I am currently working through the method of induced representations in order to calculate the unitary irreducible representations of the Poincare group.
Conventions/Notation
Metric signature $= (-,+,+,+)$
$c=\hbar=1$
Indices i,j,k run over 1,2,3 whilst any other latin indice runs over 0,1,2,3.
$x^a=(t, \vec{x})$
Context
In this thread I will consider the (positive definite) non zero mass case. I will take the standard momentum to be $k^a=(m,0,0,0)$ where m is the square root of the eigenvalue of the Casimir operator $C_1=-P^aP_a$. The corresponding little group is $H_{k}=SO(3)$.
The second casimir operator is the square of the Pauli-Lubanski vector; $C_2=W^aW_a$. For the standard momentum, $W^a=(0,mJ^i)$ where $J^i$ is the ith generator of rotations (i.e. around the ith spatial axis). Thus, $C_2=m^2\vec{J}\cdot\vec{J}$. From QM we know the eigen values of $J^2$ are $s(s+1)$. So by Schur's lemma, on any irrep of the poincare group, $C_2=m^2s(s+1)\text{Id}$. Id is the identity operator… \mathbb{1} won't work.
Thus, if I fix a value for m (and hence fix the mass shell surface) then the unitary irrep's of the Poincare group (for this m) are classified by the 'spin', s.
In this case, the method of induced reps can be expressed as:
$$U(\Lambda)|p,\sigma\rangle=\Sigma_{\sigma'}\mathcal{D}^{(s)}(h(\Lambda,p))_{\sigma'\sigma}|\Lambda p,\sigma'\rangle$$
Where $h(\Lambda,p)=L^{-1}(\Lambda p)\cdot\Lambda\cdot L(p) \in H_k$ and $L(p)$ is the standard Lorentz transformation which takes the standard momentum to p, i.e. $L(p)^a_{~~b}k^b=p^a$. Also, $\mathcal{D}^{(s)}(h(\Lambda,\sigma))$ is the unitary irrep of the little group element $h(\Lambda,\sigma)$ in the spin s representation.
So to work out how the homogeneous Lorentz transformation $U(\Lambda)$ acts on the state $|p,\sigma\rangle$, we need to work out the unitary irreps of the little group.
By convention I will take the standard Lorentz transformation, $L(p)$, to be
$$L(p)^a_{~~~b}=\begin{pmatrix} E_p/m &&p^j/m \\ p^i/m &&\delta^{ij}+\frac{p^ip^j}{m(E_p+m)} \end{pmatrix},~~~~~~~~~~~i,j=1,2,3.$$
Question/Attempt to solve
Unfortunately in my non relativistic QM class during undergrad we never talked about the spin representations of the rotation group $SO(3)$, so I am trying to bridge that gap now. I am a little confused on how we go about constructing the unitary irreps of $SO(3)$, in particular I am not so sure how to compute the matrix elements $\mathcal{D}^{(s)}(h(\Lambda,\sigma))_{\sigma'\sigma}$. This is my question, and the following is what I have so far.
In trying to translate between what I learned in undergrad QM and the above, I have come to the following conclusions:
- For the massive case (and for the standard momentum so that we are in the rest frame), the $\sigma$'s (which are supposed to represent any degrees of freedom other then 4-momentum) correspond to the 'magnetic quantum number', $m_j$.
- For a given spin s, $\sigma=m_j = \{-s,-s+1,..,s-1,s\}$. Here I am taking $j=l+s=s$ since we are in the rest frame and there is no 'orbital angular momentum' ($l=0$). Hence for a given mass, m, the dimension of the irrep of the little group is $2s+1$.
Define the 3-vector $S^i=\frac{1}{m}W^i=\frac{1}{2}\varepsilon_{ijk}J^{jk} \implies \vec{S}\cdot\vec{S}=s(s+1)\text{Id}$ which obeys the following algebra:
$$[S_i,S_j]=i\varepsilon_{ijk}S_k$$
Thus $[S_i,S^2]=[\frac{1}{2}\varepsilon_{ijk}J^{jk},\frac{1}{m^2}C_2]=0$ (since $C_2$ is a Casimir operator of ISO$^\uparrow (3,1)$). Hence we can simultaneously diagonalize $S^2$ and one of $S_i$, say, $S_3$.
In the usual way we can work out that:
$$S^2|s,\sigma\rangle=s(s+1)|s,\sigma\rangle~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$
$$S_3|s,\sigma\rangle=\sigma|s,\sigma\rangle ~~~~~~~~~~~~~~~~(\text{where } \sigma=m_s)~~~(2)$$
And define raising/lowering operators:
$$S_{\pm}=S_1\pm iS_2$$
$$\implies S_{\pm}|s,\sigma\rangle=\sqrt{s(s+1)-\sigma(\sigma\pm 1)}~|s,\sigma\pm 1\rangle~~~~(3)$$
According to Weinberg, for an infinitesimal rotation, $R_{ik}=\delta_{ik}+\Theta_{ik}$, we have that $$\mathcal{D}^{(s)}(1+\Theta)_{\sigma'\sigma}=\delta_{\sigma'\sigma}+\frac{i}{2}\Theta_{ik}(J^{(s)~ik})_{\sigma'\sigma}$$
To obtain a representation of a finite rotation we need to exponentiate the infinitesimal case:
$$\implies \mathcal{D}^{(s)}(R)=\exp\{i\Theta_{ik}J^{ik}\}$$
Which, I think, translates in my notation to:
$$\mathcal{D}^{(s)}(R(\vec{\theta}))=\exp\{i\vec{\theta}\cdot
\vec{S}\}$$
For a rotation by an angle $|\vec{\theta}|$ around $\vec{\theta}$, along with the observation that $S_i=\frac{1}{2}\varepsilon_{ijk}J^{jk}\implies \vec{S}=(J^{23},J^{31},J^{12})=(J^1,J^2,J^3)=\vec{J}$.
So we can compute the unitary representations of the little group $SO(3)$ by:
$$\mathcal{D}^{(s)}(R(\vec{\theta}))_{\sigma'\sigma}=\langle s,\sigma'|e^{i\vec{\theta}\cdot\vec{S}}|s,\sigma\rangle~~~~~~~~~~~~(4)$$
We can compute this explicitly using equations (2) and (3), and so for a fixed s, the size of the matrix D will be (2s+1)(2s+1).
I think what I have said so far is mostly correct (but I would like some confirmation). If what I have said is indeed correct, then the problem I am having is that I do not know how to translate between $\mathcal{D}^{(s)}(h(\Lambda,p))_{\sigma'\sigma}$ and $\mathcal{D}^{(s)}(R(\vec{\theta}))_{\sigma'\sigma}$. For example, given (4), how can I compute the corresponding matrix elements of $\mathcal{D}^{(s)}(h(\Lambda,p))_{\sigma'\sigma}$? I know that there must be a relation because they are both just rotations. So I can infer that $\vec{\theta}=\vec{\theta}(\Lambda,p)$. But it seems that the standard way to compute the desired matrix elements is to use the standard Lorentz transformations which I provided in the context section… but to me it seems that this is only useful if we have an explicit method to compute the elements of the representation of $h(\Lambda,p)$, like equation (4). Is there such an explicit expression? Cheers.
Best Answer
This answer is based on this article by A. Ungar.
Ungar computed the Thomas rotation formula which is almost what you need. I'll describe the general procedure, and in some cases, I'll refer you to Ungar for the proof. I'll express (just like Ungar), the Boosts in terms of velocities rather than momenta. If you wish you can repeat the exercise with the momentum parametrization. From Wikipedia we have $$B(\mathbf{v}) = \begin{bmatrix} \gamma_{\mathbf{v}}& \frac{\gamma_{\mathbf{v}}}{c}\mathbf{v}^t \\ \frac{\gamma_{\mathbf{v}}}{c}\mathbf{v} & 1_{(3\times3)}+ \frac{\gamma_{\mathbf{v}}^2}{1+\gamma_{\mathbf{v}}} \frac{\mathbf{v}\mathbf{v}^t}{c^2} \end{bmatrix}$$
The key point is finding the Wigner rotation is the observation that every Lorentz transformation can be decomposed (nonuniquely) as a product of a Boost and a rotation: $$\Lambda = B(\mathbf{u}) R$$ Any choice of the decomposition method will do, but we need to work in a fixed method of decomposition. I'll describe to you a possible method at the end of the answer Now, when we multiply two boosts, we get a boost with the relativistic addition velocity + a rotation (often referred to as the Thomas rotation): $$ B(\mathbf{u}) B(\mathbf{v}) = B(\mathbf{u}\oplus \mathbf{v}) \mathrm{Tom}(\mathbf{u},\mathbf{v})$$ Where $\mathrm{Tom}$ is a rotation matrix Ungar found the general solution for the Thomas rotation (Equation : (13) in the article) $$\mathrm{Tom}(\mathbf{u},\mathbf{v}) = B(-\mathbf{u}\oplus \mathbf{v}) B(\mathbf{u}) B(\mathbf{v})$$ Now, we are in a position to solve the equation for the Wigner rotation. We need to solve: $$ \Lambda B(\mathbf{v}) = B(\Lambda \mathbf{v}) W$$ for $W$. We parametrize $\Lambda$, we get for the left handside:
$$\begin{align*} \Lambda B(\mathbf{v}) & = B(\mathbf{u}) R B(\mathbf{v}) \\ & = B(\mathbf{u}) R B(\mathbf{v}) R^{-1} R \\ & = B(\mathbf{u}) B(\mathbf{Rv}) R \\ & = B(\mathbf{u}\oplus R\mathbf{v}) \mathrm{Tom}(\mathbf{u},R\mathbf{v}) R \end{align*} $$
and for the right hand side $$ B(\Lambda \mathbf{v}) W = B(B(\mathbf{u}) R\mathbf{v}) W = B(\mathbf{u}\oplus R\mathbf{v}) W$$
Thus: $$W = \mathrm{Tom}(\mathbf{u},R\mathbf{v}) R$$ What is left is to describe a specific parametrization of a general Lorentz matrix into a boost and a rotation: We need to find $B(\mathbf{u})$ and $R$ such that: $$\begin{bmatrix} \lambda_0 &\mathbf{\xi}^t \\ \mathbf{\eta} & \Lambda_1 \end{bmatrix} = \begin{bmatrix} \gamma_{\mathbf{v}}& \frac{\gamma_{\mathbf{v}}}{c}\mathbf{v}^t \\ \frac{\gamma_{\mathbf{v}}}{c}\mathbf{v} & 1_{(3\times3)}+ \frac{\gamma_{\mathbf{v}}^2}{1+\gamma_{\mathbf{v}}} \frac{\mathbf{v}\mathbf{v}^t}{c^2} \end{bmatrix} \begin{bmatrix} 1&0 \\ 0 &R_1 \end{bmatrix}$$
We observe that $R_1$ needs to satisfy the following relation $$\mathbf{\xi} = R_1 \mathbf{\eta}$$ Thus $R_1$ needs to rotate the 3-vector $\mathbf{\mathbf{\eta}}$ into the 3-vector $\mathbf{\xi}$ The solution (using a similar method as Ungar) can be written as: $$R_1 = 1_{(3\times3)}+ \sin(\theta) \Omega + (\cos(\theta-1)) \Omega^2$$
Where $\theta$ is the angle between the vectors $\mathbf{\eta}$ and $\mathbf{\xi}$ $$\sin(\theta) = \frac{\mathbf{\eta} \times \mathbf{\xi}}{|\mathbf{\eta}||\mathbf{\mathbf{\xi}}|}$$ and $$\Omega_{ij} =\frac{ \eta_i \xi_j - \xi_i \eta_j}{|\mathbf{\eta}||\mathbf{\xi}|}$$