Irreducible Representation Matrix Elements for Spin j

The Standard Angular Momentum Basis

Let's begin by formally setting up our workspace. For a given irreducible representation labeled by spin $j$, we work in a $(2j+1)$-dimensional complex vector space. The standard orthonormal basis for this space consists of the kets $|j,m⟩$, where $m\in \{j,j-1,\dots ,-j\}$. These are, by definition, the simultaneous eigenvectors of the Casimir operator $J^2$ and the generator $J_3$:

Our objective is to find the matrix elements of the algebra generators in this basis: $⟨j,m^{\prime }|\hat{O}|j,m⟩$, where $\hat{O}$ can be $J^2,J_3,J_{+},J_{-}$. Note that since these operators are constructed from the algebra generators, they cannot change the total spin $j$. An operator acting within a given irrep cannot connect states of different $j$; hence, matrix elements like $⟨j^{\prime },m^{\prime }|\hat{O}|j,m⟩$ are zero unless $j^{\prime }=j$.

Matrix Elements of $J^2$ and $J_3$

Constructing the matrices for $J^2$ and $J_3$ is straightforward, as our basis states are their eigenvectors. The matrix elements are:

This means that in the $|j,m⟩$ basis, the matrix for $J^2$ is simply the identity matrix multiplied by $j(j+1){\hslash }^2$. The matrix for $J_3$ is a diagonal matrix whose entries are the possible values of $m\hslash$. If we order our basis vectors from $m=j$ down to $m=-j$, the matrix for $J_3$ will have the entries $(j\hslash ,(j-1)\hslash ,\dots ,-j\hslash )$ along its diagonal.

Matrix Elements of the Ladder Operators

The most crucial part of the construction is finding the matrix elements for the ladder operators, $J_{+}$ and $J_{-}$. We know from the previous lesson that they act on a state $|j,m⟩$ to produce a state proportional to $|j,m\pm 1⟩$:

Our task is to determine the coefficients $C_{\pm }(j,m)$. The standard approach is to compute the squared norm of the state $J_{+}|j,m⟩$.

\begin{align*}
| J_+ |j,m\rangle |^2 &= \langle j,m | J_+^\dagger J_+ | j,m \rangle \
&= \langle j,m | J_- J_+ | j,m \rangle \
&= |C_{+}(j,m)|^2 \langle j,m+1 | j,m+1 \rangle = |C_{+}(j,m)|^2
\end{align*}

To evaluate this, we use the identity $J_{-}{J}_{+}=J^2-J_3^2-\hslash J_3$, which you can derive from the fundamental commutation relations. Applying this to our state:

\begin{align*}
\langle j,m | J_- J_+ | j,m \rangle &= \langle j,m | (J^2 - J_3^2 - \hbar J_3) | j,m \rangle \
&= [j(j+1)\hbar^2 - (m\hbar)^2 - \hbar(m\hbar)] \langle j,m|j,m \rangle \
&= \hbar^2 [j(j+1) - m(m+1)] \
&= \hbar^2 (j-m)(j+m+1)
\end{align*}

This gives us the magnitude of the coefficient: $|C_{+}(j,m){|}^2={\hslash }^2(j-m)(j+m+1)$. The coefficient $C_{+}(j,m)$ is, in general, a complex number. However, the relative phases of the basis states $|j,m⟩$ for a fixed $j$ are not physically determined. We can use this freedom to make a convenient choice. The Condon-Shortley phase convention sets these coefficients to be real and positive.

Thus, we have our final expressions for the action of the ladder operators:

where the second equation can be derived in a completely analogous way using the identity $J_{+}{J}_{-}=J^2-J_3^2+\hslash J_3$.

The following video provides an excellent, detailed walkthrough of this derivation.

The corresponding matrix elements are found by taking the inner product with $⟨j,m^{\prime }|$:

Notice that $J_{+}$ only has non-zero matrix elements one position above the main diagonal (the superdiagonal), and $J_{-}$ only has them one position below (the subdiagonal). Since $J_{+}^{†}=J_{-}$, the matrix for $J_{-}$ is the conjugate transpose of the matrix for $J_{+}$.

For a more formal and complete treatment of this derivation, the following notes are an excellent resource.

Examples: Constructing the Matrices

With the general formulae in hand, let's construct the matrices for the two most important cases in physics: spin-1/2 and spin-1.

Spin $j=1/2$ (Dimension = 2)

The basis is $\{|1/2,1/2⟩,|1/2,-1/2⟩\}$.

• $J_3$: The matrix is diagonal with eigenvalues $m\hslash$. $$J_3=\hslash \left(\begin{array}{cc}1/2& 0\\ 0& -1/2\end{array}\right)=\frac{\hslash }{2}{\sigma }_3$$
• $J_{+}$: The only non-zero element is $⟨1/2,1/2|J_{+}|1/2,-1/2⟩$. Using our formula with $j=1/2,m=-1/2$:
  $⟨1/2,1/2|J_{+}|1/2,-1/2⟩=\hslash \sqrt{(1/2-(-1/2))(1/2+(-1/2)+1)}=\hslash \sqrt{1\cdot 1}=\hslash$. $$J_{+}=\hslash \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$$
• $J_{-}$: This is the conjugate transpose of $J_{+}$. $$J_{-}=\hslash \left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$$
• $J_1$ and $J_2$: We can now find the matrices for the Cartesian generators using $J_1=(J_{+}+J_{-})/2$ and $J_2=(J_{+}-J_{-})/(2i)$. $$J_1=\frac{\hslash }{2}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\frac{\hslash }{2}{\sigma }_1\text{and}{J}_2=\frac{\hslash }{2}\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)=\frac{\hslash }{2}{\sigma }_2$$

As expected, the generators of the spin-1/2 representation are proportional to the Pauli matrices.

Spin $j=1$ (Dimension = 3)

The basis is $\{|1,1⟩,|1,0⟩,|1,-1⟩\}$.

• $J_3$: $$J_3=\hslash \left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right)$$
• $J_{+}$: We have two non-zero elements:
  • $⟨1,1|J_{+}|1,0⟩=\hslash \sqrt{(1-0)(1+0+1)}=\hslash \sqrt{2}$
  • $⟨1,0|J_{+}|1,-1⟩=\hslash \sqrt{(1-(-1))(1+(-1)+1)}=\hslash \sqrt{2}$
  $$J_{+}=\hslash \left(\begin{array}{ccc}0& \sqrt{2}& 0\\ 0& 0& \sqrt{2}\\ 0& 0& 0\end{array}\right)$$
• $J_{-}$: Taking the conjugate transpose: $$J_{-}=\hslash \left(\begin{array}{ccc}0& 0& 0\\ \sqrt{2}& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right)$$
• $J_1$ and $J_2$: $$J_1=\frac{\hslash }{\sqrt{2}}\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right)\text{and}{J}_2=\frac{\hslash }{\sqrt{2}}\left(\begin{array}{ccc}0& -i& 0\\ i& 0& -i\\ 0& i& 0\end{array}\right)$$

This procedure can be systematically applied for any $j$. The eigenchris video also walks through these constructions.

This construction reveals a deep truth. The abstract commutation relations of the $\mathfrak{su}(2)$ algebra contain all the information needed to build the entire family of angular momentum matrices that form the foundation of quantum mechanics.

Conclusion

In this lesson, we have successfully bridged the gap from the abstract classification of $\mathfrak{su}(2)$ representations to their concrete construction. By systematically evaluating the matrix elements of the algebra generators in the standard $|j,m⟩$ basis, we have developed a universal algorithm to build the matrix for any arbitrary spin $j$.

Key Takeaways:

• The matrix elements of the $\mathfrak{su}(2)$ generators can be fully determined from the algebra's commutation relations and the definition of the $|j,m⟩$ basis.
• The matrices for $J^2$ and $J_3$ are diagonal in this basis, with eigenvalues $j(j+1){\hslash }^2$ and $m\hslash$, respectively.
• The crucial step is determining the matrix elements of the ladder operators, $J_{+}$ and $J_{-}$, which are found to be $⟨j,m^{\prime }|J_{\pm }|j,m⟩=\hslash \sqrt{(j\mp m)(j\pm m+1)}{\delta }_{m^{\prime },m\pm 1}$.
• The Condon-Shortley phase convention is a standard choice that simplifies these matrices by making their non-zero elements real and positive.
• This procedure yields the familiar Pauli matrices for $j=1/2$ and provides a blueprint for constructing the matrices for $j=1,3/2,2,\dots$.

We now possess the explicit matrices that represent the action of angular momentum operators. In the next lesson, we will solidify the physical interpretation of this mathematical machinery, explicitly connecting the irreducible representations of SU(2) to the quantum theory of angular momentum and the behavior of particles with spin.