Solutions: Quantum Computation and Quantum Information by Nielsen and Chuang

Solutions: Quantum Computation and Quantum Information by Nielsen and Chuang

Chapter 2

2.1

$$ \begin{pmatrix} 1 \\\
-1 \end{pmatrix} + \begin{pmatrix} 1 \\\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\\ 1 \end{pmatrix} =0 $$


2.2

$$ \begin{eqnarray} &A&:V\rightarrow V,\,\, \mathrm{basis}:\left| 0 \right\rangle, \left| 1 \right\rangle \\ &A&\left| 0 \right\rangle=\left| 0 \right\rangle,\,\, A\left| 1 \right\rangle = \left| 1 \right\rangle \\ \end{eqnarray} $$

that is,

$$ \begin{eqnarray} &A&\left| 0 \right\rangle = \left| 0 \right\rangle = 1\left| 0 \right\rangle + 0\left| 1 \right\rangle \\ &A&\left| 1 \right\rangle = \left| 1 \right\rangle = 0\left| 0 \right\rangle + 1\left| 1 \right\rangle. \end{eqnarray} $$

Thus,

$$ \begin{equation} A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation} $$

2.3

$$ \begin{equation} A:V\rightarrow W,\,\,B:W\rightarrow X \end{equation} $$

From eq.(2.12),

$$ \begin{equation} A\left| v_j \right\rangle = \sum_i A_i {}_j \left| w_i \right\rangle, \,\, B\left| w_i \right\rangle = \sum_k B_k {}_i \left| x_k \right\rangle . \end{equation} $$

Then,

$$ \begin{eqnarray} B\left( A\left| v_j \right\rangle \right) &=& B \left( \sum_i A_i {}_j \left| w_i \right\rangle \right) \\ &=& \sum_i A_i {}_j \sum_k B_k {}_i \left| x_k \right\rangle \\ &=& \sum_i \sum_k B_k {}_i A_i {}_j \left| x_k \right\rangle \\ &=& \sum_k \left( \sum_i B_k {}_i A_i {}_j \right) \left| x_k \right\rangle \\ &=& \sum_k \left( BA \right)_k {}_j \left| x_k \right\rangle \end{eqnarray} $$

2.4

$$ \begin{eqnarray} &A&:V\rightarrow V,\,\, \mathrm{basis}:\left| 0 \right\rangle, \left| 1 \right\rangle \\ &A&\left| 0 \right\rangle=\left| 0 \right\rangle,\,\, A\left| 1 \right\rangle = \left| 1 \right\rangle \\ \end{eqnarray} $$

that is,

$$ \begin{eqnarray} &A&\left| 0 \right\rangle = \left| 0 \right\rangle = 1\left| 0 \right\rangle + 0\left| 1 \right\rangle \\ &A&\left| 1 \right\rangle = \left| 1 \right\rangle = 0\left| 0 \right\rangle + 1\left| 1 \right\rangle. \end{eqnarray} $$

Thus,

$$ \begin{equation} A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation} $$

2.5

Put

$$ \begin{eqnarray} &\left| v \right\rangle = \left( v_1\; v_2\; \cdots\; v_n \right)^T \\ &\left| w \right\rangle = \left( w_1\; w_2\; \cdots\; w_n \right)^T \end{eqnarray} $$

For (1),

$$ \begin{eqnarray} \left( \left| v \right\rangle,\, \sum_i \lambda_i \left| w_i \right\rangle \right) &=& \sum_j v^*_j \left( \sum_i \lambda_iw_i {}_j \right) \\ &=& \sum_i \lambda_i \left( \sum_j v^{*}_j w_i {}_j \right) \\ &=& \sum_i \lambda_i \left( \left| v \right\rangle,\, \left| w_i \right\rangle \right). \end{eqnarray} $$

For (2),

$$ \begin{eqnarray} \left( \left| v \right\rangle,\, \left| w \right\rangle \right) &=& \sum_i v^*_i w_i \\ &=& \sum_i \left( w^{*}_i v_i\right)^{*} \\ &=& \left( \left| w \right\rangle,\, \left| v \right\rangle \right)^{*}. \end{eqnarray} $$

For (3),

$$ \begin{eqnarray} \left( \left| v \right\rangle,\, \left| v \right\rangle \right) &=& \sum_i v^*_i v_i \\ &=& \sum_i |v_i|^2 \ge 0. \end{eqnarray} $$

2.6

$$ \begin{eqnarray} \left( \sum_i \lambda_i \left| w_i \right\rangle,\, \left| v \right\rangle \right) &=& \left( \left| v \right\rangle,\, \sum_i \lambda_i \left| w_i \right\rangle \right)^* \\ &=& \left( \sum_i \lambda_i \left( \left| v \right\rangle,\, \left| w_i \right\rangle \right) \right)^* \\ &=& \sum_i \lambda^*_i \left( \left| v \right\rangle,\, \left| w_i \right\rangle \right)^* \\ &=& \sum_i \lambda^*_i \left( \left| w_i \right\rangle,\, \left| v \right\rangle \right) \end{eqnarray} $$

2.7

$$ \begin{equation} \left( \left| v \right\rangle,\, \left| w \right\rangle \right) = \left\langle v | w \right\rangle = 1\times 1 + 1\times (-1) = 0. \end{equation} $$

Normalized forms of $\left| v \right\rangle$ and $\left| w \right\rangle$ are,

$$ \begin{eqnarray} &{}&\frac{\left| v \right\rangle}{\left| \left| v \right\rangle \right|} = \frac{\left| v \right\rangle}{\sqrt{\left\langle v | v \right\rangle}} = \frac{\left| v \right\rangle}{\sqrt{2}},\\ &{}&\frac{\left| w \right\rangle}{\left| \left| w \right\rangle \right|} = \frac{\left| w \right\rangle}{\sqrt{\left\langle w | w \right\rangle}} = \frac{\left| w \right\rangle}{\sqrt{2}}. \end{eqnarray} $$

2.8

Use mathematical induction to prove it.

For k = 1, $$ \begin{eqnarray} \left| v_1 \right\rangle &=& \frac{\left| w_1 \right\rangle}{\left| \left| w_1 \right\rangle \right|}\\\
\left| v_2 \right\rangle &=& \frac{\left| w_2 \right\rangle - \left\langle v_1 | w_2 \right\rangle \left| v_1 \right\rangle}{\left| \left| w_2 \right\rangle - \left\langle v_1 | w_2 \right\rangle \left| v_1 \right\rangle \right|} \\\
\left\langle v_1 | v_2 \right\rangle &=& \frac{\left\langle v_1 | w_2 \right\rangle - \left\langle v_1 | w_2 \right\rangle \left\langle v_1 | v_1 \right\rangle}{\left| \left| w_1 \right\rangle \right| \left| \left| w_2 \right\rangle - \left\langle v_1 | w_2 \right\rangle \left| v_1 \right\rangle \right|} = 0 \end{eqnarray} $$ Thus, $\left| v_1 \right\rangle \perp \left| v_2 \right\rangle$.

Assume it is true for $k = k’,( k’ \le d-1)$. Then, for $k=k'+1$,

$$ \begin{eqnarray} \left| v_{k'+1} \right\rangle &=& \frac{\left| w_{k'+1} \right\rangle - \sum^{k'}_i \left\langle v_i | w_{k'+1} \right\rangle \left| v_i \right\rangle}{\left| \left| w_{k'+1} \right\rangle - \sum^{k'}_i \left\langle v_i | w_{k'+1} \right\rangle \left| v_i \right\rangle \right|} \\ \left\langle v_{k'} | v_{k'+1} \right\rangle &=& \frac{\left\langle v_{k'} | w_{k'+1} \right\rangle - \sum^{k'}_i \left\langle v_i | w_{k'+1} \right\rangle \left\langle v_k' | v_i \right\rangle}{\left| \left| w_{k'} \right\rangle - \sum^{k'-1}_i \left\langle v_i | w_{k'} \right\rangle \left| v_i \right\rangle \right| \left| \left| w_{k'+1} \right\rangle - \sum^{k'}_i \left\langle v_i | w_{k'+1} \right\rangle \left| v_i \right\rangle \right|} = 0. \end{eqnarray} $$

Thus, $\left| v_{k’} \right\rangle \perp\left| v_{k'+1} \right\rangle$.

Therefore, the Gram-Schmidt procedure produces an orthonormal basis for $V$.


2.9

$$ \begin{eqnarray} \sigma_0 &=&= \sum^1_{i,j=0} \left\langle j | \sigma_0 | i \right\rangle \left| j \right\rangle \left\langle i \right| = \left| 0 \right\rangle \left\langle 0 \right| + \left| 1 \right\rangle \left\langle 1 \right| \\ \sigma_1 &=& \left| 0 \right\rangle \left\langle 1 \right| + \left| 1 \right\rangle \left\langle 0 \right| \\ \sigma_2 &=& -i \left| 0 \right\rangle \left\langle 1 \right| + i\left| 1 \right\rangle \left\langle 0 \right| \\ \sigma_3 &=& \left| 0 \right\rangle \left\langle 0 \right| - \left| 1 \right\rangle \left\langle 1 \right| \end{eqnarray} $$

2.10

Define $T$ as

$$ \left| v_j \right\rangle \left\langle v_k \right| \left( v_i \right) \equiv T\left( \left| v_i \right\rangle \right). $$

Then, $T\left( \left| v_1 \right\rangle \right), T\left( \left| v_2 \right\rangle \right), \cdots, T\left( \left| v_d \right\rangle \right)$ is describes as

$$ \left( T\left( \left| v_1 \right\rangle \right),\, T\left( \left| v_2 \right\rangle \right),\, \cdots,\, T\left( \left| v_d \right\rangle \right) \right) = \left( \left( \left| v_1 \right\rangle \right),\, \left( \left| v_2 \right\rangle \right),\, \cdots,\, \left( \left| v_d \right\rangle \right) \right)A $$

where $A$ is the matrix representation of $T$.

Therefore,

$$ A_{lm} = \delta_{lj} \delta_{mk} $$

2.11

(1): X

$$ \mathrm{det}\left( X - \lambda I \right) = \begin{vmatrix} -\lambda & 1 \\ 1 & -\lambda \end{vmatrix} = \lambda^2-1 = 0\;\;\;\;\therefore \lambda = \pm 1 $$

For $\lambda = -1$,

$$ \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

This yields an eigenvector

$$ \left| \lambda_{-1} \right\rangle = \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle - \left| 1 \right\rangle \right). $$

For $\lambda = 1$,

$$ \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

This yields an eigenvector

$$ \left| \lambda_1 \right\rangle = \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right). $$

Therefore,

$$ \begin{eqnarray} X &=& \lambda_{-1} \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| + \lambda_1 \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| \\ &=& \frac{1}{2} \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right) \left( \left\langle 0 \right| + \left\langle 1 \right|\right) - \frac{1}{2} \left( \left| 0 \right\rangle - \left| 1 \right\rangle \right) \left( \left\langle 0 \right| - \left\langle 1 \right| \right) \end{eqnarray} $$

(2): Y

$$ \mathrm{det}\left( Y - \lambda I \right) = \begin{vmatrix} -\lambda & -i \\ i & -\lambda \end{vmatrix} = \lambda^2-1 = 0\;\;\;\;\therefore \lambda = \pm 1 $$

For $\lambda = -1$,

$$ \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix} \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

This yields an eigenvector

$$ \left| \lambda_{-1} \right\rangle = \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i \end{pmatrix} = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle - i \left| 1 \right\rangle \right). $$

For $\lambda = 1$,

$$ \begin{pmatrix} -1 & -i \\ i & -1 \end{pmatrix} \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

This yields an eigenvector

$$ \left| \lambda_1 \right\rangle = \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix} = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle + i \left| 1 \right\rangle \right). $$

Therefore,

$$ \begin{eqnarray} Y &=& \lambda_{-1} \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| + \lambda_1 \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| \\ &=& \frac{1}{2} \left( \left| 0 \right\rangle + i \left| 1 \right\rangle \right) \left( \left\langle 0 \right| + i \left\langle 1 \right| \right) - \frac{1}{2} \left( \left| 0 \right\rangle -i \left| 1 \right\rangle \right) \left( \left\langle 0 \right| -i \left\langle 1 \right| \right) \end{eqnarray} $$

(3): Z

$$ \mathrm{det} \left( Z - \lambda I \right) = \begin{vmatrix} 1-\lambda & 0 \\ 0 & -1-\lambda \end{vmatrix} = -\left( \lambda^2 -1 \right) = 0\;\;\;\;\therefore \lambda = \pm 1. $$

For $\lambda = -1$,

$$ \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

This yields an eigenvector

$$ \left| \lambda_{-1} \right\rangle = \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \left| 1 \right\rangle. $$

For $\lambda = 1$,

$$ \begin{pmatrix} 0 & 0 \\ 0 & -2 \end{pmatrix} \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

This yields an eigenvector

$$ \left| \lambda_1 \right\rangle = \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \left| 0 \right\rangle. $$

Therefore,

$$ \begin{eqnarray} Z &=& \lambda_{-1} \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| + \lambda_1 \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| \\ &=& \left| 0 \right\rangle \left\langle 0 \right| - \left| 1 \right\rangle \left\langle 1 \right|. \end{eqnarray} $$

2.12

Put $A = \begin{pmatrix} 1 & 0 \\\ 1& 1\end{pmatrix}$. Then,

$$ \mathrm{det} \left( A- \lambda I \right)= \begin{pmatrix} 1-\lambda & 0 \\ 1 & 1-\lambda \end{pmatrix} = \left(1-\lambda \right)^2 = 0\;\;\;\;\therefore \lambda = 1. $$

Now find the eigenvector $\left| \lambda_1 \right\rangle$.

$$ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

Thus,

$$ \left| \lambda_1 \right\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}. $$

Since $A = \begin{pmatrix} 1 & 0 \\ 1& 1\end{pmatrix}$ has only one eigenvector, it’s not diagonalizable.


2.13

$$ \begin{eqnarray} \left( \left| l \right\rangle,\, \left( \left| w \right\rangle \left\langle v \right| \right)^{\dagger} \left| m \right\rangle \right) &=& \left( \left( \left| w \right\rangle \left\langle v \right| \right) \left| l \right\rangle,\, \left| m \right\rangle \right) \\ &=& \left( \left\langle v | l \right\rangle \left| w \right\rangle,\, \left| m \right\rangle \right) \\ &=& \left( \left\langle v | l \right\rangle \right)^* \left\langle w | m \right\rangle \\ &=& \left\langle l | v \right\rangle \left\langle w | m \right\rangle \\ &=& \left\langle l \right| \left( \left| v \right\rangle \left\langle w \right| \right) \left| m \right\rangle \\ &=& \left( \left| l \right\rangle,\, \left( \left| v \right\rangle \left\langle w \right| \right) \left| m \right\rangle \right) \end{eqnarray} $$

2.14

$$ \begin{eqnarray} \left( \left( \sum_i a_i A_i \right)^{\dagger} \left| v \right\rangle, \left| w \right\rangle \right) &=& \left( \left| v \right\rangle, \sum_i a_i A_i \left| w \right\rangle \right) \\ &=& \sum_i a_i \left( A^{\dagger}_i \left| v \right\rangle, \left| w \right\rangle \right) \\ &=& \left( \sum_i a^*_i A^{\dagger}_i \left| v \right\rangle, \left| w \right\rangle \right) \end{eqnarray} $$

2.15

$$ \begin{eqnarray} \left( \left( A^{\dagger} \right)^{\dagger} \left| v \right\rangle, \left| w \right\rangle \right) &=& \left( \left| v \right\rangle, A^{\dagger} \left| w \right\rangle \right) \\ &=& \left( A^{\dagger} \left| w \right\rangle, \left| v \right\rangle \right)^* \\ &=& \left( \left| w \right\rangle, A \left| v \right\rangle \right)^* \\ &=& \left( A\left| v \right\rangle, \left| w \right\rangle \right) \end{eqnarray} $$

2.16

$$ \begin{eqnarray} P^2 &=& \left( \sum^k_{i=1} \left| i \right\rangle \left\langle i \right| \right) \left( \sum^k_{j=1} \left| j \right\rangle \left\langle j \right| \right) \\ &=& \sum^k_{i=1} \sum^k_{j=1} \left| i \right\rangle \left\langle i | j \right\rangle \left\langle j \right| \\ &=& \sum^k_{i=1} \sum^k_{j=1} \left| i \right\rangle \left\langle j \right| \delta_{ij} \\ &=& \sum^k_{i=1} \left| i \right\rangle \left\langle i \right| \\ &=& P \end{eqnarray} $$

2.17

A normal matrix A is diagonalized by an unitary matrix U, such that

$$ A = U^{\dagger} D U, $$
where $D$ is a diagonal matrix.

Hermitian conjugate of A is,

$$ A^{\dagger} = U^{\dagger}D^{\dagger}U. $$

If all eigenvalues are real, $D=D^{\dagger}$. Then,

$$ A^{\dagger} = U^{\dagger}D^{\dagger}U = U^{\dagger}DU = A. $$

2.18

$$ \begin{eqnarray} U \left| v \right\rangle &=& \lambda \left| v \right\rangle \;\;\;\; &(1) \\ \left\langle v \right| U^{\dagger} &=& \lambda^* \left\langle v \right| \;\;\;\; &(2) \end{eqnarray} $$

$(2) \times (1)$,

$$ \left\langle v \middle| U^{\dagger} U \middle| v \right\rangle = \left\langle v \middle| v \right\rangle = \lambda^* \lambda \left\langle v \middle| v \right\rangle\;\;\;\; \therefore \left| \lambda \right|^2 = 1 \rightarrow \lambda = \mathrm{e}^{i\theta} $$

2.19

$$ \begin{eqnarray} \sigma^{\dagger}_0 &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \sigma_0 \\ \sigma^{\dagger}_0 \sigma_0 &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \end{eqnarray} $$
$$ \begin{eqnarray} \sigma_1^{\dagger} &=& \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \sigma_1 \\ \sigma^{\dagger}_1 \sigma_1 &=& \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \end{eqnarray} $$
$$ \begin{eqnarray} \sigma^{\dagger}_2 &=& \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \sigma_2 \\ \sigma^{\dagger}_2 \sigma_2 &=& \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \end{eqnarray} $$
$$ \begin{eqnarray} \sigma^{\dagger}_3 &=& \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \sigma_3 \\ \sigma^{\dagger}_3 \sigma_3 &=& \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \end{eqnarray} $$

2.20

$$ \begin{eqnarray} A^{'}_{ij} &=& \left\langle v_i \middle| A \middle| v_j \right\rangle \\ &=& \sum_k \left\langle v_i \middle| w_k \right\rangle \left\langle w_k \middle| A \middle| v_j \right\rangle \\ &=& \sum_{k,l} \left\langle v_i \middle| w_k \right\rangle \left\langle w_k \middle| A \middle| w_l \right\rangle \left\langle w_l \middle| v_j \right\rangle \\ &=& \sum_{k,l} \left\langle v_i \middle| U \middle| v_k \right\rangle \left\langle w_k \middle| A \middle| w_l \right\rangle \left\langle v_l \middle| U^{\dagger} \middle| v_j \right\rangle \\ &=& \sum_{k,l} U_{ik} A^{''}_{kl} U^{\dagger}_{lj} \end{eqnarray} $$

where $U \equiv \sum_m \left| w_m \right\rangle \left\langle v_m \right|$.


2.22

$$ \begin{eqnarray} H \left| \lambda_i \right\rangle &=& \lambda_i \left| \lambda_i \right\rangle \;\;\;\; &(1)\\ H \left| \lambda_j \right\rangle &=& \lambda_j \left| \lambda_j \right\rangle \;\;\;\; &(2)\\ \end{eqnarray} $$

$\left\langle \lambda_j \right| \times$ (1) and $\left\langle \lambda_i \right| \times$ (1)

$$ \begin{eqnarray} \left\langle \lambda_j \middle| H \middle| \lambda_i \right\rangle &=& \lambda_i \left\langle \lambda_j \middle| \lambda_i \right\rangle \;\;\;\; &(3) \\ \left\langle \lambda_i \middle| H \middle| \lambda_j \right\rangle &=& \lambda_j \left\langle \lambda_i \middle| \lambda_j \right\rangle \;\;\;\; &(4) \\ \left\langle \lambda_j \middle| H^{*} \middle| \lambda_i \right\rangle &=& \left\langle \lambda_j \middle| H \middle| \lambda_i \right\rangle = \lambda_j^{*} \left\langle \lambda_j \middle| \lambda_i \right\rangle = \lambda_j \left\langle \lambda_j \middle| \lambda_i \right\rangle \;\;\;\; &(4)^{'} \end{eqnarray} $$

$(3) - (4)^{'}$

$$ \begin{eqnarray} 0 = \left( \lambda_i - \lambda_j \right) \left\langle \lambda_j \middle| \lambda_i \right\rangle \end{eqnarray} $$

Therefore, $\lambda_i \neq \lambda_j \rightarrow \left\langle \lambda_i \middle| \lambda_j \right\rangle = 0$.


2.23

$$ \begin{eqnarray} P &\equiv& \sum_i \left| i \right\rangle \left\langle i \right| \\ P^2 &=& \sum_{i,j} \left| i \right\rangle \left\langle i \middle| j \right\rangle \left\langle j \right| \\ &=& \sum_i \left| i \right\rangle \left\langle i \right| = P. \end{eqnarray} $$
$$ \begin{eqnarray} P \left| \lambda \right\rangle &=& \lambda \left| \lambda \right\rangle \\ P^2 \left| \lambda \right\rangle &=& \lambda P \left| \lambda \right\rangle = \lambda^2 \left| \lambda \right\rangle \end{eqnarray} $$

Since $P = P^2$,

$$ \lambda = \lambda^2 $$
$$ \Leftrightarrow \lambda \left( \lambda - 1 \right) = 0 \;\;\; \therefore \lambda = 0,1. $$

2.24

Let $A$ be a positive operator. $A$ can be written as,

$$ \begin{eqnarray} A &=& \frac{A+A^{\dagger}}{2}+i\frac{A-A^{\dagger}}{2i} \\ &=& B + iC, \;\text{where}\; B=\frac{A+A^{\dagger}}{2},\;C=\frac{A-A^{\dagger}}{2i}. \end{eqnarray} $$ Then, $$ \begin{eqnarray} \left\langle v \middle| A \middle| v\right\rangle &=& \left\langle v \middle| B+iC \middle| v \right\rangle \\ &=& \left\langle v \middle| B \middle| v \right\rangle + i \left\langle v \middle| C \middle| v \right\rangle \end{eqnarray} $$

Since $A$ is a positive operator, $\left\langle v \middle| C \middle| v \right\rangle = 0$. So, $C=0$. Therefore, A is Hermitian.


2.25

$$ \left\langle v \middle| \left( A^{\dagger}A \right) \middle| v \right\rangle = \left\langle v \middle| A^{\dagger}A \middle| v \right\rangle = \left| A\left| v \right\rangle \right|^2 \ge 0. $$

2.26

$$ \begin{eqnarray} \left| \psi \right\rangle^{\otimes 2} &=& \left( \frac{1}{\sqrt{2}} \right)^2 \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right) \otimes \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right) \\ &=& \frac{1}{2} \left( \left| 0 \right\rangle \otimes \left| 0 \right\rangle + \left| 0 \right\rangle \otimes \left| 1 \right\rangle + \left| 1 \right\rangle \otimes \left| 0 \right\rangle + \left| 1 \right\rangle \otimes \left| 1 \right\rangle \right) \\ &=& \frac{1}{2} \left( \left| 00 \right\rangle + \left| 01 \right\rangle + \left| 10 \right\rangle + \left| 11 \right\rangle \right) \\ \left| \psi \right\rangle^{\otimes 3} &=& \left( \frac{1}{\sqrt{2}} \right)^{3} \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right) \otimes \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right) \otimes \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right) \\ &=& \frac{1}{2\sqrt{2}} \left( \left|000\right\rangle + \left|001\right\rangle + \left|010\right\rangle + \left|011\right\rangle + \left|100\right\rangle + \left|101\right\rangle + \left|110\right\rangle + \left|110\right\rangle + \left|111\right\rangle \right) \end{eqnarray} $$

Kronecker Product:

$$ \begin{eqnarray} \left| \psi \right\rangle &=& \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \\ \left| \psi \right\rangle^{\otimes 2} &=& \frac{1}{2} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} \\ \left| \psi \right\rangle^{\otimes 3} &=& \frac{1}{2\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{2\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} \end{eqnarray} $$

2.27

(a)

$$ \begin{eqnarray} X \otimes Z &=& \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix} \end{eqnarray} $$

(b)

$$ I \otimes X = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} $$

(c)

$$ X \otimes I = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} $$

Since $I\otimes X \ne X\otimes I$, tensor product is not commutative.


2.28

$$ \begin{eqnarray} \left(A\otimes B\right)^* &=& \begin{pmatrix} a_{11}^*B^* & \cdots & a_{1N}^*B^* \\ \vdots & \ddots & \vdots \\ a_{N1}^*B^* & \cdots & a_{NN}^*B^* \end{pmatrix} = A^* \otimes B^* \\ \left(A \otimes B\right)^T &=& \begin{pmatrix} a_{11}B^T & \cdots & a_{N1}B^T \\ \vdots & \ddots & \vdots \\ a_{1N}B^T & \cdots & a_{NN}B^T \end{pmatrix} = A^T \otimes B^T \\ \left( A \otimes B \right) ^{\dagger} &=& \left( \left( A \otimes B \right) ^* \right) ^T = \left( A^* \right)^T \otimes \left( B^* \right)^T = A^{\dagger} \otimes B^{\dagger} \end{eqnarray} $$

2.29

Let $U_1$ and $U_2$ be unitary operators.

$$ \left( U_1 \otimes U_2 \right) \left( U_1 \otimes U_2 \right)^{\dagger} = \left( U_1 \otimes U_2 \right) \left( U_1^{\dagger} \otimes U_2^{\dagger} \right) = U_1 U_1^{\dagger} \otimes U_2 U_2^{\dagger} = I \otimes I $$

2.30

Let $H_1$ and $H_2$ be Hermitian operators.

$$ \left( H_1 \otimes H_2 \right)^{\dagger} = H_1^{\dagger} \otimes H_2^{\dagger} = H_1 \otimes H_2 $$

2.31

Let $A_1$ and $A_2$ be positive operators. Then,

$$ \begin{eqnarray} \left( \left| v \right\rangle \otimes \left| w \right\rangle, A_1 \otimes A_2 \left| v \right\rangle \otimes \left| w \right\rangle \right) &=& \left\langle w \right| \otimes \left\langle v \right|A_1 \otimes A_2 \left| v \right\rangle \otimes \left| w \right\rangle = \left\langle v \middle| A_1 \middle| v \right\rangle \left\langle w \middle| A_2 \middle| w \right\rangle \ge 0. \end{eqnarray} $$

2.32

Let $P_1$ and $P_2$ be projectors. Then,

$$ \left( P_1 \otimes P_2 \right)^2 = \left( P_1 \otimes P_2 \right) \left( P_1 \otimes P_2 \right) = P_1^2 \otimes P_2^2 = P_1 \otimes P_2 $$

2.33

$$ \begin{eqnarray} H &=& \frac{1}{\sqrt{2}}\left[ \left( \left| 0 \right\rangle + \left| 1 \right\rangle \right) \left\langle 0 \right| + \left( \left| 0 \right\rangle - \left| 1 \right\rangle \right) \left\langle 1 \right| \right] \\ &=& \frac{1}{\sqrt{2}} \left[ \left| 0 \right\rangle \left\langle 0 \right| + \left| 1 \right\rangle \left\langle 0 \right| + \left| 0 \right\rangle \left\langle 1 \right| - \left| 1 \right\rangle \left\langle 1 \right| \right] \\ &=& \frac{1}{\sqrt{2}} \sum_{x,y} \left( -1 \right)^{x\cdot y} \left| x \right\rangle \left\langle y \right| \end{eqnarray} $$

Then,

$$ \begin{eqnarray} H^{\otimes n} &=& \frac{1}{\sqrt{2^n}} \sum_{x1,y1} \left( -1 \right)^{x_1 \cdot y_1} \left| x_1 \right\rangle \left\langle y_1 \right| \otimes \sum_{x_2,y_2} \left( -1 \right)^{x_2 \cdot y_2} \left| x_2 \right\rangle \left\langle y_2 \right| \otimes \cdots \\ &=& \frac{1}{\sqrt{2^n}} \sum_{\boldsymbol{x},\boldsymbol{y}} \left( -1 \right)^{\boldsymbol{x} \cdot \boldsymbol{y}} \left| \boldsymbol{x} \right\rangle \left\langle \boldsymbol{y} \right|. \end{eqnarray} $$
$$ \begin{eqnarray} H^{\otimes 2} &=& \frac{1}{2} \left[ \left| 00 \right\rangle \left\langle 00 \right| + \left| 01 \right\rangle \left\langle 00 \right| + \left| 00 \right\rangle \left\langle 01 \right| - \left| 01 \right\rangle \left\langle 01 \right| \\ + \left| 10 \right\rangle \left\langle 00 \right| + \left| 11 \right\rangle \left\langle 00 \right| + \left| 10 \right\rangle \left\langle 01 \right| - \left| 11 \right\rangle \left\langle 01 \right| \\ + \left| 00 \right\rangle \left\langle 10 \right| + \left| 01 \right\rangle \left\langle 10 \right| + \left| 00 \right\rangle \left\langle 11 \right| - \left| 01 \right\rangle \left\langle 11 \right| \\ -\left| 10 \right\rangle \left\langle 10 \right| - \left| 11 \right\rangle \left\langle 10 \right| - \left| 10 \right\rangle \left\langle 11 \right| + \left| 11 \right\rangle \left\langle 11 \right| \right] \\ &=& \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{pmatrix} \end{eqnarray} $$

2.34

Let

$$ A = \begin{pmatrix} 4 & 3 \\ 3 & 4 \end{pmatrix}. $$

Then,

$$ {\rm det}\left( A - \lambda I \right) = \begin{vmatrix} 4-\lambda & 3 \\ 3 & 4-\lambda \end{vmatrix} = \left( 4-\lambda \right)^2 - 9 = 0 $$ $$ \therefore \lambda = 1,\;7 $$

For $\lambda = 1$,

$$ \begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix} \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

This yields an eigenvector

$$ \left| \lambda_1 \right\rangle = \begin{pmatrix} a_1 \\ b_1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}. $$

For $\lambda = 7$,

$$ \begin{pmatrix} -3 & 3 \\ 3 & -3 \end{pmatrix} \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$

This yields an eigenvector

$$ \left| \lambda_7 \right\rangle = \begin{pmatrix} a_2 \\ b_2 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}. $$

Then $A$ is written as

$$ A = \lambda_1 \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| + \lambda_7 \left| \lambda_7 \right\rangle \left\langle \lambda_7 \right| = \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| + 7 \left| \lambda_7 \right\rangle \left\langle \lambda_7 \right|. $$

Thus,

$$ \begin{eqnarray} \sqrt{A} &=& \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| + \sqrt{7} \left| \lambda_7 \right\rangle \left\langle \lambda_7 \right| \\ &=& \frac{1}{2} \begin{pmatrix} 1 \\ -1 \end{pmatrix} \begin{pmatrix} 1 & -1 \end{pmatrix} + \frac{\sqrt{7}}{2} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} \\ &=& \frac{1}{2} \begin{pmatrix} 1+ \sqrt{7} & -1 + \sqrt{7} \\ -1 + \sqrt{7} & 1 + \sqrt{7} \end{pmatrix},\\ \ln A &=& \ln 1 \left| \lambda_1 \right\rangle \left\langle \lambda_7 \right| + \ln 7 \left| \lambda_7 \right\rangle \left\langle \lambda_7 \right| \\ &=& \frac{\ln 7}{2} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} \\ &=& \frac{\ln 7}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}. \end{eqnarray} $$

2.35

Let $\vec{v} = \left(v_1, v_2, v_3\right)$. Then,

$$ \begin{eqnarray} \vec{v}\cdot \vec{\sigma} &=& v_1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + v_2 \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} + v_3 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ &=& \begin{pmatrix} v_3 & v_1-iv_2 \\ v_1 + iv_2 & -v_3 \end{pmatrix}. \end{eqnarray} $$

Find eigenvalues by solving characteristic equation.

$$ \begin{eqnarray} \det \left( \vec{v}\cdot \vec{\sigma} - \lambda I \right) &=& \begin{vmatrix} v_3-\lambda & v_1-iv_2 \\ v_1+iv_2 & -v_3-\lambda \end{vmatrix} \\ &=& -\left( v_3-\lambda \right)\left( v_3+\lambda \right) - \left( v_1-iv_2 \right)\left( v_1+iv_2 \right) \\ &=& \lambda^2 -\left( v_1^2 + v_2^2 + v_3^2 \right) \\ &=& \lambda^2 -1 \;\;\;\;\;\; \left( \because \vec{v} \text{ is a unit vector} \right) \\ &=& 0 \;\;\;\;\;\; \therefore \lambda=\pm 1 \end{eqnarray} $$

So, $\vec{v}\cdot\vec{\sigma}$ has a spectral decomposition

$$ \vec{v}\cdot\vec{\sigma} = \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| - \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| . $$

Therefore,

$$ \begin{eqnarray} \exp \left( i\theta \vec{v} \cdot \vec{\sigma} \right) &=& \exp \left( i\theta \right) \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| + \exp \left( -i\theta \right) \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| \\ &=& \left( \cos\theta + i\sin\theta \right) \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| + \left( \cos\theta - i\sin\theta \right) \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| \\ &=& \cos\theta \left( \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| + \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| \right) + i\sin\theta \left( \left| \lambda_1 \right\rangle \left\langle \lambda_1 \right| - \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| \right) \\ &=& \cos\theta + i\sin\left( \theta \right) \vec{v}\cdot\vec{\sigma} \end{eqnarray} $$

2.36

$$ \begin{eqnarray} \mathrm{tr} \left( X \right) &=& \mathrm{tr} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = 0 \\ \mathrm{tr} \left( Y \right) &=& \mathrm{tr} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = 0 \\ \mathrm{tr} \left( Z \right) &=& \mathrm{tr} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = 0 \end{eqnarray} $$

2.37

Suppose $\dim \left( A \right) = \dim \left( B \right) = n$. Then,

$$ \begin{eqnarray} \mathrm{tr} \left( AB \right) = \sum_i^n \left( \sum_j^n A_{ij}B_{ji} \right) = \sum_j^n \left( \sum_i^n B_{ji}A_{ij} \right) = \mathrm{tr} \left( BA \right) \end{eqnarray} $$

2.38

Suppose $\dim \left( A \right) = \dim \left( B \right) = n$. Then,

$$ \begin{eqnarray} \mathrm{tr} \left( A + B \right) &=& \sum_i^n \left( A + B \right)_{ii} = \sum_i^n \left( A_{ii} + B_{ii} \right) = \mathrm{tr} \left( A \right) + \mathrm{tr} \left( B \right) \\ \mathrm{tr} \left( zA \right) &=& \sum_i^n zA_{ii} = z\sum_i^n A_{ii} = z\,\mathrm{tr} \left( A \right) \end{eqnarray} $$

2.39

(1) We show that this operator satisfies (1), (2) and (3) in “2.1.4 Inner products” (p.65).

$$ \begin{align} \left( A, \sum_i \lambda_i B_i \right) &= \mathrm{tr} \left( A^{\dagger}\sum_i \lambda_i B_i \right) \\ &= \mathrm{tr} \left( \sum_i \lambda_i A^{\dagger}B_i \right) \\ &= \sum_i \lambda_i \mathrm{tr} \left( A^{\dagger}B_i \right) \\ &= \sum_i \lambda_i \left( A, B_i \right). \end{align} $$
$$ \begin{align} \left( A, B \right)^* &= \left( \mathrm{tr} \left( A^{\dagger}B \right) \right)^* \\ &= \mathrm{tr} \left( \left( A^{\dagger}B \right)^{\dagger} \right) \\ &= \mathrm{tr} \left( B^{\dagger}A \right) \\ &= \left( B, A \right). \end{align} $$
$$ \begin{align} \left( A,A \right) &= \mathrm{tr} \left( A^{\dagger}A \right) \\ &= \sum_i \left| A_{ii} \right|^2 \ge 0. \end{align} $$

So, the function $(\cdot,\cdot)$ on $L_V\times L_V$ defined by $(A,B)\equiv \mathrm{tr}(A^{\dagger}B)$ is an inner product function.

(2) A linear transformation $T:;V\rightarrow V;(\mathrm{dim}V=d)$ can be represented as a $d\times d$ matrix. Since there are $d\times d=d^2$ matrices that are linearly independent, the dimension of $L_V$ is $d^2$.


2.40

$$ \begin{align} \left[ X,Y \right] &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} - \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2i & 0 \\ 0 & -2i \end{pmatrix} = 2iZ \\ \left[ Y,Z \right] &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2i \\ 2i & 0 \end{pmatrix} = 2iX \\ \left[ Z,X \right] &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} = 2iY \end{align} $$

2.41

$$ \begin{align} \left\{ \sigma_1,\sigma_2 \right\} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} + \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = 0 \\ \left\{ \sigma_1,\sigma_3 \right\} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} =0 \\ \left\{ \sigma_2,\sigma_1 \right\} &= \left\{ \sigma_1,\sigma_2 \right\} = 0 \\ \left\{ \sigma_2,\sigma_3 \right\} &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = 0 \\ \left\{ \sigma_3,\sigma_1 \right\} &= \left\{ \sigma_1,\sigma_3 \right\} = 0 \\ \left\{ \sigma_3,\sigma_2 \right\} &= \left\{ \sigma_2,\sigma_3 \right\} = 0 \\ \sigma_1^2 &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbb{I} \\ \sigma_2^2 &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbb{I} \\ \sigma_3^2 &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbb{I} \end{align} $$

2.42

$$ \begin{equation} \frac{\left[ A,B \right]+\left\{ A,B \right\}}{2} = \frac{AB-BA+AB+BA}{2} = AB \end{equation} $$

2.43

$$ \begin{align} \sigma_j\sigma_k &= \frac{\left[ \sigma_j,\sigma_k \right] + \left\{ \sigma_j,\sigma_k \right\}}{2}\;\;\;\left( \because \mathrm{Excercise}\; 2.42 \right) \\ &= \frac{2i\sum_{l=1}^3\epsilon_{jkl}\sigma_l + 2\delta_{j,k}\mathbb{I}}{2} \\ &= \delta_{j,k}\mathbb{I} + i\sum_{l=1}^3\epsilon_{jkl}\;\;\;\left( \because \mathrm{Excercise}\;2.40,\;2.41 \right) \end{align} $$

2.44

$$ \begin{align} \left[ A,B \right] &= AB-BA = 0\;\;\; \therefore AB=BA \\ \left\{ A,B \right\} &= AB+BA = 2AB=0\;\;\; \left( \because AB=BA \right) \end{align} $$

Since $A$ is invertible, there exists $A^{-1}$ such that $AA^{-1}=A^{-1}A=\mathbb{I}$. Then,

$$ \begin{equation} 2A^{-1}AB = A^{-1}0 = 0\;\longrightarrow\;B=0 \end{equation} $$

2.45

$$ \begin{equation} \left[ A,B \right]^{\dagger} = \left( AB-BA \right)^{\dagger} = \left( B^{\dagger}A^{\dagger} - A^{\dagger}B^{\dagger} \right) = \left[ B^{\dagger},A^{\dagger} \right] \end{equation} $$

2.46

$$ \begin{equation} \left[ A,B \right] = AB-BA = -\left( BA-AB \right) = -\left[ B,A \right] \end{equation} $$

2.47

$$ \begin{align} \left\{ i\left[ A,B \right] \right\}^{\dagger} &= -i\left[ B^{\dagger},A^{\dagger} \right]\;\;\; \left( \because \mathrm{Excercise}\;2.45 \right) \\ &= -i\left[ B,A \right]\;\;\; \left( \because A,B\;\mathrm{are\;Hermitian} \right) \\ &= i\left[ A,B \right]\;\;\; \left( \because \;\mathrm{Excercise}\;2.46 \right) \end{align} $$

2.48

Polar decomposition (left): $A = UJ$.
(i) Positive matrix $P$

Since $P$ is positive matrix, we obtain the spectral decomposition of $P$,

$$ \begin{equation} P = \sum_{i} \lambda_i \left| i \right\rangle \left\langle i \right|. \end{equation} $$

Then,

$$ \begin{align} J = \sqrt{P^{\dagger}P} &= \sqrt{\sum_{i,j}\lambda_i \lambda_j \left| i \right\rangle \left\langle i \middle| j \right\rangle \left\langle j \right|}\\ &= \sqrt{\sum_{i,j}\lambda_i\lambda_j \delta_{i,j} \left| i \right\rangle \left\langle j \right|}\\ &= \sqrt{\sum_i \lambda_i^2 \left| i \right\rangle \left\langle i \right|} = \sqrt{P^2} = P. \end{align} $$

So, $U = I$ since $P=UP$ for all $P$.

(ii) Unitary matrix $U'$

$$ \begin{equation} J = \sqrt{U^{\dagger}U} = I \end{equation} $$

Then, $U=U'$ since $U’ = UI$ for all $U$.

(iii) Hermitian matrix $H$

The spectral decomposition of $H$ is,

$$ \begin{equation} H = \sum_i \lambda_i \left| i \right\rangle \left\langle i \right|. \end{equation} $$

Then,

$$ \begin{align} J = \sqrt{H^{\dagger}H} &= \sqrt{H^2}\;\;\; \left( \because H^{\dagger} = H \right)\\ &= \sqrt{\sum_{i,j}\lambda_i \lambda_j \left| i \right\rangle \left\langle i \middle| j \right\rangle \left\langle j \right|}\\ &= \sqrt{\sum_{i,j}\lambda_i \lambda_j \delta_{i,j} \left| i \right\rangle \left\langle j \right|}\\ &= \sqrt{\sum_i \lambda_i^2 \left| i \right\rangle \left\langle i \right|} = \sum_i |\lambda_i| \left| i \right\rangle \left\langle i \right|. \end{align} $$

So, $H = U\sum_i |\lambda_i| \left| i \right\rangle \left\langle i \right|$.


2.49

The spectral decomposition of normal matrix $A$ is,

$$ \begin{equation} A = \sum_i \lambda_i \left| i \right\rangle \left\langle i \right|. \end{equation} $$

Then,

$$ \begin{align} J = \sqrt{A^{\dagger}A} &= \sqrt{\sum_{i,j}\lambda_i^{*} \lambda_j \left| i \right\rangle \left\langle i \middle| j \right\rangle \left\langle j \right|}\\ &= \sqrt{\sum_{i,j}\lambda_i^{*} \lambda_j \delta_{i,j} \left| i \right\rangle \left\langle j \right|}\\ &= \sqrt{\sum_i \left|\lambda_i\right|^2 \left| i \right\rangle \left\langle i \right|} = \sum_i |\lambda_i | \left| i \right\rangle \left\langle i \right|. \end{align} $$

So, $A=U\sum_i |\lambda_i | \left| i \right\rangle \left\langle i \right|$.


2.50

$$ \begin{equation} A^{\dagger}A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \end{equation} $$

where

$$ \begin{equation} A \equiv \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}. \end{equation} $$

Then,

$$ \begin{align} \det\left( A^{\dagger}A-\lambda I \right) = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 1-\lambda \end{vmatrix} = \lambda^2 - 3\lambda + 1 = 0 \;\;\; \therefore \lambda = \frac{3\pm\sqrt{5}}{2}. \end{align} $$

For $\lambda_- = \frac{3-\sqrt{5}}{2}$,

$$ \begin{align} \begin{pmatrix} 2-\frac{3-\sqrt{5}}{2} & 1 \\ 1 & 1-\frac{3-\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \;\;\; \therefore \left| \lambda_- \right\rangle = \frac{1}{\sqrt{10+2\sqrt{5}}} \begin{pmatrix} 2 \\ -1-\sqrt{5} \end{pmatrix}. \end{align} $$

For $\lambda_+ = \frac{3+\sqrt{5}}{2}$,

$$ \begin{equation} \begin{pmatrix} 2-\frac{3+\sqrt{5}}{2} & 1 \\ 1 & 1-\frac{3+\sqrt{5}}{2} \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \;\;\; \therefore \left| \lambda_+ \right\rangle = \frac{1}{\sqrt{10-2\sqrt{5}}} \begin{pmatrix} 2 \\ -1+\sqrt{5} \end{pmatrix}. \end{equation} $$

Here,

$$ \begin{align} \left\langle \lambda_- \middle| \lambda_- \right\rangle &= \frac{1}{10+2\sqrt{5}} \left( 4+\left( -1-\sqrt{5} \right)^2 \right) = 1 \\ \left\langle \lambda_+ \middle| \lambda_+ \right\rangle &= \frac{1}{10-2\sqrt{5}} \left( 4+\left( -1+\sqrt{5} \right)^2 \right) = 1 \\ \left\langle \lambda_- \middle| \lambda_+ \right\rangle &= \frac{1}{\sqrt{10+2\sqrt{5}}\sqrt{10-2\sqrt{5}}}\left( 4+\left( -1-\sqrt{5} \right)\left( -1+\sqrt{5} \right) \right) = 0. \end{align} $$

Thus,

$$ \begin{align} J &= \sqrt{A^{\dagger}A} = \sqrt{\lambda_-} \left| \lambda_- \right\rangle \left\langle \lambda_- \right| + \sqrt{\lambda_+} \left| \lambda_+ \right\rangle \left\langle \lambda_+ \right| \\ J^{-1} &= \frac{1}{\sqrt{\lambda_-}} \left| \lambda_- \right\rangle \left\langle \lambda_- \right| + \frac{1}{\sqrt{\lambda_+}} \left| \lambda_+ \right\rangle \left\langle \lambda_+ \right| \\ &= \sqrt{\frac{2}{3-\sqrt{5}}} \frac{1}{10+2\sqrt{5}} \begin{pmatrix} 4 & -2-2\sqrt{5} \\ -2-2\sqrt{5} & 6+2\sqrt{5} \end{pmatrix} + \sqrt{\frac{2}{3+\sqrt{5}}}\frac{1}{10-2\sqrt{5}} \begin{pmatrix} 4 & -2+2\sqrt{5} \\ -2+2\sqrt{5} & 6-2\sqrt{5} \end{pmatrix} \\ &= \sqrt{\frac{2}{3-\sqrt{5}}} \frac{1}{5+\sqrt{5}} \begin{pmatrix} 2 & -1-\sqrt{5} \\ -1-\sqrt{5} & 3+\sqrt{5} \end{pmatrix} + \sqrt{\frac{2}{3+\sqrt{5}}}\frac{1}{5-\sqrt{5}} \begin{pmatrix} 2 & -1+\sqrt{5} \\ -1+\sqrt{5} & 3-\sqrt{5} \end{pmatrix} \\ &= \frac{1}{\sqrt{5}} \begin{pmatrix} 2 & -1 \\ -1 & 3 \end{pmatrix} \\ U &= AJ^{-1} = \frac{1}{\sqrt{5}} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -1 & 3 \end{pmatrix} = \frac{1}{\sqrt{5}} \begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix} \end{align} $$

2.51

$$ \begin{equation} HH^{\dagger} = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{equation} $$

2.52

$$ \begin{equation} H^2 = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \end{equation} $$

2.53

$$ \begin{align} \det \left( H - \lambda I \right) &= \begin{pmatrix} \frac{1}{\sqrt{2}}-\lambda & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}-\lambda \end{pmatrix}\\ &= \lambda^2-1 = 0\;\;\; \therefore \lambda = \pm 1. \end{align} $$

For $\lambda_{-1} = -1$,

$$ \begin{align} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = - \begin{pmatrix} a \\b \end{pmatrix} \;\;\; \therefore \left| \lambda_{-1} \right\rangle = \frac{1}{\sqrt{4+2\sqrt{2}}} \begin{pmatrix} 1 \\ -1-\sqrt{2} \end{pmatrix}. \end{align} $$

For $\lambda_{+1} = 1$,

$$ \begin{align} \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix} \;\;\; \therefore \left| \lambda_{+1} \right\rangle = \frac{1}{\sqrt{4-2\sqrt{2}}} \begin{pmatrix} 1 \\ -1+\sqrt{2} \end{pmatrix}. \end{align} $$

2.54

Solution 1:
$A$ and $B$ are simultaneously diagonalizable since $A$ commutes with $B$, i.e. $A = \sum_i a_i \left| i \right\rangle \left\langle i \right|,; B = \sum_j b_j \left| j \right\rangle \left\langle j \right|$. Then,

$$ \begin{align} \exp \left( A \right)\exp \left( B \right) &= \sum_i \exp\left( a_i \right)\left| i \right\rangle \left\langle i \right| \sum_j \exp\left( b_j \right) \left| j \right\rangle \left\langle j \right| \\ &= \sum_{i,j} \exp\left(a_i + b_j\right) \delta_{i,j} \left| i \right\rangle \left\langle j \right| \\ &= \sum_i \exp\left(a_i + b_i\right) \left| i \right\rangle \left\langle i \right| \\ &= \exp \left( A + B \right). \end{align} $$

Solution 2:
Using Baker–Campbell–Hausdorff formula, which is frequently used in quantum mechanics, $\exp(A)\exp(B)$ is written as follows:

$$ \begin{equation} \exp(A)\exp(B) = \exp(C) \end{equation} $$

where $C$ is,

$$ \begin{equation} C = A + B + \frac{1}{2}\left[ A,B \right] + \frac{1}{12}\left[ A, \left[ A,B \right] \right] - \frac{1}{12} \left[ B, \left[ A,B \right] \right] + \cdots. \end{equation} $$

Since $A$ commutes with $B$, $\exp(A)\exp(B) = \exp(A+B)$.


2.55

Solution 1:
From eq.(2.90),

$$ \begin{align} U\left(t_1,t_2 \right) &= \exp\left[ \frac{-iH\left( t_2-t_1 \right)}{\hbar} \right],\\ U^{\dagger}\left( t_1,t_2 \right) &= \exp\left[ \frac{iH\left( t_2-t_1 \right)}{\hbar} \right]. \end{align} $$

Then,

$$ \begin{align} U\left( t_1,t_2 \right)U^{\dagger}\left( t_1,t_2 \right) &= \exp\left[ \frac{-iH\left( t_2-t_1 \right)}{\hbar} \right]\exp\left[ \frac{iH\left( t_2-t_1 \right)}{\hbar} \right] \\ &= \sum_E\exp\left[\frac{-iE\left( t_2-t_1 \right)}{\hbar} \right] \left| E \right\rangle \left\langle E \right| \sum_{E'}\exp\left[ \frac{iE'\left( t_2-t_1 \right)}{\hbar} \right] \left| E' \right\rangle \left\langle E' \right| \;\;\; \left( \because \mathrm{eq.}\;(2.87) \right) \\ &= \sum_{E,E'} \exp\left[ \frac{-iE\left( t_2-t_1 \right)+iE'\left( t_2-t_1 \right)}{\hbar} \right] \delta_{E,E'} \left| E \right\rangle \left\langle E' \right| \\ &= \sum_E \exp\left( 0 \right) \left| E \right\rangle \left\langle E \right| \\ &= \sum_E \left| E \right\rangle \left\langle E \right| = 1. \end{align} $$

Solution 2:

$$ \begin{align} U\left( t_1,t_2 \right)U^{\dagger}\left( t_1,t_2 \right) &= \exp\left[ \frac{-iH\left( t_2-t_1 \right)}{\hbar} \right]\exp\left[ \frac{iH\left( t_2-t_1 \right)}{\hbar} \right] \\ &= \sum_k \frac{ \left[ -iH\left( t_2-t_1 \right)/\hbar \right]^{k}} {k!} \sum_l \frac{ \left[ iH\left( t_2-t_1 \right)/\hbar \right]^{l}} {l!} \\ &= \sum_{N=0}^{\infty}\sum_{k=0}^{N} \frac{\left[ -iH\left( t_2-t_1 \right)/\hbar \right]^{N-k}}{\left( N-k \right)!} \frac{\left[ iH\left( t_2-t_1 \right)/\hbar \right]^k}{k!} \\ &= \sum_{N=0}^{\infty} \left[ \frac{-iH\left( t_2-t_2 \right)}{\hbar} \right]^N \sum_{k=0}^N \frac{\left[ -iH\left( t_2-t_1 \right)/\hbar \right]^{-k}}{\left( N-k \right)!} \frac{\left[ iH\left( t_2-t_1 \right) \right]^k}{k!} \\ &= \sum_{N=0}^{\infty} \left[ \frac{-iH\left( t_2-t_2 \right)}{\hbar} \right]^N \sum_{k=0}^N \left( -1 \right)^{-k} \frac{\left[ iH\left( t_2-t_1 \right)/\hbar \right]^{-k}}{\left( N-k \right)!} \frac{\left[ iH\left( t_2-t_1 \right) \right]^k}{k!} \\ &= \sum_{N=0}^{\infty} \left[ \frac{-iH\left( t_2-t_2 \right)}{\hbar} \right]^N \sum_{k=0}^N \frac{\left( -1 \right)^k}{\left( N-k \right)!\;k!} \end{align} $$

When $N>0$, remembering the Binomial theorem, which is,

$$ \begin{equation} (x+y)^N = \sum_{k=0}^N \begin{pmatrix} N \\ k \end{pmatrix}x^{N-k}y^k \end{equation} $$

the latter part will be

$$ \begin{equation} \sum_{k=0}^N \frac{\left( -1 \right)^k}{\left( N-k \right)!\;k!} = \frac{\left( 1-1 \right)^N}{N!} = 0. \end{equation} $$

Thus,

$$ \begin{equation} U\left( t_1,t_2 \right)U^{\dagger}\left( t_1,t_2 \right) = 1. \end{equation} $$

2.56

Since unitary operator $U$ is normal, it has spectral decomposition. Then,

$$ \begin{align} K &\equiv -i \log(U) \\ &= -i\sum_j\log\left( a_j \right) \left| j \right\rangle \left\langle j \right| \\ &= -i\sum_j\log\left( e^{i\theta_j} \right) \left| j \right\rangle \left\langle j \right| \;\;\; \left( \because \mathrm{Exercise\; 2.18} \right) \\ &= -i\sum_j i\theta_j \left| j \right\rangle \left\langle j \right| \\ &= \sum_j \theta_j \left| j \right\rangle \left\langle j \right| \\ \therefore K^{\dagger} &= K. \end{align} $$

2.57

The state of the system $\left| \psi_1 \right\rangle$ after the measurement ${L_l}$ is,

$$ \begin{equation} \left| \psi_1 \right\rangle = \frac{L_l \left| \psi \right\rangle}{\sqrt{\left\langle \psi \middle| L_l^{\dagger}L_l \middle| \psi \right\rangle}}. \end{equation} $$

Then, the state of the system $\left| \psi_2 \right\rangle$ after the measurement ${M_m}$ is,

$$ \begin{align} \left| \psi_2 \right\rangle &= \frac{M_m \left| \psi_1 \right\rangle}{\sqrt{\left\langle \psi_1 \middle| M_m^{\dagger}M \middle| \psi_1 \right\rangle}} \\ &= \frac{M_mL_l\left| \psi \right\rangle}{\sqrt{\left\langle \psi \middle| L_l^{\dagger}L_l \middle| \psi \right\rangle}} \cdot \frac{\sqrt{\left\langle \psi \middle| L_l^{\dagger}L_l \middle| \psi \right\rangle}}{\left\langle \psi \middle| L_l^{\dagger}M_m^{\dagger}M_mL_l \middle| \psi \right\rangle} \\ &= \frac{M_mL_l \left| \psi \right\rangle}{\sqrt{\left\langle \psi \middle| L_l^{\dagger}M_m^{\dagger}M_mL_l \middle| \psi \right\rangle}}. \end{align} $$

On the other hand, the state of the system $\left| \psi_3 \right\rangle$ after the measurement $N_l {}_m = M_m L_l$ is,

$$ \begin{equation} \left| \psi_3 \right\rangle = \frac{M_mL_l\left| \psi \right\rangle}{\sqrt{\left\langle \psi \middle| L_l^{\dagger}M_m^{\dagger}M_mL_l \middle| \psi \right\rangle}} = \left| \psi_2 \right\rangle. \end{equation} $$

2.58

The average observed value of M is,

$$ \begin{align} \left\langle \psi \middle| M \middle| \psi \right\rangle &= \left\langle \psi \middle| M \middle| \psi \right\rangle \\ &= \sum_{m'} m' \left\langle \psi \middle| P_{m'} \middle| \psi \right\rangle \\ &= m \left\langle \psi \middle| \psi \right\rangle = m. \end{align} $$

The standard deviation is,

$$ \begin{align} \left[ \Delta \left( M \right) \right]^2 &= \left\langle M^2 \right\rangle - \left\langle M \right\rangle^2 \;\;\; \left( \because \mathrm{eq. (2.115)} \right) \\ &= \left\langle \psi \middle| M^2 \middle| \psi \right\rangle - \left( \left\langle \psi \middle| M \middle| \psi \right\rangle \right)^2 \\ &= \sum_{m'} m'^2 \left\langle \psi \middle| P_{m'} \middle| \psi \right\rangle - \left( \sum_{m''} m'' \left\langle \psi \middle| P_{m''} \middle| \psi \right\rangle \right)^2 \\ &= m^2 - m^2 = 0. \end{align} $$

2.59

$$ \begin{align} \left\langle X \right\rangle &= \left\langle 0 \middle| X \middle| 0 \right\rangle = \left\langle 0 \middle| 1 \right\rangle = 0 \\ \left\langle X^2 \right\rangle &= \left\langle 0 \middle| X^2 \middle| 0 \right\rangle = \left\langle 0 \middle| I \middle| 0 \right\rangle = 1. \end{align} $$

Therefore, the standard deviation $\Delta \left( X \right)$ is,

$$ \begin{equation} \Delta\left( X \right) = \sqrt{\left\langle X^2 \right\rangle - \left( \left\langle X \right\rangle \right)^2} = 1. \end{equation} $$

2.60

Similar to Exercise 2.35,

$$ \begin{align} \vec{v}\cdot\vec{\sigma} &= v_1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + v_2 \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} + v_3 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ &= \begin{pmatrix} v_3 & v_1-iv_2 \\ v1+iv_2 & -v_3. \end{pmatrix} \end{align} $$

Now find eigenvalues by solving the characteristic equation.

$$ \begin{align} \det \left( \vec{v}\cdot\vec{\sigma} - \lambda I \right) &= \begin{vmatrix} v_3 - \lambda & v_1-iv_2 \\ v_1+iv_2 & -v_3-\lambda \end{vmatrix} \\ &= \lambda^2 - (v_1^2+v_2^2+v_3^2) \\ &= \lambda^2 - 1 \;\;\; \left( \because \vec{v} \mathrm{\; is\; a\; unit\; vector} \right) \\ &= 0 \;\;\; \therefore \lambda = \pm 1 \end{align} $$

For $\lambda=-1$,

$$ \begin{align} \begin{pmatrix} v_3+1 & v_1-iv_2 \\ v_1+iv_2 & -v_3+1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \end{align} $$

This yields an eigenvector

$$ \begin{equation} \left| \lambda_{-1} \right\rangle = \frac{1}{\sqrt{2\left( 1+v_3 \right)}} \begin{pmatrix} v_1-iv_2 \\ -v_3-1 \end{pmatrix}. \end{equation} $$

Thus, $P_-$ is

$$ \begin{align} P_- = \left| \lambda_{-1} \right\rangle \left\langle \lambda_{-1} \right| &= \frac{1}{2\left( 1+v_3 \right)} \begin{pmatrix} v_1-iv_2 \\ -v_3-1 \end{pmatrix} \begin{pmatrix} v_1+iv_2 & -v_3-1 \end{pmatrix} \\ &= \frac{1}{2\left( 1+v_3 \right)} \begin{pmatrix} v_1^2+v_2^2 & -v_1\left( v_3+1 \right) +iv_2\left( v_3+1 \right) \\ -v_1\left( v_3+1 \right)-iv_2\left( v_3+1 \right) & \left( v_3+1 \right)^2 \end{pmatrix} \\ &= \frac{1}{2} \begin{pmatrix} \frac{v_1^2+v_2^2}{1+v_3} & -v_1+iv_2 \\ -v_1-iv_2 & 1+v_3 \end{pmatrix} \\ &= \frac{1}{2} \left(I-\vec{v}\cdot\vec{\sigma} \right). \end{align} $$

For $\lambda=+1$,

$$ \begin{align} \begin{pmatrix} v_3-1 & v_1-iv_2 \\ v_1+iv_2 & -v_3+1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \end{align} $$

This yields an eigenvector

$$ \begin{equation} \left| \lambda_{+1} \right\rangle = \frac{1}{\sqrt{2\left( 1-v_3 \right)}} \begin{pmatrix} v_1-iv_2 \\ 1-v_3 \end{pmatrix}. \end{equation} $$

Thus, $P_+$ is

$$ \begin{align} P_+ = \left| \lambda_{+1} \right\rangle \left\langle \lambda_{+} \right| &= \frac{1}{2\left( 1-v_3 \right)} \begin{pmatrix} v_1-iv_2 \\ 1-v_3 \end{pmatrix} \begin{pmatrix} v_1+iv_2 & 1-v_3 \end{pmatrix} \\ &= \frac{1}{2\left( 1-v_3 \right)} \begin{pmatrix} v_1^2+v_2^2 & v_1\left( 1-v_3 \right) -iv_2\left( 1-v_3 \right) \\ v_1 \left( 1-v_3 \right)+iv_2\left( 1-v_3 \right) & \left( 1-v_3 \right)^2 \end{pmatrix} \\ &= \frac{1}{2} \begin{pmatrix} \frac{v_1^2+v_2^2}{1-v_3} & v_1-iv_2 \\ v_1+iv_2 & 1-v_3 \end{pmatrix} \\ &= \frac{1}{2}\left( I+\vec{v}\cdot\vec{\sigma} \right) \end{align} $$

2.61

The probability of obtaining the result $+1$ is,

$$ \begin{align} p(+) &= \left\langle 0 \middle| P_+^{\dagger}P_+ \middle| 0 \right\rangle \\ &= \left\langle 0 \middle| P_+ \middle| 0 \right\rangle \\ &= \frac{1}{2}\left[ \left\langle 0 \middle| I \middle| 0 \right\rangle + \left\langle 0 \middle| \vec{v}\cdot\vec{\sigma} \middle| 0 \right\rangle \right] \\ &= \frac{1}{2}\left[ \left\langle 0 \middle| 0 \right\rangle + \left\langle 0 \middle| v_1 \middle| 1 \right\rangle + \left\langle 0 \middle| iv_2 \middle| 1 \right\rangle + \left\langle 0 \middle| v_3 \middle| 0 \right\rangle \right] \\ &= \frac{1+v_3}{2}. \end{align} $$

The state of the system after the measurement is,

$$ \begin{align} \frac{P_+\left| 0 \right\rangle}{\sqrt{p(+)}} &= \frac{\left( I+\vec{v}\cdot\vec{\sigma}\right) \left| 0 \right\rangle}{\sqrt{2\left( 1+v_3 \right)}} \\ &= \frac{1}{\sqrt{2\left( 1+v_3 \right)}} \begin{pmatrix} 1+v_3 \\ v_1+iv_2 \end{pmatrix} \\ &= \frac{1}{\sqrt{2\left( 1+v_3 \right)}}\frac{1+v_3}{v_1-iv_2} \begin{pmatrix} v_1-iv_2 \\ \frac{v_1^2+v_2^2}{1+v_3} \end{pmatrix} \\ &= \frac{1}{\sqrt{2\left( 1-v_3 \right)}} \begin{pmatrix} v_1-iv_2 \\ 1-v_3 \end{pmatrix} = \left| \lambda_+ \right\rangle. \end{align} $$

2.62

Let $M_m$ be measurement operators. $M_m$ correspond to the POVM elements, that is,

$$ \begin{equation} E_m = M_m^{\dagger}M = M_m. \end{equation} $$

So, $M_m$ are positive operators. Since positive operators are Hermitian (Exercise 2.24),

$$ \begin{equation} M_m^{\dagger}M = M_m^2 = M_m. \end{equation} $$

Therefore, $M_m$ are projective operators.


2.63

The left polar decomposition of $M_m$

$$ \begin{equation} M_m = U_m J_m = U_m \sqrt{M_m^{\dagger}M} = U_m \sqrt{E_m}. \end{equation} $$

2.64

We can construct $\left| \psi’_j \right\rangle$ that is orthogonal to all states except $\left| \psi_j \right\rangle$. That is,

$$ \begin{equation} \left| \psi'_j \right\rangle = \left| \psi_j \right\rangle - \sum_{k=1,k\neq j}^m\frac{\left\langle \psi_j \middle| \psi_k \right\rangle \left| \psi_k \right\rangle}{\left| \left| \psi_k \right\rangle \right|^2} \end{equation} $$

Then, $E_m$ is,

$$ \begin{equation} E_m = A\left| \psi'_m \right\rangle \left\langle \psi'_m \right|, \end{equation} $$

where $A$ is chosen such that

$$ \begin{equation} E_{m+1} = I-\sum_{j=1}^m E_j \end{equation} $$

is positive.


2.65

$$ \begin{align} \left| + \right\rangle &\equiv \frac{\left| 0 \right\rangle + \left| 1 \right\rangle}{\sqrt{2}} \\ \left| - \right\rangle &\equiv \frac{\left| 0 \right\rangle - \left| 1 \right\rangle}{\sqrt{2}} \end{align} $$

2.66

Let $\left| \psi \right\rangle \equiv \left( \left| 00 \right\rangle + \left| 11 \right\rangle \right)/\sqrt{2}$. Then,

$$ \begin{align} E\left( X_1Z_2 \right) &= \left\langle \psi \middle| X_1Z_2 \middle| \psi \right\rangle \\ &= \frac{1}{2} \left( \left\langle 00 \middle| X_1Z_2 \middle| 00 \right\rangle + \left\langle 00 \middle| X_1Z_2 \middle| 11 \right\rangle + \left\langle 11 \middle| X_1Z_2 \middle| 00 \right\rangle + \left\langle 11 \middle| X_1Z_2 \middle| 11 \right\rangle \right) \\ &= \frac{1}{2}\left( \left\langle 00 \middle| 10 \right\rangle - \left\langle 00 \middle| 01 \right\rangle + \left\langle 11 \middle| 10 \right\rangle - \left\langle 11 \middle| 01 \right\rangle \right) \\ &= 0 \end{align} $$

2.67

It seems intuitively easy (I wouldn’t say it’s trivial). However, it’s too much to prove…


2.68

Suppose one of the Bell states $\left| \psi \right\rangle \equiv \left( \left| 00 \right\rangle + \left| 11 \right\rangle \right)/\sqrt{2}$ can be written as a product pf single particle states

$$ \begin{align} \left| a \right\rangle &= \alpha_a \left| 0 \right\rangle + \beta_a \left| 1 \right\rangle \\ \left| b \right\rangle &= \alpha_b \left| 0 \right\rangle + \beta_b \left| 1 \right\rangle. \end{align} $$

Then,

$$ \begin{equation} \left| a \right\rangle \left| b \right\rangle = \alpha_a\alpha_b \left| 00 \right\rangle + \alpha_a\beta_b \left| 01 \right\rangle + \beta_a\alpha_b \left| 10 \right\rangle + \beta_a\beta_b \left| 11 \right\rangle. \end{equation} $$

Comparing this with a bell state $\left| \psi \right\rangle = \left( \left| 00 \right\rangle + \left| 11 \right\rangle \right)/\sqrt{2}$,

$$ \begin{align} \alpha_a\beta_b &= 0 \\ \beta_a\alpha_b &= 0. \end{align} $$

This contradicts $\left| \psi \right\rangle = \left| a \right\rangle \left| b \right\rangle$. So, $\left| \psi \right\rangle \neq \left| a \right\rangle \left| b \right\rangle$ for all single qubit states $\left| a \right\rangle$ and $\left| b \right\rangle$


2.69

Define the Bell basis as

$$ \begin{align} \left| \Phi^+ \right\rangle &= \frac{\left( \left| 00 \right\rangle + \left| 11 \right\rangle \right)}{\sqrt{2}} \\ \left| \Phi^- \right\rangle &= \frac{\left( \left| 00 \right\rangle - \left| 11 \right\rangle \right)}{\sqrt{2}} \\ \left| \Psi^+ \right\rangle &= \frac{\left( \left| 01 \right\rangle + \left| 10 \right\rangle \right)}{\sqrt{2}} \\ \left| \Psi^- \right\rangle &= \frac{\left( \left| 01 \right\rangle - \left| 10 \right\rangle \right)}{\sqrt{2}}. \end{align} $$

Then,

$$ \begin{align} \left\langle \Phi^+ \middle| \Phi^+ \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 00 \right\rangle + \left\langle 00 \middle| 11 \right\rangle + \left\langle 11 \middle| 00 \right\rangle + \left\langle 11 \middle| 11 \right\rangle \right) = 1 \\ \left\langle \Phi^+ \middle| \Phi^- \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 00 \right\rangle - \left\langle 00 \middle| 11 \right\rangle + \left\langle 11 \middle| 00 \right\rangle - \left\langle 11 \middle| 11 \right\rangle \right) = 0 \\ \left\langle \Phi^+ \middle| \Psi^+ \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 01 \right\rangle + \left\langle 00 \middle| 10 \right\rangle + \left\langle 11 \middle| 01 \right\rangle + \left\langle 11 \middle| 10 \right\rangle \right) = 0 \\ \left\langle \Phi^+ \middle| \Psi^- \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 01 \right\rangle - \left\langle 00 \middle| 10 \right\rangle + \left\langle 11 \middle| 01 \right\rangle - \left\langle 11 \middle| 10 \right\rangle \right) = 0 \\ \left\langle \Phi^- \middle| \Phi^- \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 00 \right\rangle - \left\langle 00 \middle| 11 \right\rangle - \left\langle 11 \middle| 00 \right\rangle + \left\langle 11 \middle| 11 \right\rangle \right) = 1 \\ \left\langle \Phi^- \middle| \Psi^+ \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 01 \right\rangle + \left\langle 00 \middle| 10 \right\rangle - \left\langle 11 \middle| 01 \right\rangle - \left\langle 11 \middle| 10 \right\rangle \right) = 0 \\ \left\langle \Phi^- \middle| \Psi^- \right\rangle &= \frac{1}{2}\left( \left\langle 00 \middle| 01 \right\rangle - \left\langle 00 \middle| 10 \right\rangle - \left\langle 11 \middle| 01 \right\rangle + \left\langle 11 \middle| 10 \right\rangle \right) = 0 \\ \left\langle \Psi^+ \middle| \Psi^+ \right\rangle &= \frac{1}{2}\left( \left\langle 01 \middle| 01 \right\rangle + \left\langle 01 \middle| 10 \right\rangle + \left\langle 10 \middle| 01 \right\rangle + \left\langle 10 \middle| 10 \right\rangle \right) = 1 \\ \left\langle \Psi^+ \middle| \Psi^- \right\rangle &= \frac{1}{2}\left( \left\langle 01 \middle| 01 \right\rangle - \left\langle 01 \middle| 10 \right\rangle + \left\langle 10 \middle| 01 \right\rangle - \left\langle 10 \middle| 10 \right\rangle \right) = 0 \\ \left\langle \Psi^- \middle| \Psi^- \right\rangle &= \frac{1}{2}\left( \left\langle 01 \middle| 01 \right\rangle - \left\langle 01 \middle| 10 \right\rangle - \left\langle 10 \middle| 01 \right\rangle + \left\langle 10 \middle| 10 \right\rangle \right) = 1. \end{align} $$

So, the Bell basis forms an orthonormal basis.


2.70

Any single qubit operation can be expressed as a linear combination of the Pauli matrices, i.e.,

$$ \begin{align} E = c_I I + c_x X +c_y Y + c_z Z. \end{align} $$

Using the relations

$$ \begin{align} \left( Z\otimes I \right) \left| \Phi^+ \right\rangle &= \frac{1}{\sqrt{2}}\left( \left| 00 \right\rangle - \left| 11 \right\rangle \right) = \left| \Phi^- \right\rangle \\ \left( X\otimes I \right)\left| \Phi^+ \right\rangle &= \frac{1}{\sqrt{2}}\left( \left| 01 \right\rangle + \left| 10 \right\rangle \right) = \left| \Psi^+ \right\rangle \\ \left( iY\otimes I \right)\left| \Phi^+ \right\rangle &= \frac{1}{\sqrt{2}}\left( \left| 01 \right\rangle - \left| 10 \right\rangle \right) = \left| \Psi^- \right\rangle \end{align} $$

and since these are orthogonal to each other,

$$ \begin{align} \left\langle \Phi^+ \middle| \left( E\otimes I \right) \middle| \Phi^+ \right\rangle &= \left\langle \Phi^+ \middle| c_I I\otimes I \middle| \Phi^+ \right\rangle + \left\langle \Phi^+ \middle| c_x X\otimes I \middle| \Phi^+ \right\rangle + \left\langle \Phi^+ \middle| c_y Y\otimes I \middle| \Phi^+ \right\rangle + \left\langle \Phi^+ \middle| c_z Z\otimes I \middle| \Phi^+ \right\rangle \\ &= c_I \left\langle \Phi^+ \middle| \Phi^+ \right\rangle + c_x \left\langle \Phi^+ \middle| \Psi^+ \right\rangle -ic_y \left\langle \Phi^+ \middle| \Psi^- \right\rangle + c_z \left\langle \Phi^+ \middle| \Phi^- \right\rangle \\ &= c_I. \end{align} $$

Then,

$$ \begin{align} \left\langle \Phi^- \middle| E\otimes I \middle| \Phi^- \right\rangle &= \left\langle \Phi^+ \middle| \left( Z\otimes I \right)^{\dagger} \left( E\otimes I \right) \left( Z\otimes I \right) \middle| \Phi^+ \right\rangle \\ &= \left\langle \Phi^+ \middle| \left( Z\otimes I \right) \left[ \left( c_I I + c_x X + c_y Y + c_z Z \right) \otimes I \right] \left( Z\otimes I \right) \middle| \Phi^+ \right\rangle \\ &= \left\langle \Phi^+ \middle| \left( Z\otimes I \right) \left[ \left( c_I Z + -ic_x Y + ic_y X + c_z I \right) \otimes I \right] \middle| \Phi^+ \right\rangle \\ &= \left\langle \Phi^+ \middle| \left[ \left( c_I I - c_x X - c_y Y + c_z Z \right) \otimes I \right] \middle| \Phi^+ \right\rangle \\ &= c_I \\ \left\langle \Psi^+ \middle| \left( E\otimes I \right) \middle| \Psi^+ \right\rangle &= \left\langle \Phi^+ \middle| \left( X\otimes I \right)^{\dagger}\left( E\otimes I \right)\left( X\otimes I \right) \middle| \Phi^+ \right\rangle \\ &= \left\langle \Phi^+ \middle| \left( X\otimes I \right)\left[ \left( c_I I + c_x X + c_y Y + c_z Z \right) \otimes I \right] \left( X\otimes I \right) \middle| \right\rangle \\ &= \left\langle \Phi^+ \middle| \left( X\otimes I \right)\left[ \left( c_I X + c_x I -i c_y Z + ic_z Y \right)\otimes I \right] \middle| \Phi^+ \right\rangle \\ &= \left\langle \Phi^+ \middle| \left[ \left( c_I I + c_x X - c_y Y - c_z Z \right)\otimes I \right] \middle| \Phi^+ \right\rangle \\ &= c_I \\ \left\langle \Psi^- \middle| \left( E\otimes I \right) \middle| \Psi^- \right\rangle &= \left\langle \Phi^+ \middle| \left( iY\otimes I \right)^{\dagger}\left( E\otimes I \right)\left( iY\otimes I \right) \middle| \Phi^+ \right\rangle \\ &= \left\langle \Phi^+ \middle| \left( -iY\otimes I \right)\left[ \left( c_I I + c_x X + c_y Y + c_z Z \right)\otimes I \right]\left( iY\otimes I \right) \middle| \Phi^+ \right\rangle \\ &= \left\langle \Phi^+ \middle| \left( Y\otimes I \right)\left[ \left( c_I Y + ic_x Z + c_y I -ic_z X \right)\otimes I \right] \middle| \Phi^+ \right\rangle \\ &= \left\langle \Phi^+ \middle| \left[ \left( c_I I - c_x X + c_y Y - c_z Z \right)\otimes I \right] \middle| \Phi^+ \right\rangle \\ &= c_I. \end{align} $$

If Eve gets the qubit that Alice sent to Bob and performs the measurement $M_m$, the result Eve gets is

$$ \begin{equation} \left\langle \psi \middle| \left( M_m^{\dagger}M_m\otimes I \right) \middle| \psi \right\rangle. \end{equation} $$

Here $M_m^{\dagger}M_m$ is positive operator. As confirmed above, the measurement result doesn’t depend on the state Alice prepared. Therefore, Eve cannot infer anything about the information Alice sent.


2.71

$$ \begin{align} \mathrm{tr} \left( \rho^2 \right) &= \mathrm{tr}\left( \sum_{j,k}p_jp_k \left| j \right\rangle \left\langle j \middle| k \right\rangle \left\langle k \right| \right) \\ &= \mathrm{tr}\left( \sum_{j,k}p_jp_k \delta_{j,k} \left| j \right\rangle \left\langle k \right| \right) \\ &= \mathrm{tr}\left( \sum_j p_j^2 \left| j \right\rangle \left\langle j \right| \right) \\ &= \sum_j p_j^2. \end{align} $$

Since $\rho$ is positive matrix and $\sum_j p_j=1$, $;0 \le p_j \le 1$. So, $p_j^2 \le p_j$ for all $p_j$. Therefore, $\sum_j p_j^2 \le 1$. Equality occurs if and only if $\rho$ is a pure state, that is, $\rho = \left| \psi \right\rangle \left\langle \psi \right|$.


2.72

(1)

Since the Pauli matrices along with the identity matrix form an orthogonal basis for the complex 2-dimensional Hilbert space, an arbitrary density matrix for a mixed state qubit is written as a real linear combination of $I,\sigma_1,\sigma_2,\sigma_3$, i.e.

$$ \begin{align} \rho &= v_0 \sigma_0 + v_1 \sigma_1 + v_2 \sigma_2 + v_3 \sigma_3 \\ &= \begin{pmatrix} v_0 + v_3 & v_1-iv_2 \\ v_1+iv_2 & v_0-v_3 \end{pmatrix}. \end{align} $$

Since $\mathrm{tr} \left( \rho \right) = 1,; v_0 = \frac{1}{2}$.

Now, find eigenvalues by solving characteristic equation

$$ \begin{align} \det \left( \rho -\lambda I \right) &= \begin{vmatrix} v_0 + v_3 - \lambda & v_1 -iv_2 \\ v_1+iv_2 & v_0 -v_3 -\lambda \end{vmatrix} \\ &= v_0^2 - v_3^2 -2v_0\lambda + \lambda^2 - v_1^2 - v_2^2 \\ &= \lambda^2 -2v_0\lambda + v_0^2 - (v_1^2 + v_2^2 + v_3^2) = 0. \end{align} \\ \therefore \lambda = v_0 \pm \sqrt{v_1^2 + v_2^2 + v_3^2} = \frac{1}{2} \left( 1\pm \sqrt{\left( 2v_1 \right)^2+\left( 2v_2 \right)^2+\left( 2v_3 \right)^2} \right) $$

Since $\rho$ is a positive semidefinite matrix,

$$ \begin{equation} \sqrt{\left( 2v_1 \right)^2+\left( 2v_2 \right)^2+\left( 2v_3 \right)^2} \le 1 \end{equation} $$

Therefore,

$$ \begin{equation} \rho = \frac{I+\vec{r}\cdot\vec{\sigma}}{2} \end{equation} $$

where $|\vec{r}|\le 1$.

(2)

$\rho=\frac{I}{2}$ is at the origin of the Bloch sphere (the maximally mixed state).

(3)

As we did in (1), the eigenvalues of $\rho=\frac{I+\vec{r}\cdot\vec{\sigma}}{2}$ are $\lambda=\frac{1}{2}\left( 1\pm \left| \vec{r} \right| \right)$. Since $\mathrm{tr} \left( \rho^2 \right)=\lambda_1^2+\lambda_2^2$, $\rho$ is pure if and only if $|\vec{r}|=1$.

(4)

Since $\rho$ is pure, $\rho = \left| \psi \right\rangle \left\langle \psi \right|$ where $\left| \psi \right\rangle = \alpha\left| 0 \right\rangle + \beta\left| 1 \right\rangle$. Here $\alpha$ and $\beta$ are constrained by the equation $|\alpha|^2+|\beta|^2=1$. So,

$$ \begin{equation} \left| \psi \right\rangle = \mathrm{e}^{i\gamma}\left( \cos\frac{\theta}{2}\left| 0 \right\rangle + \mathrm{e}^{i\varphi}\sin\frac{\theta}{2}\left| 1 \right\rangle \right). \end{equation} $$

2.73

$$ \begin{align} \left|\psi_i\right\rangle &= \rho\rho^{-1}\left|\psi_i\right\rangle \\ &= \sum_j p_j \left|\psi_j\right\rangle\left\langle\psi_j\middle|\rho^{-1}\middle|\psi_i\right\rangle \\ &= \sum_j p_j \left|\psi_j\right\rangle\left\langle\psi_j\middle|\rho^{-1}\middle|\psi_i\right\rangle\delta_{i,j} \\ &= p_i\left|\psi_i\right\rangle\left\langle\psi_i\middle|\rho^{-1}\middle|\psi_i\right\rangle \\ &\therefore p_i=\frac{1}{\left\langle\psi_i\middle|\rho^{-1}\middle|\psi_i\right\rangle} \end{align} $$

2.74

$$ \begin{align} \rho^{AB} &= \left| a \right\rangle \left| b \right\rangle\left\langle a \right|\left\langle b \right| = \left| a\right\rangle\left\langle a \right| \otimes \left| b \right\rangle\left\langle b \right|. \\ \rho^A &= \left| a\right\rangle\left\langle a \right| \mathrm{tr}\left( \left| b \right\rangle\left\langle b \right| \right) = \left| a \right\rangle\left\langle a \right|.\\ \mathrm{tr}\left( \left( \rho^A \right)^2 \right) &= \mathrm{tr}\left( \left| a\right\rangle\left\langle a \middle| a \right\rangle\left\langle a \right| \right) = \mathrm{tr} \left( \left| a\right\rangle\left\langle a \right| \right) = 1. \end{align} $$

2.75

$$ \begin{align} \rho^{\Phi^+} &= \left|\Phi^+\right\rangle\left\langle\Phi^+\right| = \frac{\left| 00\right\rangle\left\langle 00\right| + \left| 00\right\rangle\left\langle 11\right| + \left| 11\right\rangle\left\langle 00\right| + \left| 11\right\rangle\left\langle 11\right|}{2} \\ \rho^{\Phi^-} &= \frac{\left| 00\right\rangle\left\langle 00\right| - \left| 00\right\rangle\left\langle 11\right| - \left| 11\right\rangle\left\langle 00\right| + \left| 11\right\rangle\left\langle 11\right|}{2} \\ \rho^{\Psi^+} &= \frac{\left| 01\right\rangle\left\langle 01\right| + \left| 01\right\rangle\left\langle 10\right| + \left| 10\right\rangle\left\langle 01\right| + \left| 10\right\rangle\left\langle 10\right|}{2} \\ \rho^{\Psi^-} &= \frac{\left| 01\right\rangle\left\langle 01\right| - \left| 01\right\rangle\left\langle 10\right| - \left| 10\right\rangle\left\langle 01\right| + \left| 10\right\rangle\left\langle 10\right|}{2} \\ \rho^{\Phi^+,1} &= \frac{\left| 0\right\rangle\left\langle 0\right| + \left| 1\right\rangle\left\langle 1\right|}{2} = \frac{I}{2} \\ \rho^{\Phi^+,2} &= \frac{\left| 0\right\rangle\left\langle 0\right| + \left| 1\right\rangle\left\langle 1\right|}{2} = \frac{I}{2} \\ \rho^{\Phi^-,1} &= \frac{\left| 0\right\rangle\left\langle 0\right| + \left| 1\right\rangle\left\langle 1\right|}{2} = \frac{I}{2} \\ \rho^{\Phi^-,2} &= \frac{\left| 0\right\rangle\left\langle 0\right| + \left| 1\right\rangle\left\langle 1\right|}{2} = \frac{I}{2} \\ \rho^{\Psi^+,1} &= \frac{\left| 0\right\rangle\left\langle 0\right| + \left| 1\right\rangle\left\langle 1\right|}{2} = \frac{I}{2} \\ \rho^{\Psi^+,2} &= \frac{\left| 1\right\rangle\left\langle 1\right| + \left| 0\right\rangle\left\langle 0\right|}{2} = \frac{I}{2} \\ \rho^{\Psi^-,1} &= \frac{\left| 0\right\rangle\left\langle 0\right| + \left| 1\right\rangle\left\langle 1\right|}{2} = \frac{I}{2} \\ \rho^{\Psi^,2} &= \frac{\left| 0\right\rangle\left\langle 0\right| + \left| 1\right\rangle\left\langle 1\right|}{2} = \frac{I}{2} \\ \end{align} $$

2.76

See Wikipedia .


2.77

$$ \begin{equation} \frac{\left| 001\right\rangle + \left| 010\right\rangle + \left| 100\right\rangle}{\sqrt{3}} \end{equation} $$

References:
Higher order Schmidt decompositions ( arXiv )
Existence of the Schmidt decomposition for tripartite systems ( arXiv )


2.78

(i) product state $\Leftrightarrow$ Schmidt rank is 1
Suppose $\left|\psi\right\rangle$ is a product state. Then

$$ \begin{equation} \left|\psi\right\rangle = \left|\psi_A\right\rangle\otimes\left|\psi_B\right\rangle . \end{equation} $$

Obviously, the Schmidt rank of $\left|\psi\right\rangle$ is 1.
Suppose the Schmidt rank of $\left|\psi\right\rangle$ is 1. Then $\left|\psi\right\rangle = \left|\psi_A\right\rangle\otimes\left|\psi_B\right\rangle$.

(ii) product state $\Leftrightarrow$ $\rho^A$ and $\rho^B$ are pure states.
Using the Schmidt decomposition, $\left|\psi\right\rangle$ can be written as

$$ \begin{equation} \left|\psi\right\rangle = \sum_i \lambda_i\left| i_A\right\rangle\left| i_B\right\rangle . \end{equation} $$

Then,

$$ \begin{align} \rho^{AB} &= \sum_i\lambda_i^2\left| i_Ai_B\right\rangle\left\langle i_Ai_B\right|\\ \rho^{A} &= \sum_i\lambda_i^2\left| i_A\right\rangle\left\langle i_A\right|\\ \rho^{B} &= \sum_i\lambda_i^2\left| i_B\right\rangle\left\langle i_B\right| . \end{align} $$

Now, suppose the Schmidt rank of $\left|\psi\right\rangle$ is 1 ($\Leftrightarrow$ product state). Then

$$ \begin{equation} \rho^A = \left| i_A\right\rangle\left\langle i_A\right| . \end{equation} $$

Therefore, $\rho^A$ and $\rho^B$ are pure states.
Suppose $\rho^A$ and $\rho^B$ are purestates. Then

$$ \begin{equation} \rho^A = \left| i_A\right\rangle\left\langle i_A\right| . \end{equation} $$

Therefore, the Schmidt rank of $\left|\psi\right\rangle$ is 1 ($\Leftrightarrow$ product state).


2.79

$$ \begin{align} \frac{\left| 00\right\rangle + \left| 11\right\rangle}{\sqrt{2}} &= \sum_{i=1}^{2} \frac{1}{\sqrt{2}}\left| i \right\rangle\left| i\right\rangle \\ \frac{\left| 00\right\rangle + \left| 01\right\rangle + \left| 10\right\rangle + \left| 11\right\rangle}{2} &= \frac{\left| 0\right\rangle + \left| 1\right\rangle}{\sqrt{2}}\otimes\frac{\left| 0\right\rangle + \left| 1\right\rangle}{\sqrt{2}} \\ &= \left| +\right\rangle\left| +\right\rangle \end{align} $$

To find the Schmidt decomposition of $\left|\psi\right\rangle\equiv\frac{\left| 00\right\rangle + \left| 01\right\rangle + \left| 10\right\rangle}{\sqrt{3}}$, we construct the density matrix $\rho$ and find the reduced density matrices $\rho^1$ and $\rho^2$.

$$ \begin{align} \rho &= \frac{1}{\sqrt{3}}\left( \left| 00\right\rangle + \left| 01\right\rangle + \left| 10\right\rangle \right)\left\langle\psi\right| = \frac{1}{3} \begin{pmatrix} 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} \\ \rho^1 &= \rho^2 = \frac{1}{3} \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \\ \end{align} $$

Now we find the eigenvalues and eigenvectors of $\rho^1$.

$$ \begin{align} \mathrm{det}\left(\rho^1-\lambda I\right) &= \begin{vmatrix} \frac{2}{3}-\lambda & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3}-\lambda \end{vmatrix} \\ &= 9\lambda^2-9\lambda +1 = 0 \;\;\; \therefore \lambda_{\pm} = \frac{3\pm \sqrt{5}}{6} \end{align} $$

For $\lambda_+$,

$$ \begin{equation} \begin{pmatrix} \frac{2}{3}-\lambda_+ & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3}-\lambda_+ \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \;\;\; \therefore \left|\lambda_+\right\rangle = \sqrt{\frac{2}{5+\sqrt{5}}} \begin{pmatrix} \frac{1+\sqrt{5}}{2} \\ 1 \end{pmatrix} \end{equation} $$

For $\lambda_-$,

$$ \begin{equation} \begin{pmatrix} \frac{2}{3}-\lambda_- & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3}-\lambda_- \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \;\;\; \therefore \left|\lambda_-\right\rangle = \sqrt{\frac{2}{5-\sqrt{5}}} \begin{pmatrix} \frac{1-\sqrt{5}}{2} \\ 1 \end{pmatrix} \end{equation} $$

So,

$$ \begin{equation} \frac{\left| 00\right\rangle +\left| 01\right\rangle + \left| 10\right\rangle}{\sqrt{3}} = \sum_{i=\{ +,-\} } \sqrt{\lambda_i} \left|\lambda_i\right\rangle\left|\lambda_i\right\rangle, \end{equation} $$

where

$$ \begin{equation} \lambda_{\pm} = \frac{2}{\sqrt{5\pm\sqrt{5}}},\;\;\; \left|\lambda_{\pm}\right\rangle = \sqrt{\frac{2}{5\pm\sqrt{5}}} \begin{pmatrix} \frac{1\pm\sqrt{5}}{2} \\ 1 \end{pmatrix} \end{equation} $$