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31 Jul 2017 07:18
in discussion Hidden / Per page discussions » Ex4 14

It appears to be easier to do part b) first and then use it to produce answers for part a).

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 $\frac{1}{2}$.

by , 31 Jul 2017 07:18
30 Jul 2017 07:28
in discussion Hidden / Per page discussions » Ex4 13

In the above explanation, we intentionally said "the result of a measurement of $\ket{\phi}$ according to $O$" instead of "the result of applying $O$ to $\ket {\phi}$". An observable only describes how a measurement is performed, but is not by itself applied to a state vector to produce the measured state. In other words, it would be incorrect to attempt to express that result as

(1)
\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 $\bra{\psi} P_i \ket{\psi}$.

b) by , 30 Jul 2017 07:28
30 Jul 2017 07:10
in discussion Hidden / Per page discussions » Ex4 13

Let's try to express this more formally.

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}
a) by , 30 Jul 2017 07:10
29 Jul 2017 06:11
in discussion Hidden / Per page discussions » Ex4 12

Since a projector onto a direct sum of spaces is a sum of projectors onto the individual spaces, the projectors onto $S_5$ and $S_6$ are $P_5 = P_1 + P_2$ and $P_6 = P_3 + P_4$. Thus, a measurement operator for $S_5 \oplus S_6$ can be written as

(1)
\begin{align} O' = P_1 + P_2 - \big(P_3 + P_4 \big) \end{align}
by , 29 Jul 2017 06:11
29 Jul 2017 05:52
in discussion Hidden / Per page discussions » Ex4 11

Using the number of bits as the eigenvalues:

(1)
\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}
by , 29 Jul 2017 05:52
29 Jul 2017 05:50
in discussion Hidden / Per page discussions » Ex4 10

Assigning the eigenvalue of $1$ to the "even" outcome and $0$ to "odd":

(1)
\begin{align} M = \ket{000}\bra{000} + \ket{011}\bra{011} + \ket{110}\bra{110} \end{align}
by , 29 Jul 2017 05:50
29 Jul 2017 05:42
in discussion Hidden / Per page discussions » Ex4 9

It could be argued that the simple

(1)
\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.

by , 29 Jul 2017 05:42
28 Jul 2017 06:46
in discussion Hidden / Per page discussions » Errata

p67, Exercise 4.7 b: the state $\ket{\Psi^+}$ as specified is actually $\ket{\Phi^+}$ from the Bell state set.

by , 28 Jul 2017 06:46
27 Jul 2017 21:07
in discussion Hidden / Per page discussions » Exercise 2.2

For these and many other exercises throughout the book, it helps to first work out the following cheat sheet of basis vector identities:

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

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

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

(an independent correctness check is welcome).

A Cheat Sheet by , 27 Jul 2017 21:07
27 Jul 2017 06:31
in discussion Hidden / Per page discussions » Ex4 4

Let's rewrite these in LaTex for better readability, and remembering that $a$ and $b$ are complex. Given a ket $a\ket{\beta}$ where $\ket{\beta}$ is a standard basis vector, the corresponding bra is $\overline{a}\bra{\beta}$. So these would be:

(1)
\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}
by , 27 Jul 2017 06:31
26 Jul 2017 02:50
in discussion Hidden / Per page discussions » Ex4 3

I can't readily "see" the inferences in the above reasoning, so I'm going to work out what the matrix of a projection operator looks like.

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 $i = k$.

by , 26 Jul 2017 02:50
20 Jul 2017 17:39
in discussion Hidden / Per page discussions » Exercise 2.14 Discussion

b)

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.

by , 20 Jul 2017 17:39
20 Jul 2017 15:24
in discussion Hidden / Per page discussions » Exercise 2.14 Discussion

a)

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:

### $\frac{1- \mid\alpha_2\mid^2}{\mid\alpha_2\mid^2 + 1} = \frac{\mid\alpha\mid^2 - 1}{\mid\alpha\mid^2 + 1} \ \ (3)$

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:

### $\frac{2S_2\mid\alpha\mid^2}{\mid\alpha\mid^2 + 1} = \frac{-2S}{\mid\alpha\mid^2 + 1}$

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

### $\frac{2T_2\mid\alpha\mid^2}{\mid\alpha\mid^2 + 1} = \frac{-TS}{\mid\alpha\mid^2 + 1}$

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

(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

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

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

by , 20 Jul 2017 15:24
19 Jul 2017 17:46
in discussion Hidden / Per page discussions » Exercise 2.8 Discussion

In the first case the coefficient "-1" represent a global phase. this coefficient does not alter in any way the probability of this qubit in any basis.

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

by , 19 Jul 2017 17:46
19 Jul 2017 16:14
in discussion Hidden / Per page discussions » Exercise 2.5 Discussion

a)

### $\frac{1}{\sqrt{2}}(\ket + + e^{i\theta}\ket -) \Rightarrow \frac{1 + eî\theta}{2}\ket 0 + \frac{1-e^{}i\theta}{2}\ket 1$

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

### $e^{i\theta} = cos(\theta) + i sen(\theta) = -1$

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

anwser to 2.5 a) by , 19 Jul 2017 16:14
19 Jul 2017 14:11
in discussion Hidden / Per page discussions » Exercise 2.3 Discussion

I agree with FlippingBits. Just complementing the anwser

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

by , 19 Jul 2017 14:11
19 Jul 2017 13:38
in discussion Hidden / Per page discussions » Exercise 2.2

### $\ket v = (\frac{e^{\frac{-i\pi}{4}}}{\sqrt{2}},\frac{1}{\sqrt{2}}) \rightarrow \bra v = (\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}},\frac{1}{\sqrt{2}})$

(1)
\begin{align} P_\psi = {\mid \braket v \psi \mid}^2 = {\mid \sum_{j,k}a_j^*b_k \mid}^2 \end{align}

### $P_\psi = {\mid \frac{e^{\frac{i\pi}{4}}}{2} + \frac{e^{\frac{i\pi}{4}}}{2} \mid}^2 = {\mid e^{\frac{i\pi}{4}} \mid}^2$

by eulers formula

so

### $P_\psi = {\mid \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \mid}^2 = 1$

anwser to 2.2 j) by , 19 Jul 2017 13:38
19 Jul 2017 13:05
in discussion Hidden / Per page discussions » Exercise 2.2

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

por 2.2 g) $\Rightarrow \frac{1}{\sqrt{2}}(\ket + + \ket -) = \ket 0$

Both are the same state with global phase 1

anwser to 2.2 i) by , 19 Jul 2017 13:05
19 Jul 2017 13:01
in discussion Hidden / Per page discussions » Exercise 2.2

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

Both are the same state with global phase $-i$

anwser to 2.2 h) by , 19 Jul 2017 13:01
19 Jul 2017 12:55
in discussion Hidden / Per page discussions » Exercise 2.2

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

Both are the same state with global phase 1

anwser to 2.2 g) by , 19 Jul 2017 12:55
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