Exercise 2.7 Discussion

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 2.7}$

For each of the following states, describe all orthonormal bases that includes that state.

a) $\frac{1}{\sqrt 2}\left(\ket{0} + \i\ket 1 \right)$

b) $\frac{1 + \i}{2}\ket{0} - \frac{1 - \i}{2}\ket 1$

c) $\frac{1}{\sqrt 2}\left(\ket{0} + e^{\i\pi/6}\ket 1 \right)$

d) $\frac{1}{2}\ket{+} - \frac{\i\sqrt 3}{2}\ket{-}$

page revision: 1, last edited: 12 Nov 2011 05:44

The wording of the question makes it sound as though, for any given unit vector, there are many possible states that would provide the second component of an orthonormal basis. But in fact there is only one (albeit with many equivalent representations); we can see this from the discussion of antipodal points on the Bloch sphere in section 2.5.2 of the book.

Answers to questions are as follows:

a.In column vector form, the given state is $\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ \mathbf{i} \end{pmatrix}$. The complex conjugate of this is $\frac{1}{\sqrt{2}}\begin{pmatrix} 1&-\mathbf{i} \end{pmatrix}$. A state $\alpha \left|0\right> + \beta\left|1\right>$ is orthogonal to the given state if $\frac{1}{\sqrt{2}}\begin{pmatrix}1&-\mathbf{i}\end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = 0$, i.e. if $\alpha = \mathbf{i}\beta$. The (unit length) vector we seek is thus $\frac{1}{\sqrt{2}}\left(\left|0\right> - \mathbf{i}\left|1\right>\right)$. For any $\theta$, an equivalent representation of this state is $\frac{1}{\sqrt{2}}e^{\mathbf{i}\theta}\left(\left|0\right> - \mathbf{i}\left|1\right>\right)$.Shorter answer: the given state is $\left|\mathbf{i}\right>$, and the orthonormal basis containing it is $\{\left|\mathbf{i}\right>, \left|-\mathbf{i}\right>\}$.

b.The state given in questionais $\frac{1}{\sqrt{2}}\left(\left|0\right>+\mathbf{i}\left|1\right>\right)$. If we multiply this by the constant value $\frac{1+\mathbf{i}}{\sqrt{2}}$, we get $\frac{1+\mathbf{i}}{2}\left|0\right>-\frac{1-\mathbf{i}}{2}\left|1\right>$, which is the state given in questionb. Thus these two states are equivalent, and so the answer tobis the same as the answer toa.c.In Extended Complex Plane representation, the given vector is $e^{\frac{\mathbf{i}\pi}{6}} = \frac{\sqrt{3}}{2} + \frac{1}{2}\mathbf{i}$. In Block Sphere representation, this becomes $\left(\frac{\sqrt{3}}{2},\frac{1}{2},0\right)$. The antipodal point on the Bloch Sphere is $\left(\frac{-\sqrt{3}}{2},\frac{-1}{2},0\right)$, which in Extended Complex Plane representation is $\frac{-\sqrt{3}}{2} - \frac{1}{2}\mathbf{i} = e^{\frac{7\,\mathbf{i}\,\pi}{6}}$. Translating this back into standard representation we have $\frac{1}{\sqrt{2}}\left(\left|0\right> + e^{\frac{7\,\mathbf{i}\,\pi}{6}}\left|1\right>\right)$, and this is the vector that forms an orthonormal basis together with the given vector.d. Throughout this solution we will work with respect to the Hadamard basis. In column vector form, the given state is $\begin{pmatrix} \frac{1}{2} \\ \frac{\mathbf{i}\sqrt{3}}{2} \end{pmatrix}$. The complex conjugate of this is $\begin{pmatrix} \frac{1}{2} & \frac{-\mathbf{i}\sqrt{3}}{2} \end{pmatrix}$. A state $\alpha \left|+\right> + \beta\left|-\right>$ is orthogonal to the given state if $\begin{pmatrix} \frac{1}{2} & \frac{-\mathbf{i}\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = 0$, i.e. if $\alpha = (\mathbf{i}\sqrt{3})\beta$. The (unit length) vector we seek is thus $\left(\frac{\mathbf{i}\sqrt{3}}{2}\left|+\right> + \frac{1}{2}\left|-\right>\right)$. This and the given state form an orthonormal basis.ReplyOptions