quantum mechanics - Matrix representation of rotation operators in QM

The usual matrix form of an operator $X$ is given by matrix components

$$\langle a''| X | a' \rangle$$ where $|a' \rangle$ forms a basis for the ket space. In the case where we define matrix of rotation (where $|j,m \rangle$ are eigenstates of $J^2$ and $J_z$) $$\mathcal{D}_{m',m}^{(j)}(R) = \langle j, m' |\text{exp} \bigg(\frac{-i \mathbf{J} \cdot \hat{n} \phi}{\hbar} \bigg) |j,m \rangle$$ we go through all of $j$'s and $m$'s and get a block matrix representation of the rotation $\text{exp}\bigg(\frac{-i \mathbf{J} \cdot \hat{n} \phi}{\hbar} \bigg)$. That fits with the above definition of matrix representation of a rotation operator. But in a text (Sakurai's 'Modern Quantum Mechanics' pages 196-197) it states that for specific $j$, we still get a matrix representation of a rotation operator, would this not imply that we are not using the full set of the basis $|j,m \rangle$ for this representation, how is that a valid matrix representation of an operator?

Answer

I hope I understand you right.

In general the game is to take the infinite-dimensional Hilbert space and break it into a sum of blocks. This partitioning of course is not unique but quite clearly there is advantage in choosing a partitioning where the operators mix the blocks as little as possible. In particular, if your operator does not mix the blocks and the blocks are finite dimensional, then by linearity you can work in each finite-dimensional block, an obvious advantage over working in an infinite-dimensional space. (This is clearly the optimal situation, not always realized in practice.)

In the case of rotations, the partitioning will involve basis states of the form $\vert \alpha j m\rangle$ where $\alpha$ is any other quantum number that is not $j$ or $m$. In this basis your Hilbert space now looks (cartoon version) like

$$\left(\begin{array}{c|c|c|c|c} \ldots&\hbox{$ \begin{array}{c} \phantom{\hbox{a}}\\ (2j_1+1)\times (2j_1+1)\\ \phantom{\hbox{a}}\end{array}$}& 0 & 0&\ldots \\ \hline \ldots&0&\hbox{$ \begin{array}{c} \phantom{\hbox{a}}\\ (2j_2+1)\times (2j_2+1)\\ \phantom{\hbox{a}}\end{array}$}& 0&\ldots \\ \hline \vdots & \vdots & 0 & \hbox{$ \begin{array}{c} \phantom{\hbox{a}}\\ \ddots\\ \phantom{\hbox{a}}\end{array}$} \end{array}\right)$$ with the blocks spanned by states $\vert\alpha \ell m\rangle$, one pair $(\alpha,\ell)$ for each block.

In this basis, the rotation operators connects states not only with the same $j$ but also with the same $\alpha$, i.e. $\vert\alpha jm\rangle$ and $\alpha j m'\rangle$: even if there are multiple states with the same $j$ value (this could occur for instance in the hydrogen atom or the 3d oscillator, where some specific $\ell$ values will occur for different energies), the rotation operator will not connect those, i.e.

$$\langle \beta j m\vert R\vert \alpha jm \rangle =\delta_{\alpha\beta}D^{j}_{mm'}(R)\, .$$

Thus, if I understand your question well, the statement about getting a matrix representation of the rotation operator for a specific $j$ really means that the rotation operators acting in an infinite dimensional space can be broken up in $(2j+1)\times (2j+1)$ pieces (obviously there is more than one possible value of $j$), and that inside each $j$ subspace you have a representation of the rotation operator.

Your kets $\vert a\rangle$ and $\vert a'\rangle$ will live in the infinite-dimensional Hilbert space. Since by assumption $\vert \alpha \ell m\rangle$ form a complete set of orthonormal states in this infinite-dimensional space, one should first decompose each state as

$$\begin{align} \vert a\rangle&=\sum_{\alpha \ell m}\vert \alpha \ell m\rangle\langle \alpha \ell m\vert a\rangle \, ,\\ \vert a'\rangle&=\sum_{\beta \ell' m'}\vert \beta \ell' m'\rangle\langle \beta \ell' m'\vert a'\rangle\, , \end{align}$$ so that $$\begin{align} \langle a'\vert R \vert a\rangle&= \sum_{\alpha\beta \ell\ell' m m'} \langle a'\vert \beta \ell' m'\rangle\langle \beta \ell' m' R\vert \alpha \ell m\rangle\langle \alpha \ell m\vert a\rangle\, ,\\ &=\sum_{\alpha\beta \ell\ell' m m'} \langle a'\vert \beta \ell' m'\rangle\langle \beta \ell' m' R\vert \alpha \ell m\rangle\langle \alpha \ell m\vert a\rangle\delta_{\alpha\beta}\delta_{\ell\ell'}\, ,\\ &=\sum_{\alpha\ell m m'} \langle a'\vert \alpha \ell m'\rangle D^\ell_{m'm}(R) \langle \alpha \ell m\vert a\rangle \end{align}$$ so that the rotation operator operates within a subblock $\alpha\ell$, but the there is a sum of such subblocks.

As a specific example, consider the states of a 3d harmonic oscillator. The only quantum number needed in addition to $\ell$ and $m$ is $n$ so $\alpha=n$.

Labelling states as $\vert n\ell m$, we have (for instance), the state $\vert 000\rangle$, $\vert 11M$, with $M=-1,0,1$, $\vert 200\rangle$, $\vert 22m\rangle$ with $m=-2,\ldots,2$ etc. The list goes on since the Hilbert space is infinite-dimensional.

The Hilbert space will contain the following blocks:

1. one $1\times 1$ block spanned by $\vert 000\rangle$,


2. one $1\times 1$ block spanned by $\vert 200\rangle$,


3. one $3\times 3$ block spanned by $\{\vert 11M\rangle, M=-1,0,1\}$,


4. one $5\times 5$ block spanned by $\{\vert 22m\rangle, m=-2, \ldots, 2\}$,


5. etc.

Thus, if you have the general states

$$\begin{align} \vert a\rangle = \sum_{n\ell m} b_{n\ell m} \vert n\ell m\rangle\, , \qquad \vert a'\rangle = \sum_{n\ell m} c_{n'\ell' m'} \vert n'\ell' m'\rangle\, , \end{align}$$ then $$\begin{align} \langle a'\vert R\vert a\rangle&=\sum_{n\ell mm'} b_{n\ell m}c^*_{n\ell m'} D^\ell_{m'm}(R)\, . \end{align}$$