Understanding Dirac Equation Spinor Boosts

[Josh1079]

Hi,

I'm recently learning the Dirac equation and we're following the more historical approaching working in the Dirac basis. At first it seems OK that the upper two components are interpreted as positive energy and the lower two negative. However, when I learned that after a boost the spinor (say initially at (1000)) becomes something like this [ tex ] \begin{array}{cc}1\\0\\P_z\\P_+\end{array} [ /tex ] with some constant factors.

So I'm now really curious that what is the spin direction (the direction in which this spinor is the eigenvector of the spin operator) now. I can see that it's definitely not in the z direction anymore, but I don't quite see how I can get the spin direction. And actually how do people interpret the lower indices in this case.

Thanks!

 

[PeterDonis]

I'm recently learning the Dirac equation and we're following the more historical approaching working in the Dirac basis.

Can you give the textbook or other reference that you are learning from?

At first it seems OK that the upper two components are interpreted as positive energy and the lower two negative.

That "interpretation" is not really correct and can easily lead to misunderstandings.

after a boost the spinor (say initially at (1000)) becomes something like this

Again, it would really help if you would give a specific reference where this is coming from.

 

[Josh1079]

Hi @PeterDonis ,
Thanks for replying! I'm actually just using my lecturer's notes, but he told us that most of his stuff come from Greiner. So I'm really curious about how you interpret the indices of Dirac basis. How could I see the spin orientation after boosting?

And I know I neglected a E + mc^2 beneath the P_z and P_+, just in case you wonder.

 

[vanhees71]

The physical meaning is that the spin is defined in the restframe of the particle. Then you use the Wigner basis as the most simple way to realize the unitary irreps of the orthochronous Poincare transformation (that's why you get a bispinor, i.e., a Dirac spinor and not a Weyl spinor) in a most simple way. That means for the boosted vectors the meaning is still that the spin is defined in the particle's rest frame, but in the boosted frame, where the particle it's moving it's not so easy to direcly interpret the spin degrees of freedom.

For details, see Appendix B in

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[Josh1079]

Hi @vanhees71 ,

Thank you for your reply! So are you saying that since the dirac basis is not eigen to the lorentz boost, it just becomes less sensible to speak of the spin orientation in the boosted frame?

And your notes are brilliant! In appendix B I saw something called the Pauli-Lubanski vector which find very interesting. I think I've bumped into that expression before in chapter 11.11 of Jackson "classical electrodynamics" when he's introducing the so called four-spins which was never clear to me why it was defined that way. So if you don't mind may I also ask a few question on this topic?

First of all, I'm really curious about how people come up with this definition of this vector and associate it to the spin? I've read the wiki page about it but I don't think I can see how all of that come into being. Moreover, this might be a silly one, what does the contraction with the anti-symmetric tensor mean? With three indicies I know it's the cross product, but with four I get a bit confused.

Thank you so much!

 

[vanhees71]

Although it's a bit effort in the beginning, I find the study of elementary group and representation theory the key to understand a lot about why the physical theories look the way they look. It's all about symmetry, most importantly to begin with, the symmetry of spacetime, i.e., Galilei-Newton spacetime for non-relativistic physics and Einstein-Minkowski spacetime for Special Relativity.

To understand spin, you have to study the unitary representations of the special orthochronous Poincare group, i.e., the group which is a semidirect product of space-time translations and Lorentz transformations that leaves the direction of time invariant and doesn't contain spatial reflections either. This is the space-time symmetry group that's strictly realized in Nature, as far as we know according to the Standard Model. You find all the formalities in Sect. B.4, but let's look at the implied physics for massive particles (massless particles deserve a special treatment, because there the physics is slightly different as is also indicated by the formal mathematical analysis).

The difficulty with spin in the relativistic case compared to the non-relativistic case comes from the fact that boosts, i.e., the rotation-free change from one inertial frame to another, do not build a subgroup of the Poincare group. In other words two boosts with non-collinear velocities are not only non-commuting but also lead to a boost followed by a spatial rotation, the socalled Wigner rotation. This mathematical subtlety gives rise to important phenomena as the Thomas-precession of spins for moving particles and thus the famous gyrofactor 2 for electrons (modulo radiative corrections of QED). That's why spin is not easily defined in a Lorentz-covariant way. Here, Wigner's analysis of the unitary irreps of the Poincare group helps. As shown in Sect. B.4 of my manuscript, to built the representations you first fix the mass of the particle, in our case we make $m^2>0$. Thus the one-particle states are restricted to the momentum shell, $p_{\mu} p^{\mu}=m^2>0$, and the task to find the irreps of the Poincare group is reduced to find the unitary representations of the socalled little group. You define it by choosing an in principle arbitrary standard four-momentum $p_0$ on the mass shell. It's clear that you can get any other four-momentum on the mass shell by a Lorentz boost, and since the most convenient choice to define covariantly intrinsic properties of massive particles is to define the corresponding observables in their rest frame you choose $p_0^{\mu}=(m,0,0,0)$, which is the four-vector of the particle in its rest frame. Now you can transport this four-vector to any place on the mass shell by, e.g., a rotation free Lorentz boost.

As shown in the manuscript, a particularly important role is played by that subgroup of the proper orthochronous Lorentz group that leaves this "standard momentum" invariant, the socalled little group. That are of course the spatial rotations, and as is also shown in the manuscript, you can construct all unitary irreps of the Poincare group by finding all irreps of the little group. Then the irrep of the full representation is defined by defining a convenient one-particle basis (the socalled Wigner basis), where you can define easily the group action of the Lorentz group, and then it's defined via the representation of the little group. Now the irreps of the rotation group, which is the little group for the case of massive particles, are well known and determined by the well-known angular-momentum algebra, i.e., the representations of the corresponding Lie algebra (which also includes the half-integer spin realizations, i.e., you substitute the rotation group SO(3) by its covering group SU(2), implying the substitution of the full orthochronous proper Lorentz group by its covering group $\mathrm{SL}(2,\mathbb{C})$).

Now the Pauli-Lubanski vector is defined as the generators of this little group. In the rest frame of the particle it's simply given by the three angular-momentum operators $\vec{J}$, because angular momenta are generators of rotations. Now you can define in the rest frame the angular momentum operators simply as $(J_0^{\mu})=(0,\vec{J})$. In an arbitrary frame then you can define this fourvector by the appropriate boost, and it's most convenient to choose rotation-free boosts to transform from the rest frame to the particle to the particle, where it has an arbitrary momentum, but that's a pretty cumbersome procedure, and it's much easier to define the angular momentum operators in a manifestly covariant way. This you can argue as follows:

The infinitesimal Lorentz transformations are parametrized by a set of 6 infinitesimal parameters $\delta \omega_{\mu \nu}=-\delta \omega_{\nu \mu}$, and the corresponding generators of the Lorentz transformations are thus 6 self-adjoint operators $\hat{M}_{\mu \nu}$. The little group's Lie algebra is defined by the subset that leaves the standard momentum invariant, i.e.,
$\delta \omega_{\mu \nu} p_0^{\nu}=m \delta \omega_{m0}=0,$
and indeed the remaining three $\delta \omega_{jk}$ with $j,k \in \{1,2,3\}$ are the infinitesimal rotations, which can be mapped to the angular-momentum vectors. In the rest frame the associated generators can be written as
$$\hat{J}_{0\rho}=\frac{1}{2m} \epsilon_{\mu \nu \rho \sigma} \hat{M}^{\mu \nu} \hat{p}_0^{\sigma},$$
but this is a covariant expression, and we can define it in any frame as
$$\hat{J}_{\rho} = \frac{1}{2m} \epsilon_{\mu \nu \rho \sigma} \hat{M}^{\mu \nu} \hat{p}_0^{\sigma},$$
which up to a factor $m$ is the Pauli-Lubanski vector.

To leave out this factor $1/m$ in (B.74) has the simple reason that you can generalize the construction of the irreducible reps of the proper orthochronous Poincare group also to all other possible representations, particularly the physically relevant case providing the description of massless particles with arbitrary spin.

 

[Josh1079]

Thank you @vanhees71 ! This is the clearest comment on this subject I've ever seen! I'll try to work with the details.

 

[DrDu]

You can also show by direct computation that the operators $P^2$ and the square of the Pauli Lubanski vector are two operators which can be constructed from the algebra of the generators of the Poincaré group and commute with this group and among. Operators of this kin are called Casimir operators. Hence you can use them to define an irreducible basis of this algebra. In an irreducible representation, operators commuting with all of the algebra will be represented by ordinary numbers, corresponding to $m^2$ and $m^2s^2$ in this case.