$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 3.14}$

a) For the standard basis, the Hadamard basis, and the basis $B = \{\frac{1}{\sqrt 2}(\ket 0 + \i\ket 1, \ket 0 - \i\ket 1 \}$, determine the probability of each outcome when the second qubit of a two-qubit system in the state $\ket{00}$ is measured in each of the bases.

b) Determine the probability of each outcome when the second qubit of the state $\ket{00}$ is first measured in the Hadamard basis and then in the

basis $B$ of part a).

c) Determine the probability of each outcome when the second qubit of the state $\ket{00}$ is first measured in the Hadamard basis and then in the standard basis.

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:We can compute

(1)and

(2)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)and

(4)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:We can compute

(5)and

(6)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)and

(8)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 $\frac{1}{2}$.

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