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Appendix D Solutions

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Table of Contents

Basic

Medium

Basic

B-1. Additive Group of Integers

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Problem:

Verify that integer addition \((\mathbb{Z}, +)\) satisfies the four group axioms.

For \((\mathbb{Z}, +)\):

  1. Closure: For integers \(a, b\), \(a + b\) is an integer ✓
  2. Associativity: \((a + b) + c = a + (b + c)\) ✓ (associativity of addition)
  3. Identity element: \(0\) (\(a + 0 = 0 + a = a\)) ✓
  4. Inverse element: The inverse of \(a\) is \(-a\) (\(a + (-a) = 0\)) ✓

Since all four conditions are satisfied, \((\mathbb{Z}, +)\) is a group.

B-2. Multiplicative Group of Positive Real Numbers

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Problem:

Verify that the set of all positive real numbers with multiplication \((\mathbb{R}^+, \times)\) forms a group. What are the identity element and the inverse element?

For \((\mathbb{R}^+, \times)\):

  1. Closure: For positive real numbers \(a, b\), \(a \times b\) is a positive real number ✓
  2. Associativity: \((a \times b) \times c = a \times (b \times c)\) ✓
  3. Identity element: \(1\) (\(a \times 1 = 1 \times a = a\)) ✓
  4. Inverse element: The inverse of \(a\) is \(1/a\) (\(a \times (1/a) = 1\)). If \(a > 0\) then \(1/a > 0\) ✓

Since all four conditions are satisfied, \((\mathbb{R}^+, \times)\) is a group.

B-3. Multiplication of Integers Does Not Form a Group

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Problem:

Show that \((\mathbb{Z}, \times)\), the set of all integers \(\mathbb{Z}\) equipped with ordinary multiplication \(\times\) as the operation, does not form a group, by checking the group axioms (closure, associativity, identity element, inverse element) one by one.

For \((\mathbb{Z}, \times)\):

  1. Closure: ✓ (integer × integer = integer)
  2. Associativity: ✓
  3. Identity element: \(1\) ✓
  4. Inverse: The inverse of \(a = 2\) is \(1/2\), but \(1/2 \notin \mathbb{Z}\). ✗

Since the inverse condition is violated, \((\mathbb{Z}, \times)\) is not a group. (Only \(a = \pm 1\) have inverses)

B-4. Group Conditions of \(U(1)\)

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Problem:

Verify that the elements \(e^{i\theta}\) of \(U(1)\) satisfy the group axioms.

For \(U(1) = \{e^{i\theta} \mid \theta \in \mathbb{R}\}\) under multiplication:

  1. Closure: \(e^{i\theta_1} \cdot e^{i\theta_2} = e^{i(\theta_1 + \theta_2)} \in U(1)\) ✓
  2. Associativity: Trivial from the associativity of complex number multiplication ✓
  3. Identity: \(e^{i \cdot 0} = 1\) ✓
  4. Inverse: \((e^{i\theta})^{-1} = e^{-i\theta} \in U(1)\) ✓

B-5. 3-Dimensional \(z\)-Axis Rotation

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Problem:

The rotation matrix about the \(z\)-axis in 3-dimensional space is given by

\[ R_z(\theta) = \begin{pmatrix}\cos\theta & -\sin\theta & 0\\\sin\theta & \cos\theta & 0\\0 & 0 & 1\end{pmatrix} \]

Show that \(R_z(\theta_1) R_z(\theta_2) = R_z(\theta_1 + \theta_2)\) holds by explicitly computing the matrix product.

\(R_z(\theta) = \begin{pmatrix}\cos\theta & -\sin\theta & 0\\\sin\theta & \cos\theta & 0\\0 & 0 & 1\end{pmatrix}\)

We compute \(R_z(\theta_1) R_z(\theta_2)\). Focusing on the upper-left \(2 \times 2\) block:

\(\begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\\sin\theta_1 & \cos\theta_1\end{pmatrix}\begin{pmatrix}\cos\theta_2 & -\sin\theta_2\\\sin\theta_2 & \cos\theta_2\end{pmatrix}\)

\((1,1)\) component: \(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 = \cos(\theta_1 + \theta_2)\)

\((1,2)\) component: \(-\cos\theta_1\sin\theta_2 - \sin\theta_1\cos\theta_2 = -\sin(\theta_1 + \theta_2)\)

\((2,1)\) component: \(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2 = \sin(\theta_1 + \theta_2)\)

\((2,2)\) component: \(-\sin\theta_1\sin\theta_2 + \cos\theta_1\cos\theta_2 = \cos(\theta_1 + \theta_2)\)

Therefore \(R_z(\theta_1) R_z(\theta_2) = R_z(\theta_1 + \theta_2)\) ✓

B-6. Commutation Relation of Pauli Matrices \([\sigma_1, \sigma_2]\)

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Problem:

For the Pauli matrices \(\sigma_1, \sigma_2, \sigma_3\), verify that \([\sigma_1, \sigma_2] = 2i\sigma_3\) by explicitly computing the matrix products.

\(\sigma_1\sigma_2 = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix} = \begin{pmatrix}i&0\\0&-i\end{pmatrix}\)

\(\sigma_2\sigma_1 = \begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}-i&0\\0&i\end{pmatrix}\)

\([\sigma_1, \sigma_2] = \sigma_1\sigma_2 - \sigma_2\sigma_1 = \begin{pmatrix}2i&0\\0&-2i\end{pmatrix} = 2i\begin{pmatrix}1&0\\0&-1\end{pmatrix} = 2i\sigma_3\) ✓

B-7. Commutation Relations of Pauli Matrices (All Pairs)

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Problem:

Compute \([\sigma_2, \sigma_3]\) and \([\sigma_3, \sigma_1]\) as well, and verify that \([\sigma_i, \sigma_j] = 2i\varepsilon_{ijk}\sigma_k\) holds for all combinations.

\([\sigma_2, \sigma_3]\):

\(\sigma_2\sigma_3 = \begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0&i\\i&0\end{pmatrix}\)

\(\sigma_3\sigma_2 = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix} = \begin{pmatrix}0&-i\\-i&0\end{pmatrix}\)

\([\sigma_2, \sigma_3] = \begin{pmatrix}0&2i\\2i&0\end{pmatrix} = 2i\begin{pmatrix}0&1\\1&0\end{pmatrix} = 2i\sigma_1\) ✓

\([\sigma_3, \sigma_1]\):

\(\sigma_3\sigma_1 = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}0&1\\-1&0\end{pmatrix}\)

\(\sigma_1\sigma_3 = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0&-1\\1&0\end{pmatrix}\)

\([\sigma_3, \sigma_1] = \begin{pmatrix}0&2\\-2&0\end{pmatrix} = 2i\begin{pmatrix}0&-i\\i&0\end{pmatrix} = 2i\sigma_2\) ✓

Summary: \([\sigma_i, \sigma_j] = 2i\varepsilon_{ijk}\sigma_k\) holds for all cyclic combinations.

B-8. Antisymmetry of the Commutator

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Problem:

Show from the definition that \([A, B] = -[B, A]\) (antisymmetry of the commutator).

From the definition:

\([A, B] = AB - BA\)

\([B, A] = BA - AB = -(AB - BA) = -[A, B]\)

Therefore \([A, B] = -[B, A]\). In particular, from \([A, A] = -[A, A]\) it follows that \([A, A] = 0\).

B-9. Eigenvalues and Eigenvectors of Spin 1/2

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Problem:

The state of a spin \(1/2\) particle is represented by a two-component vector \(|\psi\rangle = \begin{pmatrix}\alpha\\\beta\end{pmatrix}\). Find the eigenvalues and eigenvectors of \(\sigma_3\), and confirm that they correspond to "spin up" and "spin down."

\(\sigma_3 = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\)

Eigenvalue equation \(\sigma_3 |\psi\rangle = \lambda |\psi\rangle\):

\(\det(\sigma_3 - \lambda I) = (1-\lambda)(-1-\lambda) = 0\)

Eigenvalues: \(\lambda = +1, -1\)

\(\lambda = +1\): \(\sigma_3 |\psi\rangle = |\psi\rangle\) → \(|\psi\rangle = \begin{pmatrix}1\\0\end{pmatrix} \equiv |\uparrow\rangle\) (spin up)

\(\lambda = -1\): \(\sigma_3 |\psi\rangle = -|\psi\rangle\) → \(|\psi\rangle = \begin{pmatrix}0\\1\end{pmatrix} \equiv |\downarrow\rangle\) (spin down)

The eigenvalues of \(\sigma_3/2\) are \(\pm 1/2\) (in units of \(\hbar\)), corresponding to the "up" and "down" states of spin \(1/2\).

Medium

M-1. Exponential Representation of \(SU(2)\) Elements

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Problem:

Explicitly compute the \(2 \times 2\) matrix \(U = e^{i\theta\sigma_3/2}\) (using \(e^{iA} = I + iA + (iA)^2/2! + \cdots\); since \(\sigma_3\) is a diagonal matrix, this is straightforward). Verify that \(U^\dagger U = I\) and \(\det U = 1\).

Since \(\sigma_3 = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\) is a diagonal matrix:

\(\frac{i\theta\sigma_3}{2} = \begin{pmatrix}i\theta/2&0\\0&-i\theta/2\end{pmatrix}\)

The exponential of a diagonal matrix is the exponential of each component:

\(U = e^{i\theta\sigma_3/2} = \begin{pmatrix}e^{i\theta/2}&0\\0&e^{-i\theta/2}\end{pmatrix}\)

Verification of unitarity:

\(U^\dagger = \begin{pmatrix}e^{-i\theta/2}&0\\0&e^{i\theta/2}\end{pmatrix}\)

\(U^\dagger U = \begin{pmatrix}e^{-i\theta/2} \cdot e^{i\theta/2}&0\\0&e^{i\theta/2} \cdot e^{-i\theta/2}\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I\) ✓

Verification of the determinant:

\(\det U = e^{i\theta/2} \cdot e^{-i\theta/2} = e^0 = 1\) ✓

M-2. Jacobi Identity (Verification with Pauli Matrices)

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Problem:

Using the Pauli matrices \(\sigma_1, \sigma_2, \sigma_3\), verify by explicit calculation that the Jacobi identity

\[ [A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0 \]

holds for the case \(A = \sigma_1,\; B = \sigma_2,\; C = \sigma_3\).

Setting \(A = \sigma_1, B = \sigma_2, C = \sigma_3\):

\([A, [B, C]] = [\sigma_1, 2i\sigma_1] = 2i[\sigma_1, \sigma_1] = 0\)

\([B, [C, A]] = [\sigma_2, 2i\sigma_2] = 2i[\sigma_2, \sigma_2] = 0\)

\([C, [A, B]] = [\sigma_3, 2i\sigma_3] = 2i[\sigma_3, \sigma_3] = 0\)

\(0 + 0 + 0 = 0\) ✓

(In this case each term individually vanishes, but this is not true in general. The Jacobi identity states that the sum of the three terms equals zero.)

M-3. Parameter Count of \(SU(N)\)

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Problem:

Show that the number of parameters of \(SU(N)\) is \(N^2 - 1\) by the argument "subtract 1 from the \(N^2\) parameters of an \(N \times N\) unitary matrix due to the condition \(\det = 1\)." Verify this for the cases \(N = 2, 3\).

We count the number of parameters of an \(N \times N\) unitary matrix \(U\) (\(U^\dagger U = I\)).

An \(N \times N\) complex matrix has \(2N^2\) real parameters. The unitarity condition \(U^\dagger U = I\) imposes \(N^2\) real conditions (the number of independent components of an \(N \times N\) Hermitian matrix). Therefore, the number of parameters of the unitary matrix group \(U(N)\) is \(2N^2 - N^2 = N^2\).

\(SU(N)\) further imposes the condition \(\det U = 1\) (one real condition), so:

\(\dim SU(N) = N^2 - 1\)

Verification: - \(N = 2\): \(4 - 1 = 3\) (\(SU(2)\) has 3 parameters) ✓ - \(N = 3\): \(9 - 1 = 8\) (\(SU(3)\) has 8 parameters) ✓