Transverse-plus, transverse minus, and longitudinal polarization of spin 1 particle

What is difference in transverse-plus, transverse minus, and longitudinal polarization of spin 1 particle, and how are this related to its three spin projections states? What is difference in spin 1/2 case? If spin 1/2 particle has two projections of spin (i.e two transverse), why it is often longitudinal one introduced, and what is its meaning?

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Well, first thing to note is that components of a field of arbitrary spin $j$ have nothing to do with space (e.g. there are not $2j + 1$ independent directions), so that your discussion of spin 1/2 just doesn't make any sense here. The reason we can interpret spin 1 field as vectors is that it transforms not only in the usual $SU(2)$ representation but also in the $SO(3)$ adjoint three-dimensional representation.

Now, since the vector has three components we can decompose it into arbitrary base of three vectors and it is obviously convenient to choose one vector in the longitudinal direction and the other two in the transverse direction. Since you explicitly ask, let me show you how this is done. Denote the longitudinal direction as the $z$-axis. Then (as always when dealing with spin) we can require that the basis of the representation be composed of eigenstates of the $L_z$ operator with eigenvalues $-1$, $0$ and $1$ so that the matrix $D_z$ is diagonal in this basis. Matrices $D_x$ and $D_y$ can be determined from the relation of $L_x$ and $L_y$ to ladder operators $L_{\pm} = L_x \pm i L_y$ acting on the $L_z$ eigenstates. This is the standard representation $\mathbf D$. Now, there is an unitarily equivalent adjoint representation of $SO(3)$ given by the matrices $M^i_{jk} = i \epsilon_{ijk}$. It is now a routine exercise to find a unitary matrix that transforms the representation $\mathbf D$ to the representation $\mathbf M$. Doing this is in reverse, this is also how you pass from the standard orthonormal $x,y,z$ components of the field to the basis of the $-1, 0, 1$ polarization states.

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@Newman: I was talking about usual vectorial realization of the representation. This is in general not possible. For half-integer spin even tensorial representations aren't possible, there are just spinors. Spin 1 of course transforms in SU(2) representation, as does any other spin. On the other it has nothing to do with the Lorentz group since this group is non-compact and does not have any non-trivial unitary representations (that's why boosts are represented as non-hermitian matrices). Only the spin subgroup Spin(3) = SU(2) has these representations and it's here that spin comes from – Marek Aug 18 '11 at 6:14
@Marek, I think you have some misunderstanding here. Spin 1 particle is by definition a particle with wavefunction components transform under $SO(3)$. It has nothing to do with spin $1/2$ particle which is indeed associated with $Spin(3)=SU(2)$. The fact that you may map $SU(2)$ to $SO(3)$ allows us to operate with spin of spin-$1/2$ particle as a vector to some extent by constructing $\vec{s}=\psi^{+}\vec{\sigma}\psi$. But you should be really careful here because wavefunction transforms as a spinor and not as a vector contrary to spin-1 particle. – Misha Aug 19 '11 at 6:19
@Misha: I don't agree. Spin has always to do with Spin(3), not SO(3), except when people are lazy and forget to mention it. SO(3) only gives you orbital momentum, never spin. The reason one can pretend integer spins come from SO(3) (and exploit the nice tensorial structure thereof) is that we have canonical projection from Spin(3) to SO(3) which induces map of representations for integer spin reps. This map is used implicitly all over the physics and only people who are not careful or have no idea what's going on can claim that spin comes from SO(3). – Marek Aug 19 '11 at 6:27
Other way to say the same thing is that Spin group distinguishes bosons and fermions. Bosons have special property that they represent in a spinorial representation that is trivial on the kernel of the $\pi: Spin(3) \to SO(3)$ map and therefore bosons can be identified simply as representations of $\pi(Spin(3)) = SO(3)$. But it's important not to forget that Spin group is the fundamental one here (if nothing else, the name should ring a bell...). – Marek Aug 19 '11 at 6:32
And another remark is that your ubiquitous use of SU(2) is also due to accidental coincidence here. Again it's important to realize that spin doesn't come from SU(2). It comes from Spin(3) and one is implicitly using the isomorphism of these groups given by identification of quaternions with hermitian matrices. In general, Spin groups have nothing to with SU groups since they are just double covers of SO groups. – Marek Aug 19 '11 at 6:36