# Matrix of SU(2) representation

I'm reading D. Gross' Lecture Notes on QFT. In finding the representations of the Poincare Group he finds the represenations of SU(2), which we know are classified by $$j=0, \frac12,1,...$$ and are $$2j+1$$ dimensional, with vector space $$H=span \{ |\lambda \rangle , \lambda=-j,-j+1,...,j \}$$. Each element $$R$$ of the group will then be represented by an operator on H : $$U(R)|λ\rangle =\sum_{\lambda'} D^j_{\lambda'\lambda}(R) |\lambda'\rangle$$

Thus to fully define the representation we need the matrix elements $$D^j_{\lambda'\lambda}(R)$$.

Now Gross moves on to write that $$D^j_{\lambda'\lambda}(R)= \langle \lambda'| e^{-i\alpha J_3} e^{-i\beta J_2} e^{-i\gamma J_3}| \lambda \rangle$$ Where $$\alpha,\beta,\gamma$$ are the Euler Angles. However I can't derive this expression for $$U(R)$$.

What I know is that if $$X$$ is an element of the Lie Algebra then $$e^{x}$$ is an element of the group, and since SU(2) is simply connected every element $$R$$ can be written that way. Now the Lie algebra of SU(2) is 3 dimensional with basis $$J_1$$, $$J_2$$,$$J_3$$ thus what I would say that $$U(R)= e^{-i (\phi J_1 + \theta J_2+ \omega J_3)}$$ Where $$\phi,\theta,\omega$$ are some angles. Gross' expression should follow from the one I wrote, right? I can't see how that could be done, so I'd appreciate some help.

PS On second thoughts, does this belong to mathstackexchange with "mathematical physics" tag?

• Actually, the latter, axis-angle representation of rotations you may be aiming for is $U(R)= e^{-i (\phi J_1 +\theta J_2 + \omega J_3)}$, that is the exponential of an arbitrary Lie-algebra element. Conversions between axis-angle and Euler angles are quite messy. Commented Oct 31, 2018 at 19:33
• You are correct, in fact that's what I meant to write. I corrected it. Commented Oct 31, 2018 at 19:35
• Conversion of explicit axis-angle to Euler angles can be seen here. But Euler angles are superior above, on J_3 eigenspaces.... Commented Oct 31, 2018 at 19:43
• As written your $U(R)$ is not an operator. Presumably you are missing some $\langle \lambda\vert$ to make it an operator. Commented Oct 31, 2018 at 19:56

## 1 Answer

What you are interested in are the Wigner D-matrices, which take the generic form $$D^j_{m'm}(\alpha,\beta,\gamma)\equiv \langle jm'\vert R(\alpha,\beta,\gamma)\vert jm\rangle = e^{-im'\alpha}d^j_{mm'}(\beta)e^{-im\gamma}$$ with $$d^j_{mm'}(\beta)= \langle jm'\vert R_y(\beta)\vert jm\rangle\, .$$ There are several ways of obtaining the $$d^j_{mm'}(\beta)$$ and the wikipage linked above gives various closed forms expressions as sums; the $$d^j_{m'm}$$ are related to Jacobi polynomials.

Lengthy derivations are found in many textbooks on quantum angular momentum, or online resources. A good (older) paper on techniques to find these is Wolters, G. F. "Simple method for the explicit calculation of d-functions." Nuclear Physics B 18.2 (1970): 625-653.

Possibly the most explicit derivation uses the map \begin{align} L_{+} &\rightarrow \xi \frac{\partial }{% \partial \eta }, \\ L_{-} &\rightarrow \eta \frac{\partial }{% \partial \xi }, \\ L_{z} &\rightarrow \frac{1}{2}\left( \xi \frac{\partial }{\partial \xi }-\eta \frac{\partial }{\partial \eta }\right) . \end{align} The states $$\vert LM\rangle$$ are then mapped to fucctions of $$\xi ,\eta :$$ \begin{align} \vert LM\rangle &\rightarrow &\frac{1}{\sqrt{(L+M)!(L-M)!}}\xi ^{L+M}\eta ^{L-M}, \\ \left\langle LM\right\vert &\rightarrow &\frac{1}{\sqrt{(L+M)!(L-M)!}}\left( \frac{\partial }{\partial \xi }\right) ^{L+M}\left( \frac{\partial }{% \partial \eta }\right) ^{L-M}. \end{align}

For this, we observe that, because of the identification $$\vert \textstyle \frac{1}{2} , \textstyle\frac{1}{2} \rangle \leftrightarrow \xi , \qquad \vert \textstyle\frac{1}{2} ,- \textstyle\frac{1}{2} \rangle \leftrightarrow \eta ,$$ the transformations of the kets $$\begin{eqnarray} R_{y}(\beta )\vert \textstyle\frac{1}{2} , \textstyle\frac{1}{2} \rangle &=&\cos \left( \textstyle\frac{\beta }{2}\right) \vert \textstyle\frac{1}{2},\textstyle\frac{1}{2} \rangle +\sin \left( \textstyle\frac{\beta }{2}\right) \vert \textstyle\frac{1}{2} ,-\textstyle\frac{1}{2} \rangle , \\ R_{y}(\beta )\vert \textstyle\frac{1}{2} ,-\textstyle\frac{1}{2} \rangle &=&-\sin \left( \textstyle\frac{\beta }{2}\right) \vert \textstyle\frac{1}{2} ,\textstyle\frac{1}{2} \rangle +\cos \left( \textstyle\frac{\beta }{2}\right) \vert \textstyle\frac{1}{2},- \textstyle\frac{1}{2} \rangle \end{eqnarray}$$ follows from direct exponentiation of the Pauli matrix $$e^{-i\beta\sigma_y}$$ and imply the transformation of the dummy variables $$\begin{eqnarray} R_{y}(\beta )\xi R_{y}^{-1}(\beta ) &=&\cos \left( \textstyle\frac{\beta }{2}\right) \xi +\sin \left( \textstyle\frac{\beta }{2}\right) \eta , \\ R_{y}(\beta )\eta R_{y}^{-1}(\beta ) &=&-\sin \left( \textstyle\frac{\beta }{2}\right) \xi +\cos \left( \textstyle\frac{\beta }{2}\right) \eta . \end{eqnarray}$$

Hence, from the identification $$\vert LM\rangle \rightarrow \frac{% \xi ^{L+M}\eta ^{L-M}}{\sqrt{(L+M)!(L-M)!}}$$ we infer $$\begin{eqnarray} R_{y}(\beta )\vert LM\rangle &\rightarrow &R_{y}(\beta )\frac{% \xi ^{L+M}\eta ^{L-M}}{\sqrt{(L+M)!(L-M)!}}R_{y}^{-1}(\beta )=\frac{% R_{y}(\beta )\xi ^{L+M}R_{y}^{-1}(\beta )R_{y}(\beta )\eta ^{L-M}R_{y}^{-1}(\beta )}{\sqrt{(L+M)!(L-M)!}} \\ &=&\frac{\left( R_{y}(\beta )\xi R_{y}^{-1}(\beta )\right) ^{L+M}\left( R_{y}(\beta )\eta R_{y}^{-1}(\beta )\right) ^{L-M}}{\sqrt{(L+M)!(L-M)!}} \\ &=&\frac{\left( \cos \left( \frac{\beta }{2}\right) \xi +\sin \left( \frac{% \beta }{2}\right) \eta \right) ^{L+M}\left( -\sin \left( \frac{\beta }{2}% \right) \xi +\cos \left( \frac{\beta }{2}\right) \eta \right) ^{L-M}}{\sqrt{% (L+M)!(L-M)!}} \\ &=&\frac{1}{\sqrt{(L+M)!(L-M)!}}\nonumber \\ &&\times \sum_{x,y}\left( \cos \left( \frac{\beta }{2}% \right) \xi \right) ^{L+M-x}\left( -\sin \left( \frac{\beta }{2}\right) \xi \right) ^{L-M-y}\left( \sin \left( \frac{\beta }{2}\right) \eta \right) ^{x}\left( \sin \left( \frac{\beta }{2}\right) \eta \right) ^{y} \\ &=&\frac{1}{\sqrt{(L+M)!(L-M)!}}\sum_{x,y}(-1)^{L-M-y}\cos \left( \frac{% \beta }{2}\right) ^{L+M-x+y}\sin \left( \frac{\beta }{2}\right) ^{L-M-y+x}\xi ^{2L-x-y}\eta ^{x+y}. \end{eqnarray}$$ Now $$$$\left\langle LM^{\prime }\right\vert \rightarrow \frac{1}{\sqrt{(L+M^{\prime })!(L-M^{\prime })!}}\left( \frac{\partial }{\partial \xi }\right) ^{L+M^{\prime }}\left( \frac{\partial }{\partial \eta }\right) ^{L-M^{\prime }}$$$$ so the matrix element $$\left\langle LM^{\prime }\right\vert R_{y}(\beta )\vert LM\rangle$$ will be non-zero only when there are precisely $$L+M^{\prime }$$ powers of $$\xi$$ in $$R_{y}(\beta )\vert LM\rangle$$ and $$L-M^{\prime }$$ powers of $$\eta$$ in $$R_{y}(\beta )\vert LM\rangle$$. In this case, the multiple derivatives will produce a factor of $$\left( L+M^{\prime }\right) !\left( L-M^{\prime }\right) !$$ and, after tedious but straightforward algebra, we obtain (one possible version of) the final form $$\begin{eqnarray*} d^L_{M^{\prime }M}(\beta ) &=&\sum_{x}(-1)^{M^{\prime }-M+x}\frac{\sqrt{% (L+M^{\prime })!(L-M^{\prime })!(L+M)!(L-M)!}}{(L+M-x)!x!(L-M^{\prime }-x)!(M^{\prime }-M+x)!} \\ &&\times \cos \left( \frac{\beta }{2}\right) ^{2L+M-M^{\prime }-2x}\sin \left( \frac{\beta }{2}\right) ^{M^{\prime }-M+2x}. \end{eqnarray*}$$