = Alpha · p + βm

[L,HD] = i( alpha × p).

This shows that the orbital angular momentum is not a constant of the motion.

Now consider the matrix operator

[S,HD] = −i( a × p).

Combining this with our previous result for L we see that the operator

J := L + S

is conserved because [J,HD] = 0, and furthermore it is an angular momentum

operator because

So the average of L = -S.  Here they do not even take into account if p = 0 or not.

 So if the electron is not going around an electron, it is may be still moving straight or in som helical way with p not equal to 0.  

So either the Electron S is changing between S = + 1/2 and -1/2 or the Electron though not in Orbit, or under zero Magnetic/Electric Potential, would somehow still going helical - but that need Magnetic Field. But B  = Dell Operator x A.

Hence the confusion is not gone yet, Sir.

first: HD = ap + βm = Alpha · p + βm. I don't know how to write greek alpha here. In details: ap = a1p1+ a2p2+a3p3, or a = alpha.

YOU write: i( a × p), and  i( alpha × p). ---> i( a × p) =  i( alpha × p). Next: [L,HD]? What is YOUR L?, my L is rxp, and [rxp,HD] doesn't exist. Note: HD is a 4x4 matrix, S is a contraction of 4x4 matrices. Then:[S,HD] = −i( a × p)X = 0, if and only if p = 0, particle at rest. Spin = 1/2, and conserved. (X is another tripel of 4x4 matrices, contracted with a x p)) All this s clumsy, because this HD is not a covariant formulation.

 It is about L = -S on average.

Will L = 0 when p = 0? Yes  Will S = 0 when p = 0? No.

So how does one explain 

explected L = i( alpha × p). = - expected S = - i( alpha × p).

your L is wrong: explected L = i( alpha × p) ??? What means expected"??? This crossproduct is not defined, but even if you FORMALLY multiply the matrix tripel alpha (dimensonless) with the tripel p you get nonsense: take p = (0,0,1), then alpha × p = (alphay, - alphax, 0) ??? this is never an angular momentum, with and without "i".

###############

You still work with not defined terms. First write down IN ALL DETAILS your L! Next: what is your S? You use this symbol in many different ways. Is it a 4x4 matrix? If yes, define it! or the expectation value?, Or the eigenvalue of the Dirac state?

##############'#

(ap + βm)ψ = Eψ: in momentum space: ψ = |a,b,-b bar,a bar>, and |a|²+ |b|² = 1, expt S = = 1/2, and  = 0. You find this in every textbook.

 Thanks for your patience, what you say I understand but not able to connect with what I am reading in a Text.

The Dirac Equation and the Lorentz Group, Joel G. Broida.  You can google. 

Page 10.

Let us first look at the orbital angular momentum L = r × p. Using the

commutator identity

[ab, c] = a[b, c] + [a, c]b

along with the fundamental commutation relations [pi, pj ] = 0 and [xi, pj ] =

iδij , we compute (using a sloppy summation convention)

[Li,HD] = εijk[xjpk,HD] = εijk(xj [pk,HD] + [xj ,HD]pk).

But

[pk,HD] = [pk, αlpl + βm] = 0

while

εijk[xj ,HD]pk = εijk[xj , αlpl + βm]pk = iεijkδj

l αlpk = i( × p)i

so that

[L,HD] = i( × p).

This shows that the orbital angular momentum is not a constant of the motion.

Hence we find that

[σ′i ,HD] = [σ′i , αjpj + βm] = 2iεijkαkpj = 2i(p × )i

or, alternatively,

[S,HD] = −i( × p).

Combining this with our previous result for L we see that the operator

J := L + S

is conserved because [J,HD] = 0, and furthermore it is an angular momentum

operator because.

I understand he is skipping the Spinor on right,

But as a commen sense, if p   = 0, then L = r x p and it is zero. But Spin is always considered intrinsic, but the above arguments imply that = and that baffles me.

The commutator [.,.] is defined on a Lie algebra. HD = HD+ is a generator of the Lie group (actually SU(2,2)xf(p,q)) acting on complex 4-space, or space of Dirac spinors. alpha generates the Lorentz boosts, beta is the zero component of a Lorentz-4-vector (=current). So far, HD is a 4x4 matrix, and a observable (=energy). We have to consider the Lie algebra of observavles, 

it says: L = r x p (= LE, E= unit 4x4 matrix) , a triple of functions on phase space times unit matrix., and HD = ap + bm = HD(p), also defined on phase space.

Now S: S is ALSO a 4-contraction of the 4-spinor s and the 4 spin matrices,

The Dirac system of the FREE particle is covariant, so we are allowed to go into the rest system: P = (m;0,0,0), or p = 0. L = 0, s = (0;0,0,1), S = S(3), and S(3) commutes with beta:

So L = 0 and S is NOT 0.  L commutes trivially with HD in the rest system,  so they will commute for arbtrary r, p (however, L HERE is not covariant). INDEPENDENTLY. J + S is true, but missleading or trivial.

So it s enough to consider the DIRAC spinor at rest, spin in 3-direction: |1,0,00> and this is an eigenvector of beta x m, (p=0), and of S(3): S(3) |1,0,00> = ½ |1,0,00>.

Similar for arbitrary p and s. S(s) |abcd> = ½ |abcd>.

The situation is different if a potential is added in the HD, for instance the Coulomb potential: but then J = S + L is conserved.

Resume: the free case is almost trivial, and you have to take it as an exercise to check that you understand gamma gymnastics. A real exercise is L + S with Coulomb potential. This author is …. say,… , I would not recommend this book.  Take f.i. Bjoren and Drell, rel. quantum mechanics.

 ########

The commutator [.,.] is defined on a Lie algebra. HD = HD+ is a generator of the Lie group (actually SU(2,2)xf(p,q)) acting on complex 4-space, or space of Dirac spinors. alpha generates the Lorentz boosts, beta is the zero component of a Lorentz-4-vector (=current). So far, HD is a 4x4 matrix, and a observable (=energy). We have to consider the Lie algebra of observavles, 

it says: L = r x p (= LE, E= unit 4x4 matrix) , a triple of functions on phase space times unit matrix., and HD = ap + bm = HD(p), also defined on phase space.

Now S: S is ALSO a 4-contraction of the 4-spinor sand the 4 spin matrices,

The Dirac system of the FREE particle is covariant, so we are allowed to go into the rest system: P = (m;0,0,0), or p = 0. L = 0, s = (0;0,0,1), S = S(3), and S(3) commutes with beta:

So L = 0 and S is NOT 0.  L commutes with HD in the rest system,  so they will commute for arbtrary r, p (however, L HERE is not covriant). INDEPENDENTLY. J + S is true, but missleading or tricial.

So it s enough to consider the DIRAC spinor at rest, spin in 3-direction: |1,0,00> and this is an eigenvector of beta x m, (p=0), and of S(3): S(3) |1,0,00> = ½ |1,0,00>.

Similar for arbitrary p and s. S(s) |abcd> = ½ |abcd>.

The situation is different if a potential is added in the HD, for instance the Coulomb potential: but then J = S + L is conserved.

Resume: the free case is almost trivial, and you have to take it as an exercise to check that you understand gamma gymnastics. A real exercise is L + S with Coulomb potential. This author is …. say,… ,I would not recommend this book.  Take f.i. Bjoren and Drell, rel. quantum mechanics.

The commutator [.,.] is defined on a Lie algebra. HD = HD+ is a generator of the Lie group (actually SU(2,2)xf(p,q)) acting on complex 4-space, or space of Dirac spinors. alpha generates the Lorentz boosts, beta is the zero component of a Lorentz-4-vector (=current). So far, HD is a 4x4 matrix, and a observable (=energy). We have to consider the Lie algebra of observavles, 

it says: L = r x p (= LE, E= unit 4x4 matrix) , a triple of functions on phase space times unit matrix., and HD = ap + bm = HD(p), also defined on phase space.

Now S: S is ALSO a 4-contraction of the 4-spinor sand the 4 spin matrices,

The Dirac system of the FREE particle is covariant, so we are allowed to go into the rest system: P = (m;0,0,0), or p = 0. L = 0, s = (0;0,0,1), S = S(3), and S(3) commutes with beta:

So L = 0 and S is NOT 0.  L commutes with HD in the rest system,  so they will commute for arbtrary r, p (however, L HERE is not covriant). INDEPENDENTLY. J + S is true, but missleading or tricial.

So it s enough to consider the DIRAC spinor at rest, spin in 3-direction: |1,0,00> and this is an eigenvector of beta x m, (p=0), and of S(3): S(3) |1,0,00> = ½ |1,0,00>.

Similar for arbitrary p and s. S(s) |abcd> = ½ |abcd>.

The situation is different if a potential is added in the HD, for instance the Coulomb potential: but then J = S + L is conserved.

Resume: the free case is almost trivial, and you have to take it as an exercise to check that you understand gamma gymnastics. A real exercise is L + S with Coulomb potential. This author is …. say,… ,I would not recommend this book.  Take f.i. Bjorken and Drell, rel. quantum mechanics.

##########

" ...but the above arguments imply that = and that baffles me.". better:  =  is wrong

Thanks.

Elementary Particle Physics in Nutshell by Christopher Tully says the same thing.

Page 14, 15.  I will re read your post.

http://www.phy.ohiou.edu/~elster/lectures/advqm_4.pdf  

Section 4.8 does a very good job. :-)  It is related to Zitterbewegung the change in L and S but expected values of each L and S are zero.  Which is again strange, as we are taught expected S is the Intrinsic S  = + - 1/2.  So expected  S = 0, only when it is changing from -1/2 to + 1/2.  And he says this is independent of positive Energy and uses random p.

in this book, S is NOT spin, (4.113) : triple S are the generators of the rotations , the same in (4.231). (4.237) above are the (correct) commutators of boosts and rotators, below: beta is a Lorentz scalar. L is not the covariant angular momentum, take the Pauli Lubjanski 4-vector! (4.241) is wrong. It may be tru in a kind of nonrelativistic limit.

(4.252), (4.253) is tru: the angular momentum of a free Dirac particle is zero, the spin is + or - 1/2, both are constant. One may develope a free plane wave by spherical waves, each partial wave carries angular momentum quantum numbers l,m. 

################

in (4.113): S is magnetic moment. Put on the second (lower) sigma matrix a minus sign: that is the spin (Bjorken/Drell). Then you see: in |1,0.0.0> and  |0,1,0.0> spin and magn. moment are parallel (electron), in  |0,0.1.0> and  |0,0.0.1> they are opposite: positron.

page 79 "We introduced here the spin matrices ....

, which describe the spin in a Dirac theory.

Thus the rst term on the right-hand side of (4.91) contains directly the spin. The second

term is of relativistic nature, since here spatial and time components are mixed."

this is wrong and badly missleading: These 6 matrices are the generators of the Lorentz group, the upper three rotate the state, the lower three are the generators of the boosts. Further, they act  as tensor operator of the electric and magnetic moment. Spin is a space like 4-vector s = (s0, s) , s2 = -1, in the rest system f.i. (0;001), pointing in z-direction.

replace in 4.96 p by s and m by +- 1, then you have the "spiin" equation which commutes with the p equation.

the representation there of the 15 gamma matrices do not reveal their simplicity.

section 4.8: (4.231), the S is wrong, it is NOT the spin as explained above. (4.237) are above Lorentz algebra commutators (rotator/boost) and below says, beta is a Lorentz scalar. (4.238) : d/dt S = 0, S is a constant matrix. of course, tru for p=0.

Thanks for your time. BTW, I am not that young as it may look in the picture. I am doing as Hobby. My main job is something else.  What do you do? Please tell more about yourself.  BTW, I have also 22 Patents.

 >>in (4.113): S is magnetic moment ?

Understand L and S are operators and he getting expected value of operators by doing Integration and taking multiplication of Conjugate Spinor and Spinor.

>>replace in 4.96 p by s and m by +- 1, then you have the "spiin" equation which commutes with the p equation.?  

You mean this shows the Helicity / Chirality, depending on which two parts of Spinors you are looking.

What is Spin Equation?  Where did you get this from and what basis?  p and s com-mutate?

You mean, Pauli-Lubanski spin vector. [Wσ, Pμ] = 0

I am also reading Eq 148 from 

http://physics.ucsd.edu/students/courses/fall2009/physics130b/Dirac_Lorentz.pdf

>>d/dt S = 0, S is a constant matrix. of course, tru for p=0.

Also true for any p 0 ?

>>(4.241) is wrong

He says the equation is wrong and corrects it later.