Worked Problems in Physics

Nielsen and Chuang Exercise 2.13

Posted in Nielsen/Chuang by rpmuller on February 27, 2010

Exercise 2.13 Show that {\left(\left|w\right>\left<v\right|\right)^\dagger=\left(\left|v\right>\left<w\right|\right)}.

Using the result of Exercise 2.10, we obtain

\displaystyle \left(\left|w\right>\left<v\right|\right)^\dagger= \left[ \begin{array}{rrr} w_1^*v_1 & \dots & w_m^*v_1 \\ \vdots & \ddots & \vdots \\ w_1^*v_n & \dots & w_m^*v_n \end{array} \right]^\dagger \ \ \ \ \ (20)

\displaystyle = \left[ \begin{array}{rrr} w_1v_1^* & \dots & w_1v_n^* \\ \vdots & \ddots & \vdots \\ w_mv_1^* & \dots & w_mv_n^* \end{array} \right] =\left(\left|v\right>\left<w\right|\right) \ \ \ \ \ (21)

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Nielsen and Chuang Exercise 2.12

Posted in Nielsen/Chuang by rpmuller on February 27, 2010

Exercise 2.12 Prove that the matrix

\displaystyle \left[ \begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right] \ \ \ \ \ (17)

is not diagonalizable.

We first solve

\displaystyle 0 = \det\left( \begin{array}{rr} 1-\lambda & 0 \\ 1 & 1-\lambda \end{array} \right) \ \ \ \ \ (18)

to yield the root {\lambda=1} as a doubly degenerate eigenvalue.

To find the corresponding eigenvectors, we solve

\displaystyle \left[ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right] = 0 \ \ \ \ \ (19)

to yield the coupled equations

\displaystyle \begin{array}{rrr} 0 v_1 + 0 v_2 &=& 0 \\ 1 v_1 + 0 v_2 &=& 0 \end{array}. \ \ \ \ \ (20)

Since the equation doesn’t have solutions, the matrix is not diagonalizable.

Alternatively, we can show that the matrix is not normal.

\displaystyle AA^\dagger = \left[ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right] \neq \left[ \begin{array}{rr} 2 & 1 \\ 1 &1 \end{array} \right] = A^\dagger A \ \ \ \ \ (21)

Since the two are different, the matrix is not normal and thus not diagonalizable.

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Nielsen and Chuang Exercise 2.11

Posted in Nielsen/Chuang by rpmuller on February 27, 2010

Exercise 11 Eigendecomposition of the Pauli matrices: Find the eigenvectors, eigenvalues, and diagonal representations of the Pauli matrices {X}, {Y}, and {Z}.

The eigenvalues are {\mp 1}. The corresponding matrix with column eigenvectors are:

\displaystyle U_x = \left[\begin{array}{rr}-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right] \ \ \ \ \ (14)

\displaystyle U_y = \left[\begin{array}{rr}-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\\frac{j}{\sqrt{2}} & -\frac{j}{\sqrt{2}} \end{array}\right] \ \ \ \ \ (15)

\displaystyle U_z = \left[\begin{array}{rr}0 & 1 \\1 & 0 \end{array}\right] \ \ \ \ \ (16)

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