Exercise 3.14 Discussion / Discussion

This is the discussion related to the wiki page Exercise 3.14 Discussion.

Solution to exercise 3.14

shb65 17 Aug 2018 12:13

a. When measured with respect to the standard basis, the second bit of the state $\ket{00}$ is clearly equal to 0 with probability 1.

With respect to the Hadamard basis, the state $\ket{00}$ is equal to $\frac{1}{2}\left(\ket{+}\ket{+} + \ket{+}\ket{-} +\ket{-}\ket{+} +\ket{-}\ket{-}\right)$. Thus, when measured with respect to the Hadamard basis, the second bit equals $\ket{+}$ with probability $\frac{1}{2}$ and equals $\ket{-}$ with probability $\frac{1}{2}$.

As a shorthand for the vectors of the basis $B$, let us write $\ket{u} = \frac{1}{\sqrt{2}}\left(\ket{0} + \mathbf{i}\ket{1}\right)$ and $\ket{v} = \frac{1}{\sqrt{2}}\left(\ket{0} - \mathbf{i}\ket{1}\right)$. Then $\ket{0} = \frac{1}{\sqrt{2}}\left(\ket{u} + \ket{v}\right)$, and $\ket{00} = \frac{1}{2}\left(\ket{u}\ket{u} + \ket{u}\ket{v} +\ket{v}\ket{u} +\ket{v}\ket{v}\right)$. Much as in the previous answer, therefore, we see that when measured with respect to the basis $B$, the second bit equals $\ket{u}$ with probability $\frac{1}{2}$ and equals $\ket{v}$ with probability $\frac{1}{2}$.

b. We noted above that, with respect to the Hadamard basis, the state $\ket{00}$ is equal to $\frac{1}{2}\left(\ket{+}\ket{+} + \ket{+}\ket{-} +\ket{-}\ket{+} +\ket{-}\ket{-}\right)$. Thus, when its second bit is measured with respect to the Hadamard basis, one of two things can happen with equal probability:

• Case 1: the second bit is measured as $\ket{+}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{+} +\ket{-}\ket{+}\right)$ – let’s call this state $X$.
• Case 2: the second bit is measured as $\ket{-}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{-} +\ket{-}\ket{-}\right)$ – let’s call this state $Y$.

We can compute

(1)

\begin{align} \ket{+}\ket{+} = \frac{1}{2}\left(-\mathbf{i}\ket{u}\ket{u} + \ket{u}\ket{v} + \ket{v}\ket{u} + \mathbf{i}\ket{v}\ket{v}\right) \end{align}

and

(2)

\begin{align} \ket{-}\ket{+} = \frac{1}{2}\left(\ket{u}\ket{u} + \mathbf{i}\ket{u}\ket{v} - \mathbf{i}\ket{v}\ket{u} + \ket{v}\ket{v}\right) \end{align}

and so state $X = \frac{1}{2\sqrt{2}}\left((1-\mathbf{i})\ket{u}\ket{u} + (1+\mathbf{i})\ket{u}\ket{v} + (1-\mathbf{i})\ket{v}\ket{u} + (1+\mathbf{i})\ket{v}\ket{v}\right)$. Thus, if Case 1 occurred on the first measurement, then the second measurement yields $\ket{u}$ with probability $\frac{1}{8}\left({\left|1-\mathbf{i}\right|}^2 + {\left|1-\mathbf{i}\right|}^2\right) = \frac{1}{2}$, and yields $\ket{v}$ with probability $\frac{1}{8}\left({\left|1+\mathbf{i}\right|}^2 + {\left|1+\mathbf{i}\right|}^2\right) = \frac{1}{2}$.

Similarly we can compute

(3)

\begin{align} \ket{+}\ket{-} = \frac{1}{2}\left(\ket{u}\ket{u} - \mathbf{i}\ket{u}\ket{v} + \mathbf{i}\ket{v}\ket{u} + \ket{v}\ket{v}\right) \end{align}

and

(4)

\begin{align} \ket{-}\ket{-} = \frac{1}{2}\left(\mathbf{i}\ket{u}\ket{u} + \ket{u}\ket{v} + \ket{v}\ket{u} - \mathbf{i}\ket{v}\ket{v}\right) \end{align}

and so state $Y = \frac{1}{2\sqrt{2}}\left((1+\mathbf{i})\ket{u}\ket{u} + (1-\mathbf{i})\ket{u}\ket{v} + (1+\mathbf{i})\ket{v}\ket{u} + (1-\mathbf{i})\ket{v}\ket{v}\right)$. Thus, if Case 2 occurred on the first measurement, then the second measurement yields $\ket{u}$ with probability $\frac{1}{8}\left({\left|1+\mathbf{i}\right|}^2 + {\left|1+\mathbf{i}\right|}^2\right) = \frac{1}{2}$, and yields $\ket{v}$ with probability $\frac{1}{8}\left({\left|1-\mathbf{i}\right|}^2 + {\left|1-\mathbf{i}\right|}^2\right) = \frac{1}{2}$.

c. As in question b, when the second bit of $\ket{00}$ is measured with respect to the Hadamard basis, one of two things can happen with equal probability:

• Case 1: the second bit is measured as $\ket{+}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{+} +\ket{-}\ket{+}\right)$ – we called this state $X$.
• Case 2: the second bit is measured as $\ket{-}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{-} +\ket{-}\ket{-}\right)$ – we called this state $Y$.

We can compute

(5)

\begin{align} \ket{+}\ket{+} = \frac{1}{2}\left(\ket{0}\ket{0} + \ket{0}\ket{1} + \ket{1}\ket{0} + \ket{1}\ket{1}\right) \end{align}

and

(6)

\begin{align} \ket{-}\ket{+} = \frac{1}{2}\left(\ket{0}\ket{0} + \ket{0}\ket{1} - \ket{1}\ket{0} - \ket{1}\ket{1}\right) \end{align}

and so state $X = \frac{1}{\sqrt{2}}\left(\ket{0}\ket{0} + \ket{1}\ket{1}\right)$. Thus, if Case 1 occurred on the first measurement, then the second measurement yields $\ket{0}$ with probability $\frac{1}{2}$ and yields $\ket{1}$ with probability $\frac{1}{2}$.

Similarly we can compute

(7)

\begin{align} \ket{+}\ket{-} = \frac{1}{2}\left(\ket{0}\ket{0} - \ket{0}\ket{1} + \ket{1}\ket{0} - \ket{1}\ket{1}\right) \end{align}

and

(8)

\begin{align} \ket{-}\ket{-} = \frac{1}{2}\left(\ket{0}\ket{0} - \ket{0}\ket{1} - \ket{1}\ket{0} + \ket{1}\ket{1}\right) \end{align}

and so state $Y = \frac{1}{\sqrt{2}}\left(\ket{0}\ket{0} - \ket{0}\ket{1}\right)$. Thus, if Case 2 occurred on the first measurement, then again the second measurement yields $\ket{0}$ with probability $\frac{1}{2}$ and yields $\ket{1}$ with probability $\frac{1}{2}$.

Re: Solution to exercise 3.14

HornetMaX 17 Mar 2020 18:11

I think there's a minor typo in the above, on the line after your equation (6), X should be $\frac{1}{\sqrt{2}}(\ket{0}\ket{0}+\ket{0}\ket{1})$.