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HephaestisHephaestis 14 Jan 2017 23:07
in discussion Hidden / Per page discussions » Ex5 8

iY has the following eignevalue, eigenvector pairs

1, <ix, x|-transpose and -1, <-ix, x|-transpose, for any non-zero x. (Sorry, no Tex on my computer, hence my use of transpose of bras when I'd really rather have typed kets.)

The eigenvectors also can be written i|0> + |1> and -i|0> + |1>

by HephaestisHephaestis, 14 Jan 2017 23:07

Let M be an arbitrary 2x2 matrix, with entries A and B in the top row and entries C and D in the second row.

Set M = aI + bX + cY + dZ, where I, X, Y, and Z are the Pauli operators and a, b, c, and d are any scalars.

Then,
A = a + d. B = b + c.
C = b - c. D = a - d.

Solving simultaneously the equations involving a and d, we get a = (A+D)/2 and d = (A-D)/2.
Similarly solving for b and c, we get b = (B+C)/2 and c = (B-C)/2.

This shows that any 2x2 matrix is a linear combination of the 4 Pauli matrices. Since there are 4 degrees of freedom in choosing the 4 entries of M, the basis must have 4 elements.

Easy Simultaneous Equations by HephaestisHephaestis, 14 Jan 2017 00:26
HephaestisHephaestis 13 Jan 2017 23:58
in discussion Hidden / Per page discussions » Ex5 5

As Laker4 said, except for f. this is straight-forward. Just set up a matrix-vector product M|v> and solve for |v> = M|v>. (I'm using |v> in place of |psi> in the problem statement. I'm also occasionally writing the answers as transposes because I don't have Tex.) We get
a. |v> = a(|0> + |1>), for any scalar a. [I.e., the 2-D vector with both coordinates equal to each other.]
b. Only the null vector, because the product shows you that the first coordinate of |v> must equal the second coordinate and it must also equal the negative of the second.
c. |v> = a|0>, for any scalar a. [I.e., any 2-D vector with second coordinate = zero.]
d. <v| = <u|<w|, where <u| = <a,a| and <w|= <b,b|, for any coordinates a and b.
e. <v| = <u|<w|, where <w| = <a,0| and <u| = <b,b| for any coordinates a and b
f. |v> = |0>|u> for any vector |u> = a|0> + b|1>, for any scalars a and b, because C-not with control |0> does not change either the control or the target. In Laker4's answer for f., there is a typo, I think. I believe the last term should be a|00> + b|01>

by HephaestisHephaestis, 13 Jan 2017 23:58
sperytsperyt 10 May 2016 01:56
in discussion Hidden / Per page discussions » Errata

p. 15, 4 line from bottom there is: $v=a\lvert0\rangle+b\lvert1\rangle \rightarrow$ should be $\lvert v\rangle=a\lvert0\rangle+b\lvert1\rangle$

by sperytsperyt, 10 May 2016 01:56
Michel BonnardeauMichel Bonnardeau 02 Oct 2015 09:12
in discussion Hidden / Per page discussions » Ex4 20

a) Let be the two pairs of othogonal vectors $|v_{\theta_{i}}\rangle=\cos({\theta_{i}})|0\rangle+\sin({\theta_{i})}|1\rangle$ and $|v^{\bot}_{\theta_{i}}\rangle=-\sin({\theta_{i}})|0\rangle+\cos({\theta_{i}})|1\rangle$ , $i\in\{1,2\}$ corresponding to photon 1 and photon 2. An observable for only one photon $i$ is:

$O_{\theta_{i}}=|v_{\theta_{i}}\rangle \langle v_{\theta_{i}}|-|v^{\bot}_{\theta_{i}}\rangle \langle v^{\bot}_{\theta_{i}}|$

and an observable for observing photon 1 without observing photon 2 is:

$O_{\theta_{1}}\otimes I=|v_{\theta_{1}}\rangle \langle v_{\theta_{1}}|\otimes|I\rangle \langle I|-|v^{\bot}_{\theta_{1}}\rangle \langle v^{\bot}_{\theta_{1}}|\otimes|I\rangle \langle I|$

with $|I\rangle=(1/\sqrt{2})(|0\rangle +|1\rangle)$ for photon 2.

The first term of $O_{\theta_{1}}\otimes I$ corresponds to the eigenvalue +1, the second one to the eigenvalue -1 (see Exercice 4.19).

The projector for photon 1 for the eigenvalue +1 is then:

$P_{1}^{+1}=|v_{\theta_{1}}\rangle \langle v_{\theta_{1}}|\otimes|I\rangle \langle I|$

$P_{1}^{+1}=|v_{\theta_{1}}\rangle |I\rangle (\cos({\theta_{1}})\langle 0|+\sin({\theta_{1})}\langle 1|)(1/\sqrt{2})(\langle 0|+\langle 1|)$

$P_{1}^{+1}=|v_{\theta_{1}}\rangle |I\rangle (1/\sqrt{2})(\cos({\theta_{1}})\langle 00|+\cos({\theta_{1}})\langle 01|+\sin({\theta_{1})}\langle 10|+\sin({\theta_{1})}\langle 11|)$

When this projector is applied to the entangled state $|\psi\rangle =(1/\sqrt{2})(|00\rangle+|11\rangle)$ the result is:

$P_{1}^{+1}|\psi\rangle=|v_{\theta_{1}}\rangle |I\rangle (1/2) (\cos(\theta_{1})+\sin(\theta_{1}))$

with the probability (for photon 1 to have the outcome corresponding to the +1 eigenvalue):

$\Pi_{1}^{+1}=|P_{1}^{+1}|\psi\rangle|^2=(1/4)(\cos(\theta_{1})+\sin(\theta_{1}))^2$

In the same way, the probability for photon 1 to have the outcome corresponding to the -1 eigenvalue is:

$\Pi_{1}^{-1}=(1/4)(-\sin(\theta_{1})+\cos(\theta_{1}))^2$

An observable for photon 2 without observing photon 1 is:

$I\otimes O_{\theta_{2}}=|I\rangle \langle I|\otimes|v_{\theta_{2}}\rangle \langle v_{\theta_{2}}|-|I\rangle \langle I|\otimes|v^{\bot}_{\theta_{2}}\rangle \langle v^{\bot}_{\theta_{2}}|$

In the same way as for photon 1, the probability for photon 2 to have the outcome corresponding to the +1 eigenvalue is:

$\Pi_{2}^{+1}=(1/4)(\cos(\theta_{2})+\sin(\theta_{2}))^2$

and for the outcome corresponding to the -1 eigenvalue:

$\Pi_{2}^{-1}=(1/4)(-\sin(\theta_{2})+\cos(\theta_{2}))^2$

The sum of all these probabilities is 1:

$\Pi_{1}^{+1}+\Pi_{1}^{-1}+\Pi_{2}^{+1}+\Pi_{2}^{-1}=1$

An observable to observe both photon 1 and photon 2 is:

$O_{\theta_{1}}\otimes O_{\theta_{2}}=|v_{\theta_{1}}\rangle|v_{\theta_{1}}\rangle \langle v_{\theta_{1}}|\langle v_{\theta_{2}}|-|v_{\theta_{1}}\rangle |v^{\bot}_{\theta_{2}}\rangle \langle v_{\theta_{1}}| \langle v^{\bot}_{\theta_{2}}|-|v^{\bot}_{\theta_{1}}\rangle |v_{\theta_{2}}\rangle \langle v^{\bot}_{\theta_{1}}| \langle v_{\theta_{2}}|+|v^{\bot}_{\theta_{1}}\rangle |v^{\bot}_{\theta_{2}}\rangle \langle v^{\bot}_{\theta_{1}}| \langle v^{\bot}_{\theta_{2}}|$

Using the projectors this becomes:

$O_{\theta_{1}}\otimes O_{\theta_{2}}=P_{12}^{+1}-P_{12}^{+1-1}-P_{12}^{-1+1}+P_{12}^{-1}$

where $P_{12}^{+1}$ is the projector for both photons having an outcome corresponding to eigenvalues +1, $P_{12}^{+1-1}$ the projector for photon 1 with eigenvalue +1 and photon 2 with -1, and so on. One has for the projector $P_{12}^{+1}$:

$P_{12}^{+1}=|v_{\theta_{1}}\rangle|v_{\theta_{1}}\rangle \langle v_{\theta_{1}}|\langle v_{\theta_{2}}|$

$P_{12}^{+1}=|v_{\theta_{1}}\rangle|v_{\theta_{2}}\rangle (\cos(\theta_{1})\langle 0|+\sin(\theta_{1})\langle 1|)(\cos(\theta_{2})\langle 0|+\sin(\theta_{2})\langle 1|)$

$P_{12}^{+1}=|v_{\theta_{1}}\rangle|v_{\theta_{2}}\rangle (\cos(\theta_{1})\cos(\theta_{2})\langle 00|+\cos(\theta_{1})\sin(\theta_{2})\langle 01|+\sin(\theta_{1})\cos(\theta_{2})\langle 10|+\sin(\theta_{1})\sin(\theta_{2})\langle 11|)$

Applied to the entangled state $|\psi\rangle =(1/\sqrt{2})(|00\rangle +|11\rangle)$ this gives:

$P_{12}^{+1}|\psi\rangle =|v_{\theta_{1}}\rangle|v_{\theta_{2}}\rangle (1/\sqrt{2}) (\cos(\theta_{1}) \cos(\theta_{2})+\sin(\theta_{1}) \sin(\theta_{2}))$

$P_{12}^{+1}|\psi\rangle =|v_{\theta_{1}}\rangle|v_{\theta_{2}}\rangle (1/\sqrt{2})\cos(\theta_{1}-\theta_{2})$

The probability for both photons to have the outcome corresponding to the +1 eigenvalue is:

$\Pi_{12}^{+1}=|P_{12}^{+1}|\psi\rangle|^2=(1/2)\cos(\theta_{1}-\theta_{2})^2$

The same way, the probability for both photons to have the outcome corresponding to the -1 eigenvalue is:

$\Pi_{12}^{-1}=|P_{12}^{-1}|\psi\rangle|^2=(1/2)\cos(\theta_{1}-\theta_{2})^2$

So, the probability for both photons to have the same outcome is:

$\Pi_{12}^{same}=\Pi_{12}^{+1}+\Pi_{12}^{-1}=\cos(\theta_{1}-\theta_{2})^2$

The same way, the probability for both photons to have different outcomes is:

$\Pi_{12}^{different}=\Pi_{12}^{+1-1}+\Pi_{12}^{-1+1}=\sin(\theta_{1}-\theta_{2})^2$

and, of course:

$\Pi_{12}^{same}+\Pi_{12}^{different}=1$

b) The results always agree for $\theta_{1}=\theta_{2}$ (modulo $180\,^{\circ}$).

c) The results never agree for $\theta_{1}-\theta_{2}=90\,^{\circ}$ (modulo $180\,^{\circ}$).

d) The results agree half the time for $\theta_{1}-\theta_{2}=45\,^{\circ}$ (modulo $180\,^{\circ}$).

e) When $\theta_{1}=0$, which happens 1/3 of the time, then $\theta_{2}=\pm60\,^{\circ}\neq\theta_{1}$ and $\Pi_{12}^{same}=1/4$.
When $\theta_{1}=60\,^{\circ}$, which happens 1/3 of the time, then $\theta_{2}=0$ or $-60\,^{\circ}$ and $\Pi_{12}^{same}=1/4$.
The same when $\theta_{1}=-60\,^{\circ}$.

So, the probability that the results agree (with $\theta_{1}\neq\theta_{2}$) is 3x(1/3)x(1/4)=1/4 of the time.

(If $\theta_{1}$ is allowed to be equal to $\theta_{2}$, the results agree 1/2 the time).

by Michel BonnardeauMichel Bonnardeau, 02 Oct 2015 09:12
Michel BonnardeauMichel Bonnardeau 29 Sep 2015 11:42
in discussion Hidden / Per page discussions » Ex4 19

One has the two orthogonal vectors $|v_{\theta}\rangle=\cos({\theta})|0\rangle+\sin({\theta)}|1\rangle$ and $|v^{\bot}_{\theta}\rangle=-\sin({\theta})|0\rangle+\cos({\theta})|1\rangle$, then the two projectors $P_{i}=|v_{i}\rangle \langle v_{i}|$. With the eigenvalues $\lambda=1$ and $\lambda^{\bot}=-1$ one defines the observable $O_{\theta}=\sum\limits_{i}\lambda_{i}.P_{i}$:
$O_{\theta}=|v_{\theta}\rangle \langle v_{\theta}|-|v^{\bot}_{\theta}\rangle \langle v^{\bot}_{\theta}|$

$O_{\theta}=(\cos({\theta})|0\rangle+\sin({\theta)}|1\rangle)(\cos({\theta}) \langle 0|+\sin({\theta}) \langle 1|)-(-\sin({\theta})|0\rangle+\cos({\theta})|1\rangle)(-\sin({\theta}) \langle 0|+\cos({\theta}) \langle 1|)$

$O_{\theta}=(\cos(\theta)^2-\sin(\theta)^2)|0\rangle \langle0|+2\cos(\theta)\sin(\theta)|0\rangle \langle1|+2\cos(\theta)\sin(\theta)|1\rangle \langle0|+(\sin(\theta)^2-\cos(\theta)^2)|1\rangle \langle1|$

$O_{\theta}=\cos(2\theta)|0\rangle \langle0|+\sin(2\theta)|0\rangle \langle1|+\sin(2\theta)|1\rangle \langle0|-\cos(2\theta)|1\rangle \langle1|$

In matrix notation:

$O_{\theta}=\begin{pmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{pmatrix}$

Another way is to write $O_{\theta}=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and to consider that $|v_{\theta}\rangle$ and $|v^{\bot}_{\theta}\rangle$ are eigenvectors with +1 and -1 as the eigenvalues, respectively, i.e.

$\begin{pmatrix} a & b \\ c & d \end{pmatrix}.\begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix}=\begin{pmatrix} \cos(\theta) \\ \sin(\theta) \end{pmatrix}$ and $\begin{pmatrix} a & b \\ c & d \end{pmatrix}.\begin{pmatrix} -\sin(\theta) \\ \cos(\theta) \end{pmatrix}=-\begin{pmatrix} -\sin(\theta) \\ \cos(\theta) \end{pmatrix}$

So we have 4 equations with 4 unknowns. This can be solved to obtain again the previous result.

by Michel BonnardeauMichel Bonnardeau, 29 Sep 2015 11:42

To show the decomposition of $I$, note that $K(0) = T(0) = R(0) = I$. Thus,

(1)
\begin{equation} K(0)T(0)R(0)T(0) = IIII = I . \end{equation}

To show the decomposition of $X$, note that

(2)
\begin{align} T(\pi/2) = \left(\begin{array}{cc} \mathbf{i} & 0 \\ 0 & -\mathbf{i} \end{array}\right) \quad \text{and} \quad R(\pi/2) = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right). \end{align}

Thus,

(3)
\begin{align} -\mathbf{i}T(\pi/2)R(\pi/2)T(0) & = -\mathbf{i}\left(\begin{array}{cc} \mathbf{i} & 0 \\ 0 & -\mathbf{i} \end{array}\right) \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) I \\ \\ & = -\mathbf{i}\left(\begin{array}{cc} 0 & \mathbf{i} \\ \mathbf{i} & 0 \end{array}\right) = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \\ \\ & = X . \end{align}

To show the decomposition of $H$, note that $R(\pi/4) = \left(\begin{array}{cc} \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \end{array}\right)$. Thus,

(4)
\begin{align} -\mathbf{i}T(\pi/2)R(\pi/4)T(0) & = -\mathbf{i}\left(\begin{array}{cc} \mathbf{i} & 0 \\ 0 & -\mathbf{i} \end{array}\right) \left(\begin{array}{cc} \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \end{array}\right) I \\ \\ & = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) \left(\begin{array}{cc} \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \end{array}\right) = \left(\begin{array}{cc} \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{array}\right) \\ \\ & = H. \quad \square \end{align}
Solution to Exercise 5.4. by FlippingBitsFlippingBits, 02 Jun 2014 03:47

The measurement basis determines the outcomes, so in all cases the possible outcomes are the same as the measurement basis states. The probability that a measurement of qubit state $\ket{\psi}$ will yield a measurement basis state $\ket{b}$ as its outcome is the square of the magnitude of the component of $\ket{\psi}$ along $\ket{b}$. The probability $p_{\ket{b}}$ of obtaining outcome $\ket{b}$ is given below for each part of the exercise.

a. $p_{\ket{0}} = \tfrac{3}{4}$ and $p_{\ket{1}} = \tfrac{1}{4}$

b. $p_{\ket{0}} = \tfrac{1}{4}$ and $p_{\ket{1}} = \tfrac{3}{4}$

c. $p_{\ket{0}} = \tfrac{1}{2}$ and $p_{\ket{1}} = \tfrac{1}{2}$

d. $\ket{0} = \tfrac{1}{\sqrt{2}}\big(\ket{+}+\ket{-}\big)$, thus $p_{\ket{+}} = \tfrac{1}{2}$ and $p_{\ket{-}} = \tfrac{1}{2}$

e. $\tfrac{1}{\sqrt{2}}\big(\ket{0} - \ket{1}\big) = \tfrac{1+\mathbf{i}}{2}\ket{\mathbf{i}} + \tfrac{1-\mathbf{i}}{2}\ket{-\mathbf{i}}$, thus $p_{\ket{\mathbf{i}}} = \tfrac{1}{2}$ and $p_{\ket{-\mathbf{i}}} = \tfrac{1}{2}$

f. $\ket{1} = \tfrac{-\mathbf{i}}{\sqrt{2}}\ket{\mathbf{i}} + \tfrac{\mathbf{i}}{\sqrt{2}}\ket{-\mathbf{i}}$, thus $p_{\ket{\mathbf{i}}} = \tfrac{1}{2}$ and $p_{\ket{-\mathbf{i}}} = \tfrac{1}{2}$

g. Let $\ket{b_{1}} = \big(\tfrac{1}{2}\ket{0}+\tfrac{\sqrt{3}}{2}\ket{1}\big)$ and $\ket{b_{2}} = \big(\tfrac{\sqrt{3}}{2}\ket{0} - \tfrac{1}{2}\ket{1}\big)$.

Then $\ket{+} = \tfrac{\sqrt{3}+1}{2\sqrt{2}}\ket{b_{1}} + \tfrac{\sqrt{3}-1}{2\sqrt{2}}\ket{b_{2}}$. Thus, $p_{\ket{b_{1}}} = \tfrac{4+2\sqrt{3}}{8}$ and $p_{\ket{b_{2}}} = \tfrac{4-2\sqrt{3}}{8}$.

Let $U$ and $W$ both be vector spaces over field $\mathbb{F}$, with $B_{U} = \big\{u_{1}, \ldots, u_{m}\big\}$ a basis for $U$ and $B_{W} = \big\{w_{1}, \ldots, w_{n}\big\}$ a basis for $W$. Then, the set $B_{U\otimes W} = \big\{u_{i} \otimes w_{j} | 1 \leq i \leq m \,, 1 \leq j \leq n \big\}$ is a basis for the tensor product space $U \otimes W$.

Thus, two different bases for $V \otimes V$ can be obtained using two different bases for $V$. Let $B$ be the basis given in the problem description, with $b_{1} = (1,0,0)$, $b_{2} = (0,1,0)$, and $b_{3} = (0,0,1)$. Let a second basis for $V$ be the set $B' = \big\{b_{1}', b_{2}', b_{3}' \big\}$ where $b_{1}' = \tfrac{1}{\sqrt{2}}(1,0,1)$, $b_{2}' = \tfrac{1}{\sqrt{2}}(1,0,-1)$, and $b_{3}' = (0,1,0)$.

With these two bases for $V$, it is possible to form the two following bases for $V \otimes V$:

(1)
\begin{align} \mathcal{B}_{B,B} = \big\{& b_{1} \otimes b_{1} \,,\, b_{1} \otimes b_{2} \,,\, b_{1} \otimes b_{3} \,, \\ & b_{2} \otimes b_{1} \,,\, b_{2} \otimes b_{2} \,,\, b_{2} \otimes b_{3} \,, \\ & b_{3} \otimes b_{1} \,,\, b_{3} \otimes b_{2} \,,\, b_{3} \otimes b_{3} \big\} \\ \\ \text{and} \\ \\ \mathcal{B}_{B,B'} = \big\{& b_{1} \otimes b'_{1} \,,\, b_{1} \otimes b'_{2} \,,\, b_{1} \otimes b'_{3} \,, \\ & b_{2} \otimes b'_{1} \,,\, b_{2} \otimes b'_{2} \,,\, b_{2} \otimes b'_{3} \,, \\ & b_{3} \otimes b'_{1} \,,\, b_{3} \otimes b'_{2} \,,\, b_{3} \otimes b'_{3} \big\} \end{align}

Noting that

(2)
\begin{align} b'_{1} & = \tfrac{1}{\sqrt{2}} \big( b_{1} + b_{3} \big) \,, \\ b'_{2} & = \tfrac{1}{\sqrt{2}} \big( b_{1} - b_{3} \big) \,, \quad \textrm{and} \\ b'_{3} & = b_{2} \end{align}

it is possible to write the $\mathcal{B}_{B,B'}$ basis exclusively in terms of the $b_{i}$ basis vectors:

(3)
\begin{align} b_{1} \otimes b'_{1} & = \tfrac{1}{\sqrt{2}} \big( b_{1} \otimes b_{1} + b_{1} \otimes b_{3} \big) \,, \\ b_{1} \otimes b'_{2} & = \tfrac{1}{\sqrt{2}} \big( b_{1} \otimes b_{1} - b_{1} \otimes b_{3} \big) \,, \\ b_{1} \otimes b'_{3} & = b_{1} \otimes b_{2} \,, \\ b_{2} \otimes b'_{1} & = \tfrac{1}{\sqrt{2}} \big( b_{2} \otimes b_{1} + b_{2} \otimes b_{3} \big) \,, \\ b_{2} \otimes b'_{2} & = \tfrac{1}{\sqrt{2}} \big( b_{2} \otimes b_{1} - b_{2} \otimes b_{3} \big) \,, \\ b_{2} \otimes b'_{3} & = b_{2} \otimes b_{2} \,, \\ b_{3} \otimes b'_{1} & = \tfrac{1}{\sqrt{2}} \big( b_{3} \otimes b_{1} + b_{3} \otimes b_{3} \big) \,, \\ b_{3} \otimes b'_{2} & = \tfrac{1}{\sqrt{2}} \big( b_{3} \otimes b_{1} - b_{3} \otimes b_{3} \big) \,, \\ b_{3} \otimes b'_{3} & = b_{3} \otimes b_{2} \,. \end{align}
Solution to Exercise 3.1. by FlippingBitsFlippingBits, 31 May 2014 20:42
(1)
\begin{align} P_{1} & = \ket{+}\ket{+}\bra{+}\bra{+} + \ket{-}\ket{-}\bra{-}\bra{-} \,. \\ \\ P_{2} & = \ket{+}\ket{-}\bra{+}\bra{-} + \ket{-}\ket{+}\bra{-}\bra{+} \,. \end{align}
Solution to part (i) by FlippingBitsFlippingBits, 31 May 2014 05:34

g.

(1)
\begin{align} X \otimes Z & = \big(\ket{0}\bra{1} + \ket{1}\bra{0}\big) \otimes \big(\ket{0}\bra{0} - \ket{1}\bra{1}\big) \\ \\ & = \ket{0}\bra{1} \otimes \ket{0}\bra{0} - \ket{0}\bra{1} \otimes \ket{1}\bra{1} + \ket{1}\bra{0} \otimes \ket{0}\bra{0} - \ket{1}\bra{0} \otimes \ket{1}\bra{1} \\ \\ & = \ket{00}\bra{10} - \ket{01}\bra{11} + \ket{10}\bra{00} - \ket{11}\bra{01} \,. \end{align}

Alternatively, one can use the Kronecker product to form the matrix representation of the tensor product (in the standard basis) and then translate the matrix into bra/ket notation, as shown next.

h.

(2)
\begin{align} H \otimes H & = \frac{1}{2}\left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{array}\right) \\ \\ \\ & = \frac{1}{2}\big(\ket{00}\bra{00} + \ket{00}\bra{01} + \ket{00}\bra{10} + \ket{00}\bra{11} \\ \\ & \quad\:\:\,\, + \ket{01}\bra{00} - \ket{01}\bra{01} + \ket{01}\bra{10} - \ket{01}\bra{11} \\ \\ & \quad\:\:\,\, + \ket{10}\bra{00} + \ket{10}\bra{01} - \ket{10}\bra{10} - \ket{10}\bra{11} \\ \\ & \quad\:\:\,\, + \ket{11}\bra{00} - \ket{11}\bra{01} - \ket{11}\bra{10} + \ket{11}\bra{11} \big)\,. \end{align}

c. $Y = \left(\begin{array}{c} 1 \\ 0 \end{array}\right)\left(\begin{array}{cc}0 & 1 \end{array}\right) - \left(\begin{array}{c} 0 \\ 1 \end{array}\right)\left(\begin{array}{cc}1 & 0 \end{array}\right) = \ket{0}\bra{1} - \ket{1}\bra{0}$.

d. $Z = \left(\begin{array}{c} 1 \\ 0 \end{array}\right)\left(\begin{array}{cc}1 & 0 \end{array}\right) - \left(\begin{array}{c} 0 \\ 1 \end{array}\right)\left(\begin{array}{cc}0 & 1 \end{array}\right) = \ket{0}\bra{0} - \ket{1}\bra{1}$.

e. $23 \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array}\right)\left(\begin{array}{cccc}1 & 0 & 0 & 0 \end{array}\right) - 5 \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right)\left(\begin{array}{cccc}0 & 1 & 0 & 0 \end{array}\right) + 9 \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}\right)\left(\begin{array}{cccc}0 & 0 & 0 & 1 \end{array}\right)= 23\ket{00}\bra{00} - 5\ket{01}\bra{01} + 9\ket{11}\bra{11}$.

(1)
\begin{align} \frac{1}{\sqrt{2}}\big(\ket{\mathbf{i}}\ket{\mathbf{i}}+\ket{-\mathbf{i}}\ket{-\mathbf{i}}\big) & = \frac{1}{\sqrt{2}}\left[\frac{1}{2}\big(\ket{0}+\mathbf{i}\ket{1}\big) \otimes \big(\ket{0}+\mathbf{i}\ket{1}\big) + \frac{1}{2}\big(\ket{0}-\mathbf{i}\ket{1}\big) \otimes \big(\ket{0}-\mathbf{i}\ket{1}\big) \right] \\ \\ & = \frac{1}{2\sqrt{2}}\big(\ket{0}\ket{0} + \mathbf{i}\ket{0}\ket{1} + \mathbf{i}\ket{1}\ket{0} - \ket{1}\ket{1} + \ket{0}\ket{0} - \mathbf{i}\ket{0}\ket{1} - \mathbf{i}\ket{1}\ket{0} - \ket{1}\ket{1}\big) \\ \\ & = \frac{1}{\sqrt{2}}\big(\ket{0}\ket{0} - \ket{1}\ket{1}\big)\,. \quad \square \end{align}
Solution to part (b) by FlippingBitsFlippingBits, 31 May 2014 03:27

Using the notation for Bell states given in the text:

a. $\ket{00} = \frac{1}{\sqrt{2}} \big( \ket{\Phi^{+}} + \ket{\Phi^{-}} \big)$,

b. $\ket{+}\ket{-} = \frac{1}{2}\big( \ket{00} - \ket{01} + \ket{10} - \ket{11}\big) = \frac{1}{\sqrt{2}}\big( \ket{\Phi^{-}} - \ket{\Psi^{-}}\big)$, and

c. $\frac{1}{\sqrt{3}}\big( \ket{00} + \ket{01} + \ket{10}\big) = \frac{1}{\sqrt{6}}\ket{\Phi^{+}} + \frac{1}{\sqrt{6}}\ket{\Phi^{-}} + \sqrt{\frac{2}{3}} \ket{\Psi^{+}}$.

Solution to Exercise 3.7. by FlippingBitsFlippingBits, 31 May 2014 00:30

The states in (b), (c), and (d) are clearly superpositions with respect to the Hadamard basis. The state (e) is also a superposition since

(1)
\begin{align} \frac{1}{\sqrt{2}}\big(\ket{\mathbf{i}} - \ket{-\mathbf{i}}\big) = \mathbf{i}\ket{1} = \frac{\mathbf{i}}{\sqrt{2}}\big( \ket{+} - \ket{-} \big) \,. \end{align}

This leaves the states in (a) and (f), which are not superpositions with respect to the Hadamard basis. The state in (a) is $\ket{+}$ and the state in (f) is equal to $\ket{-}$, which are the Hadamard basis vectors.

Solution to Exercise 2.4. by FlippingBitsFlippingBits, 31 May 2014 00:10

The states in (a), (d), and (f) are superpositions with respect to the standard basis.

A single-qubit state $\ket{\psi}$ is not a superposition with respect to a basis $\{\ket{\psi}, \ket{\psi_{\perp}} \}$ that contains the state $\ket{\psi}$ as well as a unit vector $\ket{\psi_{\perp}}$ that is orthogonal to $\ket{\psi}$.

Thus, the state in (a) is not a superposition with respect to the Hadamard basis $\{ \ket{+}, \ket{-} \}$.
The state in (d) is not a superposition with respect to the basis $\{\tfrac{\sqrt{3}}{2}\ket{+} - \tfrac{1}{2}\ket{-} \,,\, \tfrac{1}{2}\ket{+} + \tfrac{\sqrt{3}}{2}\ket{-} \}$.
The state in (f) is equal to $\ket{-}$, so it is not a superposition with respect to the Hadamard basis.

Solution to Exercise 2.3 by FlippingBitsFlippingBits, 30 May 2014 23:59
enough?
Laker4Laker4 22 Jan 2014 01:59
in discussion Hidden / Per page discussions » Ex5 5

So for a) through c) this question seems pretty straight forward, basically and eigenvector problem. You put the operators into matrix form and figure out the eigenvectors. For d) and e) I did the same thing where $X \otimes X$ is

(1)
\begin{align} \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\\ 0 & 1 &0 & 0\\ 1 & 0 & 0 & 0 \end{array} \right) \end{align}

etc.
However, for the $C_{not}$ operator I had some difficulty solving for the eigenvectors, I think it's because there are essentially infinite solutions. Basically any vector of the form

(2)
\begin{align} |0>\otimes \left(a|0> + b|1>\right) = a|00> + b|11> \end{align}

is going to be stable when acted on with $C_{not}$

enough? by Laker4Laker4, 22 Jan 2014 01:59

If $U$ takes unit vectors to unit vectors it preserves the inner product because $<x|x> = 1=<x|U^\dagger U|x>$
So then if $|x>$ and $|y>$ are orthogonal then $<x|y>$ = 0 and $<x|U^\dagger U|y> = 0$ as well.

Not sure if this is sufficient but best I came up with so far.

Does this make sense? by Laker4Laker4, 22 Jan 2014 01:50
confusion
Laker4Laker4 21 Jan 2014 23:20
in discussion Hidden / Per page discussions » Ex5 2

I'm not sure what is being asked in this question. Is the point to consider if an arbitrary vector from a vector spaced spanned by the given vectors can be cloned? If each of the vectors can be cloned with the same operator? I'm pretty sure everyway I read this question the no cloning principle says that the operator doesn't exist.

confusion by Laker4Laker4, 21 Jan 2014 23:20
Laker4Laker4 13 Jan 2014 00:53
in discussion Hidden / Per page discussions » Ex4 13

SOooooo I'm not sure how to answer this. I have trouble using the correct language but basically I'm thinking this

If after a measurement of $|\psi>$ according to $O$ we obtain $|\phi>$ this means that one of the subspaces of the measurement is generated by $|\phi>$ with projector $|\phi><\phi|$. This means the operator for the measurment of observable $O$ contains $|\phi><\phi|$. Since the subspaces of any observable are orthogonal when $O$ is applied to $|\phi>$ the only non orthonal projector in the operator is $|\phi><\phi|$.

ie.

(1)
\begin{align} O = |\phi><\phi| + \sum_{i} |i><i| \end{align}

where $|i><i|$ is the projection operator of the ith subspace that $O$

(2)
\begin{align} O|\phi> = \left(|\phi><\phi| + \sum_{i}|i><i|\right)|\phi>=|\phi> \end{align}

this is true because the subspaces defined by an observable are orthogonal ie $|i><\phi|=0$ for all i.

Now for $O$ vs $O^2$ I'm really confused. Unless $O^2$ is a different measurement than two individual $O$ measurements I can't think of how they aren't the same…..thoughts anyone

by Laker4Laker4, 13 Jan 2014 00:53
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