In each of the following cases, we will suppose that one of two single qubit states $\ket{a}$ and $\ket{b}$ is selected randomly with probability $1/2$, but we are not told which. The goal is to guess which state was selected with as high a probability as we can achieve. What will be described is a distinguishing procedure in the form of a unitary operation followed by a measurement in the computational basis, and its success probability.

__CASE: (i)__

$\ket{a}=\ket{0}$ and $\ket{b}=\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$

The inner product of the two states $\ket{a}$ and $\ket{b}$ is:

\[\ip{a}{b}=\bra{0}\super{\ket{0}+\ket{1}}=\frac{1}{\sqrt{2}}(\ip{0}{0}+\ip{0}{1})=\frac{1}{\sqrt{2}},\]

since $\ip{i}{j}=\delta_{ij}$ for all $i,j\in\{0,1\}$. Then using the geometrical relation $\ip{a}{b}=\cos{\phi}$, where $\phi\in[0,2\pi)$ denotes the angle between the two states $\ket{a}$ and $\ket{b}$, it is seen that in this case

\[\phi=\arccos{\frac{1}{\sqrt{2}}}=\frac{\pi}{4}.\]

In general, two pure states can only be perfectly distinguished if they are orthogonal. That is, if $\phi=\frac{\pi}{2},\frac{3\pi}{2}$. Therefore, in this case the two states $\ket{a}$ and $\ket{b}$ cannot be perfectly distinguished.

Consider the unitary operation $R$ that rotates states by $\theta=\frac{\pi}{8}$ given by:

\[

R=\begin{pmatrix}

\cos{\frac{\pi}{8}}&-\sin{\frac{\pi}{8}} \\

\sin{\frac{\pi}{8}}&\cos{\frac{\pi}{8}}

\end{pmatrix}

\]

Then applying $R$ to the two computational basis states $\ket{0}$ and $\ket{1}$ gives:

\[\begin{array}{rcl}

R\ket{0}&=&\cos{\frac{\pi}{8}}\ket{0}+\sin{\frac{\pi}{8}}\ket{1}\\

R\ket{1}&=&-\sin{\frac{\pi}{8}}\ket{0}+\cos{\frac{\pi}{8}}\ket{1}

\end{array}\]

Therefore, applying $R$ to the states $\ket{a}$ and $\ket{b}$ yields:

\[\begin{array}{rcl}

R\ket{a}=R\ket{0}&=&\cos{\frac{\pi}{8}}\ket{0}+\sin{\frac{\pi}{8}}\ket{1},\\

R\ket{b}=\frac{1}{\sqrt{2}}R(\ket{0}+\ket{1})

&=&\frac{1}{\sqrt{2}}(\cos{\frac{\pi}{8}}\ket{0}+\sin{\frac{\pi}{8}}\ket{1}-\sin{\frac{\pi}{8}}\ket{0}+\cos{\frac{\pi}{8}}\ket{1})\\

&=&\frac{1}{\sqrt{2}}((\cos{\frac{\pi}{8}}-\sin{\frac{\pi}{8}})\ket{0}+(\cos{\frac{\pi}{8}}+\sin{\frac{\pi}{8}})\ket{1})

\end{array}\]

Since $R$ is unitary the inner product of $R\ket{a}$ and $R\ket{b}$ is preserved, and also the angle $\phi=\frac{\pi}{4}$ between the states is preserved.

When either $R\ket{a}$ or $R\ket{b}$ are measured in the computational basis what will result is one of the basis states $\ket{0}$ or $\ket{1}$, but with certain probability. If the state happened to be $R\ket{a}$, then the probability of observing $\ket{0}$ and $\ket{1}$ is $\cos^2{\frac{\pi}{8}}$ and $\sin^2{\frac{\pi}{8}}$, respectively. If instead the state $R\ket{b}$ was measured, the probability of observing $\ket{0}$ would be $\sin^2{\frac{\pi}{8}}$, and the probability of observing $\ket{1}$ would be $\cos^2{\frac{\pi}{8}}$.

Thus when either $\ket{a}$ or $\ket{b}$ is given uniformly at random, first apply $R$ to the state and then measure. If $\ket{0}$ is observed guess the state $\ket{a}=\ket{0}$, and if the state $\ket{b}$ is observed guess the state $\ket{b}=\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$.

The probability of guessing the state correctly is given by $\frac{1}{2}(1+\sin{\phi})$, where $\phi=\arccos({\ip{a}{b}})$ is the angle between the states. In this case, with $\phi=\frac{\pi}{4}$, the probability of successfully guessing the correct state is $\frac{1}{2}(1+\sin{\frac{\pi}{4}})\approx .50$.

**Case: (ii)**

$\ket{a}=\super{\ket{0}+i\ket{1}}$ and $\ket{b}=\super{i\ket{0}+\ket{1}}$

Consider the inner product between $\ket{a}$ and $\ket{b}$:

\[\ip{a}{b}=(\super{\bra{0}-i\bra{1}})(\super{i\ket{0}+\ket{1}})=\frac{1}{2}(i\ip{0}{0}+\ip{0}{1}+\ip{1}{0}-i\ip{1}{1})=\frac{i}{2}-\frac{i}{2}=0\]

Hence, the two states are orthogonal so the angle between them is $\phi=\frac{\pi}{2}$. There must exist a unitary transformation $U$ that maps these two states to the computational basis states $\ket{0}$ and $\ket{1}$. Consider the following unitary map:

\[

U=\frac{1}{\sqrt{2}}\begin{pmatrix}

1&-i \\

i&-1\end{pmatrix},\]

and observe that

\[

U\ket{a}=\frac{1}{\sqrt{2}}\begin{pmatrix}

1&-i \\

i&-1\end{pmatrix}

\begin{pmatrix}

\frac{1}{\sqrt{2}}\\

\frac{i}{\sqrt{2}}

\end{pmatrix}=

\begin{pmatrix}

1\\

0

\end{pmatrix},\]

\[U\ket{b}=\frac{1}{\sqrt{2}}\begin{pmatrix}

1&-i \\

i&-1\end{pmatrix}

\begin{pmatrix}

\frac{i}{\sqrt{2}}\\

\frac{1}{\sqrt{2}}

\end{pmatrix}=

\begin{pmatrix}

0\\

1

\end{pmatrix},

\]

Then by applying the unitary operation $U$ to either of the states $\ket{a}$ or $\ket{b}$ chosen uniformly at random, and then measuring in the computational basis, either the state $\ket{0}$ or $\ket{1}$ will be observed with certainty. If $\ket{0}$ is observed, then the original state was $\ket{a}=\super{\ket{0}+i\ket{1}}$. Otherwise, if the state $\ket{1}$ was observed, then the original state was $\ket{b}=\super{i\ket{0}+\ket{1}}$. Since these states are orthogonal, the probability of success with the procedure is $1$.

**Case: (iii)**$\ket{a}=\cos\theta\ket{0}+\sin\theta\ket{1}$ and $\ket{b}=\cos\theta\ket{0}-\sin\theta\ket{1}$, where $\theta\in[0,2\pi)$ is known.

The inner product of the states $\ket{a}$ and $\ket{b}$ is given by:

\[\begin{array}{rcl}\ip{a}{b}&=&(\cos\theta\bra{0}+\sin\theta\bra{1})(\cos\theta\ket{0}-\sin\theta\ket{1})\\

&=&\cos^2{\theta}\ip{0}{0}+\cos{\theta}\sin{\theta}\ip{0}{1}-(\sin{\theta}\cos{\theta}\ip{1}{0}+\sin^2{\theta}\ip{1}{1} \\

&=&\cos^2{\theta}-\sin^2{\theta} \\

&=&\cos{2\theta}

\end{array}\]

Therefore the angle between the two states is $\phi=2\theta$. If $\theta=0$, then $\phi=2\theta=0$ and the states are actually the same. In this case, there is no improved way of distinguishing between the two states besides merely guessing with a $50/50$ chance. Otherwise, consider the unitary operator $R$ that performs a rotation by $\pi/4$:

\[R=\begin{pmatrix}\cos{\frac{\pi}{4}}&-\sin{\frac{\pi}{4}} \\

\sin{\frac{\pi}{4}}&\cos{\frac{\pi}{4}}

\end{pmatrix}\]

Geometrical considerations show that applying $R$ to the states $\ket{a}$ and $\ket{b}$ yields:

\[\begin{array}{rcl}R\ket{a}&=&\cos{(\theta-\pi/4)}\ket{0}+\sin{(\theta-\pi/4)}\ket{1},\\

R\ket{b}&=&\cos{(\theta+\pi/4)}\ket{0}+\sin{(\theta+\pi/4)}\ket{1}.

\end{array}\]

Then if $R\ket{a}$ is measured in the computation basis,$\ket{0} $will be observed with probability $\cos^2{(\theta-\pi/4)}$ and $\ket{1}$ will be observed with probability $\sin^2{(\theta-\pi/4)}$. On the other hand, if the state $R\ket{b}$ is measured, the states $\ket{0}$ and $\ket{1}$ will be observed with probability $\cos^2{(\theta+\pi/4)}$ and $\sin^2{(\theta+\pi/4)}$, respectively.

Therefore, for $0\leq\theta\leq\pi/2$ or $\pi\leq\theta\leq3\pi/2$, if the state $\ket{0}$ is observed, guess that the original state was $\ket{a}$. For $\pi/2\leq\theta\leq\pi/2$ or $3\pi/2\leq\theta\leq2\pi$, if $\ket{1}$ is observed, then guess that the original state was $\ket{b}$. This method has a probability of success given by $P=\frac{1}{2}(1+\sin{(2\theta)})$.

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