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?

  • 1
    $\begingroup$ 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. $\endgroup$ Oct 31, 2018 at 19:33
  • $\begingroup$ You are correct, in fact that's what I meant to write. I corrected it. $\endgroup$
    – Arbiter
    Oct 31, 2018 at 19:35
  • 1
    $\begingroup$ Conversion of explicit axis-angle to Euler angles can be seen here. But Euler angles are superior above, on J_3 eigenspaces.... $\endgroup$ Oct 31, 2018 at 19:43
  • $\begingroup$ As written your $U(R)$ is not an operator. Presumably you are missing some $\langle \lambda\vert$ to make it an operator. $\endgroup$ Oct 31, 2018 at 19:56

1 Answer 1


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 \begin{equation} \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 }} \end{equation} 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*}


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