Exercise 2.2#

(1)#\[\begin{split} \begin{align} A|v_1\rangle &= A|0\rangle = A_{11}|0\rangle + A_{21}|1\rangle = |1\rangle \\ A|v_2\rangle &= A|1\rangle = A_{12}|0\rangle + A_{22}|1\rangle = |0\rangle \end{align} \end{split}\]

From eq. (1) we could conclude that

(2)#\[\begin{split} A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \end{split}\]

Suppose we have a different basis of \(V\) with

(3)#\[ |+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, |-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}} \]

To check whether they are basis, we could do

(4)#\[\begin{split} \begin{align} \langle +|+\rangle &= \frac{\langle 0 | + \langle 1|}{\sqrt{2}}\cdot\frac{|0\rangle + |1\rangle}{\sqrt{2}} = \frac{\langle 0 |0\rangle + \langle 1|1\rangle}{2} = 1 \\ \langle +|-\rangle &= \frac{\langle 0 | + \langle 1|}{\sqrt{2}}\cdot\frac{|0\rangle - |1\rangle}{\sqrt{2}} = \frac{\langle 0 |0\rangle - \langle 1|1\rangle}{2} = 0 \\ \langle -|+\rangle &= \frac{\langle 0 | - \langle 1|}{\sqrt{2}}\cdot\frac{|0\rangle + |1\rangle}{\sqrt{2}} = \frac{\langle 0 |0\rangle - \langle 1|1\rangle}{2} = 0 \\ \end{align} \end{split}\]

Then we could conclude from eq. (4) that the basis defined in eq. (3) is a valid orthonormal basis. Under the basis of eq. (3), basis vectors \(|0\rangle\) and \(|1\rangle\) becomes

(5)#\[ |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}}, |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}} \]

From eq. (5), eq. (3) becomes

(6)#\[\begin{split} \begin{align} A|0\rangle = A\left(\frac{|+\rangle + |-\rangle }{\sqrt{2}}\right) = \frac{A|+\rangle + A|-\rangle }{\sqrt{2}} = |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}}\\ A|1\rangle = A\left(\frac{|+\rangle - |-\rangle }{\sqrt{2}}\right) = \frac{A|+\rangle - A|-\rangle }{\sqrt{2}} = |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}} \\ \end{align} \end{split}\]

From eq. (6) we could find that

(7)#\[ A|+\rangle = |+\rangle, A|-\rangle = -|-\rangle \]

Using similar method with eq. (1), we obtain

(8)#\[\begin{split} \begin{align} A|v_1\rangle = A|+\rangle = A_{11}|+\rangle + A_{21}|-\rangle = |+\rangle\\ A|v_2\rangle = A|-\rangle = A_{12}|+\rangle + A_{22}|-\rangle = -|-\rangle\\ \end{align} \end{split}\]

Thus, the matrix representation of \(A\) in basis \(|+\rangle\) and \(|-\rangle\) becomes

(9)#\[\begin{split} A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} \end{split}\]

Then under basis \(|+\rangle\) and \(|-\rangle\), the matrix representation of \(A\) is different.