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14 Sep 2018 19:51
in discussion Hidden / Per page discussions » Ex5 5

Also |0>|+> and |1>|+> are stabilized by CNOT, so anything of the form a|00> + b|01> + c|1>|+> is stabilized. That's the complete three-dimensional eigenspace.

More eigenvectors for (f) by , 14 Sep 2018 19:51
17 Aug 2018 12:13
in discussion Hidden / Per page discussions » Exercise 3.14 Discussion

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:

• Case 1: the second bit is measured as $\ket{+}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{+} +\ket{-}\ket{+}\right)$ – let’s call this state $X$.
• Case 2: the second bit is measured as $\ket{-}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{-} +\ket{-}\ket{-}\right)$ – let’s call this state $Y$.

We can compute

(1)
\begin{align} \ket{+}\ket{+} = \frac{1}{2}\left(-\mathbf{i}\ket{u}\ket{u} + \ket{u}\ket{v} + \ket{v}\ket{u} + \mathbf{i}\ket{v}\ket{v}\right) \end{align}

and

(2)
\begin{align} \ket{-}\ket{+} = \frac{1}{2}\left(\ket{u}\ket{u} + \mathbf{i}\ket{u}\ket{v} - \mathbf{i}\ket{v}\ket{u} + \ket{v}\ket{v}\right) \end{align}

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)
\begin{align} \ket{+}\ket{-} = \frac{1}{2}\left(\ket{u}\ket{u} - \mathbf{i}\ket{u}\ket{v} + \mathbf{i}\ket{v}\ket{u} + \ket{v}\ket{v}\right) \end{align}

and

(4)
\begin{align} \ket{-}\ket{-} = \frac{1}{2}\left(\mathbf{i}\ket{u}\ket{u} + \ket{u}\ket{v} + \ket{v}\ket{u} - \mathbf{i}\ket{v}\ket{v}\right) \end{align}

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:

• Case 1: the second bit is measured as $\ket{+}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{+} +\ket{-}\ket{+}\right)$ – we called this state $X$.
• Case 2: the second bit is measured as $\ket{-}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{-} +\ket{-}\ket{-}\right)$ – we called this state $Y$.

We can compute

(5)
\begin{align} \ket{+}\ket{+} = \frac{1}{2}\left(\ket{0}\ket{0} + \ket{0}\ket{1} + \ket{1}\ket{0} + \ket{1}\ket{1}\right) \end{align}

and

(6)
\begin{align} \ket{-}\ket{+} = \frac{1}{2}\left(\ket{0}\ket{0} + \ket{0}\ket{1} - \ket{1}\ket{0} - \ket{1}\ket{1}\right) \end{align}

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)
\begin{align} \ket{+}\ket{-} = \frac{1}{2}\left(\ket{0}\ket{0} - \ket{0}\ket{1} + \ket{1}\ket{0} - \ket{1}\ket{1}\right) \end{align}

and

(8)
\begin{align} \ket{-}\ket{-} = \frac{1}{2}\left(\ket{0}\ket{0} - \ket{0}\ket{1} - \ket{1}\ket{0} + \ket{1}\ket{1}\right) \end{align}

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 $\frac{1}{2}$.

Solution to exercise 3.14 by , 17 Aug 2018 12:13
25 Jul 2018 13:09
in discussion Hidden / Per page discussions » Exercise 3.11 Discussion

For $n=1$, the stated theorem is clearly not true – the distance can be zero. For the rest of this answer we assume $n > 1$.

Let’s label the standard basis vectors $\ket{b_0} \ldots \ket{b_{n-1}}$, and let $\ket{b_n}$ be another name for $\ket{b_0}$ – this will simplify the proof notation.

For any state $\ket{\psi}$, the sum of the distances from $\ket{\psi}$ to the basis vectors is

(1)
\begin{align} \displaystyle \sum_{i=0}^{n-1} \left| \ket{\psi} - \ket{b_i} \right| = \frac{1}{2} \displaystyle \sum_{i=0}^{n-1} \left( \left| \ket{\psi} - \ket{b_i} \right| + \left| \ket{\psi} - \ket{b_{i+1}} \right| \right) \end{align}

We can now use the triangle inequality:

(2)
\begin{align} \left| \ket{\psi} - \ket{b_i} \right| + \left| \ket{\psi} - \ket{b_{i+1}} \right| \ge \left| \ket{b_i} - \ket{b_{i+1}} \right| \end{align}

It is easy to see that the distance between any two standard basis vectors is $\sqrt{2}$. Hence

(3)
\begin{align} \displaystyle \sum_{i=0}^{n-1} \left| \ket{\psi} - \ket{b_i} \right| \ge \frac{1}{2} \times n \times \sqrt{2} = \frac{n}{\sqrt{2}} \end{align}

This is a positive lower bound dependent on $n$. (It is not a tight lower bound.)

Solution to exercise 3.11 by , 25 Jul 2018 13:09
24 Jul 2018 12:20
in discussion Hidden / Per page discussions » Exercise 3.13 Discussion

a. Assume $\ket{\psi}$ is not entangled with respect to this decomposition. Then (still using the 0/1/2/3 representation for each two-qubit part), $\ket{\psi}$ can be written as

(1)
\begin{align} \ket{\psi} = (a_0\ket{0} + a_1\ket{0} + a_2\ket{2} + a_3\ket{3}) \otimes (b_0\ket{0} + b_1\ket{0} + b_2\ket{2} + b_3\ket{3}) \end{align}

We can see that $a_0 b_0 = \frac{1}{2}$, so $a_0$ is non-zero; and $a_1 b_1 = \frac{1}{2}$, so $b_1$ is non-zero. But we can also see that $a_0 b_1 = 0$. This is impossible. Hence $\ket{\psi}$ must in fact be entangled with respect to this decomposition.

b. Let us write $\ket{\psi}$ as $\frac{1}{2}(\ket{0000} +\ket{0101} +\ket{1010} + \ket{1111})$.

First consider the one qubit / three qubit decomposition where the one qubit is the leftmost in this representation. Suppose $\ket{\psi}$ is not entangled with respect to this decomposition. Then $\ket{\psi}$ can be written as

(2)
\begin{align} \ket{\psi} = (a_0\ket{0} + a_1\ket{1}) \otimes (b_0\ket{000} + b_1\ket{001} + b_2\ket{010} + b_3\ket{011} + b_4\ket{100} + b_5\ket{101} + b_6\ket{110} + b_7\ket{111}) \end{align}

We can see that $a_0 b_0 = \frac{1}{2}$, so $a_0$ is non-zero; and $a_1 b_2 = \frac{1}{2}$, so $b_2$ is non-zero. But we can also see that $a_0 b_2 = 0$. This is impossible. Hence $\ket{\psi}$ must in fact be entangled with respect to this decomposition.

For each of the other three possible positions for the single qubit, a similar argument can be constructed. Hence $\ket{\psi}$ is entangled with respect to any of these four decompositions.

Solution to exercises 3.13 by , 24 Jul 2018 12:20
03 Jul 2018 12:43
in discussion Hidden / Per page discussions » Exercise 3.5 Discussion

The talk of “opposite signs” seems to assume that $a_1$, $a_2$, $b_1$ and $b_2$ are real, but in fact they can be complex.

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 [[$a_1 a_2]], and we can see that this is not the case. Thus the system has no solutions and the state is indeed entangled. Comment on the answer from vb314 by , 03 Jul 2018 12:43 03 Jul 2018 12:30 in discussion Hidden / Per page discussions » Exercise 3.4 Discussion Assume$\ket{GHZ_n}$is not entangled and$n > 1$. Then$\ket{GHZ_n}can be written as (1) \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 thea_ivalues must be non-zero; and (3) \begin{align} b_1 \cdots b_n = \frac{1}{\sqrt{2}} \end{align} so all of theb_ivalues 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}$must be entangled. Solution to exercise 3.4 by , 03 Jul 2018 12:30 11 Jun 2018 12:23 in discussion Hidden / Per page discussions » Exercise 2.8 Discussion |1⟩ and −|1⟩ are different representations of the same state. When you carry out an operation on a state, you have to do so with respect to the particular representation you're using. Here's a really simple analogy. Consider (a) an isosceles triangle pointing upwards, and (b) the same size isosceles triangle rotated through 180 degrees, i.e. pointing downwards. We can say that these two images are different representations of the same shape. Now let's carry out an operation "put a circle on" to each of these shapes. Position a circle directly above the triangle on the page, just touching the triangle. Are the shapes still the same? No. To keep the shape the same after the operation, we need to make sure that the "put a circle on" operation is carried out in the right way, i.e. with respect to the particular representation we initially chose. A higher level answer by , 11 Jun 2018 12:23 08 Jun 2018 13:07 in discussion Hidden / Per page discussions » Exercise 2.7 Discussion The wording of the question makes it sound as though, for any given unit vector, there are many possible states that would provide the second component of an orthonormal basis. But in fact there is only one (albeit with many equivalent representations); we can see this from the discussion of antipodal points on the Bloch sphere in section 2.5.2 of the book. Answers to questions are as follows: a. In column vector form, the given state is$\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)$. This and the given state form an orthonormal basis. Solution to exercise 2.7 by , 08 Jun 2018 13:07 22 May 2018 15:55 in discussion Hidden / Per page discussions » Exercise 2.5 Discussion If you multiply the first state expression by$e^{- i \theta}$, you get the second state expression. Therefore the two states are identical, for any value of$\theta$. I disagree with the answer to 2.5 b) by , 22 May 2018 15:55 22 May 2018 15:41 in discussion Hidden / Per page discussions » Exercise 2.1 The question seemed ambiguous to me. "Assume that each photon generated by the laser pointer has random polarization" - does that mean 50%$\ket{\rightarrow}$and 50%$\ket{\uparrow}$, or does it mean a uniformly distributed random superposition? But either way, it's true that a fraction 0.5 of photons pass A and are then in state$\ket{\rightarrow}$, so the answer given is correct. Re: Sample answer by , 22 May 2018 15:41 11 Nov 2017 20:05 in discussion Hidden / Per page discussions » Exercise 2.13 a)$\alpha \vert 0 \rangle + \beta \vert 1 \rangle = \begin{bmatrix} \alpha\\\beta \end{bmatrix} = \begin{bmatrix} \alpha & \beta \end{bmatrix}^{T} = \alpha \vert 0 \rangle + \beta \vert 1 \rangle$b)$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))$by , 11 Nov 2017 20:05 11 Nov 2017 19:42 in discussion Hidden / Per page discussions » Exercise 2.12 a) A qubit can be represented as$\vert \psi \rangle = \alpha \vert 0 \rangle + \beta \vert 1 \rangle$, where$\vert \alpha \vert^2 + \vert \beta \vert ^2 = 1$. With$z = x + iy$on the Bloch sphere, we can convert the qubit representation s.t.$\vert \psi \rangle = r_{\alpha} e^{i \phi_{\alpha}} \vert 0 \rangle + r_{\beta}e^{i \phi_{\beta}} \vert 1 \rangle$since$z = x + iy = r e^{i \theta}$. Since we're concerned with the relative angle, we can substitute$\phi = \phi_{\beta} - \phi_{\alpha}$, so that$\vert \psi \rangle = r_{\alpha} \vert 0 \rangle + r_{\beta} e^{i \phi} \vert 1 \rangle$, which is equivalent to$\psi = r_{\alpha} \vert 0 \rangle + (x + iy) \vert 1 \rangle$. Before converting to polar coordinates its important to see that$0 \leq \theta \leq \pi$and$0 \leq \phi \lt 2 \pi$, and that the angle separating two bases is maximum at$\frac{\theta}{2}$. Using the map:$(x, y, z) \to (r sin(\theta) cos(\phi), r sin(\theta) sin(\phi), r cos(\theta))$, we can convert to polar coordinates, so that we have that$\vert \psi \rangle = cos(\frac{\theta}{2}) \vert 0 \rangle + e^{i \phi} sin(\frac{\theta}{2}) \vert 1 \rangle$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. by , 11 Nov 2017 19:42 11 Nov 2017 18:44 in discussion Hidden / Per page discussions » Exercise 2.11 a) Suppose Alice encodes a$0 \to \vert 0 \rangle$and send it to Bob. There are two possibilities: 1) Bob generates a random string,$y_i=0$, so he measures in the Hadamard basis, and has a 50% chance of measuring$\vert + \rangle$, which will be discarded and the same chance of measuring$\vert - \rangle$, which will be kept. 2) Bob generates a random string,$y_i=1$, so he measures in the standard basis, and has a 100% chance of measuring$\vert 0 \rangle$. As such, all of these bits will be discarded. Thus, when Alice encodes$\vert 0 \rangle$, the remaining bits satisfy$x' = y'$. Now, suppose Alice encodes a$1 \to \vert + \rangle = \frac{1}{\sqrt{2}}(\vert 0 \rangle + \vert 1 \rangle)$. Again, there are two possibilities: 1) Bob generates a random string,$y_i=0$, measuring in the Hadamard basis, the result has a 100% chance of being$\vert + \rangle$, and so all bits from this instance are discarded 2) Bob generates a random string,$y_i=1$, measuring in the standard basis, the result has a 50% chance of being$\vert 0 \rangle$, which are discarded or$\vert 1 \rangle$, which are kept. From the 4 possible instances, the result after discarding whenever Bob sends a 0 bit, is that$x'=y'$. 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:$1 - (\frac{3}{4})^s$by , 11 Nov 2017 18:44 11 Nov 2017 03:21 in discussion Hidden / Per page discussions » Exercise 2.10 a) Eve only learns the correct basis for measurement on the classical line after she has a chance to measure the qubits, so those that she knows for sure are those that by chance she was able to measure exactly in the basis Alice encoded them. When Eve doesn't know which two bases to choose from, there are an infinite number of possibilities for her to choose from, so that chance is approximately zero. Thus, Eve doesn't know any of the final key bit values for sure. 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 8by , 11 Nov 2017 03:21 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$P_kwith probability 1. Because (3) \begin{align} P_k = \sum_{i=1}^{m} \ket{\alpha_i}\bra{\alpha_i} \end{align} the result is (4) \begin{align} P_k \ket{\phi} = \sum_{i=1}^{m} \sum_{j=1}^{m} \big( \ket{\alpha_i}\bra{\alpha_i} \big) a_j \ket{\alpha_j} = \sum_{i=1}^{m} \sum_{j=1}^{m} a_j \ket{\alpha_i}\bra{\alpha_i} \ket{\alpha_j} = \sum_{i=1}^{m} a_i \ket{\alpha_i} = \ket{\phi} \end{align} a) by , 30 Jul 2017 07:10 29 Jul 2017 06:11 in discussion Hidden / Per page discussions » Ex4 12 Since a projector onto a direct sum of spaces is a sum of projectors onto the individual spaces, the projectors ontoS_5$and$S_6$are$P_5 = P_1 + P_2$and$P_6 = P_3 + P_4$. Thus, a measurement operator for$S_5 \oplus S_6can be written as (1) \begin{align} O' = P_1 + P_2 - \big(P_3 + P_4 \big) \end{align} by , 29 Jul 2017 06:11 29 Jul 2017 05:52 in discussion Hidden / Per page discussions » Ex4 11 Using the number of bits as the eigenvalues: (1) \begin{align} M = \ket{001}\bra{001} + \ket{010}\bra{010} + \ket{100}\bra{100} + 2 \Big( \ket{011}\bra{011} + \ket{101}\bra{101} + \ket{110}\bra{110} \Big) + 3 \ket{111}\bra{111} \end{align} by , 29 Jul 2017 05:52 29 Jul 2017 05:50 in discussion Hidden / Per page discussions » Ex4 10 Assigning the eigenvalue of1$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
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