# 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.
@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
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