# Nielsen and Chuang -- Chapter 2

Updated on December 23, 2018

### (2.11) Find the eigenvectors and eigenvalues of the Pauli matrices.

\begin{align} X &= \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \
X - \lambda I &= \begin{bmatrix} -\lambda & 1 \ 1 & -\lambda \end{bmatrix} \
\lambda^2 - 1 &= 0 \
\lambda_{\pm} &= \pm 1 \
\begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} v_+ &= 0 \
\implies v_+ &= \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \ 1 \end{bmatrix} \
\begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} v_- &= 0 \
\implies v_- = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \ -1 \end{bmatrix} \end{align}

Similarly, $Y$ has eigenvalues $\pm 1$ with respective eigenvectors $$\Big\{ \begin{bmatrix} 1 \\ i \end{bmatrix} , \begin{bmatrix} 1 \\ -i \end{bmatrix} \Big\}$$. $Z$ has eigenvalues $\pm 1$ with respective eigenvectors $$\Big\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \end{bmatrix} \Big\}$$

### (2.51) Verify that $H$ is unitary

$HH^\dagger = 1/2\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix} \\ = 1/2\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \\ = I = H^\dagger H$

### (2.52) Verify that $H^2 = I$

Because $H$ is hermitian we have that

$H^2 = HH^\dagger \\ = I$

from $H$ being unitary.

### (2.53) What are the eigenvalues and eigenvectors of $H$?

$\det \frac{1}{\sqrt{2}}\begin{pmatrix} 1 - \lambda & 1 \\ 1 & -1 - \lambda\end{pmatrix} = -(1-\sqrt{2}\lambda)(1+\sqrt{2}\lambda) - 1 = 0 \\ = -1 + 2\lambda^2 -1 \\ \lambda = \pm 1 \\ \frac{1}{\sqrt{2}}\begin{pmatrix} 1-\sqrt{2} & 1 \\ 1 & -1 +\sqrt{2}\end{pmatrix}v_1 = 0 \\ (1-\sqrt{2})v_{11} + v_{12} = 0 \\ v_{11} -(1-\sqrt{2})v_{12} = 0 \\ v_1 = \begin{pmatrix} 1 + \sqrt{2} \\ 1\end{pmatrix} \\ \frac{1}{\sqrt{2}}\begin{pmatrix} 1+\sqrt{2} & 1 \\ 1 & -1 -\sqrt{2}\end{pmatrix}v_2 = 0 \\ v_2 = \begin{pmatrix} 1 - \sqrt{2} \\ 1\end{pmatrix}$

### (2.54) Suppose $[A, B] = 0$ and $A, B$ are Hermitian. Prove that $\exp(A)\exp(B) = \exp(A + B)$

From Theorem 2.2, $A$ and $B$ are simultaneously diagonalizable. Hence, there is a common set of orthonormal eigenvectors ${ \ket{i} }$. Hence, $A = \sum_i a_i \ket{i}\bra{i}, B = \sum_i b_i \ket{i}\bra{i}$. So,

$\exp(A)\exp(B) = \sum_{k'=0}^\infty\sum_{i'} \frac{(b_{i'} \ket{i'}\bra{i'})^{k'}}{k'!}\sum_{k=0}^\infty\sum_{i} \frac{(a_i \ket{i}\bra{i})^k}{k!} \\ = \sum_{i}\Big[\sum_{k'=0}^\infty \frac{b_i^{k'} \ket{i}\bra{i}}{k'!}\sum_{k=0}^\infty \frac{a_i^k \ket{i}\bra{i}}{k!}\Big] \tag{orthonormality} \\ = \sum_{i}\sum_{k'=0}^\infty\sum_{k=0}^\infty \frac{a_i^k b_i^{k'} \ket{i}\bra{i}}{k!k'!}\\ = \sum_{i}\sum_{l=0}^\infty\sum_{k=0}^l \frac{a_i^k b_i^{l-k} \ket{i}\bra{i}}{k!(l-k)!}\\ = \sum_{i}\sum_{l=0}^\infty \frac{1}{l!}\sum_{k=0}^l \binom{k}{l} a_i^k b_i^{l-k} \ket{i}\bra{i} \\ = \sum_{i}\sum_{l=0}^\infty \frac{(a_i +b_i)^l}{l!} \ket{i}\bra{i} \\ = \exp(A + B)$

### (2.55) Prove that $U(t_1, t_2)$ is unitary

Using the result of 2.54,

$UU^\dagger = U^\dagger U = \exp[\frac{-iH(t_2 - t_1)}{\hbar}]\exp[\frac{iH(t_2 - t_1)}{\hbar}] \\ = \exp(\hat{0}) \\ = I$

### (2.56) Use the spectral decomposition to show that $K := -i\log U$ is Hermitian for any unitary $U$ and thus $U = \exp(iK)$ for some Hermitian $K$

The eigenvalues of $U$ can be given as $\exp(i\theta)$ by unitary. Furthermore, from spectral theorem, $U$ is diagonalizable as $U = V \Lambda V^\dagger$ where $V$ is unitary\footnote{Quick proof: $U$ can be written as $U=VTV^\dagger$ where $V$ is unitary and $T$ is upper triangular by Schur Decomposition. However, $UU^\dagger = U^\dagger U, VV^\dagger = I = V^\dagger V$ $\implies$ $T$ is normal $\implies$ $T$ is diagonal.}. Hence, $U = V \Lambda V^\dagger$ where diagonal matrix $\Lambda$ has elements of the form $\exp(i\theta)$ across the diagonal.

Furthermore, $(V\Lambda V^\dagger)^n = V\Lambda V^\dagger V\Lambda V^\dagger\cdots V\Lambda V^\dagger = V\Lambda^n V^\dagger$ $\implies$ $\exp(V \Lambda V^\dagger) = V \exp(\Lambda) V^\dagger$. Therefore, let $\Lambda’ = \log(\Lambda)$ which therefore has elements of the form $i\theta$. Hence, $U = \exp(V \Lambda’ V^\dagger)$

$K = -i\log U \\ = -i\log (\exp(V \Lambda' V^\dagger)) \\ = -iV \Lambda' V^\dagger \\ = V\Theta V^\dagger$

where $\Theta = -i\Lambda’$ has elements of the form $\theta$ (and hence the elements are real along the diagonal and zero elsewhere $\implies$ $\Theta^\dagger = \Theta$). Therefore, $K^\dagger = V^\dagger\Theta^\dagger V = V\Theta V^\dagger = K$.

### (2.57) Suppose ${L_l}$ and ${M_m}$ are two sets of measurement operators. Show that a measurement defined by the measurement operators ${L_l }$ followed by a measurement defined by the measurement operators ${ M_m }$ is physically equivalent to a single measurement defined by measurement operators ${N_{lm}}$ with the representation $N_{lm} = M_mL_l$.

Let $\ket{\phi}$ be our initial state and recall that if $l$ is measured then the post-measurement state is given by $\frac{L_l \ket{\psi}}{\sqrt{p(l)}}$. Furthermore, if we then measure $m$ we have $\frac{M_m(L_l \ket{\psi})}{\sqrt{p(m)}\sqrt{p(l)}} = \frac{N_{lm} \ket{\psi})}{\sqrt{p(m)p(l)}}$.

Now,

$p(m)p(l) = \bra{\psi} L_l^\dagger L_l \ket{\psi}\frac{\bra{\psi} L_l^\dagger}{\sqrt{p(m)}}M_m^\dagger M_m \frac{L_l \ket{\psi}}{\sqrt{p(m)}} \\ = p(l) \frac{\bra{\psi} L_l^\dagger M_m^\dagger M_m L_l \ket{\psi}}{p(l)}\\ = \bra{\psi} N_{lm}^\dagger N_{lm} \ket{\psi}\\ = p(lm)$

Hence, $\frac{N_{lm} \ket{\psi}}{\sqrt{p(m)p(l)}}=\frac{N_{lm} \ket{\psi}}{\sqrt{p(lm)}}$. Therefore, the representation is physically equivalent.

### (2.58) Suppose we prepare a quantum system in an eigenstate $\ket{\psi}$ of some observable $M$, with corresponding eigenvalue $m$. What is the average observed value of $m$ and the standard deviation?

First,

$\langle M \rangle = \bra{\psi} M \ket{\psi} \\ = \bra{\psi} m \ket{\psi} = m$

Furthermore,

$\langle M^2 \rangle - \langle M \rangle^2 = \bra{\psi} M^2 \ket{\psi} - m^2 \\ = \bra{\psi} M^\dagger M \ket{\psi} -m^2 \\ = m^2 - m^2 = 0$

### (2.59) Suppose we have a qubit in the state $\ket{0}$, and we measure the observable $X$. What is the average value of $X$? What is the standard deviation of $X$?

$X$ has eigenvalues $+1$ and $-1$ and eigenstates $\ket{+}$ and $\ket{-}$, respectively. Hence,

$\langle X \rangle = \bra{\psi} X \ket{\psi} \\ = \bra{\psi} (\ket{+}\bra{+} - \ket{-}\bra{-}) \ket{\psi} \\ = \bra{0}\ket{+}\bra{+}\ket{0} - \bra{0}\ket{-}\bra{-}\ket{0}\\ = \frac{1}{2} - \frac{1}{2} = 0$

Furthermore,

$\langle M^2 \rangle - \langle M \rangle^2 = \bra{\psi} M^2 \ket{\psi} - 0 \\ = \bra{\psi} (\ket{+}\bra{+} + \ket{-}\bra{-}) \ket{\psi} \\ = \frac{1}{2} + \frac{1}{2} = 1$

### (2.60) Show that $v \cdot \sigma$ has eigenvalues $\pm 1$ and that the projectors onto the corresponding eigenspaces are given by $P_{\pm} = (I \pm v \cdot \sigma)/2$.

First, $v\cdot \sigma$ is Hermitian so it’s spectral decomposition is given by $v \cdot \sigma = U \Lambda U^\dagger$ for some unitary $U$, diagonal matrix $\Lambda$. Hence, using $(v\cdot \sigma)^2 = I$ we have

$I = (v \cdot \sigma)^2 = (U \Lambda U^\dagger)^2 \\ = U \Lambda^2 U^\dagger \\ \implies U^\dagger I U = \Lambda^2 \\ I = \Lambda^2$

Therefore, $\Lambda$ must have diagonal entries $\pm 1$.

Next, $P_i P_j = \delta_{ij}P_j$ since if $i \neq j$ then $(I + v \cdot \sigma)(I - v\cdot \sigma) = I - (v \cdot \sigma)^2 = I - I = 0$. Furthermore, $P_+ + P_- = (I + v \cdot \sigma)/2 + (I - v \cdot \sigma)/2 = I$.

Finally, $(+1)P_+ + (-1)P_- = (I + v \cdot \sigma)/2 - (I - v \cdot \sigma)/2 = v \cdot \sigma$.

### (2.61) Calculate the probability of obtaining result $+1$ for a measurement of $v \cdot \sigma$, given that the state prior to measurement is $\ket{0}$. What is the state of the system after measurement if $+1$ is obtained?

First,

$p(+1) = \bra{\psi} P_+ \ket{\psi} \\ = \bra{\psi}(I + v \cdot \sigma)/2\ket{\psi} \\ = 1 + \frac{1}{2}[ v_1\bra{0} X \ket{0} + v_2\bra{0} Y \ket{0} + v_3\bra{0} Z \ket{0}] \\ = 1 + \frac{1}{2}[ v_1\bra{0}\ket{1} + iv_2\bra{0}\ket{1} + v_3\bra{0}\ket{0}] \\ = 1+ \frac{v_3}{2}$

Furthermore, after measurement of $+1$ we have

$(I + v \cdot \sigma)/2\ket{0} = \ket{0} + \frac{1}{2}[ v_1\ket{1} + iv_2\ket{1} + v_3\ket{0}] \\ = \Big[\big(\frac{v_3}{2} + 1\big)\ket{0} + \frac{v_1 + iv_2}{2}\ket{1}\Big]/\sqrt{1+ \frac{v_3}{2}}$

### (2.62) Show that any measurement where the measurement operators and the POVM elements coincide is a projective measurement

We would then have $M_m = E_m = M_m^\dagger M_m$. Furthermore, $E_m$ is a positive operator $\implies$ $M_m = M_m^\dagger M_m = M_m M_m^\dagger = M_m^\dagger$ so $M_m$ is Hermitian. Hence, $M_m = M_m^2$ so the measurement is projective.

### (2.63) Suppose a measurement is described by measurement operators $M_m$. Show that there exist unitary operators $U_m$ such that $M_m = U_m\sqrt{E_m}$ where $E_m$ is the POVM associated to the measurement.

From SVD, we have that $M_m = UDV$ for $U,V$ unitary and $D$ real, diagonal. Hence,

$\sqrt{E_m} = \sqrt{M_m^\dagger M_m} = \sqrt{V^\dagger DU^\dagger UDV} \\ = \sqrt{V^\dagger D^2V} \\ = V^\dagger DV = V^\dagger U^\dagger UDV \\ = U_m^\dagger M_m$

where $U_m := UV$. Therefore, there exists the unitary transformation of interest.

### (2.64) Suppose Bob is given a quantum state chosen from a set $S = \ket{\psi_1} , \cdots , \ket{\psi_m}$ of linearly independent states. Construct a POVM ${ E_1 , \cdots , E_{m+1} }$ such that if outcome $E_i$ occurs, $1 \leq i \leq m$, then Bob knows with certainty that he was given state $\ket{\psi_i}$.

To distinguish the states we require $\bra{\psi_i} E_j \ket{\psi_i} = p_i \delta_{ij}$ where $p_i > 0$ and $1 \leq i,j \leq m$.

So, we can use the Gram-Schmidt process using $S$ as our linearly independent set. This will give us an orthonormal set $U = \ket{\phi_1} , \cdots , \ket{\phi_m}$ that spans the same subspace as $S$. Next, we can represent each $\ket{\psi_i}$ in this orthonormal basis, $U$. Finally, for each $i$ we can find a vector $\ket{\psi_i’}$ in the span of $U$ that is orthogonal to all $\ket{\psi_j}, j \neq i$. Hence, we can define $E_i = \ket{\psi_i’} \bra{\psi_i’}, 1 \leq i \leq m$. Finally, take $E_{m+1} = I - \sum_m E_i$.

Creating an optimal POVM is much trickier (in the sense of minimizing the probability $p_{m+1}$).

### (2.65) Express the states $(\ket{0} + \ket{1})/\sqrt{2}$ and $(\ket{0} - \ket{1})/\sqrt{2}$ in a basis in which they are not the same up to relative phase shift.

Trivially, the $\ket{+}$ and $\ket{-}$ suffices as a basis where they are not the same up to relative phase shift.

### (2.66)

Show that the average value of the observable $X_1Z_2$ ($X$ acting on the first qubit and $Z$ on the second) for a two qubit system measured in the state $\frac{\ket{00} + \ket{11}}{\sqrt{2}}$ is zero.

Let observable $M = X_1Z_2$. Hence,

$\langle M \rangle = \frac{\bra{00} + \bra{11}}{\sqrt{2}} M \frac{\ket{00} + \ket{11}}{\sqrt{2}}\\ = \frac{\bra{00} + \bra{11}}{\sqrt{2}}\frac{X_1\ket{0}Z_2\ket{0} + X_1\ket{1}Z_2\ket{1}}{\sqrt{2}} \\ = \frac{\bra{00} + \bra{11}}{\sqrt{2}}\frac{\ket{1}\ket{0} - \ket{0}\ket{1}}{\sqrt{2}} \\ = 0$

### (2.68) Prove that $\ket{\psi} \neq \ket{a}\ket{b}$ for all single qubit state $\ket{a}$ and $\ket{b}$ where $\ket{\psi} = \frac{\ket{00} + \ket{11}}{\sqrt{2}}$.

First, decompose the qubit state in their basis, $\ket{a} = \alpha_0 \ket{0} + \alpha_1 \ket{1}$ and $\ket{b} = \beta_0 \ket{0} + \beta_1 \ket{1}$. Now, we prove by contradiction

$\ket{a}\ket{b} = \alpha_0\beta_0 \ket{00} + \alpha_0\beta_1 \ket{01} + \alpha_1\beta_0 \ket{10} + \alpha_1\beta_1 \ket{11} \\ = \frac{\ket{00} + \ket{11}}{\sqrt{2}}$

which would imply that either $\alpha_0$ or $\beta_1$ are zero in order to remove the $\ket{01}$ term. However, this would also remove either the $\ket{00}$ or $\ket{11}$ term, so we have a contradiction.

### (2.69) Verify that the Bell basis forms an orthonormal basis for the two qubit state space.

Two qubit state space consists of states of the form $\ket{\psi} = a\ket{00} + b\ket{01} + c \ket{10} + d\ket{11}$. Evidently, $\ket{00} = \frac{\sqrt{2}}{2}\Big[\frac{\ket{00} + \ket{11}}{\sqrt{2}} + \frac{\ket{00} - \ket{11}}{\sqrt{2}}\Big], \ket{01} = \frac{\sqrt{2}}{2}\Big[\frac{\ket{10} + \ket{01}}{\sqrt{2}} - \frac{-\ket{10} + \ket{01}}{\sqrt{2}}\Big]$ and similarly for the others. Hence, we span the same space.

Furthermore, $\bra{\beta_{00}}\ket{\beta_{00}} = \frac{\bra{00} + \bra{11}}{\sqrt{2}}\frac{\ket{00} + \ket{11}}{\sqrt{2}} = (\bra{00}\ket{00} + \bra{11}\ket{11}) /2 = 1$. Also, $\bra{\beta_{00}}\ket{\beta_{01}} = \frac{\bra{00} + \bra{11}}{\sqrt{2}}\frac{\ket{00} - \ket{11}}{\sqrt{2}} = (\bra{00}\ket{00} - \bra{11}\ket{11})/2 = 0$. The other combinations follow similarly.

Therefore, we have an orthonormal basis.

### (2.70) Suppose $E$ is any positive operator acting on Alice’s qubit. Show that $\bra{\psi} E \otimes I \ket{\psi}$ takes the same value when $\ket{\psi}$ is any of the four Bell states. Suppose some malevolent third party (‘Eve’) intercepts Alice’s qubit on the way to Bob in the superdense coding protocol. Can Eve infer anything about which of the four possible bit strings 00, 01, 10, 11 Alice is trying to send? If so, how, or if not, why not?

$\bra{00} + \bra{11} (E \otimes I) \ket{00} + \ket{11} = \bra{0}E\ket{0} + \bra{1}E\ket{1}\\ \bra{00} - \bra{11}(E \otimes I)\ket{00} - \ket{11} = \bra{0}E\ket{0} + \bra{1}E\ket{1}\\ \bra{10} + \bra{01} (E \otimes I)\ket{10} + \ket{01} = \bra{0}E\ket{0} + \bra{1}E\ket{1}\\ -\bra{10} + \bra{01}(E \otimes I)-\ket{10} + \ket{01} = \bra{0}E\ket{0} + \bra{1}E\ket{1}\\$

Hence, Eve can’t infer anything. The states are only distinguishable if one can perform a measurement that acts on both qubits.

### (2.71) Let $\rho$ be a density operator. Show that $\tr(\rho^2) \leq 1$ with equality iff $\rho$ is a pure state.

$\rho^2 = \sum_i p_i \ket{\psi_i}\bra{\psi_i} \sum_{i'} p_{i'} \ket{\psi_{i'}}\bra{\psi_{i'}} \\ = \sum_i p_i^2 \ket{\psi_i}\bra{\psi_i}$

by orthonormality. Hence,

$\tr \rho^2 = \sum_i \sum_j p_i^2 \psi_{i,jj}^2 \\ = \sum_i p_i^2 \tag{And \sum_j \psi_{i,jj}^2=1 by normalization}$

Now, we have that $\sum_i p_i = 1 \implies \sum_i p_i^2 = 1 \iff p_i = 1$. If $p_i = 1$, then there is only one index and hence we have a pure state. Otherwise, $\sum_i p_i^2 < 1$ and we have a mixed state.

### (2.74) Suppose a composite of systems $A$ and $B$ is in state $\ket{a}\ket{b}$, where $\ket{a}$ is a pure state of system $A$ and $\ket{b}$ is a pure state of system $B$. Show that the reduced density operator of system $A$ alone is a pure state.

$\rho = \ket{a}\ket{b}\bra{a}\bra{b} \\ \rho^A = \ket{a}\bra{a}\bra{b}\ket{b} = \ket{a}\bra{a}$

where we were given that $\ket{a}$ is a pure state.

## (2.75) For each of the four Bell states, find the reduced density operator for each qubit

First, $\frac{\ket{00} + \ket{11}}{\sqrt{2}}$

$\rho = \frac{\ket{00} + \ket{11}}{\sqrt{2}}\frac{\bra{00} + \bra{11}}{\sqrt{2}} \\ \rho^1 = \frac{\ket{0}\bra{0}\bra{0}\ket{0}+\ket{1}\bra{0}\bra{0}\ket{1}+ \ket{0}\bra{1}\bra{1}\ket{0}+ \ket{1}\bra{1}\bra{1}\ket{1}}{2} \\ = \frac{\ket{0}\bra{0} + \ket{1}\bra{1}}{2} = \frac{I}{2} \\ \rho^2 = \frac{\ket{0}\bra{0}\bra{0}\ket{0}+\ket{1}\bra{0}\bra{0}\ket{1}+ \ket{0}\bra{1}\bra{1}\ket{0}+ \ket{1}\bra{1}\bra{1}\ket{1}}{2} \\ = \frac{\ket{0}\bra{0} + \ket{1}\bra{1}}{2} = \frac{I}{2}$

Next, $\frac{\ket{00} - \ket{11} }{\sqrt{2}}$

$\rho = \frac{\ket{00} - \ket{11}}{\sqrt{2}}\frac{\bra{00} - \bra{11}}{\sqrt{2}} \\ \rho^1 = \frac{\ket{0}\bra{0}\bra{0}\ket{0}-\ket{1}\bra{0}\bra{0}\ket{1}- \ket{0}\bra{1}\bra{1}\ket{0}+ \ket{1}\bra{1}\bra{1}\ket{1}}{2} \\ = \frac{\ket{0}\bra{0} + \ket{1}\bra{1}}{2} = \frac{I}{2} \\ \rho^2 = \frac{\ket{0}\bra{0}\bra{0}\ket{0}-\ket{1}\bra{0}\bra{0}\ket{1}- \ket{0}\bra{1}\bra{1}\ket{0}+ \ket{1}\bra{1}\bra{1}\ket{1}}{2} \\ = \frac{\ket{0}\bra{0} + \ket{1}\bra{1}}{2} = \frac{I}{2}$

The remaining two are similar.