Does Wightman's unitary $U(\Lambda)$ really exist for Lorentz boost?

Question

This question is related to another question here. But I am asking a more fundamental question about the existence of Wightman's unitary $U(\Lambda)$ for Lorentz transformation.

Let $\psi^\alpha$ be a quantum spinor field transforming in the $(\frac{1}{2},0)$ irrep of $SL(2,\mathbb{C})$. That is, under a boost $\Lambda$, with unitary representative $U(\Lambda)$, the field transforms as:

\begin{equation} U(\Lambda)^\dagger \psi^\alpha(x) U(\Lambda) = S(\Lambda)^\alpha_{\space\space\beta} \psi^\beta(\Lambda^{-1}x) \tag{1}\end{equation}

with $S(\Lambda)\in SL(2,\mathbb{C})$.

And the unitary $U(\Lambda)$ also satisfies $$ U(\Lambda)|\Omega \rangle = |\Omega \rangle \tag{2} $$ for all Lorentz boost $\Lambda$, where $|\Omega \rangle $ is the vacuum.

My question is: given that $S(\Lambda)$ for Lorentz boost is NOT unitary, is there really a unitary $U(\Lambda)$ that satisfies both eq. (1) and eq. (2) under Lorentz boost?

The existence of unitary $U(\Lambda)$ is crucial for Wightman axioms. So please DON'T just quote the the Wightman axioms as your answer. And for that matter, please DON'T just quote relativity or Lorentz invariance as your answer. Lorentz invariance implies the existence of $S(\Lambda)$ (known as the non-unitary $K(\Lambda)$ for Lorentz boost), but not necessarily $U(\Lambda)$. Instead, please give a proof of the existence of unitary $U(\Lambda)$.

The proof does not have to be super rigorous, one concrete example for spinor field under Lorentz boost would suffice. And yes, it has to be specifically for spinor field, with details of $U(\Lambda)$ worked out, rather than a general procedural guidance of how to get $U(\Lambda)$. Note that I am NOT asking for a proof for complex/real scalar field.

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Added note:

The references given by various users are pertaining to the scalar field. The whole situation made me doubtful of the proof of unitarity for the spinor field. That is why I am asking the question.

Answer 1

A QFT is said to be relativistic invariant if it realizes a unitary representation of the Lorentz group. Not all QFTs are relativistic, so there is no general proof. Instead, this is a definition: if you have a unitary representation, then your theory is relativistic.

So, how do we construct relativistic theories? The so-called canonical approach is arguably the most useful tool in doing this. The claim is that the QFT is relativistic if you begin with a hermitian and Lorentz invariant Lagrangian. What follows is a sketch of this claim, although the "full proof" requires a whole textbook (cf.ref1).

In the canonical approach to QFT the symmetries under Lorentz transformations give rise to the Noether current (cf. ref1 §7.4) $$ M_{\mu\nu\rho}\sim\frac{\partial\mathcal L}{\partial \phi_{,\mu}}\delta_{\nu\rho}\phi $$ where $\phi$ denotes the fields of your theory and $\delta_{\nu\rho}$ denotes a Lorentz variation in the $\nu\rho$ direction.

The generator of Lorentz transformations is, then $$ J_{\mu\nu}=\int M_{\mu\nu0}\mathrm d\boldsymbol x $$ which is conserved thanks Noether's theorem.

Finally, the operator that implements Lorentz transformations is, by definition, $$ U(\Lambda)=\exp\big(\frac i2J_{\mu\nu}\omega^{\mu\nu}\big) $$ where $\Lambda=e^{\frac i2 \mathcal J_{\mu\nu}\omega^{\mu\nu}}$, with $(\mathcal J_{\mu\nu})_{\rho\sigma}\sim\eta_{\mu[\rho}\eta_{\sigma]\nu}$ the generators of the Lorentz group in the fundamental representation.

Unitarity of $U$ follows from hermiticity of $J$, which in turns follows from hermiticity of $\mathcal L$.

Whether $U(\Lambda)$ leaves $\Omega$ invariant or not is a question of whether the Lorentz symmetry is spontaneously broken or not. As a matter of principle this can happen, so again, there is no proof. We are again in a situation where things become definitions: if $\Omega$ is invariant, then we say that Lorentz is unbroken.

Usually, when constructing $M_{\mu\nu\rho}$ there are ordering ambiguities which can be exploited in order to make $\Omega$ invariant. For example, the Hamiltonian is defined up to a constant, and you can fix it by declaring that $\Omega$ has zero energy. If you can do this for all Lorentz generators then the symmetry is unbroken. If you cannot, then it is broken.

References.

1. Weinberg S. - Quantum theory of fields, Vol.1. Foundations.

Answer 2

Induced Representation

After half a year of contemplation, I finally admit that my previous answer was incorrect. Here is a new one. For a short answer, you can jump to the last section.

Here is a brief introduction to the theory of induced representation, which is crucial to the understanding of the unitary representation of the Poincaré group. Everything here can be found in the book

Theory of Group Representations and Applications

——A.O. Barut

If you are not interested in the mathematical details, please jump to the next section.

First of all, suppose we have a topological group $G$. Then, we have the following theorem by George Mackey:

Theorem 1: Let $G$ be a separable, and locally compact topological group, and $K$ be its subgroup. Then, there exists a Borel set $S\subset G$ such that $\forall g\in G$ can be uniquely represented as the product $$g=s_{g}k_{g}^{-1},$$

where $k_{g}\in K$ and $s_{g}\in S$.

Next, suppose $G$ is a Lie group. Let $K$ be a Lie subgroup of $G$, and $\mathscr{H}$ be a separable Hilbert space. Let $\tau:K\rightarrow U(\mathscr{H})$ be a unitary representation of $K$ in $\mathscr{H}$. We are interested in the coset space $X=G/K$. Suppose $X=\left\{gK|g\in G\right\}\equiv\left\{x_g|g\in G\right\}$ has a $G$-invariant measure $\mu$, and consider the set $\Phi$ of all functions $u: G\rightarrow\mathscr{H}$ satisfying the following conditions:

1. The inner-product $(u(g),v)$ is measurable wrt $dg$ for all $v\in\mathscr{H}$.
2. $u(gk^{-1})=\tau(k)u(g)$, for all $k\in K$ and all $g\in G$.
3. $\int_{X}\|u(g)\|_{\ast}^{2}d\mu(x_{g})<\infty$, where $\|\cdot\|_{\ast}=\underset{\|v\|\leq 1}{\sup}|(\cdot,v)|$ is the dual norm induced by the norm $\|\cdot\|$ in the Hilbert space $\mathscr{H}$.

Remark: By condition $2.$, one has $\|u(gk^{-1})\|_{\ast}=\|\tau(k)u(g)\|_{\ast}=\|u(g)\|_{\ast}$, hence the integrand is defined on the coset space $X$.

Then, one can prove the following theorems:

Theorem 2: The space $\Phi$ is isomorphic to the Hilbert space $L(X,d\mu,\mathscr{H})$ of ($\mathscr{H}$-valued) square integrable functions on $X$, with the isomorphism given by $$u(g)=\tau(k_{g})\tilde{u}(x_{g}),$$

where $k_{g}$ is the factor of $g\in G$ in the Mackey decomposition $g=k_{g}s_{g}$, and $\tilde{u}(x_{g})\in L(X,d\mu,\mathscr{H})$. The map $\tilde{u}(x_{g})\rightarrow u(g)$ defines an isometry from $L^{2}(X,\mu,\mathscr{H})$ into $\Phi$.

Theorem 3: Let $g$, $h\in G$. The map $g\rightarrow U(g)$ given by $$U(g)u(h)\equiv\sqrt{\rho_{g}(x_{h})}u(g^{-1}h), \tag{$\star$}$$

where $\rho_{g}(x_{h})=d\mu(x_{h}\cdot g)/d\mu(x_{h})$ is the Radon-Nikodym deritivate of the $G$-invariant measure $\mu$ on $X$, defines a unitary representatin of $G$ in $\Phi$, which is known as Mackey's induced representation of $G$ by the unitary represetation $\tau: K\rightarrow U(\mathscr{H})$.

The Lorentz Group

The Poincaré group has the semi-direct product strucuture $\mathcal{P}=\mathbb{R}^{4}\rtimes\mathcal{L}$, where $\mathcal{L}=O(3,1)$ is the Lorentz group. The Lorentz group has four connected components: \begin{align} \mathcal{L}_{+}^{\uparrow}&=\left\{\Lambda\in\mathcal{L}|\Lambda^{0}_{\,\,0}\geq 1,\det\Lambda=+1\right\} \\ \mathcal{L}_{-}^{\uparrow}&=\left\{\Lambda\in\mathcal{L}|\Lambda^{0}_{\,\,0}\geq 1,\det\Lambda=-1\right\}\equiv P\mathcal{L}_{+}^{\uparrow} \\ \mathcal{L}_{-}^{\downarrow}&=\left\{\Lambda\in\mathcal{L}|\Lambda^{0}_{\,\,0}\leq 1,\det\Lambda=-1\right\}\equiv T\mathcal{L}_{+}^{\uparrow} \\ \mathcal{L}_{+}^{\downarrow}&=\left\{\Lambda\in\mathcal{L}|\Lambda^{0}_{\,\,0}\leq 1,\det\Lambda=+1\right\}\equiv PT\mathcal{L}_{+}^{\uparrow}. \end{align}

Other than rotations $O(3)$, it contains Lie subgroups $\mathcal{L}_{+}=\mathcal{L}_{+}^{\uparrow}\cup\mathcal{L}_{+}^{\downarrow}=SO(3,1)$, and $\mathcal{L}^{\uparrow}=\mathcal{L}_{+}^{\uparrow}\cup\mathcal{L}_{-}^{\uparrow}$.

From quantum mechanics we have learnt that the projective unitary representation of $\mathcal{P}^{\uparrow}_{+}=\mathbb{R}^{4}\rtimes\mathcal{L}^{\uparrow}_{+}$ is in one-to-one correspondence with the ordinary unitary representation of the universal covering group $\widetilde{\mathcal{P}}^{\uparrow}_{+}\simeq\mathbb{R}^{4}\ltimes SL(2,\mathbb{C})$. For any vector $x\in\mathbb{R}^{4}$, we have a Hermitian $2\times 2$ matrix $\sigma(x)$ given by $$\sigma(x)=x^{\mu}\sigma_{\mu}=\begin{pmatrix} x^{0}+x^{3} & x^{1}-ix^{2}\\ x^{1}+ix^{2} & x^{0}-x^{3}\\ \end{pmatrix},$$

where $\left\{\sigma_{\mu}\right\}\equiv(𝟙_{2\times 2},\vec{\sigma})$ are the Pauli matrices, and we use the convention $\eta=\mathrm{diag}(1,-1,-1,-1)$. Up to an overall minus sign, a Lorentz transformation is determined by a matrix $A\in SL(2,\mathbb{C})$. We denote the image of $A$ under the (two-fold) canonical projection $\Pi: SL(2,\mathbb{C})\rightarrow\mathcal{L}^{\uparrow}_{+}$ by $\Lambda_{A}$. Then, the Lorentz transformation $\Lambda_{A}x$ of the four-vector $x$ is represented by $$\sigma(\Lambda_{A}x)\equiv\sigma(\Pi(A)x)=A\sigma(x)A^{\dagger}. \tag{1.a}$$

Conversely, from the given $A\in SL(2,\mathbb{C})$, the Lorentz transformation is determined by $$(\Lambda_{A}x)^{\nu}=\frac{1}{2}\sum_{\mu=0}^{3}\mathrm{Tr}(A\sigma_{\mu}A^{\dagger}\sigma_{\nu})x^{\mu}.$$

Similarly, one can associate each $x\in\mathbb{R}^{4}$ with a Hermitian matrix $$\bar{\sigma}(x)=\sum_{\mu=0}^{3}x_{\mu}\sigma_{\mu}=\begin{pmatrix} x^{0}-x^{3} & -x^{1}+ix^{2}\\ -x^{1}-ix^{2} & x^{0}+x^{3}\\ \end{pmatrix}, $$

which is related with $\sigma(x)$ via space reflection $P$. Again, for a given $B\in SL(2,\mathbb{C})$, $$\bar{\sigma}(\Lambda_{B}x)\equiv B\bar{\sigma}(x)B^{\dagger} \tag{1.b}$$

defines a Lorentz transformation. But since $\bar{\sigma}(x)$ is related with $\sigma(x)$ via the equation $\bar{\sigma}(x)=\sigma_{2}\sigma(x)^{\ast}\sigma_{2}$, one finds that $A$ and $B$ represents two inequivalent representations of $SL(2,\mathbb{C})$. They are related via the relation $$B=(A^{\dagger})^{-1}.$$

This implies that it is impossible to include the representation of space reflection in $\mathbb{C}^{2}$. As a result, one is motivated to combine $A$ and $(A^{\dagger})^{-1}$ into a $4\times 4$ matrix $$L_{A}=\begin{pmatrix} A & 0\\ 0 & (A^{\dagger})^{-1}\\ \end{pmatrix}$$

in $\mathbb{C}^{4}$. Then, the space reflection, aka the automorphism $A\rightarrow(A^{\dagger})^{-1}$, can be represented by the matrix $$L_{P}=\begin{pmatrix} 0 & 𝟙_{2\times 2}\\ 𝟙_{2\times 2} & 0\\ \end{pmatrix},$$

and one has $L_{P}L_{A}L_{P}^{-1}=L_{(A^{\dagger})^{-1}}=(L_{A}^{\dagger})^{-1}$. Then, we found the universal overing group $\widetilde{\mathcal{L}}^{\uparrow}=\left\{L_{A},L_{P}L_{A}|A\in SL(2,\mathbb{C})\right\}$, which acts on $\mathbb{C}^{4}$ irreducibly. One can generalize the Pauli matrices and define the $4\times 4$ Gamma matrices, for each $x\in\mathbb{R}^{4}$, $$\gamma(x)\equiv\begin{pmatrix} 0 & \sigma(x) \\ \bar{\sigma}(x) & 0\\ \end{pmatrix}.$$

Then, the Lorentz transformation and space reflection are represented by \begin{align} L_{A}\gamma(x)L_{A}^{-1}&=\begin{pmatrix} 0 & A\sigma(x)A^{\dagger} \\ (A^{\dagger})^{-1}\bar{\sigma}(x)A^{-1} & 0\\ \end{pmatrix}=\gamma(\Lambda_{A}x), \tag{1.c} \\ L_{P}\gamma(x)L_{P}^{-1}&=\begin{pmatrix} 0 & \bar{\sigma}(x) \\ \sigma(x) & 0\\ \end{pmatrix}=\gamma(P\cdot x) \tag{1.d} . \end{align}

By expressing the $4\times 4$ matrices as $\gamma(x)=\gamma_{\mu}x^{\mu}$, it's easy to verify that the canonical basis $\gamma_{\mu}$ satisfy the Clifford algebra $$\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}𝟙_{4\times 4},$$

and each Lorentz transformation can be represented via Gamma matrices by the formula $$(\Lambda_{A})^{\mu}_{\,\,\nu}=\frac{1}{4}\mathrm{Tr}\left[L_{A}^{-1}\gamma^{\mu}L_{A}\gamma_{\nu}\right].$$

For each $SL(2,\mathbb{C})$ matrix $A$, it has a unique polar decomposition, $$A=HU,$$ where $H=\sqrt{A^{\dagger}A}$ is Hermitian and corresponds to a pure Lorentz boost, and $U=(\sqrt{A^{\dagger}A})^{-1}A$ is unitary and corresponds to a ratation. Suppose we start from a standard momentum $\pi$, and apply a Lorentz transformation $\Lambda_{p}$ that transforms it into $p$, i.e. $\Lambda_{p}\pi=p$. Then, up to an overall sign, there's a matrix $l(p)\in SL(2,\mathbb{C})$ such that $\Pi(l(p))=\Lambda_{p}$. To achieve this transformation, we consider the following two scenarios:

1. $\pi$ is timelike: the standard momentum is $\pi=(m,0,0,0)^{T}$. For any timelike $p$, there's a Lorentz boost $$\Lambda_{p}=\frac{1}{m}\begin{pmatrix} p^{0} & (\vec{p})^{T} \\ \vec{p} & m𝟙_{3\times 3}+\frac{p^{0}-m}{(\vec{p})^{2}}\vec{p}\otimes(\vec{p})^{T} \\ \end{pmatrix}\in\mathcal{L}^{\uparrow}_{+} \tag{1.e}$$

that transforms $\pi$ into $p$, and so the corresponding $l(p)\in SL(2,\mathbb{C})$ must be Hermitian. Following (1.a), we can solve this $2\times 2$ Hermitian matrix $l(p)$ from the equation $l(p)\sigma(\pi)l(p)^{\dagger}=\sigma(p)$. One can easily check that the solution is $$l(p)=\frac{m𝟙_{2\times 2}+\sigma(p)}{\sqrt{2m(m+p^{0})}}\in SL(2,\mathbb{C}), \tag{1.f}$$

which is known as the Foldy-Wouthuysen transformation. Similarly, following (1.c) there's a $4\times 4$ Hermitian matrix $L(p)$ such that $\gamma(p)=L(p)\gamma(\pi)L(p)^{-1}$. Then, using the fact that for any $L\in\widetilde{\mathcal{L}}^{\uparrow}$, $L\gamma^{0}=\gamma^{0}(L^{\dagger})^{-1}$, one has $$L(p)=\frac{m𝟙_{4\times 4}+\gamma(p)\gamma^{0}}{\sqrt{2m(m+p^{0})}}, \tag{1.g}$$

which incorperates the Lorentz boost $l(p)$ and parity $L_{P}$.

2. $\pi$ is lightlike: the standard momentum is $\pi=(1/2,0,0,1/2)^{T}$. Then, the Lorentz transformation $\Lambda_{p}$ is achieved by a rotation $R(p)$ which rotates $\vec{\pi}$ into the $\vec{p}$ direction, followed by a pure Lorentz boost $B(p)$. i.e. $$\Lambda_{p}=R(p)B(p).$$ I don't bother to find out the specific expressions of $l(p)$ and $L(p)$ in this case. The only diffence is that in the lightlike case, they take the form of a product $U(p)H(p)$, where $U(p)$ is unitary and $H(p)$ is Hermitian.

Wigner's Classification

Our goal is to find a unitary representation $\mathbf{U}(a,A)$ of $(a,A)\in\mathbb{R}^{4}\ltimes SL(2,\mathbb{C})$. Since the Poincaré transformation should satisfy the multiplication rule $$(a,\Lambda_{A})=(a,𝟙)(0,\Lambda_{A})=(0,\Lambda_{A})(\Lambda_{A}^{-1}a,𝟙),$$

the corresponding unitary operators should satisfy $$\mathbf{U}(a,A)=\mathbf{U}(a,𝟙)\mathbf{U}(0,A)=\mathbf{U}(0,A)\mathbf{U}(\Lambda_{A}^{-1}a,𝟙). \tag{2}$$

First of all, if we restrict to the subgroup $(\mathbb{R}^{4},𝟙)$ of spacetime translations generated by four momenta $\mathbf{P}^{\mu}$, we obtain (according to the SNAG theorem) a unitary representation $$\mathbf{U}(a)=e^{i\mathbf{P}\cdot a},$$

where $a\in\mathbb{R}^{4}$. We introduce the following (infinite-dimensional) basis $$\mathbf{P}^{\mu}|p,\alpha\rangle=p^{\mu}|p,\alpha\rangle,$$

with the Lorentz covariant normalization condition $$\langle q,\alpha|p,\beta\rangle=2p^{0}\delta_{\alpha\beta}\delta(\vec{p}-\vec{q})$$

where $\alpha$ is some degeneracy parameter to be determined, and $|p^{0}|=\sqrt{m^{2}+\vec{p}^{2}}$.

For any given momentum $p$ that can be related to the given standard momentum $\pi$ via a Lorentz transformation $\Lambda_{p}$, we consider the state $\mathbf{U}(0,l(p))|\pi,\alpha\rangle$. Applying equation (2) one finds \begin{equation} \mathbf{U}(a,𝟙)\mathbf{U}(0,l(p))|\pi,\alpha\rangle=\mathbf{U}(0,l(p))\mathbf{U}(\Lambda_{p}^{-1}a,𝟙)|\pi,\alpha\rangle \\ =\exp(i\Lambda_{p}^{-1}a\cdot\pi)\mathbf{U}(0,l(p))|\pi,\alpha\rangle=\exp(ip\cdot a)\mathbf{U}(0,l(p))|\pi,\alpha\rangle. \end{equation}

Thus, the state $\mathbf{U}(0,l(p))|\pi,\alpha\rangle$ has momentum $p$. This indicates that, up to a complex phase factor, one can write $$|p,\alpha\rangle=\mathbf{U}(0,l(p))|\pi,\alpha\rangle.$$

For any $A\in SL(2,\mathbb{C})$, it can be written as $$A=l(\Lambda_{A}p)\left[l(\Lambda_{A}p)^{-1}Al(p)\right]l(p)^{-1}\equiv l(\Lambda_{A}p)\mathcal{W}(A,p)l(p)^{-1}.$$

It is easy to check that $\mathcal{W}(A,p)$ is an element of the stabilizer of $\sigma(\pi)$. It is known as Wigner's little group. Using this decomposition, one has, for an arbitrary one-particle state $|p,\alpha\rangle$, \begin{align} \mathbf{U}(0,A)|p,\alpha\rangle&=\mathbf{U}(0,l(\Lambda_{A}p))\mathbf{U}(0,l(\Lambda_{A}p)^{-1}A\,l(p))\mathbf{U}(0,l(p)^{-1})|p,\alpha\rangle \\ &=\mathbf{U}(0,l(\Lambda_{A}p))\mathbf{U}(0,l(\Lambda_{A}p)^{-1}A\,l(p))|\pi,\alpha\rangle \\ &=\mathbf{U}(0,l(\Lambda_{A}p))\mathbf{U}(0,\mathcal{W}(A,p))|\pi,\alpha\rangle. \tag{3} \end{align}

To apply the theory of induced representation on Poincaré group, we focus on the following four Lorentz invariant orbits (which should be identified as the coset spaces of $\mathcal{P}^{\uparrow}_{+}$ modulo the stabilizer subgroup of the given standard momentum):

1. $\mathcal{H}^{\pm}_{m}=\left\{p\in\mathbb{R}^{4}|p^{2}=m^{2},m>0\right\}$ for massive particles, where $p^{0}=\pm\sqrt{m^{2}+\vec{p}^{2}}$ gives two hyperboloids.
2. $\mathcal{C}^{\pm}_{0}=\left\{p\in\mathbb{R}^{4}|p^{2}=0\right\}$ for massless particles, where $p^{0}=\pm\sqrt{\vec{p}^{2}}$ gives two lightcones.

For the massive case, the stabilizer of the standard four-momentum $(m,0,0,0)^{T}\in\mathcal{H}^{\pm}_{m}$ is $O(3)$ (the actions of spacetime translations are trivial). Correspondingly, the stabilizer of the Hermitian matrix $\sigma(\pi)$ under the action (1.a) and (1.b) is $\mathrm{Pin}(3)\equiv\left\{L_{U},L_{P}L_{U}|U\in SU(2)\right\}$. The unitary irreducible representation of the rotation group is well-known in quantum mechanics: $$\mathbf{U}(0,U[R])|\pi,\alpha\rangle=\sum_{\beta}D^{(0,s)}_{\beta\alpha}(U[R])|\pi,\beta\rangle, \tag{4.a}$$

where $R\in SO(3)$, $U[R]\in SU(2)$ is the twofold covering of $SO(3)$, and $D^{(0,s)}_{\beta\alpha}$ is the matrix element of the unitary irreducible representation of $SU(2)$ in $\mathbb{C}^{2s+1}$. In a similar manner, one has the complex conjugate representation $D^{(s,0)}_{\dot{\beta}\dot{\alpha}}$ of $SU(2)$ under space reflction: $$\mathbf{U}(0,U[R])|\pi,\dot{\alpha}\rangle=\sum_{\dot{\beta}}D^{(s,0)}_{\dot{\beta}\dot{\alpha}}(U[R])|\pi,\dot{\beta}\rangle. \tag{4.b}$$

Now, apply the above calculations to equation (3), one has \begin{align} U(0,A)|p,\alpha\rangle &=\sum_{\beta}D^{(0,s)}_{\beta\alpha}(\mathcal{W}(A,p))|\Lambda_{A}p,\beta\rangle, \tag{5.a} \\ U(0,A)|p,\dot{\alpha}\rangle &=\sum_{\dot{\beta}}D^{(s,0)}_{\dot{\beta}\dot{\alpha}}(\mathcal{W}(A,p))|\Lambda_{A}p,\dot{\beta}\rangle. \tag{5.b} \end{align}

This means that for a massive particle, the degeneracy parameter is completely determined by the spin: $s=0$, $\frac{1}{2}$, $1$, ...

For the massless case, it turns out that the stabilizer subgroup is isomorphic to the twofold covering of the Euclidean group in two dimensions. For more details, please check my answer here.

Quantum Mechanical Wave Functions

Consider the wave package for a massive particle $$|\psi\rangle=\sum_{\alpha}\int_{\mathcal{H}^{\pm}_{m}}\frac{d^{3}\vec{p}}{(2\pi)^{3}2p^{0}}f_{\alpha}(p)|p,\alpha\rangle.$$

We multiply both sides of the above equation with the unitary operator $\mathbf{U}(a,A)$, then \begin{align} \mathbf{U}(a,A)|\psi\rangle&=\sum_{\alpha,\beta}\int_{\mathcal{H}^{\pm}_{m}}\frac{d^{3}\vec{p}}{(2\pi)^{3}2p^{0}}f_{\alpha}(p)\exp(i\Lambda_{A}\,p\cdot a)D^{(0,s)}_{\beta\alpha}(\mathcal{W}(A,p))|\Lambda_{A}p,\beta\rangle \\ &=\sum_{\alpha,\beta}\int_{\mathcal{H}^{\pm}_{m}}\frac{d^{3}\vec{p}}{(2\pi)^{3}2p^{0}}f_{\beta}(\Lambda_{A}^{-1}p)\exp(ip\cdot a)D^{(0,s)}_{\alpha\beta}(\mathcal{W}(A,\Lambda_{A}^{-1}p))|p,\alpha\rangle, \end{align}

where in the last line we have used the Lorentz invariance of the measure on $\mathcal{H}^{\pm}_{m}$. From the above equation, we can define the unitary operator \begin{align} \mathbf{U}^{(0,s)}(a,A)\cdot f_{\alpha}(p)&\equiv e^{ip\cdot a}\sum_{\beta}D^{(0,s)}_{\alpha\beta}(\mathcal{W}(A,\Lambda_{A}^{-1}p))f_{\beta}(\Lambda_{A}^{-1}p), \tag{6.a} \\ \mathbf{U}^{(s,0)}(a,A)\cdot f_{\dot{\alpha}}(p)&\equiv e^{ip\cdot a}\sum_{\dot{\beta}}D^{(s,0)}_{\dot{\alpha}\dot{\beta}}(\mathcal{W}(A,\Lambda_{A}^{-1}p))f_{\dot{\beta}}(\Lambda_{A}^{-1}p) \tag{6.b}, \end{align}

with the inner-product $$(f,g)=\sum_{\alpha}\int_{\mathcal{H}^{\pm}_{m}}\frac{d^{3}\vec{p}}{(2\pi)^{3}2p^{0}}f_{\alpha}^{\ast}(p)g_{\alpha}(p)\quad\mathrm{and}\quad(f,g)=\sum_{\dot{\alpha}}\int_{\mathcal{H}^{\pm}_{m}}\frac{d^{3}\vec{p}}{(2\pi)^{3}2p^{0}}f_{\dot{\alpha}}^{\ast}(p)g_{\dot{\alpha}}(p)$$

in the Hilbert-space $L^{2}(\mathcal{H}^{\pm}_{m},d^{3}\vec{p}/2p^{0},\mathbb{C}^{2s+1})$.

Covariant States and Classical Fields

In standard QFT textbooks, instead of constructing a unitary representation of $SL(2,\mathbb{C})$, the starting point is to consider a classical field which transforms under the Lorentz group in a non-unitary representation. It is natural to ask the question how these two representations are related with each other. To solve this puzzle, we consider a finite dimensional representation of $SL(2,\mathbb{C})$ such that its restriction on $\mathcal{W}(A,p)$ is unitary. The theory of induced representation tells us that if the restriction of $D$ on $\mathcal{W}(A,p)$ is unitary, then the induced representation of $SL(2,\mathbb{C})$ is also unitary. For convenience, we still denote this extension of $D$ on $SL(2,\mathbb{C})$ by $D$. We can define the so-called "spinor basis" (also known as the covariant states): $$\zeta(p)=D(l(p))f(p) \tag{7.1}.$$

Then, it's easy to show that the unitary operator $\mathbf{U}(a,A)$ acting on $f(p)$ leads to a (non-unitary) operator $\mathbf{T}(a,A)$ acting on $\zeta(p)$ in the following way: $$\mathbf{T}(a,A)\cdot\zeta(p)=e^{ip\cdot a}D(A)\zeta(\Lambda_{A}^{-1}p), \tag{7.2} $$

where each component of $\zeta(p)$ trivially satisfies the Klein-Gordon equation $$(p^{2}-m^{2})\zeta(p)=0.$$

For convenience, we denote the Hilbert space $L^{2}(\mathcal{H}^{\pm}_{m},d^{3}\vec{p}/2p^{0},\mathbb{C}^{2s+1})$ by $\mathfrak{H}_{m,s}^{\pm}$, and denote the corresponding space of covariant states in (7.1) by $\mathfrak{C}_{m,s}^{\pm}$. Then, equation (7.1) shows that there is a homomorphism: $$\mathfrak{H}_{m,s}^{\pm}\stackrel{\iota}\longrightarrow\mathfrak{C}_{m,s}^{\pm}.$$ Together with equation (7.2), they imply the following commutative diagram: $$\require{AMScd} \begin{CD} \mathfrak{H}_{m,s}^{\pm} @>{\mathbf{U}}>> \mathfrak{H}_{m,s}^{\pm}\\ @V{\iota}VV @V{\iota}VV \\ \mathfrak{C}_{m,s}^{\pm} @>{\mathbf{T}}>> \mathfrak{C}_{m,s}^{\pm} \end{CD} \tag{7.3}$$

The above equations show that the covariant states are the Fourier modes of classical fields. In general, $\mathbf{T}(a,A)$ cannot be unitary, except when the field is in a trivial representation (i.e. a scalar). Again, we have two possibilities:

1. For the massive case: suppose we have a particle with spin $j$, then its covariant state transforms in the following manner: $$\mathbf{T}(a,A)\cdot\zeta(p)=e^{ip\cdot a}D^{(0,j)}(A)\zeta(\Lambda_{A}^{-1}p).$$

If it also admits the parity operator, then we must start from a reducible representation (by Schur's lemma) $D(A)=D^{(0,j)}\oplus D^{(j,0)}(A)$ instead. However, we now would have twice as many components of a classical field of a particle with spin $j$. The condition that removes redundant component is known as wave equations in QFT. Here is an example of the spin-$1/2$ case:

Dirac Equation: we consider the representation $D^{(1/2,0)}\oplus D^{(0,1/2)}(A)$. The covariant states transform as $$\mathbf{T}(a,A)\cdot\zeta(p)=e^{ip\cdot a}D^{(1/2,0)}\oplus D^{(0,1/2)}(A)\zeta(\Lambda_{A}^{-1}p),$$

where $D^{(1/2,0)}\oplus D^{(0,1/2)}(A)$ is given by $L_{A}$. The projection operators that remove the redundant components are $$\mathrm{P}_{R}=\frac{1}{2}(𝟙_{4\times 4}+\beta),\quad\mathrm{and}\quad \mathrm{P}_{L}=\frac{1}{2}(𝟙_{4\times 4}-\beta),$$

where $\beta=L_{P}$. For $f(p)\in\mathrm{P}_{R,L}\mathfrak{H}_{m,s}^{\pm}$, one has $f(p)=\mathrm{P}_{R,L}f(p)$, because $\mathrm{P}_{R,L}$ is a projection operator. It follows that \begin{align} \zeta(p)&=D^{(1/2,0)}\oplus D^{(0,1/2)}(l(p))f(p)=L(p)f(p)=L(p)\mathrm{P}_{R,L}f(p) \\ &=\left[L(p)\mathrm{P}_{R,L}L(p)^{-1}\right]L(p)f(p)=\left[L(p)\mathrm{P}_{R,L}L(p)^{-1}\right]\zeta(p). \tag{8.1} \end{align}

Therefore, $\mathfrak{C}_{m,s}^{\pm}$ also splits into two invariant subspaces: $$\mathfrak{C}_{m,s}^{\pm}=\mathrm{P}_{L}\mathfrak{C}_{m,s}^{\pm}\oplus\mathrm{P}_{R}\mathfrak{C}_{m,s}^{\pm}.$$

A small calculation shows $$L(p)\mathrm{P}_{R,L}L(p)^{-1}=\frac{1}{2}\left(1\pm L(p)\beta L(p)^{-1}\right)=\frac{1}{2m}\left(m𝟙_{4\times 4}\pm\gamma(p)\right).$$

Then, equation (8.1) implies that $$\left(\gamma^{\mu}p_{\mu}-m\right)\zeta(p)=0, \tag{8.2}$$

for $\zeta\in\mathfrak{C}^{+}_{+m,s}$, and $\zeta\in\mathfrak{C}^{-}_{-m,s}$.

2. For the massless case: we can derive the Fourier modes of Maxwell equations in vacuum. Please look at here for more details

Hermitian Generators

We have shown the relation between the relativistic quantum mechanical wave functions and the Fourier modes of classical fields. Here, we use the Dirac field as an example and give an exact expression of the Hermitian generator of the Lorentz group.

Starting from equation (8.2), we define $u\oplus v\in\mathfrak{C}^{+}_{+m,s}\oplus\mathfrak{C}^{-}_{-m,s}$, and $$\Psi(x)=\int_{\mathcal{H}^{+}_{m}\,\cup\mathcal{H}^{-}_{m}}\frac{d^{3}\vec{p}}{(2\pi)^{3}}\frac{m}{p^{0}}\sum_{\alpha=1}^{2}\left\{e^{-ip\cdot x}b_{\alpha}(p)u^{\alpha}(p)+e^{+ip\cdot x}d^{\ast}_{\alpha}(p)v^{\alpha}(p)\right\}, \tag{9.1}$$

where $d_{\alpha}$ and $b_{\alpha}$ are two complex Grassmann numbers.

To make sure the Langrangian density is real-valued, we define the symmetrized Lagrangian density $$\mathcal{L}_{\mathrm{S}}=\frac{1}{2}(\mathcal{L}+\mathcal{L}^{\dagger})=\frac{1}{2}\bar{\Psi}(i\overset{\leftrightarrow}{\partial}\!\!\!\!\!/-2m)\Psi.$$

The canonical momentum is given by $$\Pi_{\mathrm{S}}=\mathcal{L}_{\mathrm{S}}\frac{\overset{\leftarrow}{\delta}}{\delta\dot{\Psi}}=\frac{i}{2}\Psi^{\dagger}.$$

By virtue of (7.3) and (6.a) and (6.b), we can read off the Hermitian generators of Wigner's rotation $\mathcal{W}(A,\Lambda_{A}^{-1}p)=l(p)^{-1}A\,l(\Lambda_{A}^{-1}p)$ and work out those of the classical field (9.1).

First of all, let's consider an infinitesimal rotation $R(\vec{\theta})$: \begin{align} &\vec{x}\rightarrow\vec{x}+\vec{x}\times\vec{\theta}, \\ &x^{0}\rightarrow x^{0}. \end{align}

The corresponding infinitesimal Wigner's rotation is \begin{align} &\vec{x}\rightarrow\vec{x}+\vec{x}\times\vec{\theta}, \\ &x^{0}\rightarrow x^{0}. \end{align}

Accordingly, the quantum mechanical wave function transforms as \begin{align} f_{\alpha}(p)\rightarrow&\sum_{\beta}\left(𝟙+i\vec{\sigma}\cdot\theta\right)_{\alpha\beta}f_{\beta}(p^{0},\vec{p}-\vec{p}\times\vec{\theta}) \\ &=\sum_{\beta}\left[𝟙+i\theta\cdot\left(-i\vec{p}\times\frac{\partial}{\partial\vec{p}}+\vec{\sigma}\right)\right]_{\alpha\beta}f_{\beta}(p). \tag{9.2.a} \end{align}

Next, we consider a pure Lorentz boost $B(\vec{\omega})$: \begin{align} &x^{0}\rightarrow x^{0}-\vec{\omega}\cdot\vec{x}, \\ &\vec{x}\rightarrow\vec{x}-\omega x^{0}. \end{align}

The corresponding infinitesimal Wigner's rotation is \begin{align} &x^{0}\rightarrow x^{0}, \\ &\vec{x}\rightarrow\vec{x}+\vec{x}\times\frac{\vec{p}\times\vec{\omega}}{m+p^{0}}. \end{align}

Accordingly, the quantum mechanical wave function transforms as \begin{align} f_{\alpha}(p)\rightarrow&\sum_{\beta}\left(𝟙+i\vec{\sigma}\cdot\frac{\vec{p}\times\vec{\omega}}{p^{0}+m}\right)_{\alpha\beta}f_{\beta}(p^{0},\vec{p}+p^{0}\vec{\omega}) \\ &=\sum_{\beta}\left[𝟙+ip^{0}\vec{\omega}\cdot\left(-i\frac{\partial}{\partial\vec{p}}-\frac{\vec{p}\times\vec{\sigma}}{p^{0}(m+p^{0})}\right)\right]_{\alpha\beta}f_{\beta}(p) \tag{9.2.b}. \end{align}

From (9.2) we can read off the following Hermitian generators: $$\vec{\mathscr{J}}(p)=\frac{1}{i}\left(\vec{p}\times\frac{\partial}{\partial\vec{p}}\right)+\vec{\sigma}\quad\quad\vec{\mathscr{K}}(p)=-p^{0}\left(\frac{1}{i}\frac{\partial}{\partial\vec{p}}-\frac{\vec{p}\times\vec{\sigma}}{p^{0}(m+p^{0})}\right).$$

For convenience, we denote $\vec{\mathscr{J}}$ by $\Sigma_{ij}$, and denote $\vec{\mathscr{K}}$ by $\Sigma_{0i}$. According to [2], commmutators among $\Sigma_{\mu\nu}$ satisfy the Lorentz algebra.

Then, we have the following Noether charge: $$Q_{\mu\nu}(t)=-2i\int d^{3}\mathbf{x}\Psi^{\dagger}(t,\mathbf{x})\Sigma_{\mu\nu}\Psi(t,\mathbf{x})=\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\Sigma_{\mu\nu}\Psi(t,\mathbf{x}).$$

On the phase space, we introduce the $\mathbb{Z}_{2}$-graded Poisson-super bracket $$\left\{F(t),G(t)\right\}_{PB}=\int ds\int d^{3}\mathbf{x}F(t)\left(\frac{\overset{\leftarrow}{\delta}}{\delta\Psi(s,\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\Pi(s,\mathbf{x})}+\frac{\overset{\leftarrow}{\delta}}{\delta\Pi(s,\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\Psi(s,\mathbf{x})}\right)G(t).$$

It satisfies the commuting property $$\left\{F,G\right\}_{PB}=-(-1)^{\epsilon(F)\epsilon(G)}\left\{G,F\right\}_{PB},$$

where $\epsilon$ is the parity of the Grassmann variable. Then, it's easy to verify the Lorentz algebra $$\left\{Q_{\mu\nu},Q_{\rho\sigma}\right\}_{PB}=i(g_{\sigma\mu}Q_{\rho\nu}+g_{\nu\sigma}Q_{\mu\rho}-g_{\rho\mu}Q_{\sigma\nu}-g_{\nu\rho}Q_{\mu\sigma}).$$

Since the above Poisson bracket is anti-symmetric, in the canonical quantization, it is replaced by commutator $$\left[Q_{\mu\nu},Q_{\rho\sigma}\right]_{-}=i(g_{\sigma\mu}Q_{\rho\nu}+g_{\nu\sigma}Q_{\mu\rho}-g_{\rho\mu}Q_{\sigma\nu}-g_{\nu\rho}Q_{\mu\sigma}).$$

References:

1. Realizations of the Unitary Representations of the Inhomogeneous Space-time Groups I, II——U. H. Niederer,L. O'Raifeartaigh

2. Unitary Representations of the Poincare Group and Relativistic Wave Equations——Y. Ohnuki

3. Theory of Group Representations and Applications——A.O. Barut

4. The Dirac Equation——Bernd Thaller