Question: In my lecture script it says that \begin{equation} \theta^{(j)}(\vec{\phi})=\exp(-\frac{i}{\hbar}\vec{\phi}\cdot\vec{J}^{(j)}) \end{equation} How can this be derived from \begin{equation} \theta(\vec{\phi})=\exp(-\frac{i}{\hbar}\vec{\phi}\cdot\vec{J}) \end{equation} Notation: The matrix representation of the rotation operator is defined by the matrix elements \begin{equation} \theta_{m'm}^{(j)}(\vec{\phi})=\langle j',m'|\theta(\vec{\phi})|j,m\rangle=\delta_{j,j'}\langle j,m'|\theta(\vec{\phi})|j,m\rangle \quad \textrm{where} \quad \theta(\vec{\phi}):=\exp(-\frac{i}{\hbar}\vec{\phi}\cdot\vec{J}) \end{equation} where it has been used that the elements with $j\neq j'$ vanish. Similarily the basis representation of the anuglar momentum operator $J_i$ ($i \in \{x,y,z\}$) is defined as \begin{equation} J_{i,m'm}^{(j)}=\delta_{j,j'}\langle j,m'|J_i|j,m\rangle \end{equation} $J_i^{(j)}$ is the matrix formed by $m,m'=-j,...,j$. For example we can identify the three $J_i^{(1/2)}$ matrices as the Pauli-matrices.

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  • $\begingroup$ @Frobenius While related, this doesn't adress my question. $\endgroup$
    – user224659
    Commented Jan 3, 2020 at 21:21
  • $\begingroup$ This is a special case of the problem of constructing unitary representations of a Lie group out of hermitian representations of its Lie algebra. In your case you know the representation of the rotation algebra and wants to find the unitary operators the hermitian generators give rise to. This thread (physics.stackexchange.com/questions/520014/…) and this thread (physics.stackexchange.com/questions/474050/…) might give you some insight on that. $\endgroup$
    – Gold
    Commented Jan 3, 2020 at 21:45
  • $\begingroup$ I've never heared anything about group theory before, I don't think I will be able to understand your argument in the answers linked before reading much more on the topic. I feel like the answer should be quite easy though, as it's assumed as trivially true in both my lecture notes and sakurai @user1620696 $\endgroup$
    – user224659
    Commented Jan 3, 2020 at 21:50

2 Answers 2


It is true in all cases. The group elements are obtained by exponentiating the generic algebra element. Proof can be found in most books on Lie algebras and groups. Note that in some cases not every element can be continuously obtained starting from the identity.

Exponentiation is not necessarily easy but for $2\times 2$ Pauli one can fairly easily prove this identity: $$ e^{i a \hat n\cdot \vec \sigma}=I \cos(a)+ i(\hat n\cdot \vec \sigma)\sin(a) \tag{1} $$ which follows using $\sigma_k^2=I$.

For the more general $(2j+1)\times (2j+1)$ representation, exponentiating a general element $e^{i a \hat n\cdot \vec J}$ would be near impossible so the strategy is to factorize the elements $$ O(\vec \phi)=R_z(\alpha) R_y(\beta) R_z(\gamma) \tag{2} $$ in three consecutive transformations. In a basis where $\hat L_z$ is diagonal, the first and last transformations will produce phases: $$ R_z(\gamma)\vert jm\rangle = e^{-i\gamma m}\vert jm\rangle\, , $$ so there remains $R_y(\beta)$ which acts non-trivially. The functions $d^j_{m’m}(\beta)$ are known as the Wigner $d$-functions and there are various ways of obtaining them, as in here for instance.

The canonical reference for all this is

Varshalovich, D.A., Moskalev, A.N. and Khersonskii, V.K.M., Quantum theory of angular momentum, (World Scientific, 1988)

Finite transformations for any of the classical groups are likewise obtained by exponentiating generic algebra elements (and possibly multiplying two transformations if the final element is not continuously connected to the identity), although generalizing (1) directly is not technically possible. Instead, one can usually find a convenient factorization generalizing (2) and go from there.

There are many papers on the factorization of $SU(N)$ transformations and a simple Google search will get you some hits. Getting the group functions from the factorization is not easy for arbitrary irreps of SU(N).

There are also some pretty cool factorization of the symplectic matrices, my favorite given in

Dragt, Alex J. "Lectures on nonlinear orbit dynamics." AIP conference proceedings. Vol. 87. No. 1. AIP, 1982.

for application to optics and beam optics (accelerator physics). There is a version of (1) applicable to $SU(1,1)$ which can be found in

Puri, R.R., 2001. Mathematical methods of quantum optics (Vol. 79). Springer Science & Business Media.

and also in some special relativity textbooks given the close connection between $SU(1,1)$ transformations and Lorentz transformations.

  • $\begingroup$ Thank you for taking the time! But your answer seems to focus on how to evaluate the matrix exponential. But I'm mainly interested in how to transition from the operator exponential to this matrix exponential, not so much in getting explicit matrix representations by evaluating this exponential. And a google search sadly didn't help me with that. $\endgroup$
    – user224659
    Commented Jan 3, 2020 at 14:46
  • $\begingroup$ I'm not sure I understand your question. the operator exponential IS the matrix exponential, i.e. find the operator and exponentiate to get the matrix exponential... Granted it's not terribly convenient to evaluate the matrix exponential that way... $\endgroup$ Commented Jan 3, 2020 at 14:51
  • $\begingroup$ @TheoreticalMinimum possibly the difficulty is that, as you have written it, $\langle j m'\vert \exp(O)\vert jm\rangle$ is a matrix element whereas $\exp(\langle jm'|O|jm\rangle)$ is a matrix (and thus obviously no equality here) but the $(m',m)$ entry of that matrix is $\langle j m'\vert \exp(O)\vert jm\rangle$ $\endgroup$ Commented Jan 3, 2020 at 14:57
  • $\begingroup$ Don't you differentiate between the matrix representation of an operator and the operator itself? The operator is acting on a ket, the pauli matrices for example on a $\mathbb{R}^2$ vector. $\endgroup$
    – user224659
    Commented Jan 3, 2020 at 14:57
  • $\begingroup$ I just want to show that $\theta(\vec{\phi}):=\exp(-\frac{i}{\hbar}\vec{\phi}\cdot\vec{J}) \rightarrow \theta^{(j)}(\vec{\phi})=\exp(-\frac{i}{\hbar}\vec{\phi}\cdot\vec{J}^{(j)})$ $\endgroup$
    – user224659
    Commented Jan 3, 2020 at 15:00

Turns out the answer is quite trivial. First let $\vec{\phi}=\phi\cdot\vec{e_i}$. Then \begin{equation} e^{-\frac{i}{\hbar}\vec{\phi}\cdot\vec{J}^{(j)}}=e^{-\frac{i}{\hbar}\phi\cdot J_i^{(j)}}=\sum_n\left(-\frac{i}{\hbar}\phi\right)^n\frac{1}{n!}\left(J_i^{(j)}\right)^n \end{equation} When Evalauting $\left(J_i^{(j)}\right)^n $ one notices \begin{equation} \left(\left(J_i^{(j)}\right)^2\right)_{a,c}=\sum_b \left(J_i^{(j)}\right)_{a,b} \left(J_i^{(j)}\right)_{b,c}=\sum_b \langle j,a|J_i|j,b\rangle\langle j,b|J_i|j,c\rangle=\langle j,a|J_i^2|j,c\rangle \end{equation} Because of the completness $\sum_m |j,m\rangle\langle j,m|=1$. This generalizes to \begin{equation} \left(\left(J_i^{(j)}\right)^n\right)_{a,c}=\langle j,a|J_i^n|j,c\rangle \end{equation} Plugging into a component of the first matrix equation \begin{align} \left( e^{-\frac{i}{\hbar}\phi\cdot J_i^{(j)}}\right)_{a,c}&=\sum_n\left(-\frac{i}{\hbar}\phi\right)^n\frac{1}{n!}\left(\left(J_i^{(j)}\right)^n\right)_{a,c}=\sum_n\left(-\frac{i}{\hbar}\phi\right)^n\frac{1}{n!}\langle j,a|J_i^n|j,c\rangle\\ &=\langle j,a|\sum_n\left(-\frac{i}{\hbar}\phi\right)^n\frac{1}{n!}J_i^n|j,c\rangle=\langle j,a|e^{-\frac{i}{\hbar}\phi J_i}|j,c\rangle:=\theta^{(j)}_{a,c}(\phi \vec{e_i}) \end{align} It's easy to see that this can be generalized for arbitrary $\vec{\phi}$. So in matrix notation it's indeed true that \begin{equation} e^{-\frac{i}{\hbar}\vec{\phi}\cdot\vec{J}^{(j)}}= \theta^{(j)}(\vec{\phi})\end{equation}


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