b) $a \vert 0 \rangle + b \vert 1 \rangle \to \frac{b}{a}=\alpha$, $\alpha \to \frac{1}{\sqrt{1+\vert \alpha \vert ^2}} \vert 0 \rangle + \frac{\alpha}{\sqrt{1+ \vert \alpha \vert ^2}}$, and $\infty ↔ 1$

c) $(x, y, z) = (sin(\theta) cos(\phi), sin(\theta) sin(\phi), cos(\theta))$

]]>b) For $\vert + \rangle$, $\theta$ and $\phi$ are found by first setting $cos(\frac{\theta}{2}) = \frac {1}{\sqrt{2}}$ and $e^{i \phi} sin(\frac{\theta}{2}) = \frac{1}{\sqrt{2}}$, with the result that $\theta = \frac{\pi}{2}$ and $\phi = 0$.

The same method can used to find phi and theta for the other bases in question.

b) When Bob's bit value from measurement is 0, there is some chance that what Alice encoded could be 0 or 1, but when Bob's bit value from measurement is 1, the bit encoded by Alice must be 0 in the case that Bob measures $\vert - \rangle$, and must be 1 in the case that Bob measures $\vert 1 \rangle$.

c) With the assumption that Eve knows Alice and Bob are using the B92 protocol, if Eve follows a similar procedure to Alice and Bob, and resorts to discarding all bits measured as $\vert 0 \rangle$ or $\vert + \rangle$, then in the cases where she measures $\vert - \rangle$ and $\vert 1 \rangle$, she can be certain that the bit encoded by Alice is 0 and 1 respectively. If she doesn't know what protocol Alice and Bob are using, then there is no chance for her to be certain of the results of her measurements since she doesn't know what to discard, though she will certainly have some chance of being correct.

Without Eve's presence, in the case that Alice encodes a 0, and Bob generates a 1, there is a 100% chance that he should measure 0, and when Alice encodes a 1, and Bob generates a 0, there is again a 100% chance that he should measure a 1. When Eve interferes however, there is a 25% chance for errors in each case, and as such, overall. The likelihood the likelihood that Alice and Bob will detect Eve's presence as a function of the number of compared bits s, is then: $1 - (\frac{3}{4})^s$

]]>b) The angle between bases determines the change in the probability that Eve measures the bit correctly. Suppose the basis Eve measures in is $\lvert v \rangle = cos(\theta) \lvert 0 \rangle + sin(\theta) \lvert 1 \rangle$, where $\theta$ is the angle between the basis chosen by Alice, and that chosen by Eve. Then, if we take the instance where the \levert 0 \rangle is the what Alice encodes, the probability that Eve will correctly measure this, is $\vert \langle 0 \vert v \rangle \vert ^2 = cos^2(\theta)$. Expanding to the the more general case, the average number of bits that Eve measures correctly will be $\frac{\pi}{2} \int_0^{\frac{\pi}{2}} cos^2 (\theta) d\theta = \frac{1}{2}$. So, on average, Eve is able to correctly measure 50% of the bits encoded by Alice.

c) If Alice encodes the bit as $\vert 0 \rangle$, Eve measures it as such in her randomly chosen basis, and sends it on in that basis, the probability that Bob will still measure the value originally encoded by Alice is represented by $|\langle v \vert 0 \rangle \langle v \vert 0 \rangle|^2 + |\langle v_{\bot} \vert 0 \rangle \langle v_{\bot} \vert \rangle|^2 = cos^4 (\theta) + sin^4 (\theta)$, where Eve's chance of correctly measuring the bit Alice encoded, and Bob's chance of measuring the bit resent by Eve as the same as Alice originally encoded, are represented in the two inner products in each term of the expression. Averaging this expression yields $\frac{\pi}{2} \int_0^{\frac{\pi}{2}} cos^4 (\theta) + sin^4 (\theta) d\theta = \frac{3}{4}$, which represents the chance that Eve will correctly measure a bit encoded by Alice, and go undetected. So, in order for Alice and Bob to have a 90% chance of detecting Eve's presence, they need to compare $n$ bits s.t. $(\frac{3}{4})^n \lt 0.1$, so $n \approx 8$

]]>Part b)

Projectors for the four Bell states are easily derived from the states definitions as

(1)\begin{align} P_{\Phi^+} = \frac{1}{2} \big( \ket{00}\bra{00} + \ket {11}\bra{00} + \ket{00}\bra{11} + \ket{11}\bra{11} \big) \end{align}

(2) \begin{align} P_{\Phi^-} = \frac{1}{2} \big( \ket{00}\bra{00} - \ket {11}\bra{00} - \ket{00}\bra{11} + \ket{11}\bra{11} \big) \end{align}

(3) \begin{align} P_{\Psi^+} = \frac{1}{2} \big( \ket{01}\bra{01} + \ket {10}\bra{01} + \ket{01}\bra{10} + \ket{10}\bra{10} \big) \end{align}

(4) \begin{align} P_{\Psi^-} = \frac{1}{2} \big( \ket{01}\bra{01} - \ket {10}\bra{01} - \ket{01}\bra{10} + \ket{10}\bra{10} \big) \end{align}

Choosing eigenvalues 1, -1, 2 and 2, we could write the measurement operator as a neat

(5)\begin{align} M = \ket{11}\bra{00} + \ket{00}\bra{11} + 2 \ket{10}\bra{01} + 2 \ket{01}\bra{10} \end{align}

however it's not directly useful for our needs here.

Considering the four projectors applied to a general state

$\ket{\psi} = a\ket{00} + b\ket{01} + c\ket{10} + d\ket{11}$

we get

(6)\begin{align} P_{\Phi^+} \ket{\psi} = \frac{a + d}{2} \big( \ket{00} + \ket{11} \big) \end{align}

(7) \begin{align} P_{\Phi^-} \ket{\psi} = \frac{a - d}{2} \big( \ket{00} - \ket{11} \big) \end{align}

(8) \begin{align} P_{\Psi^+} \ket{\psi} = \frac{b + c}{2} \big( \ket{01} + \ket{10} \big) \end{align}

(9) \begin{align} P_{\Psi^-} \ket{\psi} = \frac{b - c}{2} \big( \ket{01} - \ket{10} \big) \end{align}

(note that these are not normalized) and their probabilities

(10)\begin{align} \bra{\psi} P_{\Phi^+} \ket{\psi} = \frac{(a + d)^2}{2} \end{align}

(11) \begin{align} \bra{\psi} P_{\Phi^-} \ket{\psi} = \frac{(a - d)^2}{2} \end{align}

(12) \begin{align} \bra{\psi} P_{\Psi^+} \ket{\psi} = \frac{(b + c)^2}{2} \end{align}

(13) \begin{align} \bra{\psi} P_{\Psi^-} \ket{\psi} = \frac{(b - c)^2}{2} \end{align}

Part a)

For $\ket{\psi} = \ket{00}$, substituting $a=1$ and all other parameters 0 in the above (and not forgetting to normalize), we get the outcomes

$\ket{s_1} = \frac{1}{\sqrt{2}} \big( \ket{00} + \ket{11} \big) = \ket{\Phi^+}$

and

$\ket{s_2} = \frac{1}{\sqrt{2}} \big( \ket{00} - \ket{11} \big) = \ket{\Phi^-}$,

each with probability $\frac{1}{2}$.

Analogously, measurement outcomes for $\ket{01}$ are $\ket{\Psi^+}$ and $\ket{\Psi^-}$, the outcomes for $\ket{10}$ are $\ket{\Psi^+}$ and $-\ket{\Psi^-}$, and the outcomes for $\ket{11}$ are $\ket{\Phi^+}$ and $-\ket{\Phi^-}$, all with probability $\frac{1}{2}$.

]]>\begin{align} O \ket{\phi} = \lambda_1 P_1 \ket{\phi} + \lambda_2 P_2 \ket{\phi} + \dots + \lambda_n P_n \ket{\phi} \end{align}

That would suggest that the second measurement would produce $\ket{\phi'} = \lambda_k \ket{\phi}$, which in the general case wouldn't even be a unit vector, and thus not a valid state. Thus, generally $O^2 \neq O$, and that's okay because we don't apply it to states. $O$ is simply a notation allowing to summarize an observation in a single expression. The actual conversion of a state into the result is performed by one of the projectors $P_i$ (sans the associated eigenvalue), selected with probability $\bra{\psi} P_i \ket{\psi}$.

]]>Following the Hermitian operator formalism, the observable can be written as

(1)\begin{align} O = \sum_{i=1}^{n} \lambda_i P_i \end{align}

The measured state $\ket{\phi}$ by definition belongs to some $m$-dimensional $\lambda_k$-eigenspace of $O$, $1 \leq k \leq n$, with the projector $P_k$. The state can be expressed as

(2)\begin{align} \ket{\phi} = \sum_{i=1}^{m} a_i \ket{\alpha_i} \end{align}

where $\ket{\alpha_j}$ are basis vectors of the $\lambda_k$-eigenspace. Because $\ket{\phi}$ is a $\lambda_k$-eigenvector of $O$, the result of a measurement of $\ket{\phi}$ according to $O$ is by definition a $\lambda_k$-eigenvector of $O$. Alternatively, and perhaps more correctly in light of part b), we could say that $\bra{\phi} P_i \ket{\phi} = 0$ for all $i \neq k$ so the measurement will be performed by projector $P_k$ with probability 1. Because

(3)\begin{align} P_k = \sum_{i=1}^{m} \ket{\alpha_i}\bra{\alpha_i} \end{align}

the result is

(4)\begin{align} P_k \ket{\phi} = \sum_{i=1}^{m} \sum_{j=1}^{m} \big( \ket{\alpha_i}\bra{\alpha_i} \big) a_j \ket{\alpha_j} = \sum_{i=1}^{m} \sum_{j=1}^{m} a_j \ket{\alpha_i}\bra{\alpha_i} \ket{\alpha_j} = \sum_{i=1}^{m} a_i \ket{\alpha_i} = \ket{\phi} \end{align}

]]>
\begin{align} O' = P_1 + P_2 - \big(P_3 + P_4 \big) \end{align}

]]>
\begin{align} M = \ket{001}\bra{001} + \ket{010}\bra{010} + \ket{100}\bra{100} + 2 \Big( \ket{011}\bra{011} + \ket{101}\bra{101} + \ket{110}\bra{110} \Big) + 3 \ket{111}\bra{111} \end{align}

]]>
\begin{align} M = \ket{000}\bra{000} + \ket{011}\bra{011} + \ket{110}\bra{110} \end{align}

]]>
\begin{align} M = \ket{000}\bra{000} + \ket{111}\bra{111} \end{align}

which represents the comparison outcome as $1$ for true and $0$ for false is a better choice of eigenvalues, in line with the remark on p.56.

]]>$\ket{0} = \frac{1}{\sqrt{2}}(\ket{+} + \ket{-})$

$\ket{1} = \frac{1}{\sqrt{2}}(\ket{+} - \ket{-})$

$\ket{0} = \frac{1}{\sqrt{2}}(\ket{i} + \ket{-i})$

$\ket{1} = \frac{-i}{\sqrt{2}}(\ket{i} - \ket{-i})$

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

$\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$

$\ket{+} = \frac{1 - i}{2}\ket{i} + \frac{1 + i}{2}\ket{-i}$

$\ket{-} = \frac{1 + i}{2}\ket{i} + \frac{1 - i}{2}\ket{-i}$

$\ket{i} = \frac{1}{\sqrt{2}}(\ket{0} + i\ket{1})$

$\ket{-i} = \frac{1}{\sqrt{2}}(\ket{0} - i\ket{1})$

$\ket{i} = \frac{1 + i}{2}\ket{+} + \frac{1 - i}{2}\ket{-}$

$\ket{-i} = \frac{1 - i}{2}\ket{+} + \frac{1 + i}{2}\ket{-}$

(an independent correctness check is welcome).

]]>\begin{align} \bra{\psi} P_+ \ket{\psi} = (\overline{a}\bra{0} + \overline{b}\bra{1}) \frac{1}{\sqrt{2}} (\ket{0} + \ket{1}) \frac{1}{\sqrt{2}} (\bra{0} + \bra{1}) ( a\ket{0} + b\ket{1}) = \frac{\overline{(a + b)} (a + b)}{2} = \frac{| a + b |^2}{2} \end{align}

(2) \begin{align} \bra{\psi} P_- \ket{\psi} = (\overline{a}\bra{0} + \overline{b}\bra{1}) \frac{1}{\sqrt{2}} (\ket{0} - \ket{1}) \frac{1}{\sqrt{2}} (\bra{0} - \bra{1}) ( a\ket{0} + b\ket{1}) = \frac{\overline{(a - b)} (a - b)}{2} = \frac{| a - b |^2}{2} \end{align}

]]>
Consider a basis vector $\ket{\alpha_i}$ of $S$, expressed in the standard basis as

$\ket{\alpha_i} = a_0\ket{0} + a_1\ket{1} + \dots + a_n\ket{n}$

To simplify the presentation, instead of the "full" $P_s$, let's consider the matrix of each individual outer product $A = \ket{\alpha_i} \bra{\alpha_i}$. An element of $A$ has the form

$A_{jk} = a_j\overline{a_k}$

An element of its conjugate transpose $A^{\dagger}$

$A^{\dagger}_{jk} = \overline{A_{kj}} = \overline{a_k\overline{a_j}} = \overline{a_k} a_j = a_j \overline{a_k} = A_{jk}$

Thus, $A$ is its own conjugate transpose and $P_s$, which is a sum of all $A$s, is also its own conjugate transpose.

Note that this does not mean that the matrix of $P_s$ or any of the $A$s is symmetric and real. Only the diagonal elements of $A$s and $P_s$ are necessarily real because $a_i\overline{a_k}$ is real if $i = k$.

]]>We start by assuming that the two states are orthogonal thus equation (8) of the previous exercise is by definition equal zero

$\braket {\psi_2} \psi = 0 \therefore \frac{1}{\sqrt{\alpha^2 + 1}}( \overline{a_{\psi_2}} + \alpha\overline{b_{\psi_2}}) = 0$

$\overline{a_{\psi_2}} = -\alpha\overline{b_{\psi_2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (14)$

$\alpha_2 = \frac{b_{\ket {\psi_2}}}{a_{\ket {\psi_2}}} \ \ \therefore \ \ \overline{$\alpha_2} = \frac{\overline{b_{\ket {\psi_2}}}}{\overline{a_{\ket {\psi_2}}}} \ \ \therefore \ \ \overline{$\alpha_2} = \frac{-1}{\alpha}$

$\overline{\alpha_2} = \frac{-\overline{\alpha}}{\mid \alpha \mid^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (15)$

we now see that (15) is the same as (11). It is strait forward now to go from (15) to (1), (2) and (3) proving that orthogonal states generates antipodal points.

]]>To get antipodal points $\ket \psi$ and $\ket {\psi_2}$ in a sphere, you just invert the signal of the Cartesian coordinates.

So the correspondent $\alpha_2$ of antipodal state $\ket {\psi_2}$, where $\alpha_2 = S_2 + i.T_2$ is given by:

We just need to show that $\braket {\psi_2} {\psi}$ must be zero. In order to prove that $\ket \psi$ and $\ket {\psi_2}$ are orthogonal.

from $(3)$ we can get $\alpha_2$ from $\alpha$

$(\mid\alpha\mid^2 - 1)(\mid\alpha_2\mid^2 + 1)= (1 - \mid\alpha_2\mid^2)(\mid\alpha\mid^2 + 1)$

$2\mid\alpha_2\mid^2\mid\alpha\mid^2 = 2$

$\mid\alpha_2\mid^2 = \frac{1}{\mid\alpha\mid^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

we know that:

$\alpha = \frac{b}{a}\ \ \ \ \ \ \ \ \ \ \ \ \ (5)$

$a_{\ket \psi} = \frac{1}{\sqrt{\alpha^2 + 1}}\ \ (6)$

$b_{\ket \psi} = \frac{\alpha}{\sqrt{\alpha^2 + 1}}\ \ (7)$

and that

$\braket {\psi_2} {\psi} = a_{\ket \psi}\ \overline{a_{\ket {\psi_2}}} + b_{\ket \psi}\ \overline{b_{\ket {\psi_2}}}$

applying (6) and (7) we ge

$\braket {\psi_2} {\psi} = \frac{1}{\sqrt{\alpha^2 + 1}}(\overline{a_{\ket {\psi_2}}} + \alpha\overline{b_{\ket {\psi_2}}}) \ \ (8)$

applying (4) to (1) and (2) we have:

$S_2 = \frac{-S}{\mid\alpha\mid^2 } \ \ \ \ \ (9)$

$T_2 = \frac{-T}{\mid\alpha\mid^2 } \ \ \ \ \ (10)$

(4) gives the modulus of $\alpha_2$ in relation to modulus of $\alpha$, we can use the fact that $\overline{\alpha_2} = S_2 - i.T_2$ and use (9) and (10) to get $\alpha_2$ in relation to $\alpha$

$\overline{\alpha_2} = S_2 - i.T_2 \ \ \therefore \ \ \overline{\alpha_2} = \frac{-S}{\mid\alpha\mid^2 } - i. \frac{-T}{\mid\alpha\mid^2 }$

$\overline{\alpha_2} = \frac{-\overline{\alpha}}{\mid\alpha\mid^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11)$

now if we use (11), (6) and (7) it is strait forward to calculate $\overline{a_{\ket {\psi_2}}}$ and $\overline{b_{\ket {\psi_2}}}$

$\overline{a_{\ket {\psi_2}}} = \frac{1}{\sqrt{\frac{\overline{\alpha^2}}{\mid\alpha\mid^4} + 1}}\ \ \therefore\ \ \overline{a_{\ket {\psi_2}}} = \frac{\mid\alpha\mid^2}{\sqrt{\overline{\alpha^2} + \mid\alpha\mid^4}}\ \ (12)$

$\overline{b_{\ket {\psi_2}}} = \frac{-\overline{\alpha^2}}{\mid\alpha\mid^2\sqrt{\frac{\overline{\alpha^2}}{\mid\alpha\mid^4} + 1}}\ \ \therefore\ \ \overline{b_{\ket {\psi_2}}} = \frac{-\overline{\alpha^2}}{\sqrt{\overline{\alpha^2} + \mid\alpha\mid^4}}\ \ (13)$

Knowing that $\overline{\alpha}\alpha = \mid \alpha \mid^2$ for any complex number, we can use (12) e (13) to prove that (8) is zero

$\braket {\psi_2} {\psi} = \frac{1}{\sqrt{\alpha^2 + 1}}( \frac{\mid\alpha\mid^2}{\sqrt{\overline{\alpha^2} + \mid\alpha\mid^4}} - \frac{\overline{\alpha\alpha^2}}{\sqrt{\overline{\alpha^2} + \mid\alpha\mid^4}} )$

$\braket {\psi_2} {\psi} = \frac{1}{\sqrt{\alpha^2 + 1}}( \frac{\mid\alpha\mid^2}{\sqrt{\overline{\alpha^2} + \mid\alpha\mid^4}} - \frac{\mid\alpha\mid^2}{\sqrt{\overline{\alpha^2} + \mid\alpha\mid^4}} )$

$\braket {\psi_2} {\psi} = 0\ \ \ \ _{Q.E.D.}$

]]>in the second case the coefficient is altering the relative phase of those qubits hence they are not the same state.

The global phase is always cancel out during the inner product calculation when changing basis, so it carries no useful information.

The relative phase on the other hand will not.

In the above example $\frac{1}{\sqrt{2}}(\ket 0 + \ket 1) = \ket +$ and $\frac{1}{\sqrt{2}}(\ket 0 - \ket 1) = \ket -$

So:

$\frac{1}{\sqrt{2}}(\ket 0 + \ket 1)\ \rightarrow\ P_{\ket +} = 1 , \ P_{\ket -} = 0$

$\frac{1}{\sqrt{2}}(\ket 0 - \ket 1)\ \rightarrow\ P_{\ket +} = 0 , \ P_{\ket -} = 1$

Coefficient of $\ket 0$ must be zero in order for the states to be equivalent, so $\frac{1 + eî\theta}{2} = 0$.

By Euler formula

therefore $\theta = \pi + 2K\pi, K \in \mathbb{N}$

**b)**

$e^{i\theta} = 1$ and $e^{-i\theta} = 1$

$cos(\theta) + i sen(\theta) = 1$ and $cos(\theta) - i sen(\theta) = 1$

therefore $\theta = 0 + 2K\pi, K \in \mathbb{N}$

**c)**

$e^{i\theta}$ is a global phase therefore any value of $\theta$ make the two states equivalent

A single qubit state $\ket \psi$ is a superposition in respect to a given basis $\{\ket K , \ket{K_\perp}\}$ if it can be written as a **non trivial** linear combination of the basis (as stated in the second paragraph of FlippingBit)

To find a basis for which $\ket \psi$ is not a superposition, all one needs to do is find a orthogonal state $\ket {\psi_\perp}$

$\braket \psi {\psi_\perp} = 0$, and the new basis is of course $\{\ket \psi , \ket{\psi_\perp}\}$

**a)** As explained by FlippingBit

**b)**

$\frac{1}{\sqrt{2}}(\ket + + \ket -) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) + \frac{1}{\sqrt{2}}(\ket 0 - \ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + \ket 1) + \frac{1}{2}(\ket 0 - \ket 1)$

$\frac{1}{\sqrt{2}}(\ket + + \ket -) = \ket 0$

"b)" is also not a superposition (non trivial linear combination)

**c)**

$\frac{1}{\sqrt{2}}(\ket + - \ket -) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) - \frac{1}{\sqrt{2}}(\ket 0 - \ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + \ket 1) - \frac{1}{2}(\ket 0 - \ket 1)$

$\frac{1}{\sqrt{2}}(\ket + - \ket -) = \ket 1$

"c)" is not a superposition (non trivial linear combination)

**d)**

$\frac{\sqrt3}{2}\ket + - \frac{1}{2}\ket - \Rightarrow \frac{\sqrt3}{2}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) ) - \frac{1}{2} (\frac{1}{\sqrt{2}}(\ket 0 - \ket 1))$

$\frac{1}{\sqrt{2}}(\ket + - \ket -) = \frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1$ which is a trivial case,

therefore "d) is a superposition"

In the basis $\{\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1 ,\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 - \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1\}$ this state is not in superposition. as explained by FlippingBit

One may check by doing $\braket {(\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1)} {(\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 - \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1\})} = 0$

**e)**

$\frac{1}{\sqrt{2}}(\ket i - \ket{-i}) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + i\ket 1) - \frac{1}{\sqrt{2}}(\ket 0 - i\ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + i\ket 1) - \frac{1}{2}(\ket 0 - i\ket 1)$

$\frac{1}{\sqrt{2}}(\ket i - \ket{-i}) = i\ket 1$

"e)" is not a superposition (non trivial linear combination)

**f)**

as explained by FlippingBit