## Jul 13, 2013

### On the Spin in Relativistic Dynamics

Spin is an intrinsic property of a particle associated with rotational symmetry. Roughly speaking, it is the intrinsic part of the angular momentum. Measurement of angular momentum $\vec{J}$ differs in different reference frames. It is tempted to decompose it $\vec{J} = \vec{L} + \vec{S}$, where $\vec{L}$ called orbital angular momentum, $\vec{S}$ called spin angular momentum or spin for short. Orbital angular momentum should vanish in the particle rest frame $\vec{P} = 0$. In non-relativistic quantum mechanics, it is defined as $\vec{L} = \vec{X} \times \vec{P}$. But it is not clear a priori that such a decomposition is always available in relativistic dynamics. Another way of defining spin, is to measure the angular momentum in the particle rest frame. In relativistic dynamics, there is no unique Lorentz transformation that transforms a momentum state to the particle rest frame. Ambiguity thus exists for the definition of spin.

Nevertheless, spin can be defined formally as a vector operator that satisfying the following conditions: $\left[ S^i, S^j \right] = i\varepsilon^{ijk} S^k; \quad (i,j,k = 1,2,3) \\ \left[ S^i, P^j \right] = 0; \qquad \qquad \qquad \qquad \quad (\mathbf{*})\\ \vec{S} = \vec{J} \qquad \text{ if } \vec{P} = \vec{p}_c. \qquad \qquad$ The last condition should be understood within momentum states $\vec{S} \left.| p, \sigma\right> = \vec{J} \left.| p, \sigma\right> \quad \text{if } \vec{p} = \vec{p}_c$. The spin operator $\vec{S}$ described above is not covariant. The canonical example of $\vec{p}_c$ is $\vec{0}$.

Spin as an operator must also depend on the specific realization of the space-time symmetry (the Poincaré symmetry), i.e. the representation. According to Wigner theorem, a symmetry transformation in quantum mechanics should be realized either as unitary operator or as anti-unitary operator. Because Lorentz group is non-compact, all unitary representation has to be infinite dimensional, namely fields. Nevertheless, a relativistic theory still can have a finite-dimensional representation. The prominent example is the Dirac theory of relativistic electrons.

### Poincaré algebra

Consider the Poincaré algebra, $\left[ P^\mu, P^\nu \right] = 0; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ \left[ P^\lambda, M^{\mu\nu} \right] = i \left( g^{\lambda\mu} P^\nu - g^{\lambda\nu} P^\mu \right); \quad \qquad \qquad \qquad \qquad \quad \\ \left[ M^{\lambda\rho}, M^{\mu\nu} \right] = i\left( g^{\lambda\nu} M^{\rho\mu} + g^{\rho\mu} M^{\lambda\nu} - g^{\lambda\mu}M^{\rho\nu} - g^{\rho\nu} M^{\lambda\mu} \right);$ $P^\mu$ and $M^{\mu\nu}$ are the 10 generators of the Poincaré algebra. The six independent components of $M^{\mu\nu}$ are, $J^k \equiv \frac{1}{2} \epsilon^{ijk}M^{ij}, (i,j,k = 1,2,3)$ the angular momenta; $K^i \equiv M^{0i}, (i = 1,2,3)$ the boosts.

The metric tensor is $g^{\mu\nu} = g_{\mu\nu} = \begin{pmatrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \\ \end{pmatrix}$
$P^2 = P_\mu P^\mu = \mathscr{M}^2$ is a Casimir element, known as invariant mass squared. In this post, we only consider massive case $\mathscr{M}^2 > 0$ for simplicity.

#### Pauli-Lubanski Pseudo Vector

$W^\mu = -\frac{1}{2} \varepsilon^{\mu\nu\kappa\rho} M_{\nu\kappa} P_\rho,$ where $\varepsilon^{\mu\nu\kappa\lambda}$ is the Levi-Civita tensor. It can be shown, $W^0 = \vec{J}\cdot\vec{P}$, $\vec{W} = \vec{K}\times\vec{P} + P^0 \vec{J}$. Pauli-Lubanski vector satisfies,
1. $P_\mu W^\mu = 0$;
2. $\left[ P^\mu, W^\nu \right]= 0$;
3. $\left[ M^{\mu\nu}, W^\kappa \right] = i\left( g^{\kappa\mu} W^\nu - g^{\kappa\nu} W^\mu\right)$;
4. $\left[ W^\mu, W^\nu\right] = i \varepsilon^{\mu\nu\kappa\rho} W_\kappa P_\rho$;
$W^\mu$ are the generators of a stabilizer group for each $P^\mu$. For irreducible representations, $W^2 = W_\mu W^\mu$ is a Casimir operator (Casimir element). To evaluate its scalar value, we can study it in the center-of-momentum frame where $\vec{P} = 0$: $W^\mu = \mathscr{M} (0, \vec{J})$. So $W^2 = - \mathscr{M}^2 \vec{J}^2 = -\mathscr{M}^2 s(s+1)$. Note that $P^2$ and $W^2$ are the only two independent polynomial invariant operators that commute will all the operators (the Casimir elements). Here the total spin appears naturally. But $W^\mu/\mathscr{M}$ (or $\vec{W}/\mathscr{M}$) does not satisfies the angular momentum commutation relations, hence it is still not a spin operator.

It is convenient to define a Pauli-Lubanski tensor, $W^{\mu\nu} = \frac{1}{i} \left[ W^\mu, W^\nu \right] \\ = - \frac{1}{-g} \left\{ M^{\mu\nu} P^2 - M^{\mu\lambda} P_\lambda P^\nu + M^{\nu \lambda} P_\lambda P^\mu \right\}$

Before we continue our discussion on spin operator, let's see how the Casimir elements help to identify irreducible representations. Casimir elements $P^2$ and $W^2$ belong to a set of mutually commuting operators. $\{ P^2, W^2, P^\mu, W^0 \}$ is one possible set of mutually commuting operators. It is also customary to choose $h = \frac{W^0}{|\vec{P}|} = \hat{P}\cdot \vec{J}$, the helicity, instead of $W^0$. Therefore, particles (irreps.) can be identified by their invariant mass, momentum and spin, helicity, $\left.| \mathscr{M}, p^\mu, s, h \right>$. It is also possible to choose $S_z$ (spin projection in $z$ direction) which we'll defined later to identify the irreps. In fact, the $z$-direction is often chosen along the longitudinal direction $\hat{P}$. In that case, spin projection $S_z$ is the same as the helicity operator $h$.

### "Relativistic Spin Operator" via Lorentz Transformed Pauli-Lubanski Vector

As we have stated in the beginning, it seems reasonable to measure the spin operator by a Lorentz transformation to the particle rest frame. In the literature spin defined in this way is called the "relativistic spin". We shall explore the idea in this section.

Let $\vec{S}_p$ be the spin operator that depends on momentum $p$. Now we know its operator value at $\vec{p} = 0$, $\vec{S}_0 = \vec{J}$. To define its operator value at arbitrary momentum $\vec{p}$,  we require $\left< \psi|\right. \vec{S}_p \left.| \psi \right> = \left<\psi|\right. U(L_p) \vec{S}_0 U(L_p^{-1})\left.|\psi\right>$, where $L_p^{-1}$ is a Lorentz transformation that takes particle with momentum $p$ to particle rest frame: $L_p^{-1} \cdot p = (\mathscr{M},\vec{0})$. But there is a subtle technicality here: Lorentz transformation demands a covariant 4-vector whereas $\vec{J}$ is only a 3-vector. One important observation is that the Pauli-Lubanski vector has the same expectation value as $\mathscr{M}(0, \vec{J})$ at $\vec{p}= 0$. We simply put $\left< \psi|\right. (0, \vec{S}_p) \left.| \psi \right> = \left<\psi|\right. U(L_p) (0, \vec{S}_0) U(L_p^{-1})\left.|\psi\right> = \frac{1}{\mathscr{M}}\left<\psi|\right. U(L_p) W U(L_p^{-1})\left.|\psi\right>$. Therefore, $(0, \vec{S}_p) = \frac{1}{\mathscr{M}} U(L_p) W U(L_p^{-1}) = \frac{1}{\mathscr{M}}L^{-1}_p \cdot W.$
Lorentz transformation $L_p$ is not unique. Let $L_p$ and $L'_p$ be two such Lorentz transformations. $L_p^{-1}L'_p \cdot (\mathscr{M},\vec{0}) = (\mathscr{M},\vec{0})$. So $L = L_p^{-1}L'_p$ belongs to the little group $\mathscr{L} = \{L \in SO(3,1)| L\cdot (\mathscr{M},\vec{0})\}$. Conversely, let $L_p \in SO(3,1)$ be some Lorentz transformation that $L_p \cdot (\mathscr{M},\vec{0}) = p$, $\forall L \in \mathscr{L}$, $L_p L$ also takes $(\mathscr{M},\vec{0})$ to $p$. For the Lorentz group, the little group for massive states $\mathscr{M}^2 > 0$, $\mathscr{L} = SO(3)$ is the 3d rotation group. This rotation is called the (generalized) Melosh-Wigner rotation.

It is convenient to choose $L_p$ to be the standard boost (rotationaless boost): ${L_p}^0_{\;0} = p^0/\mathscr{M} \qquad \qquad \qquad \qquad \\ {L_p}^i_{\;0} = {L_p}^0_{\; i} = p^i/\mathscr{M} \qquad \qquad \quad \\ {L_p}^i_{\; j} = \delta^{ij} + p^i p^j/(\mathscr{M}(p^0+\mathscr{M}))$ To obtain $L^{-1}_p$, one simply replaces $\vec{p}$ with $-\vec{p}$. For the standard boost $L_p$, $\vec{S}_p = \frac{1}{\mathscr{M}}\left( \vec{W} - \vec{p} \frac{W^0}{p^0+\mathscr{M}}\right)$. We note that the zero-component is $\frac{1}{\mathscr{M}}\left( p^0/\mathscr{M} W^0 - p^i/\mathscr{M} W^i \right) = \frac{1}{\mathscr{M}^2} P_\mu W^\mu = 0$, which justifies the notation $(0, \vec{S}_p)$. To extend to the spin operator, we simply promote momentum $p$ to momentum operator $P$. Furthermore, $\vec{S}^2 = - (0, \vec{S})^2 = - \frac{1}{\mathscr{M}^2} ( L^{-1}_p W)^2 = - \frac{1}{\mathscr{M}^2} W^2 = s(s+1)$. It can be checked that $\vec{S}$ indeed satisfies the commutation relations for the spin operator. Therefore, $\vec{S} = \frac{1}{\mathscr{M}}\left( \vec{W} - \vec{P} \frac{W^0}{P^0+\mathscr{M}}\right)$ is a spin operator. In the literature, this spin operator is also called the canonical (relativistic) spin [3], or simply the relativistic spin.

A nice feature of the canonical spin is that its longitudinal component $\hat{P}\cdot \vec{S} = \hat{P}\cdot \vec{J} \equiv h$ is the helicity operator. This is consistent with the non-relativistic quantum mechanics, where $\vec{J} = \vec{X}\times \vec{P} + \vec{S}$ hence $h \equiv \hat{P}\cdot \vec{J} = \hat{P}\cdot\vec{S}$.

As we have stated,  there are however other valid spins resulted from rotations of the canonical spin. One prominent example is the light-cone spin. The light-cone representation of a 4-vector $a = (a^0,\vec{a})$ is defined as $a^\pm = a^0 \pm a^3, a^\perp = (a^1,a^2)$. $a\cdot b = a_+ b^+ + a_- b^- - a^\perp \cdot a^\perp = \frac{1}{2} a^- b^+ + \frac{1}{2} a^+ b^- - a^\perp\cdot b^\perp$. The standard light-cone Lorentz boost (rotationaless boost) is, ${L^{-1}_p}^+_{\;\mu} = \frac{\mathscr{M}}{p^+}\omega_\mu; \qquad \qquad \qquad \qquad \qquad \qquad \\ {L^{-1}_p}^-_{\;\mu} = 2\frac{p_\mu}{\mathscr{M}}-\frac{\mathscr{M}}{p^+}\omega_\mu; \qquad \qquad \qquad \qquad \\ {L^{-1}_p}^i_{\;+} = - \frac{p^i}{p^+}; \quad {L^{-1}_p}^i_{\;-} = 0; \quad {L^{-1}_p}^i_{\; j} = \delta^{ij}$ The corresponding spin operator is $S^+ = \frac{W^+}{P^+}, \quad S^- = -\frac{W^+}{P^+}, \quad \mathbf{S}^\perp = \frac{1}{\mathscr{M}}\left( \mathbf{W}^\perp - \mathbf{P}^\perp \frac{W^+}{P^+} \right)$  $\vec{S}_{LC} = (S^-, \mathbf{S}^\perp) \text{ or } (\mathbf{S}^\perp, S^+)$.

The light-cone spin projection $S^+ = J^3 + \varepsilon^{ij} \frac{B^i P^j}{P^+}$ is kinematic, while $\mathbf{S}^\perp$ is dynamical. It is not difficult to understand this the light-front spin projection, if one note that $M^{\mu\nu} \equiv X^\mu P^\nu - X^\nu P^\mu$. Then the transverse boost $B^i = M^{+i} = X^+ P^i - X^i P^+ = - X^i P^+$ at $x^+ = 0$, the light-front quantization surface. Therefore $X^i = - \frac{B^i}{P^+}$. And hence $S^+ = J^3 - \varepsilon^{ij} X^i P^j$. The second part $X^1 P^2 - X^2 P^1 = L_z$ is an orbital angular momentum in $z$ (or longitudinal) direction. The expression makes perfect sense by stating spin is angular momentum minus orbital angular momentum $S^+ = J^+ - L^+$, where I have replace $z$ direction with longitudinal direction.

For light-cone dynamics, one nice feature of the light-cone spin is that it incorporates the helicity along longitudinal momentum $P^+$, because the longitudinal momentum is easier to access for light-cone dynamics. Another advantages of the light-cone spin is that for massless particles, $W^\mu = s P^\mu$, the light-cone spin gives non-vanishing result $\vec{S}_{LC} = s \hat{z}$. Whereas for the canonical spin, spin vector for massless particles has to be defined separately. The light-cone spin and canonical spin can be related by a Melosh rotation. Inspired by the light-cone spin, the canonical spin may be better termed as the equal-time spin.

Ji and Mitchell have constructed a spin operator within an interpolation angular that gives the equal-time spin and light-cone spin the instant and light-front limit, respectively [7].

### Angular Momentum Decomposition

In non-relativistic quantum mechanics, the angular momentum can be decomposed into an orbital part and a spin part: $\vec{J} = \vec{X}\times\vec{P} + \vec{S}.$ This can be generalized into relativistic dynamics through the angular momentum tensor: $M^{\mu\nu} = L^{\mu\nu} + S^{\mu\nu}.$ with $[S^{\mu\nu}, P^\lambda ] = 0$ where $\mathcal{L}^{\mu\nu} = i (p^\mu \partial_p^\nu - p^\nu \partial^\mu_p)$ is the orbital angular momentum tensor. It may also be defined from some position operator $X^\mu$ by $L^{\mu\nu} = \frac{1}{2}\left\{X^\mu, P^\mu \right\} - \frac{1}{2}\left\{ X^\nu, P^\mu \right\}$. The 3-vector angular momentum is defined as $J^i = \frac{1}{2}\epsilon^{ijk} M^{jk}$. Similarly, the spin operator may be defined as $S^i = \frac{1}{2}\epsilon^{ijk} S^{jk}$. In addition, define a dipole vector $D^i = S^{0i}$. It is easy to see, $\varepsilon^{\mu\nu\kappa\lambda} L_{\nu\kappa}P_\lambda = 0$. Therefore, $W^\mu = -\frac{1}{2}\varepsilon^{\mu\nu\kappa\lambda}S_{\nu\kappa}P_\lambda$. So, $W^0 = \vec{S} \cdot \vec{P}$, $\vec{W} = \vec{D} \times \vec{P} + P^0 \vec{S}$. If $S^{\mu\nu}$ is linear in terms of $W^{\mu\nu}$, the general form of it is $S^{\mu\nu} = \varepsilon^{\mu\nu\kappa\rho} W_\kappa (a P_\rho + b \eta_\rho)$  where $\eta^\rho$ is a constant 4-vector, $a,b$ are scalars. Substitute this back to Pauli-Lubanski vector, we obtain: $a P^2 + b \eta \cdot P = 1$. The corresponding spin vector is $\vec{S} = (a P^0 + b \eta^0) \vec{W} - W^0 (a \vec{P} + b \vec{\eta}) \\ \vec{D} = \vec{W} \times (a \vec{P} + b \vec{\eta})$ The supplementary condition gives additional constraint of the spin tensor $S^{\mu\nu}$. There are three popular SSCs:
1. (Møller): $S^{\mu\nu} \eta_\nu = 0$;
2. (Fokker-Synge-Pryce, Covariant): $S^{\mu\nu} P_\nu = 0$;
3. (Newton-Wigner, Canonical): $\mathscr{M} S^{\mu\nu}\eta_\nu + S^{\mu\nu}P_\nu = 0$;
Each SSC gives a definition of the spin tensor.

1. (Møller): $S^{\mu\nu} = \varepsilon^{\mu\nu\rho\kappa}W_\rho\eta_\kappa /\eta\cdot P$,($a = 0, b = 1/\eta\cdot P$);
2. (Fokker-Synge-Pryce, Covariant): $S^{\mu\nu} = \varepsilon^{\mu\nu\rho\kappa}W_\rho P_\kappa / P^2$, ($a = 1/P^2, b = 0$);
3. (Newton-Wigner, Canonical): $S^{\mu\nu} = \varepsilon^{\mu\nu\rho\kappa}W_\rho\left( \eta_\kappa + P_\kappa/\mathscr{M} \right) /(\mathscr{M} + \eta\cdot P)$,
($a = 1/(\mathscr{M}(\mathscr{M}+P\cdot \eta)), b = 1/(\mathscr{M}+P\cdot \eta)$);
Case 1: $\vec{S}$ is the equal-time spin operator. Then $\vec{S}^2 = -W^2/\mathscr{M}^2$. We conclude that
$a = \frac{1}{\mathscr{M}(P^0 \pm \mathscr{M})}, b\eta^0 = \frac{1}{P^0 \pm \mathscr{M}}, \vec{\eta} = 0$. This corresponds to the Newton-Wigner SSC. And the resultant spin vector is just the canonical spin operator that we have obtained in the previous section.

Case 2: $\vec{S}$ is not the equal-time spin operator, but $\lambda \vec{S} + \mu \vec{D}$ is. Let's evaluate $\mathscr{S}^2 \equiv -\frac{1}{2}S^{\mu\nu}S_{\mu\nu} = \vec{S}^2 + \vec{D}^2$ for Fokker-Synge-Pryce spin tensor, because its a genuine Lorentz scalar. $\mathscr{S}^2 = -W^2/\mathscr{M}^2 = s(s+1)$. Apparently, $\vec{S}^2 \ne s(s+1)$. Let's redefine a spin operator, $\vec{S'} = \vec{S} \pm \vec{D} = \frac{1}{\mathscr{M}^2}\left( P^0 \vec{W} - W^0 \vec{P} \pm \vec{W} \times \vec{P} \right)$ then $\vec{S'}^2 = s(s+1)$. It can be checked that $\vec{S'}$ satisfies the $SO(3)$ Lie algebra.

Case 3: the light-cone spin.

#### Newton-Wigner Position Operator

$\vec{X}_{NW} = -\frac{1}{2}\left\{\vec{K}, \frac{1}{P^0} \right\} - \frac{1}{\mathscr{M}}\frac{\vec{P}\times\vec{W}}{P^0(P^0+\mathscr{M})}$ The nice part of the Newton-Wigner operator is that it satisfies $\left[ X^i_{NW}, P^j \right] = i \delta^{ij}, \quad \left[ X^i_{NW}, X^j_{NW} \right] = 0 \\ \vec{J} = \vec{X}_{NW} \times \vec{P} + \vec{S}$

#### Field Decomposition

Recall the Lorentz transformation of a quantum field is, $(\Lambda \varphi)_a(x) = \sum_{b}D_{ab}(\Lambda) \varphi_b(\Lambda^{-1} x) \\ \Rightarrow \qquad [ \varphi_a(x), M^{\mu\nu} ] = -i\left( x^\mu \partial^\nu - x^\nu \partial^\mu \right) \varphi_a(x) + \mathcal{S}_{ab}^{\mu\nu} \cdot \varphi_b(x)$ where $( \Lambda \varphi)_a (x) \equiv U(\Lambda^{-1}) \varphi(x)_a U(\Lambda), U(\Lambda) = e^{-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}}, D(\Lambda) = e^{-\frac{i}{2}\omega_{\mu\nu}\mathcal{S}^{\mu\nu}}$. It's easy to recognize that $D(\Lambda)$ is a finite-dimensional representation of the Lorentz group. According to Noether theorem, the conserved current is, $\vec{J} = \int \mathrm{d}^3x \bar\psi \gamma^0 \left( \vec{r}\times (-i\nabla) + \frac{1}{2}\vec{\Sigma} \right) \psi$It seems natural to write $\vec{J} = \vec{X} \times \vec{P} + \frac{1}{2}\vec{\Sigma}$ or $M^{\mu\nu} = \mathcal{L}^{\mu\nu} + \mathcal{S}^{\mu\nu}$, namely, $S^{\mu\nu} = \mathcal{S}^{\mu\nu}$. But this is not strictly correct. Spin tensor $S^{\mu\nu}$ is an Hermitian operator in Hilbert space whereas $\mathcal{S}^{\mu\nu}$ is only a finite-dimensional linear operator. In fact, as we have stated in the beginning, $\mathcal{S}^{\mu\nu}$ cannot be Hermitian. It will be utterly wrong to identify $\mathcal{S}^{\mu\nu}$ as the spin tensor, although we'll see below, $\mathcal{S}^{\mu\nu}$ does tell us some information about the spin. We should note that most of the trouble is caused by the boosts. As spin is closely related to the rotation property, the finite-dimensional $\mathcal{S}^{\mu\nu}$ does tell us a great deal of information.

#### Finite-Dimensional Representations

The finite-dimensional irreducible representations of the Lorentz group are identified with Casimir elements of the complex Lie algebra $N^i_\pm = \frac{1}{2}(J^i \pm i K^i), \quad i = 1,2,3 \\ \left[N^i_\pm, N^j_\pm \right] = i \epsilon^{ijk} N^k_\pm, \left[N^i_+, N^j_-\right] = 0$ The Casimirs are $N_+^2$ and $N_-^2$. For irreducible representations, $N_\pm^2 = n_\pm (n_\pm+1)$, $n_\pm = 0, \frac{1}{2}, 1, \frac{3}{2}, \cdots$. We'll use $(n_+, n_-)$ to identify each irrep. The dimension of this irrep is $(2 n_++1) (2n_-+1)$.

The simplest non-trivial irreps are the 2d spinor representations $(\frac{1}{2}, 0)$ and $(0, \frac{1}{2})$, known as the left-handed and right-handed Weyl spinors, respectively. Note that they are NOT field representations. For spinor representations, $M^{\mu\nu}_L = \frac{i}{4}\left( \sigma^\mu \bar\sigma^\nu - \sigma^\nu \bar\sigma^\mu \right) \\ M^{\mu\nu}_R = -\frac{i}{4}\left( \bar\sigma^\mu \sigma^\nu - \bar\sigma^\nu \sigma^\mu \right) \\$ where $\sigma^\mu = (I, \vec\sigma), \bar\sigma^\mu = (I, -\vec\sigma)$. Since Weyl spinors are not in field representation, all of angular momentum is intrinsic. They obviously have total spin 1/2. The spin operator is $\vec{S} = \vec{J} = \frac{1}{2} \vec{\sigma}$.

The reducible 4d representation $(\frac{1}{2}, 0)\otimes(0,\frac{1}{2})$ is called the Dirac spinor. Dirac spinor also has spin 1/2. It is worth mentioning the angular momentum (hence spin) is written in the infamous $\gamma$-matrices $M^{\mu\nu} = \frac{i}{4}\left[ \gamma^\mu, \gamma^\nu \right]$ The spin operator is $\vec{S} = \frac{1}{2} \begin{pmatrix} \vec{\sigma} & \\ & \vec{\sigma} \\ \end{pmatrix} \equiv \frac{1}{2} \vec{\Sigma}$.

The irreducible 4d representation $(\frac{1}{2}, \frac{1}{2})$ is the vector representation. The matrix elements of the group generators are $( M^{\mu\nu} )_{\alpha\beta} = -i \left( \delta^\mu_{\;\alpha} \delta^{\nu}_{\;\beta} - \delta^{\nu}_{\;\alpha} \delta^\mu_{\;\beta} \right).$ The spin operator is $\left[S^i \right]_{jk} = -i \epsilon^{ijk}$.

#### Infinite-Dimensional Representations

The Poincaré group is represented by unitary operator $U(\Lambda,a)$: $U(\Lambda, a) \left.| p, \sigma \right> = e^{-i p\cdot a} \sum_{\sigma'} C_{\sigma,\sigma'}(\Lambda, p) \left.|\Lambda\cdot p, \sigma' \right>$ where $\left.|p,\sigma\right>$ is shortcut for $\left.|\mathscr{M}, p, s, \sigma \right>$. In Wigner classification, the representation of the Lorentz group is $U(\Lambda, a) \left.| p, \sigma \right> = \left( \frac{N_p}{N_{\Lambda\cdot p}} \right) e^{-i p\cdot a} \sum_{\sigma'} D^{(s)}_{\sigma,\sigma'}(W(\Lambda, p)) \left.|\Lambda\cdot p, \sigma'\right>$ where $W(\Lambda, p) = L^{-1}(\Lambda \cdot p) \Lambda L(p)$ is the Wigner rotation with respect to the standard vector $k = (\mathscr{M}, \vec{0})$, $W\cdot k = k$, the group element of the little group  $w_k$ of vector $k$. $D(W)$ is the representation of the little group. In the massive particle case, $w_k$ is the 3d rotation group $SO(3)$. $L(p)\cdot k = p$ is some standard Lorentz transformation. $N_p$ is a normalization factor that can be chosen to be $N_p = 1$.

The infinitesimal transformation $U(1 + \delta \omega) \simeq 1 - \frac{i}{2} \delta_{\mu\nu} M^{\mu\nu}$. Therefore, $M^{\mu\nu} = i(p^\mu \partial_p^\nu - p^\nu \partial_p^\mu) + S^{\mu\nu} \\ S^{\mu\nu} = L(p)^\mu_{\;\alpha} L(p)^\nu_{\; \beta} \sigma^{\alpha\beta} - \frac{1}{2}\left( p^\mu (\partial_p^\nu L^{-1}(p))^\alpha_{\;\kappa} L(p)^\kappa_\beta - p^\nu (\partial_p^\mu L^{-1}(p))^\alpha_{\;\kappa} L(p)^\kappa_\beta \right) \sigma_\alpha^{\;\beta}$ where $\sigma^{\mu\nu}$ is the generator of the 3d rational group $SO(3)$ in representation labeled by total spin $s$. Here $M^{\mu\nu}$ has been decomposed into two parts, with $\mathcal{L}^{\mu\nu} \equiv i(p^\mu \partial_p^\nu - p^\nu \partial_p^\mu)$ apparently being the orbital angular momentum. The spin operator depends on momentum $p^\mu$ of the particle as well as on the choice of the Lorentz transformation $L(p)$ that $L(p) \cdot k = p$. That means there is not unique number of definitions of the spin operator.

### Dirac Theory - the "Relativistic Wave Equation Theory"

Finally, we have to address the Dirac Theory. I first want to quote Schwinger and Weinberg to remind the readers:

"The picture of an infinite sea of negative energy electrons is now best regarded as a historical curiosity and forgotten." - Julia Schwinger

"... quantum field theory is the way it is because it is the only way to reconcile the principles of quantum mechanics with those of special relativity." - Steven Weinberg

Nevertheless, the success of Dirac theory on describing the spin-1/2 relativistic single-particle state is worth touring the theory and finding the corresponding spin operators.

Recall that Dirac spinor is a finite-dimensional representation. It has not space-time dependence. It can be used to well describe an electron with some standard momentum $p_s$. The Dirac spinor has a spin $\vec{S} = \vec{J} = \frac{1}{2}\vec{\Sigma}$. To obtain the electron wavefunction and operators with other momentum $p^\mu$, one can simply do a Lorentz boost $L(p_s, p) \cdot p_s = p$ to a frame that the electron has a momentum $p$. It is customary to choose $p_s = (\mathscr{M}, \vec{0})$. Similar to what we have analysed above, there are infinite many definition of the spin vector corresponding to different choice of the Lorentz boost $L(p_s, p)$. We first note that the generators of the Lorentz group are $\vec{J} = \frac{1}{2} \vec{\Sigma} = \begin{pmatrix} \vec{\sigma} & \\ & \vec{\sigma} \\ \end{pmatrix} \\ \vec{K} = \frac{i}{2}\gamma^0 \vec{\gamma} = \frac{i}{2}\vec{\alpha} = -\frac{i}{2}\begin{pmatrix} \vec{\sigma} & \\ & -\vec{\sigma} \\ \end{pmatrix}$
Then the canonical spin is, $\vec{S} = \frac{1}{2\mathscr{M}}\left( P^0 \vec{\Sigma} + i \gamma^0 \vec{\gamma} \times \vec{P} - \vec{P} \frac{\vec{P}\cdot \vec\Sigma}{P^0+\mathscr{M}} \right)$

#### Foldy-Wouthuysen Spin Operator

In the Foldy-Wouthuysen representation the positive modes and negative modes of the Dirac theory decouples. The transformation is very useful for obtaining the relativistic correction of a non-relativistic theory. In free theory, the transformation that carries the Dirac theory to the Foldy-Wouthuysen representation is $F(p) = \exp\left[ -i \vec\gamma \cdot \hat p \theta \right] \\ = \cos \theta + \vec{\gamma} \cdot \hat p \sin \theta$ where $\theta = \frac{1}{2}\arctan \frac{|\vec p|}{\mathscr{M}}$.

The Hamiltonian in the FW representation becomes, $H_D = \gamma^0 \vec\gamma \cdot \vec p + \gamma^0 m \to H_{FW} = \gamma^0 p^0$
In the massless limit, FW representation becomes the chiral representation.

The FW spin is defined as the inverse-transformed spinor spin $F(p) \vec{S}_{FW} F^{-1}(p) = \frac{1}{2}\vec{\Sigma}$. Then, $\vec{S}_{FW} = \frac{1}{2 P^0}\left( \mathscr{M} \vec\Sigma - i \gamma^0 \vec{\gamma} \times \vec{P} + \vec{P} \frac{\vec{P}\cdot \vec\Sigma}{P^0+\mathscr{M}} \right)$

### references:

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