Unlike rotations, the boost transformations are non-unitary. Therefore, the boost generators are not Hermitian. When boosts induce transformations in the Hilbert space, will those transformation be unitary? I think no. If that is the case, what is the physical significance of such non-unitary transformations corresponding to boosts in the Hilbert space?

  • 1
    $\begingroup$ DEar @Roopam. I have been following your probing questions with interest. As Luboš's answer shows in the details for your specific, this question is answered by recalling that the image of a group under a particular representation is in general different from the group itself. As you know from your other question, the image of $O(1,3)$ under the adjoint representation ${\rm Ad}$ is $O(1,3)$ itself. But this is not true of a general homomorphism. Nor indeed is it generally true even of ${\rm Ad}$ itself. You marked the question "representation-theory" so it sounds as though you are ... $\endgroup$ Commented Feb 15, 2014 at 1:33
  • $\begingroup$ ... learning get your head around these ideas. You just need to keep on learning about representations and their properties. When something undergoes a spacetime transformation by a member of $O(1,3)$, the corresponding quantum state must undergo a unitary transformation (the object must end up in some state, after all!). WHen you are ready, see the Woit reference linked in my answer here or also this excellent reference by the inimitable John Baez arxiv.org/abs/0904.1556 was also a great help to me. $\endgroup$ Commented Feb 15, 2014 at 1:41
  • 1
    $\begingroup$ ... Also the Stone-von Neumann theorem, Wigner's theorem and Wigner's classification are important concepts you are walking into. $\endgroup$ Commented Feb 15, 2014 at 1:43

2 Answers 2


On the actual Hilbert space of a consistent relativistic quantum mechanical system, the Lorentz transformations including boosts actually are unitary – which also means that the generators $J_{0i}$ are as Hermitian as the generators of rotations $J_{ij}$.

We say that the Hilbert space forms a unitary representation of the Lorentz group.

What the OP must be confused by is the fact that the ordinary vector representation composed of vectors $(t,x,y,z)$ is not a unitary representation of $SO(3,1)$. The $SO(3,1)$ transformations don't preserve any positively definite quadratic invariant constructed out of the coordinates $(t,x,y,z)$. After all, we know that an indefinite form, $t^2-x^2-y^2-z^2$, is conserved by the Lorentz transformations. So on a representation like the vector space of such $(t,x,y,z)$, the generators $J_{0i}$ would end up being anti-Hermitian rather than Hermitian.

But if you take a Lorentz-invariant theory with a positive definite Hilbert space, like QED, the formula for $J_{0i}$ makes it manifest that it is a Hermitian operator, which means that $\langle \psi |\psi\rangle$ is preserved by the Lorentz boosts! The complex probability amplitudes for different states $c_i$ behave differently than the coordinates $t,x,y,z$ above.

Note that the (non-trivial) unitary transformations of $SO(3,1)$ are inevitably infinite-dimensional. Finite-dimensional reps may be constructed out of the fundamental vector representation above and they are as non-unitary as the vector representation. But that's not true for infinite-dimensional reps. For example, the space of one-scalar-particle states in a QFT is a unitary representation of the Lorentz group. For each $p^\mu$ obeying $p^\mu p_\mu=m^2$, and there are infinitely (continuously) many values of such a vector (on the mass shell), the representation contains one basis vector (which are normalized to the Dirac delta function). The boosts just "permute them" along the mass shell which makes it obvious that the positively definite form is preserved when normalized properly.

  • $\begingroup$ Which formula do you mean by the formula for $J_{0i}$? I am reading weinberg and in his exposition (pg59) he just assumes the $J_{\mu \nu}$ are hermitian so that U is unitary. I'd be curious if it can be formulated a different way. $\endgroup$ Commented Feb 8, 2019 at 11:48
  • $\begingroup$ Dear @user27084, $J_{0i}$ is schematically $x_0 p_i - x_i p_0$ but in field theory, $p_\mu$ has to be written as an integral of the energy density or momentum density over space. Let me assume that you know how to write the total momentum or the total energy as an integral involving all the fields. So to get the boost generator, you just write down the same integrals that define $p_0$ or $p_i$ and add the $x_0$ or $x_i$ extra factor inside the integral - to get both terms in the difference that I started with. $\endgroup$ Commented Feb 9, 2019 at 19:02
  • $\begingroup$ To be more precise, the formula that you got so far is just the orbital part of $J_{0i}$, the $L_{0i}$, if you wish. To get the full generator of the boost, you also need to add the terms responsible for the spin, the intrinsic angular momentum, of all the fields. All such extra terms are simple bilinear expressions in the field whose index structure is fully determined by the Lorentz covariance. About 1/2 of the QFT books contain explicit formulae for the form of $J_{\mu\nu}$ for any field theory they discuss. $\endgroup$ Commented Feb 9, 2019 at 19:04
  • $\begingroup$ Note that all the terms are some integrals and the integrands contain factors such as $x_0$, $x_i$, energy density, or momentum density. And then the spin. If you look at all these factors, there is never any imaginary unit etc. which makes it almost manifest that all the generators, regardless of the indices' being spatial or temporal, are Hermitian. $\endgroup$ Commented Feb 9, 2019 at 19:06
  • $\begingroup$ Hi professor. Why the canonical quantization of the Noether charges associated with Lorentz transformation of the photon is not Hermitian? Classically, these charges generate the infinitesimal Lorentz transformations under the Dirac bracket. So I expect under canonical quantization they should generate the unitary representation of the Lorentz transformation. But they are non-Hermitian. $\endgroup$
    – Valac
    Commented Jul 1, 2022 at 18:57

Chapter I: The Lorentz Group

First of all, 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 $L_{P}$.

  1. $\pi$ 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.

Chapter II: 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 $\widetilde{O}(3)\equiv\left\{L_{U},L_{P}L_{U}|U\in SU(2)\right\}$. 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, a small calculation shows that members in its stabilizer group take the form $$\begin{pmatrix} e^{i\theta} & z \\ 0 & e^{-i\theta} \\ \end{pmatrix},\quad\mathrm{where}\,\,\,\theta\in[0,2\pi),\,\,\,\mathrm{and}\,\,\,z\in\mathbb{C}.$$

By the isomorphism $\theta\rightarrow\varphi/2$, $z\rightarrow ze^{-i\varphi/2}$, $\varphi\in[0,4\pi)$, one finds that the stabilizer group has a semi-direct product structure $$(z_{1},\varphi_{1})\cdot(z_{2},\varphi_{2})=(z_{1}+z_{2}e^{i\varphi_{1}},\varphi_{1}+\varphi_{2}).$$

In other words, the stabilizer group is isomorphic to the twofold covering group of the Euclidean group in two dimensions, i.e $\widetilde{\mathbb{E}}(2)=\mathbb{R}^{2}\rtimes\mathrm{Spin}(2)$. Its elements take the form $$\epsilon(z,\varphi)=\begin{pmatrix} e^{i\varphi/2} & ze^{-i\varphi/2}\\ 0 & e^{-i\varphi/2}\\ \end{pmatrix}\in SL(2,\mathbb{C}).$$

To find its unitary irreducible representations, first notice that the real Lie algebra $\mathfrak{sl}(2,\mathbb{C})_{\mathbb{R}}$ is generated by $\left\{\vec{J},\vec{K}\right\}\equiv\left\{\sigma_{1},\sigma_{2},\sigma_{3},i\sigma_{1},i\sigma_{2},i\sigma_{3}\right\}$, with the commutation relations $$[J_{a},J_{b}]=i\epsilon_{abc}J_{c},\quad[J_{a},K_{b}]=i\epsilon_{abc}K_{c},\quad[K_{a},K_{b}]=-i\epsilon_{abc}J_{c}.$$

Clearly, the two dimensionsal subgroup $\mathbb{R}^{2}$ of translations is generated by $T_{1}=J_{1}-K_{2}$ and $T_{2}=J_{2}+K_{1}$. The generator of $\mathrm{Spin}(2)$ is $J_{3}$. We have the unitary irreducible representation $$J_{3}|\lambda\rangle=\lambda|\lambda\rangle,\quad\mathrm{and}\quad \mathbf{U}^{\lambda}(\varphi)|\lambda\rangle=e^{-i\lambda\varphi}|\lambda\rangle,$$

with the identification $e^{-i4\pi\lambda}=1$. This implies that the degeneracy parameter (helicity) takes values $\lambda=0,\pm\frac{1}{2},\pm 1\cdots$.

In general, we can consider the simultaneous eigenstates of $\mathbb{R}^{2}$ and $J_{3}$, $$T^{2}|t,\lambda\rangle=t^{2}|t,\lambda\rangle,\quad J_{3}|t,\lambda\rangle=\lambda|t,\lambda\rangle,\quad\mathrm{where}\,\,\,t^{2}\geq0,\,\,\,\lambda=0,\pm\frac{1}{2},\pm1,\cdots.$$

For the $t^{2}>0$ case, it is clear that the representation is infinite dimensonal which corresponds to infinite degrees of freedom. Thus, we demand that the translation subgroup acts trivially in the Hilbert space. Since the two lightlike vectors $\pi=(1/2,0,0,1/2)^{T}$ and $p=(|\vec{p}|,\vec{p})^{T}$ are related via a Lorentz transformation, by following a similar procedure as in the massive case, we obtain, for any $A\in SL(2,\mathbb{C})$, $$\mathbf{U}(0,A)|p,\lambda\rangle=e^{i\lambda\varphi(A,\Lambda_{A}^{-1}\,p)}|\Lambda_{A}p,\lambda\rangle. \tag{5.c}$$

Notice that the representations $\varphi\rightarrow e^{i\lambda\varphi}$ and $\varphi\rightarrow e^{-i\lambda\varphi}$ are related by a complex conjugation. If the theory has space inversion symmetry, then such a particle and its antiparticle are deemed to be the same.

Chapter III: 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 $L^{2}(\mathcal{H}^{\pm}_{m},d^{3}\vec{p}/2p^{0},\mathbb{C}^{2s+1})$.

Similarly, in the massless case, we consider the wave-package $$|\psi\rangle=\sum_{\lambda}\int_{\mathcal{C}^{\pm}_{0}}\frac{d^{3}\vec{p}}{(2\pi)^{3}2p^{0}}f_{\lambda}(p)|p,\lambda\rangle.$$

Following similar steps, one define the unitary operator $$\mathbf{U}^{\lambda}(a,A)\cdot f_{\lambda}(p)=e^{ip\cdot a}e^{i\lambda\varphi(A,\Lambda_{A}^{-1}\,p)}f_{\lambda}(\Lambda_{A}^{-1}p),\tag{6.c} $$

with the inner product $$(f,g)=\sum_{\lambda}\int_{\mathcal{C}^{\pm}_{0}}\frac{d^{3}\vec{p}}{(2\pi)^{3}2p^{0}}f_{\lambda}^{\ast}(p)g_{\lambda}(p)$$

defined in the Hilbert space $L^{2}(\mathcal{C}^{\pm}_{0},d^{3}\vec{p}/2p^{0},\mathbb{C})$.

Chapter IV: 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 are two examples:

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 $L_{A}$, which is introduced in chapter I. 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 $\zeta\in\mathfrak{C}^{-}_{-m,s}$.

Proca Equations: we consider the representation $D^{\frac{1}{2}}\otimes D^{\frac{1}{2}}(A)=D^{1}\oplus D^{0}(A)$. In this case, we have a spin-$1$ particle and a scalar. The projection onto the three-vector is thus $$\mathrm{P}=\begin{pmatrix} 0 & 0\\ 0 & 𝟙_{3\times 3}\\ \end{pmatrix}.$$

i.e. $\mathrm{P}_{\mu\nu}=\frac{1}{2}\left(\delta_{\mu\nu}-g_{\mu\nu}\right)$. The expression of $D^{1}\oplus D^{0}(A)$ is given by (1.e). We have $$\left(\Lambda_{p}\mathrm{P}\Lambda_{p}^{-1}\right)_{\mu\nu}=\left(\Lambda_{p}\mathrm{P}\Lambda_{-p}\right)_{\mu\nu}=\frac{1}{2}\left(\frac{p_{\mu}p_{\nu}}{m^{2}}-g_{\mu\nu}\right).$$

Hence, we obtain $$\frac{1}{2}\left(\frac{p_{\mu}p_{\nu}}{m^{2}}-g_{\mu\nu}\right)\zeta^{\mu}(p)=\zeta_{\nu}(p).$$

Multiplying both sides by $p^{\nu}$, we have $$p^{\nu}\zeta_{\nu}(p)=0,\quad\mathrm{and}\quad (p^{2}-m^{2})\zeta_{\nu}(p)=0.$$

Then, by introducing $\Phi_{\mu\nu}=p_{\mu}\zeta_{\nu}-p_{\nu}\zeta_{\mu}$, we find $$p^{\mu}\Phi_{\mu\nu}-m^{2}\zeta_{\nu}=0, \tag{8.3}$$

which is the usual form of Proca equations.

  1. For the massless case: we denote the generators of $\mathfrak{su}(2)_{\mathbb{C}}\oplus\mathfrak{su}(2)_{\mathbb{C}}$ by $$\vec{N}=\frac{1}{2}(\vec{J}+i\vec{K})\quad\mathrm{and}\quad\vec{M}=\frac{1}{2}(\vec{J}-i\vec{K}),$$

with the ladder operators $N_{\pm}=N_{1}\pm iN_{2}$ and $M_{\pm}=M_{1}\pm iM_{2}$. We have shown that in order to have a finite degrees of freedom, the two dimensional translation group generated by $T_{1}=J_{1}-K_{2}$ and $T_{2}=J_{2}+K_{1}$ must acts trivially. This means $$N_{+}\zeta(\pi)=0,\quad\mathrm{and}\quad M_{-}\zeta(\pi)=0,$$

where $\pi$ is the standard momentum of lightlike particles. By definition, the covariant state $\zeta(p)$ is also an eigenstate of $J_{3}$. Therefore, we can write $$N_{3}\zeta(\pi)=n\zeta(\pi),\quad M_{3}\zeta(p)=-m\zeta(\pi),$$

and so the helicity is determined by $$J_{3}\zeta(\pi)=\lambda\zeta(\pi)=(n-m)\zeta(\pi). \tag{8.4}$$

The above equations imply that the covariant state $\zeta(\pi)$ is the highest weight of $\vec{N}$, and is the lowest weight of $\vec{M}$. From equation (8.4), we have $$\pi_{3}J_{3}\zeta(\pi)=|\vec{\pi}|\lambda\zeta(\pi)=\vec{\pi}\cdot\vec{J}\zeta(\pi).$$

Then, using the fact that $\frac{\vec{J}\cdot\vec{p}}{|\vec{p}|}$ is Lorentz invariant for any lightlike $p$, one finds $$\frac{\vec{J}\cdot\vec{p}}{|\vec{p}|}\zeta(p)=\lambda\zeta(p)=(n-m)\zeta(p). \tag{8.5}$$

The above equation is the most generic form of the EOM of a lightlike particle. It also implies that when parity is defined, it necessarily doubles the dimension of the space of covariant states, and so massless particles have two states of polarizations. Here is an example:

Maxwell Equation: we consider the representation $(D^{(1,0)}\oplus D^{(0,1)})(A)$. The space of covariant states splits into two parts: $\zeta=\xi\oplus\eta$. The representations $D^{(1,0)}$ and $D^{(0,1)}$ are three dimensional, and the corresponding $\vec{J}^{(1,0)}=\vec{J}^{(0,1)}$ are in the adjoint representation: $$J_{1}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \\ \end{pmatrix}\quad J_{2}=\begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ 0 & -i & 0 \\ \end{pmatrix}\quad J_{3}=\begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}.$$

i.e. $i(J^{(0,1)}_{a})_{bc}=i(J^{(1,0)}_{a})_{bc}=\epsilon_{abc}$. Thus, equation (8.5) reduces to $$\vec{p}\times\vec{\xi}(p)=-i\omega\vec{\xi}(p),\quad\vec{p}\times\vec{\eta}(p)=i\omega\vec{\eta}(p),$$

where $\omega=|\vec{p}|$. Introducing the field strength $\vec{\xi}=\vec{B}+i\vec{E}$ and $\vec{\eta}=\vec{B}-i\vec{E}$, we obtain \begin{equation} \vec{p}\times\vec{E}=\omega\vec{B},\quad\vec{p}\cdot\vec{B}=0, \\ \vec{p}\cdot\vec{E}=0,\quad\vec{p}\times\vec{B}=-\omega\vec{E}, \end{equation}

which are the Fourier modes of Maxwell equations in vacuum.

Chapter V: 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. 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 $\mathscr{L}_{ij}$, and denote $\vec{\mathscr{K}}$ by $\mathscr{L}_{0i}$. Then, we have the following Noether charge: $$Q_{\mu\nu}(t)=\int d^{3}\mathbf{x}\Psi^{\dagger}(t,\mathbf{x})\hat{\mathscr{L}}_{\mu\nu}\Psi(t,\mathbf{x})=-i\int d^{3}\mathbf{x}\Pi(t,\mathbf{x})\hat{\mathscr{L}}_{\mu\nu}\Psi(t,\mathbf{x}).$$

where $\hat{\mathscr{L}}_{\mu\nu}$ is the operator in configuration space, and $\Pi(t,\mathbf{x})=i\Psi(t,\mathbf{x})^{\dagger}$ is the the canonical momentum.

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 $$\left\{Q_{\mu\nu},Q_{\rho\sigma}\right\}_{PB}=i(g_{\sigma\mu}Q_{\rho\nu}+g_{\nu\sigma}Q_{\mu\rho}-g_{\rho\mu}Q_{\sigma\nu}-g_{\nu\rho}Q_{\mu\sigma}).$$

  • $\begingroup$ References: ----------- 1. Realizations of the Unitary Representations of the Inhomogeneous Space-time Groups I, II——U. H. Niederer,L. O'Raifeartaigh 2. Theory of Group Representations and Applications——A.O. Barut 3. The Dirac Equation——Bernd Thaller 4. Unitary Representations of the Poincare Group and Relativistic Wave Equations——Y. Ohnuki $\endgroup$
    – Valac
    Commented Nov 27, 2022 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.