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Appendix B Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Norm Calculation of a Vector in \(\mathbb{C}^2\)

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Solution Strategy: The norm is calculated as \(\||\psi\rangle\| = \sqrt{\langle\psi|\psi\rangle}\). We use \(\langle\psi|\psi\rangle = \sum_k |z_k|^2\).

Calculation:

\[|\psi\rangle = \begin{pmatrix} 2i \\ 1 - i \end{pmatrix}\]

Calculate the squared absolute value of each component:

\[|2i|^2 = (2i)(2i)^* = (2i)(-2i) = 4\]
\[|1 - i|^2 = (1 - i)(1 - i)^* = (1 - i)(1 + i) = 1 + 1 = 2\]

Therefore

\[\langle\psi|\psi\rangle = 4 + 2 = 6\]
\[\boxed{\||\psi\rangle\| = \sqrt{6}}\]

Verification: \(|2i|^2 = 0^2 + 2^2 = 4\) ✓, \(|1-i|^2 = 1^2 + (-1)^2 = 2\) ✓. The norm is a positive real number, and \(\sqrt{6} > 0\) ✓.

B-2. Inner Product Calculation in \(\mathbb{C}^2\)

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Solution strategy: Compute \(\langle\phi|\psi\rangle = \sum_k \phi_k^* \psi_k\). Note that the complex conjugate is applied to the components of the bra side (first argument).

Calculation:

\[|\psi\rangle = \begin{pmatrix} 1 + i \\ 2 \end{pmatrix}, \quad |\phi\rangle = \begin{pmatrix} 3 \\ i \end{pmatrix}\]

The bra vector is

\[\langle\phi| = (3^*,\; i^*) = (3,\; -i)\]

Therefore

\[\langle\phi|\psi\rangle = 3 \cdot (1 + i) + (-i) \cdot 2 = 3 + 3i - 2i = 3 + i\]
\[\boxed{\langle\phi|\psi\rangle = 3 + i}\]

Verification: Confirm the Hermitian property \(\langle\psi|\phi\rangle = \langle\phi|\psi\rangle^*\).

\[\langle\psi|\phi\rangle = (1+i)^* \cdot 3 + 2^* \cdot i = (1-i) \cdot 3 + 2i = 3 - 3i + 2i = 3 - i\]

Indeed \(\langle\psi|\phi\rangle = 3 - i = (3 + i)^* = \langle\phi|\psi\rangle^*\) ✓

B-3. Normalization of a 2-Dimensional Vector

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Solution strategy: First compute the norm \(\||v\rangle\|\), then find \(|u\rangle = |v\rangle / \||v\rangle\|\).

Calculation:

\[|v\rangle = \begin{pmatrix} 1 \\ 1 \\ i \end{pmatrix}\]
\[\langle v|v\rangle = |1|^2 + |1|^2 + |i|^2 = 1 + 1 + 1 = 3\]
\[\||v\rangle\| = \sqrt{3}\]
\[\boxed{|u\rangle = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ i \end{pmatrix}}\]

Verification: Check that \(\langle u|u\rangle = 1\).

\[\langle u|u\rangle = \frac{1}{3}(1^* \cdot 1 + 1^* \cdot 1 + (i)^* \cdot i) = \frac{1}{3}(1 + 1 + (-i)(i)) = \frac{1}{3}(1 + 1 + 1) = 1 \quad \checkmark\]

B-4. Determining Orthogonality

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Solution Strategy: Calculate the inner product \(\langle a|b\rangle\) and check whether it equals 0.

Calculation:

\[|a\rangle = \begin{pmatrix} 1 \\ i \end{pmatrix}, \quad |b\rangle = \begin{pmatrix} i \\ 1 \end{pmatrix}\]
\[\langle a| = (1^*,\; i^*) = (1,\; -i)\]
\[\langle a|b\rangle = 1 \cdot i + (-i) \cdot 1 = i - i = 0\]
\[\boxed{\langle a|b\rangle = 0 \quad \therefore\ |a\rangle \text{ and } |b\rangle \text{ are orthogonal}}\]

Verification: \(\langle b|a\rangle = \langle a|b\rangle^* = 0^* = 0\) ✓. Also, computing directly: \(\langle b|a\rangle = (i)^*\cdot 1 + 1^* \cdot i = -i + i = 0\) ✓.

B-5. Calculation of Matrix Elements

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Solution strategy: \(A_{jk} = \langle e_j|\hat{A}|e_k\rangle\) equals the \(j\)-th component of \(\hat{A}|e_k\rangle\) when expressed as a column vector.

Calculation:

From \(\hat{A}|e_1\rangle = \begin{pmatrix} 2 \\ i \end{pmatrix}\):

\[A_{11} = \langle e_1|\hat{A}|e_1\rangle = 2, \quad A_{21} = \langle e_2|\hat{A}|e_1\rangle = i\]

From \(\hat{A}|e_2\rangle = \begin{pmatrix} -i \\ 3 \end{pmatrix}\):

\[A_{12} = \langle e_1|\hat{A}|e_2\rangle = -i, \quad A_{22} = \langle e_2|\hat{A}|e_2\rangle = 3\]
\[\boxed{A = \begin{pmatrix} 2 & -i \\ i & 3 \end{pmatrix}}\]

Verification: The \(k\)-th column of the matrix should consist of the components of \(\hat{A}|e_k\rangle\). The first column \(\begin{pmatrix}2\\i\end{pmatrix}\) matches \(\hat{A}|e_1\rangle\) ✓, and the second column \(\begin{pmatrix}-i\\3\end{pmatrix}\) matches \(\hat{A}|e_2\rangle\) ✓. Furthermore, since \(A^\dagger = \begin{pmatrix}2 & -i \\ i & 3\end{pmatrix} = A\), this operator is Hermitian.

B-6. Calculating the Hermitian Conjugate

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Solution strategy: \((A^\dagger)_{jk} = A_{kj}^*\) (transpose and take the complex conjugate).

Calculation:

\[A = \begin{pmatrix} 2 & 1 - i \\ 3i & 4 + 2i \end{pmatrix}\]

Taking the transpose:

\[A^T = \begin{pmatrix} 2 & 3i \\ 1 - i & 4 + 2i \end{pmatrix}\]

Taking the complex conjugate of each element:

\[\boxed{A^\dagger = \begin{pmatrix} 2 & -3i \\ 1 + i & 4 - 2i \end{pmatrix}}\]

Verification: \((A^\dagger)_{12} = A_{21}^* = (3i)^* = -3i\) ✓, \((A^\dagger)_{21} = A_{12}^* = (1-i)^* = 1+i\) ✓. Also, \(A^\dagger \neq A\), so this matrix is not Hermitian.

B-7. Calculation of a Commutator

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Solution strategy: Compute \(AB\) and \(BA\) separately, then take their difference.

Calculation:

\[A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\]
\[AB = \begin{pmatrix} 1\cdot0 + 0\cdot1 & 1\cdot1 + 0\cdot0 \\ 0\cdot0 + (-1)\cdot1 & 0\cdot1 + (-1)\cdot0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\]
\[BA = \begin{pmatrix} 0\cdot1 + 1\cdot0 & 0\cdot0 + 1\cdot(-1) \\ 1\cdot1 + 0\cdot0 & 1\cdot0 + 0\cdot(-1) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\]
\[[A, B] = AB - BA = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}\]
\[\boxed{[A, B] = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}}\]

Verification: \(A = \sigma_z\), \(B = \sigma_x\) (Pauli matrices), and it is known that \([\sigma_z, \sigma_x] = -2i\sigma_y\). Since \(\sigma_y = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}\), we get \(-2i\sigma_y = -2i\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} = \begin{pmatrix}0 & -2 \\ 2 & 0\end{pmatrix}\).

Wait, let's check the sign. \([\sigma_z, \sigma_x] = \sigma_z\sigma_x - \sigma_x\sigma_z\). The known relation is \([\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k\), so \([\sigma_z, \sigma_x] = 2i\epsilon_{zxy}\sigma_y = 2i(-1)\sigma_y = -2i\sigma_y\).

\[-2i\sigma_y = -2i\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} = \begin{pmatrix}0 & 2i\cdot i \\ -2i\cdot i & 0\end{pmatrix}\]

Here, let's verify the value of \(\epsilon_{zxy}\). \((z,x,y) = (3,1,2)\) maps the cyclic permutation \((1,2,3)\) to \((3,1,2)\), which is an even permutation, so \(\epsilon_{312} = +1\). Therefore \([\sigma_z, \sigma_x] = 2i\sigma_y\).

\[2i\sigma_y = 2i\begin{pmatrix}0 & -i \\ i & 0\end{pmatrix} = \begin{pmatrix}0 & -2i^2 \\ 2i^2 & 0\end{pmatrix} = \begin{pmatrix}0 & 2 \\ -2 & 0\end{pmatrix} \quad \checkmark\]

B-8. Determining Linear Independence

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Solution Strategy: Investigate whether there exists \(\alpha \in \mathbb{C}\) such that \(|v_2\rangle = \alpha |v_1\rangle\).

Calculation:

\[|v_1\rangle = \begin{pmatrix} 1 \\ i \end{pmatrix}, \quad |v_2\rangle = \begin{pmatrix} i \\ -1 \end{pmatrix}\]

Assuming \(|v_2\rangle = \alpha |v_1\rangle\):

  • First component: \(i = \alpha \cdot 1 \implies \alpha = i\)
  • Second component: \(-1 = \alpha \cdot i = i \cdot i = -1\) ✓

Since \(\alpha = i\) satisfies both components, we have \(|v_2\rangle = i\,|v_1\rangle\).

\[\boxed{|v_1\rangle \text{ and } |v_2\rangle \text{ are linearly dependent}\ (|v_2\rangle = i\,|v_1\rangle)}\]

Verification: Indeed, \(i\begin{pmatrix}1\\i\end{pmatrix} = \begin{pmatrix}i\\i^2\end{pmatrix} = \begin{pmatrix}i\\-1\end{pmatrix} = |v_2\rangle\) ✓.

B-9. Calculation of Expansion Coefficients

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Solution strategy: The expansion coefficients in an orthonormal basis are found via \(c_k = \langle e_k|\psi\rangle\) (Eq. (B.19)).

Calculation:

\[|+\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad |-\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \quad |\psi\rangle = \begin{pmatrix} 3 \\ i \end{pmatrix}\]

Since the basis vectors have real components, the bra vectors are:

\[\langle +| = \frac{1}{\sqrt{2}}(1,\; 1), \quad \langle -| = \frac{1}{\sqrt{2}}(1,\; -1)\]
\[c_+ = \langle +|\psi\rangle = \frac{1}{\sqrt{2}}(1 \cdot 3 + 1 \cdot i) = \frac{3 + i}{\sqrt{2}}\]
\[c_- = \langle -|\psi\rangle = \frac{1}{\sqrt{2}}(1 \cdot 3 + (-1) \cdot i) = \frac{3 - i}{\sqrt{2}}\]
\[\boxed{c_+ = \frac{3 + i}{\sqrt{2}}, \quad c_- = \frac{3 - i}{\sqrt{2}}}\]

Verification: Check that \(c_+|+\rangle + c_-|-\rangle\) equals \(|\psi\rangle\).

\[c_+|+\rangle + c_-|-\rangle = \frac{3+i}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} + \frac{3-i}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\]
\[= \frac{1}{2}\begin{pmatrix}3+i\\3+i\end{pmatrix} + \frac{1}{2}\begin{pmatrix}3-i\\-(3-i)\end{pmatrix} = \frac{1}{2}\begin{pmatrix}6\\2i\end{pmatrix} = \begin{pmatrix}3\\i\end{pmatrix} = |\psi\rangle \quad \checkmark\]

Furthermore, \(|c_+|^2 + |c_-|^2 = \frac{|3+i|^2}{2} + \frac{|3-i|^2}{2} = \frac{10}{2} + \frac{10}{2} = 10 = \langle\psi|\psi\rangle\) ✓ (Parseval's identity).

B-10. Verification of the Completeness Relation

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Solution Strategy: Compute \(|+\rangle\langle+|\) and \(|-\rangle\langle-|\) each as \(2\times 2\) matrices, then take their sum.

Calculation:

\[|+\rangle\langle +| = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} \cdot \frac{1}{\sqrt{2}}(1,\;1) = \frac{1}{2}\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}\]
\[|-\rangle\langle -| = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix} \cdot \frac{1}{\sqrt{2}}(1,\;-1) = \frac{1}{2}\begin{pmatrix}1 & -1\\-1 & 1\end{pmatrix}\]
\[|+\rangle\langle +| + |-\rangle\langle -| = \frac{1}{2}\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix} + \frac{1}{2}\begin{pmatrix}1 & -1\\-1 & 1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}2 & 0\\0 & 2\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\]
\[\boxed{|+\rangle\langle +| + |-\rangle\langle -| = \hat{1} = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}}\]

Verification: The completeness relation holding is a consequence of \(\{|+\rangle, |-\rangle\}\) forming an orthonormal basis. Indeed, \(\langle+|-\rangle = \frac{1}{2}(1\cdot1 + 1\cdot(-1)) = 0\) ✓, \(\langle+|+\rangle = \frac{1}{2}(1+1) = 1\) ✓, \(\langle-|-\rangle = \frac{1}{2}(1+1) = 1\) ✓. Since two orthonormal vectors in a 2-dimensional space form a basis, the completeness relation is satisfied.

Medium

M-1. Gram–Schmidt Orthogonalization

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Strategy: Following equations (B.22)–(B.24), normalize \(|v_1\rangle\) to construct \(|e_1\rangle\), then remove the \(|e_1\rangle\) component from \(|v_2\rangle\) and normalize to construct \(|e_2\rangle\).

Step 1: Constructing \(|e_1\rangle\)

\[|v_1\rangle = \begin{pmatrix} 1 \\ i \end{pmatrix}\]
\[\langle v_1|v_1\rangle = |1|^2 + |i|^2 = 1 + 1 = 2\]
\[|e_1\rangle = \frac{|v_1\rangle}{\sqrt{2}} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}\]

Step 2: Constructing \(|e_2\rangle\)

First, compute \(\langle e_1|v_2\rangle\). Since \(\langle e_1| = \frac{1}{\sqrt{2}}(1^*,\; i^*) = \frac{1}{\sqrt{2}}(1,\; -i)\):

\[\langle e_1|v_2\rangle = \frac{1}{\sqrt{2}}(1 \cdot 1 + (-i) \cdot 1) = \frac{1 - i}{\sqrt{2}}\]

Remove the \(|e_1\rangle\) component:

\[|w_2\rangle = |v_2\rangle - \langle e_1|v_2\rangle\,|e_1\rangle = \begin{pmatrix}1\\1\end{pmatrix} - \frac{1-i}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}\]
\[= \begin{pmatrix}1\\1\end{pmatrix} - \frac{1-i}{2}\begin{pmatrix}1\\i\end{pmatrix}\]

Compute the product of \(\frac{1-i}{2}\) with each component:

\[\frac{(1-i) \cdot 1}{2} = \frac{1-i}{2}\]
\[\frac{(1-i) \cdot i}{2} = \frac{i - i^2}{2} = \frac{i + 1}{2} = \frac{1+i}{2}\]

Therefore

\[|w_2\rangle = \begin{pmatrix}1\\1\end{pmatrix} - \begin{pmatrix}\frac{1-i}{2}\\\frac{1+i}{2}\end{pmatrix} = \begin{pmatrix}1 - \frac{1-i}{2}\\1 - \frac{1+i}{2}\end{pmatrix} = \begin{pmatrix}\frac{2 - 1 + i}{2}\\\frac{2 - 1 - i}{2}\end{pmatrix} = \begin{pmatrix}\frac{1+i}{2}\\\frac{1-i}{2}\end{pmatrix}\]

Compute the norm of \(|w_2\rangle\):

\[\langle w_2|w_2\rangle = \left|\frac{1+i}{2}\right|^2 + \left|\frac{1-i}{2}\right|^2 = \frac{2}{4} + \frac{2}{4} = 1\]

Since the norm is 1, no normalization is needed:

\[|e_2\rangle = |w_2\rangle = \begin{pmatrix}\frac{1+i}{2}\\\frac{1-i}{2}\end{pmatrix}\]
\[\boxed{|e_1\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}, \quad |e_2\rangle = \frac{1}{2}\begin{pmatrix}1+i\\1-i\end{pmatrix}}\]

Verification of Orthonormality

\(\langle e_1|e_1\rangle\):

\[\langle e_1|e_1\rangle = \frac{1}{2}(|1|^2 + |i|^2) = \frac{1}{2}(1 + 1) = 1 \quad \checkmark\]

\(\langle e_2|e_2\rangle\):

\[\langle e_2|e_2\rangle = \frac{1}{4}(|1+i|^2 + |1-i|^2) = \frac{1}{4}(2 + 2) = 1 \quad \checkmark\]

\(\langle e_1|e_2\rangle\):

\[\langle e_1|e_2\rangle = \frac{1}{\sqrt{2}}(1,\;-i) \cdot \frac{1}{2}\begin{pmatrix}1+i\\1-i\end{pmatrix} = \frac{1}{2\sqrt{2}}\bigl[1\cdot(1+i) + (-i)\cdot(1-i)\bigr]\]
\[= \frac{1}{2\sqrt{2}}\bigl[(1+i) + (-i + i^2)\bigr] = \frac{1}{2\sqrt{2}}\bigl[(1+i) + (-i - 1)\bigr] = \frac{1}{2\sqrt{2}} \cdot 0 = 0 \quad \checkmark\]

Orthonormality has been verified.

M-2. Proof that Eigenvalues of Hermitian Matrices are Real

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Solution Strategy: Multiply both sides of the eigenvalue equation by a bra, and use Hermiticity to show that \(\lambda = \lambda^*\).

Proof:

Multiplying both sides of \(\hat{A}|\lambda\rangle = \lambda|\lambda\rangle\) (\(|\lambda\rangle \neq \mathbf{0}\)) from the left by \(\langle\lambda|\):

\[\langle\lambda|\hat{A}|\lambda\rangle = \lambda\langle\lambda|\lambda\rangle \tag{1}\]

Next, consider the complex conjugate of the left-hand side of (1). From the Hermitian property of the inner product (Eq. (B.7)):

\[\langle\lambda|\hat{A}|\lambda\rangle^* = \langle\hat{A}\lambda|\lambda\rangle^{*\,*} \text{ — rather than this, we proceed directly.}\]

Proceeding more carefully. Taking the complex conjugate of (1):

\[\langle\lambda|\hat{A}|\lambda\rangle^* = \lambda^*\langle\lambda|\lambda\rangle^* = \lambda^*\langle\lambda|\lambda\rangle \tag{2}\]

where the last equality holds because \(\langle\lambda|\lambda\rangle\) is real (non-negative).

On the other hand, using \(\hat{A}^\dagger = \hat{A}\):

\[\langle\lambda|\hat{A}|\lambda\rangle^* = \langle\lambda|\hat{A}^\dagger|\lambda\rangle = \langle\lambda|\hat{A}|\lambda\rangle \tag{3}\]

where the first equality follows from the general relation \(\langle\psi|\hat{O}|\phi\rangle^* = \langle\phi|\hat{O}^\dagger|\psi\rangle\) (in the case \(|\psi\rangle = |\phi\rangle = |\lambda\rangle\)).

From (3), \(\langle\lambda|\hat{A}|\lambda\rangle\) is real.

Comparing (1) and (2):

\[\lambda\langle\lambda|\lambda\rangle = \lambda^*\langle\lambda|\lambda\rangle\]

Since \(|\lambda\rangle \neq \mathbf{0}\), we have \(\langle\lambda|\lambda\rangle > 0\), so dividing both sides by \(\langle\lambda|\lambda\rangle\):

\[\boxed{\lambda = \lambda^*}\]

Therefore \(\lambda\) is real. \(\blacksquare\)

Verification: The matrix \(A = \begin{pmatrix}2 & -i \\ i & 3\end{pmatrix}\) obtained in D5 was Hermitian. Finding the eigenvalues, the characteristic equation \((2-\lambda)(3-\lambda) - (-i)(i) = 0\) gives \(\lambda^2 - 5\lambda + 6 - 1 = 0\), \(\lambda^2 - 5\lambda + 5 = 0\), \(\lambda = \frac{5 \pm \sqrt{5}}{2}\). Indeed real ✓.

M-3. Eigenvectors belonging to different eigenvalues of a Hermitian matrix are orthogonal

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Solution Strategy: Derive an equality involving \(\langle\lambda_1|\lambda_2\rangle\) from the two eigenvalue equations, and conclude \(\langle\lambda_1|\lambda_2\rangle = 0\) from \(\lambda_1 \neq \lambda_2\).

Proof:

Eigenvalue equations:

\[\hat{A}|\lambda_1\rangle = \lambda_1|\lambda_1\rangle \tag{i}\]
\[\hat{A}|\lambda_2\rangle = \lambda_2|\lambda_2\rangle \tag{ii}\]

Multiplying both sides of (ii) from the left by \(\langle\lambda_1|\):

\[\langle\lambda_1|\hat{A}|\lambda_2\rangle = \lambda_2\langle\lambda_1|\lambda_2\rangle \tag{iii}\]

Taking the Hermitian conjugate of (i) (since \(\hat{A}^\dagger = \hat{A}\), and from S2, \(\lambda_1\) is real so \(\lambda_1^* = \lambda_1\)):

\[\langle\lambda_1|\hat{A}^\dagger = \lambda_1^*\langle\lambda_1| \implies \langle\lambda_1|\hat{A} = \lambda_1\langle\lambda_1|\]

Multiplying this from the right by \(|\lambda_2\rangle\):

\[\langle\lambda_1|\hat{A}|\lambda_2\rangle = \lambda_1\langle\lambda_1|\lambda_2\rangle \tag{iv}\]

Comparing (iii) and (iv):

\[\lambda_2\langle\lambda_1|\lambda_2\rangle = \lambda_1\langle\lambda_1|\lambda_2\rangle\]
\[(\lambda_1 - \lambda_2)\langle\lambda_1|\lambda_2\rangle = 0\]

Since \(\lambda_1 \neq \lambda_2\), we have \(\lambda_1 - \lambda_2 \neq 0\), therefore:

\[\boxed{\langle\lambda_1|\lambda_2\rangle = 0}\]

\(\blacksquare\)

Verification: We can confirm orthogonality by finding the eigenvectors corresponding to the eigenvalues \(\lambda_\pm = \frac{5 \pm \sqrt{5}}{2}\) of the matrix \(A = \begin{pmatrix}2 & -i \\ i & 3\end{pmatrix}\) used in the verification of S2. For \(\lambda_+ = \frac{5+\sqrt{5}}{2}\), from \((2 - \lambda_+)v_1 - iv_2 = 0\) we get \(v_2 = \frac{2-\lambda_+}{i} \cdot v_1 = i(\lambda_+ - 2)v_1\). Similarly, for \(\lambda_-\) we get \(v_2 = i(\lambda_- - 2)v_1\). Computing the inner product gives \(1 + i(\lambda_+ - 2) \cdot (-i)(\lambda_- - 2) = 1 + (\lambda_+ - 2)(\lambda_- - 2)\). By Vieta's formulas, \((\lambda_+ - 2)(\lambda_- - 2) = \lambda_+\lambda_- - 2(\lambda_+ + \lambda_-) + 4 = 5 - 10 + 4 = -1\). Therefore the inner product \(= 1 + (-1) = 0\) ✓.

M-4. Derivation of Commutator Identity

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Solution Strategy: Expand each commutator according to its definition, write out all 12 terms, and verify the cancellations.

Proof:

Using the definition of the commutator \([\hat{X}, \hat{Y}] = \hat{X}\hat{Y} - \hat{Y}\hat{X}\), we expand each term.

First term:

\[[\hat{A}, [\hat{B}, \hat{C}]] = [\hat{A}, \hat{B}\hat{C} - \hat{C}\hat{B}] = \hat{A}\hat{B}\hat{C} - \hat{A}\hat{C}\hat{B} - \hat{B}\hat{C}\hat{A} + \hat{C}\hat{B}\hat{A}\]

Second term: Cyclic permutation \(\hat{A} \to \hat{B}\), \(\hat{B} \to \hat{C}\), \(\hat{C} \to \hat{A}\):

\[[\hat{B}, [\hat{C}, \hat{A}]] = \hat{B}\hat{C}\hat{A} - \hat{B}\hat{A}\hat{C} - \hat{C}\hat{A}\hat{B} + \hat{A}\hat{C}\hat{B}\]

Third term: Another cyclic permutation:

\[[\hat{C}, [\hat{A}, \hat{B}]] = \hat{C}\hat{A}\hat{B} - \hat{C}\hat{B}\hat{A} - \hat{A}\hat{B}\hat{C} + \hat{B}\hat{A}\hat{C}\]

Summing all 12 terms:

Term First Second Third Total
\(\hat{A}\hat{B}\hat{C}\) \(+1\) \(0\) \(-1\) \(0\)
\(\hat{A}\hat{C}\hat{B}\) \(-1\) \(+1\) \(0\) \(0\)
\(\hat{B}\hat{C}\hat{A}\) \(-1\) \(+1\) \(0\) \(0\)
\(\hat{C}\hat{B}\hat{A}\) \(+1\) \(0\) \(-1\) \(0\)
\(\hat{B}\hat{A}\hat{C}\) \(0\) \(-1\) \(+1\) \(0\)
\(\hat{C}\hat{A}\hat{B}\) \(0\) \(-1\) \(+1\) \(0\)

All terms cancel:

\[\boxed{[\hat{A}, [\hat{B}, \hat{C}]] + [\hat{B}, [\hat{C}, \hat{A}]] + [\hat{C}, [\hat{A}, \hat{B}]] = 0}\]

\(\blacksquare\)

Verification: This can be confirmed concretely using \(A = \sigma_z\), \(B = \sigma_x\) from D7, together with \(C = \sigma_y\). Using \([\sigma_z, \sigma_x] = 2i\sigma_y\), \([\sigma_x, \sigma_y] = 2i\sigma_z\), \([\sigma_y, \sigma_z] = 2i\sigma_x\), we get \([\sigma_z, 2i\sigma_z] + [\sigma_x, 2i\sigma_x] + [\sigma_y, 2i\sigma_y] = 2i[\sigma_z, \sigma_z] + 2i[\sigma_x, \sigma_x] + 2i[\sigma_y, \sigma_y] = 0 + 0 + 0 = 0\) ✓ (any operator commutes with itself).

M-5. Proof of the Schwarz Inequality

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Solution Strategy: Introduce an auxiliary vector \(|w\rangle = |\psi\rangle - t|\phi\rangle\) and derive the inequality from \(\langle w|w\rangle \geq 0\).

Proof:

Case 1: When \(|\phi\rangle = \mathbf{0}\).

Right-hand side \(= \langle\phi|\phi\rangle \cdot \langle\psi|\psi\rangle = 0\), left-hand side \(= |\langle\phi|\psi\rangle|^2 = |0|^2 = 0\). Therefore \(0 \leq 0\) and the inequality holds trivially.

Case 2: When \(|\phi\rangle \neq \mathbf{0}\).

For any complex number \(t\), let \(|w\rangle = |\psi\rangle - t|\phi\rangle\). By the positive-definiteness of the inner product:

\[\langle w|w\rangle \geq 0\]
\[\langle w|w\rangle = \langle\psi|\psi\rangle - t\langle\psi|\phi\rangle - t^*\langle\phi|\psi\rangle + |t|^2\langle\phi|\phi\rangle \geq 0 \tag{*}\]

We now choose \(t\) optimally. To find the \(t\) that minimizes \(\langle w|w\rangle\), set

\[t = \frac{\langle\phi|\psi\rangle}{\langle\phi|\phi\rangle}\]

(this is well-defined since \(\langle\phi|\phi\rangle > 0\)). Then:

\[t^* = \frac{\langle\phi|\psi\rangle^*}{\langle\phi|\phi\rangle} = \frac{\langle\psi|\phi\rangle}{\langle\phi|\phi\rangle}\]
\[|t|^2 = \frac{|\langle\phi|\psi\rangle|^2}{\langle\phi|\phi\rangle^2}\]

Substituting into (\(*\)):

\[\langle\psi|\psi\rangle - \frac{\langle\phi|\psi\rangle}{\langle\phi|\phi\rangle}\langle\psi|\phi\rangle - \frac{\langle\psi|\phi\rangle}{\langle\phi|\phi\rangle}\langle\phi|\psi\rangle + \frac{|\langle\phi|\psi\rangle|^2}{\langle\phi|\phi\rangle^2}\langle\phi|\phi\rangle \geq 0\]

Noting that \(\langle\psi|\phi\rangle \cdot \langle\phi|\psi\rangle = |\langle\phi|\psi\rangle|^2\) and simplifying:

\[\langle\psi|\psi\rangle - \frac{|\langle\phi|\psi\rangle|^2}{\langle\phi|\phi\rangle} - \frac{|\langle\phi|\psi\rangle|^2}{\langle\phi|\phi\rangle} + \frac{|\langle\phi|\psi\rangle|^2}{\langle\phi|\phi\rangle} \geq 0\]
\[\langle\psi|\psi\rangle - \frac{|\langle\phi|\psi\rangle|^2}{\langle\phi|\phi\rangle} \geq 0\]

Multiplying both sides by \(\langle\phi|\phi\rangle > 0\):

\[\boxed{|\langle\phi|\psi\rangle|^2 \leq \langle\phi|\phi\rangle \cdot \langle\psi|\psi\rangle}\]

\(\blacksquare\)

Equality condition: Equality holds when \(\langle w|w\rangle = 0\), i.e., when \(|w\rangle = \mathbf{0}\). This means

\[|\psi\rangle = t|\phi\rangle = \frac{\langle\phi|\psi\rangle}{\langle\phi|\phi\rangle}|\phi\rangle\]

That is, equality holds when \(|\psi\rangle\) and \(|\phi\rangle\) are linearly dependent (one is a scalar multiple of the other).

Verification: Let us check with a concrete example. For \(|\psi\rangle = \begin{pmatrix}1\\0\end{pmatrix}\), \(|\phi\rangle = \begin{pmatrix}1\\1\end{pmatrix}\): - Left-hand side: \(|\langle\phi|\psi\rangle|^2 = |1|^2 = 1\) - Right-hand side: \(\langle\phi|\phi\rangle \cdot \langle\psi|\psi\rangle = 2 \cdot 1 = 2\) - \(1 \leq 2\) ✓ (strict inequality since linearly independent)

For \(|\psi\rangle = \begin{pmatrix}2\\2\end{pmatrix} = 2|\phi\rangle/\sqrt{?}\)... \(|\psi\rangle = 2\begin{pmatrix}1\\1\end{pmatrix} = 2|\phi\rangle\): - Left-hand side: \(|\langle\phi|\psi\rangle|^2 = |2 \cdot 2|^2 = |4|^2 = 16\) - Right-hand side: \(2 \cdot 8 = 16\) - \(16 = 16\) ✓ (equality holds since linearly dependent)

Advanced

A-1. Basis Transformation by Unitary Matrices and Transformation Rules for Matrix Representations

→ Back to problem

(a) Proof that \(U\) is a unitary matrix

Solution strategy: Compute \((U^\dagger U)_{jk}\) and insert the completeness relation for \(\{|e_l\rangle\}\).

Proof:

From \(U_{jk} = \langle e_j|f_k\rangle\), we have \((U^\dagger)_{jk} = U_{kj}^* = \langle e_k|f_j\rangle^* = \langle f_j|e_k\rangle\).

Computing \((U^\dagger U)_{jk}\):

\[(U^\dagger U)_{jk} = \sum_{l=1}^{N} (U^\dagger)_{jl}\, U_{lk} = \sum_{l=1}^{N} \langle f_j|e_l\rangle\langle e_l|f_k\rangle\]

Using the completeness relation for \(\{|e_l\rangle\}\), \(\sum_{l=1}^{N}|e_l\rangle\langle e_l| = \hat{1}\):

\[(U^\dagger U)_{jk} = \langle f_j|\left(\sum_{l=1}^{N}|e_l\rangle\langle e_l|\right)|f_k\rangle = \langle f_j|\hat{1}|f_k\rangle = \langle f_j|f_k\rangle = \delta_{jk}\]

The last equality follows from the orthonormality of \(\{|f_k\rangle\}\). Therefore \(U^\dagger U = \hat{1}\).

Similarly, computing \((UU^\dagger)_{jk}\):

\[(UU^\dagger)_{jk} = \sum_{l=1}^{N} U_{jl}\,(U^\dagger)_{lk} = \sum_{l=1}^{N} \langle e_j|f_l\rangle\langle f_l|e_k\rangle\]

Using the completeness relation for \(\{|f_l\rangle\}\), \(\sum_{l=1}^{N}|f_l\rangle\langle f_l| = \hat{1}\):

\[(UU^\dagger)_{jk} = \langle e_j|\hat{1}|e_k\rangle = \delta_{jk}\]
\[\boxed{U^\dagger U = UU^\dagger = \hat{1}}\]

Therefore \(U\) is a unitary matrix. \(\blacksquare\)

(b) Transformation rule for matrix representations

Solution strategy: Insert the completeness relation for \(\{|f_l\rangle\}\) into \(A^{(e)}_{jk} = \langle e_j|\hat{A}|e_k\rangle\).

Proof:

\[A^{(e)}_{jk} = \langle e_j|\hat{A}|e_k\rangle = \langle e_j|\hat{1}\,\hat{A}\,\hat{1}|e_k\rangle\]

Inserting the completeness relation for \(\{|f_l\rangle\}\) on both sides of \(\hat{A}\):

\[= \sum_{l=1}^{N}\sum_{m=1}^{N} \langle e_j|f_l\rangle\langle f_l|\hat{A}|f_m\rangle\langle f_m|e_k\rangle\]
\[= \sum_{l,m} U_{jl}\, A^{(f)}_{lm}\, (U^\dagger)_{mk}\]

This is precisely the definition of matrix multiplication, so:

\[\boxed{A^{(e)} = U\, A^{(f)}\, U^\dagger}\]

\(\blacksquare\)

(c) Basis independence of the trace

Solution strategy: Use the cyclic property of the trace.

Proof:

\[\mathrm{Tr}(A^{(e)}) = \mathrm{Tr}(U\, A^{(f)}\, U^\dagger)\]

Using the cyclic property of the trace, \(\mathrm{Tr}(XYZ) = \mathrm{Tr}(ZXY)\):

\[= \mathrm{Tr}(U^\dagger U\, A^{(f)}) = \mathrm{Tr}(\hat{1}\, A^{(f)}) = \mathrm{Tr}(A^{(f)})\]
\[\boxed{\mathrm{Tr}(A^{(e)}) = \mathrm{Tr}(A^{(f)})}\]

Therefore, the trace of an operator is independent of the choice of basis. \(\blacksquare\)

Verification (direct confirmation of the cyclic property): We compute directly using indices.

\[\mathrm{Tr}(A^{(e)}) = \sum_k A^{(e)}_{kk} = \sum_k \langle e_k|\hat{A}|e_k\rangle\]

Using the completeness relation \(\sum_k |e_k\rangle\langle e_k| = \hat{1}\), this is the representation of \(\mathrm{Tr}(\hat{A})\) in the basis \(\{|e_k\rangle\}\). Similarly, in the basis \(\{|f_k\rangle\}\):

\[\sum_k \langle f_k|\hat{A}|f_k\rangle = \sum_k \langle f_k|\hat{A}\hat{1}|f_k\rangle = \sum_{k,l} \langle f_k|\hat{A}|e_l\rangle\langle e_l|f_k\rangle = \sum_l \langle e_l|\left(\sum_k |f_k\rangle\langle f_k|\right)\hat{A}|e_l\rangle = \sum_l \langle e_l|\hat{A}|e_l\rangle\]

Indeed they agree ✓.

A-2. Tensor Product Space and Construction of the Bell Basis

→ Back to problem

(a) Proof that the Bell basis forms an orthonormal basis

Solution strategy: Using the inner product in the tensor product space \(\langle jk|lm\rangle = \delta_{jl}\delta_{km}\), compute the inner product for all combinations of the four Bell states.

Norm of each vector:

\[\langle\Phi^+|\Phi^+\rangle = \frac{1}{2}(\langle 00| + \langle 11|)(|00\rangle + |11\rangle) = \frac{1}{2}(\langle 00|00\rangle + \langle 00|11\rangle + \langle 11|00\rangle + \langle 11|11\rangle)\]
\[= \frac{1}{2}(1 + 0 + 0 + 1) = 1 \quad \checkmark\]

By similar calculations (since the squares of all signs yield \(+1\)):

\[\langle\Phi^-|\Phi^-\rangle = \frac{1}{2}(1 + 1) = 1, \quad \langle\Psi^+|\Psi^+\rangle = 1, \quad \langle\Psi^-|\Psi^-\rangle = 1 \quad \checkmark\]

Inner products between different vectors:

Since \(|\Phi^\pm\rangle\) are linear combinations of \(|00\rangle, |11\rangle\) and \(|\Psi^\pm\rangle\) are linear combinations of \(|01\rangle, |10\rangle\), the inner products between the \(\Phi\) family and the \(\Psi\) family are automatically 0:

\[\langle\Phi^\pm|\Psi^\pm\rangle = 0, \quad \langle\Phi^\pm|\Psi^\mp\rangle = 0\]

(Because the inner products of distinct standard basis vectors are all 0.)

The remaining combinations:

\[\langle\Phi^+|\Phi^-\rangle = \frac{1}{2}(\langle 00| + \langle 11|)(|00\rangle - |11\rangle) = \frac{1}{2}(1 - 0 + 0 - 1) = 0 \quad \checkmark\]
\[\langle\Psi^+|\Psi^-\rangle = \frac{1}{2}(\langle 01| + \langle 10|)(|01\rangle - |10\rangle) = \frac{1}{2}(1 - 0 + 0 - 1) = 0 \quad \checkmark\]

From the above, the four Bell states are mutually orthogonal, each with norm 1. Since \(\mathbb{C}^2 \otimes \mathbb{C}^2\) is a 4-dimensional space, four orthonormal vectors form a basis.

\[\boxed{\{|\Phi^+\rangle, |\Phi^-\rangle, |\Psi^+\rangle, |\Psi^-\rangle\} \text{ is an orthonormal basis for } \mathbb{C}^2 \otimes \mathbb{C}^2.}\]

(b) Proof that \(|\Phi^+\rangle\) is an entangled state

Solution strategy: Use proof by contradiction. Assume \(|\Phi^+\rangle = |a\rangle \otimes |b\rangle\) and derive a contradiction.

Proof:

Assume \(|\Phi^+\rangle = |a\rangle \otimes |b\rangle\) can be written as a product state. Let

\[|a\rangle = a_0|0\rangle + a_1|1\rangle, \quad |b\rangle = b_0|0\rangle + b_1|1\rangle\]

Then:

\[|a\rangle \otimes |b\rangle = a_0 b_0|00\rangle + a_0 b_1|01\rangle + a_1 b_0|10\rangle + a_1 b_1|11\rangle\]

Setting this equal to \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) = \frac{1}{\sqrt{2}}|00\rangle + 0 \cdot |01\rangle + 0 \cdot |10\rangle + \frac{1}{\sqrt{2}}|11\rangle\) and comparing coefficients:

\[a_0 b_0 = \frac{1}{\sqrt{2}} \tag{I}\]
\[a_0 b_1 = 0 \tag{II}\]
\[a_1 b_0 = 0 \tag{III}\]
\[a_1 b_1 = \frac{1}{\sqrt{2}} \tag{IV}\]

From (I), \(a_0 \neq 0\) and \(b_0 \neq 0\).

From (II), since \(a_0 \neq 0\), we have \(b_1 = 0\).

From (IV), since \(b_1 = 0\), we get \(a_1 b_1 = 0 \neq \frac{1}{\sqrt{2}}\). Contradiction.

\[\boxed{|\Phi^+\rangle \text{ is non-separable (an entangled state).}}\]

\(\blacksquare\)

Verification: We confirm that a contradiction also arises via an alternative path. From (IV), \(a_1 \neq 0\) and \(b_1 \neq 0\). But from (II), either \(a_0 = 0\) or \(b_1 = 0\). Since \(b_1 \neq 0\), we have \(a_0 = 0\). Then from (I), \(a_0 b_0 = 0 \neq \frac{1}{\sqrt{2}}\). Contradiction ✓. Both paths lead to a contradiction.

(c) Expansion of an arbitrary state in the Bell basis

Solution strategy: Express the standard basis in terms of the Bell basis by inverting the definitions, then substitute. Alternatively, find each expansion coefficient via inner products.

Calculation:

First, invert the Bell basis definitions to express the standard basis in terms of the Bell basis:

\[|\Phi^+\rangle + |\Phi^-\rangle = \sqrt{2}\,|00\rangle \implies |00\rangle = \frac{1}{\sqrt{2}}(|\Phi^+\rangle + |\Phi^-\rangle)\]
\[|\Phi^+\rangle - |\Phi^-\rangle = \sqrt{2}\,|11\rangle \implies |11\rangle = \frac{1}{\sqrt{2}}(|\Phi^+\rangle - |\Phi^-\rangle)\]
\[|\Psi^+\rangle + |\Psi^-\rangle = \sqrt{2}\,|01\rangle \implies |01\rangle = \frac{1}{\sqrt{2}}(|\Psi^+\rangle + |\Psi^-\rangle)\]
\[|\Psi^+\rangle - |\Psi^-\rangle = \sqrt{2}\,|10\rangle \implies |10\rangle = \frac{1}{\sqrt{2}}(|\Psi^+\rangle - |\Psi^-\rangle)\]

Substituting into \(|\psi\rangle = \alpha|00\rangle + \beta|01\rangle + \gamma|10\rangle + \delta|11\rangle\):

\[|\psi\rangle = \frac{\alpha}{\sqrt{2}}(|\Phi^+\rangle + |\Phi^-\rangle) + \frac{\beta}{\sqrt{2}}(|\Psi^+\rangle + |\Psi^-\rangle) + \frac{\gamma}{\sqrt{2}}(|\Psi^+\rangle - |\Psi^-\rangle) + \frac{\delta}{\sqrt{2}}(|\Phi^+\rangle - |\Phi^-\rangle)\]

Grouping by Bell basis states:

\[|\psi\rangle = \frac{\alpha + \delta}{\sqrt{2}}|\Phi^+\rangle + \frac{\alpha - \delta}{\sqrt{2}}|\Phi^-\rangle + \frac{\beta + \gamma}{\sqrt{2}}|\Psi^+\rangle + \frac{\beta - \gamma}{\sqrt{2}}|\Psi^-\rangle\]
\[\boxed{|\psi\rangle = \frac{\alpha + \delta}{\sqrt{2}}|\Phi^+\rangle + \frac{\alpha - \delta}{\sqrt{2}}|\Phi^-\rangle + \frac{\beta + \gamma}{\sqrt{2}}|\Psi^+\rangle + \frac{\beta - \gamma}{\sqrt{2}}|\Psi^-\rangle}\]

Summarizing the expansion coefficients:

\[c_{\Phi^+} = \frac{\alpha + \delta}{\sqrt{2}}, \quad c_{\Phi^-} = \frac{\alpha - \delta}{\sqrt{2}}, \quad c_{\Psi^+} = \frac{\beta + \gamma}{\sqrt{2}}, \quad c_{\Psi^-} = \frac{\beta - \gamma}{\sqrt{2}}\]

Verification: Check the normalization condition.

\[\sum |c_i|^2 = \frac{|\alpha+\delta|^2 + |\alpha-\delta|^2 + |\beta+\gamma|^2 + |\beta-\gamma|^2}{2}\]

By the parallelogram law, \(|\alpha + \delta|^2 + |\alpha - \delta|^2 = 2(|\alpha|^2 + |\delta|^2)\), and similarly \(|\beta + \gamma|^2 + |\beta - \gamma|^2 = 2(|\beta|^2 + |\gamma|^2)\), so:

\[\sum |c_i|^2 = \frac{2(|\alpha|^2 + |\delta|^2) + 2(|\beta|^2 + |\gamma|^2)}{2} = |\alpha|^2 + |\beta|^2 + |\gamma|^2 + |\delta|^2 = 1 \quad \checkmark\]

Also, as a special case, when \(|\psi\rangle = |\Phi^+\rangle\) (\(\alpha = \delta = 1/\sqrt{2}\), \(\beta = \gamma = 0\)): \(c_{\Phi^+} = \frac{1/\sqrt{2} + 1/\sqrt{2}}{\sqrt{2}} = 1\), and all others are 0 ✓.