Here is an alternative argument. We see that $a_1 a_2 = a_1 b_2$, and these are non-zero; hence $a_2 = b_2$. Similarly $a_1 = b_1$. But then $b_1 b_2$ must be equal to [[$a_1 a_2]], and we can see that this is not the case. Thus the system has no solutions and the state is indeed entangled.

]]>\begin{align} \ket{GHZ_n} = (a_1\ket{0} + b_1\ket{1}) \otimes (a_2\ket{0} + b_2\ket{1}) \otimes \cdots \otimes (a_n\ket{0} + b_n\ket{1}) \end{align}

From the definition of $\ket{GHZ_n}$, we must therefore have

(2)\begin{align} a_1 \cdots a_n = \frac{1}{\sqrt{2}} \end{align}

so all of the $a_i$ values must be non-zero; and

(3)\begin{align} b_1 \cdots b_n = \frac{1}{\sqrt{2}} \end{align}

so all of the $b_i$ values must be non-zero. But we also have

(4)\begin{align} a_1 (b_2 \cdots b_n) = 0 \end{align}

which is clearly impossible. Thus our assumption was wrong and $\ket{GHZ_n}$ must be entangled.

]]>Here's a really simple analogy. Consider (a) an isosceles triangle pointing upwards, and (b) the same size isosceles triangle rotated through 180 degrees, i.e. pointing downwards. We can say that these two images are different representations of the same shape.

Now let's carry out an operation "put a circle on" to each of these shapes. Position a circle directly above the triangle on the page, just touching the triangle. Are the shapes still the same? No.

To keep the shape the same after the operation, we need to make sure that the "put a circle on" operation is carried out in the right way, i.e. with respect to the particular representation we initially chose.

]]>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 question **a** is $\frac{1}{\sqrt{2}}\left(\left|0\right>+\mathbf{i}\left|1\right>\right)$. 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 question **b**. Thus these two states are equivalent, and so the answer to **b** is the same as the answer to **a**.

**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.

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}

]]>