Lie Groups III Examples Bottom Up

$\mathfrak{so}(3)$ Lie Algebra¶

Step 1: Define the so(3) Lie Algebra¶

The Lie algebra $\mathfrak{so}(3)$ is a three-dimensional real Lie algebra generated by basis elements $J_x, J_y, J_z$ with the defining commutation relations:

where $i, j, k = x, y, z$, $\epsilon_{ijk}$ is the Levi-Civita symbol (totally antisymmetric with $\epsilon_{xyz} = 1$), and summation over repeated indices is implied. Explicitly, these are:

• $[J_x, J_y] = i J_z$,

• $[J_y, J_z] = i J_x$,

• $[J_z, J_x] = i J_y$.

This convention, common in physics, ensures that in unitary representations, the operators are Hermitian, which is suitable for the ladder operator approach. Alternatively one can absorb the $i$ into the generators and obtain $[J_i, J_j] = \epsilon_{ijk} J_k$.

Step 2: Introduce Ladder Operators¶

In the ladder operator method, we define the raising and lowering operators as:

These operators are complex linear combinations of the generators, facilitating the analysis of representations. To proceed, we compute their commutation relations using the $\mathfrak{so}(3)$ algebra.

Thus, the ladder operator algebra is:

Step 3: Representations of the Lie Algebra¶

A representation of $\mathfrak{so}(3)$ is a homomorphism from the Lie algebra to the linear operators on a complex vector space $V$, preserving the commutation relations. Since SO(3) is compact, its finite-dimensional representations are completely reducible, so we focus on irreducible representations.

Assume $V$ is a finite-dimensional complex vector space where $J_z$ is diagonalizable (possible in complex representations). Let $|m\rangle$ be an eigenvector of $J_z$ with eigenvalue $m$:

• Action of $J_+$:

Thus, $J_+ |m\rangle$ is either zero or an eigenvector of $J_z$ with eigenvalue $m + 1$.

• Action of $J_-$:

Thus, $J_- |m\rangle$ is either zero or an eigenvector with eigenvalue $m - 1$.

Since $V$ is finite-dimensional, the eigenvalues cannot increase or decrease indefinitely. There must exist a highest eigenvalue $j$ such that:

and a lowest eigenvalue $m_{\text{min}}$ such that $J_- |m_{\text{min}}\rangle = 0$.

Step 4: Construct the Representation¶

Starting from the highest weight vector $|j\rangle$, apply $J_-$ repeatedly:

• $J_- |j\rangle$ has eigenvalue $j - 1$,

• $J_-^2 |j\rangle$ has eigenvalue $j - 2$,

• and so on, until $J_-^k |j\rangle \neq 0$ but $J_-^{k+1} |j\rangle = 0$.

Define $|m\rangle \propto J_-^{j - m} |j\rangle$ (with appropriate normalization), where $m = j, j-1, \ldots$, until $J_- |m_{\text{min}}\rangle = 0$. Suppose $m_{\text{min}} = -j$, so:

The eigenvalues range from $j$ to $-j$ in steps of 1: $j, j-1, \ldots, -j+1, -j$. The number of states is:

implying $2j$ is an integer, so $j$ is a non-negative integer or half-integer ($j = 0, \frac{1}{2}, 1, \frac{3}{2}, \ldots$).

Step 5: Use the Casimir Operator¶

Define the Casimir operator:

Express it using ladder operators:

Compute:

Thus:

Since $[J_+, J_-] = 2 J_z$, we have $J_+ J_- = J_- J_+ + 2 J_z$, so:

For the highest weight vector:

Since $J^2$ commutes with $J_z, J_+, J_-$, it is a scalar in an irreducible representation: $J^2 = j (j + 1) I$. This confirms the eigenvalues $m = -j, -j+1, \ldots, j$.

Step 6: Determine Coefficients¶

In a unitary representation, $J_x, J_y, J_z$ are Hermitian, so $J_+^\dagger = J_-$. Assume orthonormal basis $\langle m | m' \rangle = \delta_{mm'}$:

Since $\langle m | J_- | m+1 \rangle = \langle m+1 | J_+ | m \rangle^*$, compute:

Thus, $|c_m|^2 = c_{m-1} d_m - 2 m$. For $m = j$, $J_+ |j\rangle = 0$, so $c_j = 0$. For $m = -j$, $J_- |-j\rangle = 0$, so $d_{-j} = 0$. Solving recursively (standard in quantum mechanics) gives:

Step 7: Relate to SO(3) Group¶

The Lie algebra so(3) is isomorphic to su(2), and its representations are labeled by $j = 0, \frac{1}{2}, 1, \ldots$, with dimension $2j + 1$. Exponentiating to the group, SU(2) (simply connected) admits all $j$, but SO(3) = SU(2)/$\mathbb{Z}_2$ requires single-valued representations. For integer $j$, the representation is single-valued on SO(3); for half-integer $j$, it is double-valued, corresponding to SU(2). Thus, SO(3) representations have $j = 0, 1, 2, \ldots$.

Final Answer¶

The irreducible representations of the SO(3) group, derived from the so(3) Lie algebra via the ladder operator approach, are labeled by integers $j = 0, 1, 2, \ldots$, with dimension $2j + 1$, where $J_z$ has eigenvalues $m = -j, \ldots, j$, and $J^2 = j (j + 1) I$.

Angular Momentum Operators and Their Transformation for $j = 1$¶

In quantum mechanics, for a system with angular momentum quantum number $j = 1$, the angular momentum operators are represented in the basis $|1, 1\rangle$, $|1, 0\rangle$, and $|1, -1\rangle$. We start with the ladder operators and $J_z$, compute $J_x$ and $J_y$, and then apply a unitary transformation to obtain transformed operators.

1. Ladder Operators and $J_z$¶

The ladder operators $J_+$ (raising) and $J_-$ (lowering), along with the $z$-component $J_z$, are defined as follows:

• Raising operator $J_+$ raises the magnetic quantum number $m$ by 1:

  $$J_+ = \begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{pmatrix}$$

  (28)

• Lowering operator $J_-$ lowers the magnetic quantum number $m$ by 1:

  $$J_- = \begin{pmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{pmatrix}$$

  (29)

• $z$-component $J_z$ is diagonal with eigenvalues corresponding to $m$:

  $$J_z = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$

  (30)

2. Computation of $J_x$ and $J_y$¶

Using the ladder operators, we compute $J_x$ and $J_y$ as follows:

• $J_x$ is the average of the raising and lowering operators: Using the relation $J_x = \frac{1}{2} (J_+ + J_-)$,

  $$J_x = \begin{pmatrix} 0 & \frac{\sqrt{2}}{2} & 0 \\ \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\ 0 & \frac{\sqrt{2}}{2} & 0 \end{pmatrix}$$

  (31)

• $J_y$ involves the imaginary unit: Using $J_y = \frac{1}{2i} (J_+ - J_-)$, where $\frac{1}{i} = -i$,

  $$J_y = \begin{pmatrix} 0 & -\frac{\sqrt{2} i}{2} & 0 \\ \frac{\sqrt{2} i}{2} & 0 & -\frac{\sqrt{2} i}{2} \\ 0 & \frac{\sqrt{2} i}{2} & 0 \end{pmatrix}$$

  (32)

These matrices, along with $J_z$, satisfy the angular momentum commutation relations, such as $[J_x, J_y] = i J_z$.

3. Unitary Transformation Matrix $U$¶

To transform the operators into a basis where certain properties (e.g., real rotation matrices) are emphasized, we use the unitary matrix $U$:

This matrix is unitary, satisfying $U^\dagger U = I$, where $U^\dagger$ is the conjugate transpose of $U$.

4. Transformed Angular Momentum Operators¶

The transformed operators are computed using the similarity transformation $J_i' = U^\dagger J_i U$ for $i = x, y, z$:

• Transformed $J_x'$:

  $$J_x' = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{pmatrix}$$

  (34)

• Transformed $J_y'$:

  $$J_y' = \begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \end{pmatrix}$$

  (35)

• Transformed $J_z'$:

  $$J_z' = \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

  (36)

5. Verification with $i J_i'$¶

To explore the properties of the transformed operators, consider $i J_i'$:

• $i J_x' = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}$

• $i J_y' = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}$

• $i J_z' = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

Each $i J_i'$ is real and skew-symmetric (i.e., $(i J_i')^T = -i J_i'$), which is consistent with generators of real rotation matrices in SO(3).

Computing the rotation matrices¶

In this section we use:

• $J_x = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}$

• $J_y = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}$

• $J_z = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

We have:

and:

We have $J_x^3=J_x J_x^2 = -J_x$ and $J_x^4=J_x^2 J_x^2 = -J_x^2$ and $J_x^5=-J_x^3=J_x$.

Now we compute:

That is the rotation matrix around the $x$-axis.

The Homogeneous Lorentz Group from the Ground Up¶

The $so(1,3)$ Lie Algebra¶

Step 1: Define the $so(1,3)$ Lie Algebra¶

The Lie algebra $so(1,3)$ is the Lie algebra of the Lorentz group, which preserves the Minkowski metric in four-dimensional spacetime with signature $(1, -1, -1, -1)$. It is a six-dimensional real Lie algebra generated by three rotation generators $J_x, J_y, J_z$ (or $J_i$ for $i = 1, 2, 3$) and three boost generators $K_x, K_y, K_z$ (or $K_i$), corresponding to rotations in spatial planes and boosts mixing time and space coordinates, respectively.

In the physics convention, where generators are Hermitian in unitary representations, the commutation relations are:

where $i, j, k = 1, 2, 3$ (with $J_1 = J_x$, etc.), $\epsilon_{ijk}$ is the Levi-Civita symbol ($\epsilon_{123} = 1$), and summation over repeated indices is implied. Explicitly:

• $[J_x, J_y] = i J_z$, cyclic permutations,

• $[J_x, K_y] = i K_z$, etc.,

• $[K_x, K_y] = -i J_z$, etc.

These relations reflect that $J_i$ generate an $so(3)$ subalgebra (rotations), while $K_i$ introduce the non-compact nature of $so(1,3)$ due to the negative sign in the boost commutator.

Step 2: Separate the Algebra Using a Complex Trick¶

To simplify the structure, define new generators:

for $i = 1, 2, 3$. These are complex linear combinations of the real generators $J_i$ and $K_i$. Compute their commutation relations to reveal the algebra’s structure.

Commutator $[A_i, A_j]$:¶

Expand:

• $[J_i, J_j] = i \epsilon_{ijk} J_k$,

• $[J_i, K_j] = i \epsilon_{ijk} K_k$,

• $[K_i, J_j] = -[J_j, K_i] = -i \epsilon_{jik} K_k = i \epsilon_{ijk} K_k$ (since $\epsilon_{jik} = -\epsilon_{ijk}$),

• $[K_i, K_j] = -i \epsilon_{ijk} J_k$.

Substitute:

so:

Commutator $[B_i, B_j]$:¶

Commutator $[A_i, B_j]$:¶

Thus:

Each set $\{A_i\}$ and $\{B_i\}$ satisfies the $su(2)$ commutation relations, and they commute with each other, suggesting $so(1,3)$ is related to $su(2) \oplus su(2)$ via complexification.

Step 3: Interpret the Algebra Structure¶

Invert the definitions:

The $so(1,3)$ algebra, with real generators $J_i, K_i$, maps to two commuting $su(2)$ algebras via complex generators $A_i, B_i$. The complexification of $so(1,3)$, denoted $so(1,3)_ℂ$, is isomorphic to $sl(2,ℂ)$, and as a real Lie algebra, $so(1,3)$ has dimension 6, matching $sl(2,ℂ)$ over the reals. This trick reveals that representations can be constructed from $su(2)$ representations.

Representations of $so(1,3)$¶

Step 4: Label Representations Using $su(2)$¶

Since $A_i$ and $B_i$ generate two independent $su(2)$ algebras, representations of $so(1,3)$ are labeled by pairs $(j_1, j_2)$, where $j_1$ is the spin of the $su(2)$ algebra generated by $A_i$, and $j_2$ is the spin of the $B_i$ algebra. From $su(2)$, $j_1, j_2 = 0, \frac{1}{2}, 1, \frac{3}{2}, \ldots$, and the representation space is the tensor product of a $(2j_1 + 1)$-dimensional space and a $(2j_2 + 1)$-dimensional space, with dimension:

In this representation:

• $A_i$ acts as $J_i^{(j_1)} \otimes I$,

• $B_i$ acts as $I \otimes J_i^{(j_2)}$,

• $J_i = A_i + B_i = J_i^{(j_1)} \otimes I + I \otimes J_i^{(j_2)}$,

• $K_i = -i (A_i - B_i) = -i (J_i^{(j_1)} \otimes I - I \otimes J_i^{(j_2)})$,

where $J_i^{(j)}$ are the $su(2)$ generators in the spin-$j$ representation.

Step 5: Identify the Vector Representation¶

The homogeneous Lorentz group $SO(1,3)$ acts on 4D Minkowski space via the four-vector representation, which is four-dimensional. Find $(j_1, j_2)$ such that:

• If $j_1 = 0, j_2 = \frac{3}{2}$: $(1)(4) = 4$ (but typically symmetric in physics),

• If $j_1 = \frac{1}{2}, j_2 = \frac{1}{2}$: $(2)(2) = 4$.

The $(1/2, 1/2)$ representation is the standard four-vector representation of $so(1,3)$, corresponding to fields transforming as four-vectors (e.g., the electromagnetic potential).

The Homogeneous Lorentz Group $SO(1,3)$¶

Step 6: Relate to the Group¶

The Lie algebra $so(1,3)$ corresponds to the Lorentz group $SO(1,3)$, which is non-compact, and its double cover is $SL(2,ℂ)$. Finite-dimensional representations are not unitary, but the $(j_1, j_2)$ labeling applies, with $SO(1,3)$ transformations obtained by exponentiating the generators.

Transformation Matrices in the $(1/2, 1/2)$ Representation¶

Step 7: Construct Generators in $(1/2, 1/2)$¶

For $j_1 = 1/2$, $j_2 = 1/2$, the basis is $|m_1, m_2\rangle$ with $m_1, m_2 = \pm 1/2$, so:

• $|1/2, 1/2\rangle$, $|1/2, -1/2\rangle$, $|-1/2, 1/2\rangle$, $|-1/2, -1/2\rangle$.

Using $J_z^{(1/2)} = \sigma_z / 2$, etc., where $\sigma_i$ are Pauli matrices:

• $A_z = (\sigma_z / 2) \otimes I = \text{diag}(1/2, 1/2, -1/2, -1/2)$,

• $B_z = I \otimes (\sigma_z / 2) = \text{diag}(1/2, -1/2, 1/2, -1/2)$,

• $J_z = A_z + B_z = \text{diag}(1, 0, 0, -1)$,

• $K_z = -i (A_z - B_z) = -i \text{diag}(0, 1, -1, 0)$.