Nielsen and Chuang -- Chapter 4

Updated on January 11, 2019

(4.1) Find the points on the Bloch sphere which correspond to the normalized eigenvectors of the different Paul matrices.

Recall that, from Exercise 2.11, $X$ has eigenvalues $\pm 1$ with respective eigenvectors

$$\{\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right]\}$$

Similarly, $Y$ has eigenvalues $\pm 1$ with respective eigenvectors

$$\{\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ i\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -i\end{array}\right]\}$$

Finally, $Z$ has eigenvalues $\pm 1$ with respective eigenvectors

$$\{\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]\}$$

First, we solve for $X$.

$$\{\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right]\}⟺\{\frac{1}{\sqrt{2}}(|0⟩+|1⟩),\frac{1}{\sqrt{2}}(|0⟩-|1⟩)\}$$

So, for $\frac{1}{\sqrt{2}}(|0⟩+|1⟩)$, we have that $\cos \left(\theta /2\right)=\frac{1}{\sqrt{2}}$. Hence, $\theta =\pi /2$. Now, $e^{i\varphi }\sin \left(\theta /2\right)=e^{i\varphi }/\sqrt{2}=1/\sqrt{2}$. Hence, $\varphi =0$.

Similarly, for the second eigenvector, $\theta =\pi /2$ but $\varphi =-\pi$.

Therefore, for the first eigenvector,

$$\begin{gathered} (\cos \varphi \sin \theta ,\sin \varphi \sin \theta ,\cos \theta )=(\cos \left(0\right)\sin \left(\pi /2\right),\sin \left(0\right)\sin \left(\pi /2\right),\cos \left(\pi /2\right)) \\ =(1,0,0) \end{gathered}$$

And for the second we have,

$$\begin{gathered} (\cos \left(\pi \right)\sin \left(\pi /2\right),\sin \left(\pi \right)\sin \left(\pi /2\right),\cos \left(\pi /2\right)) \\ =(-1,0,0) \end{gathered}$$

Similarly, we find the Bloch vectors $(0,\pm 1,0)$ for $Y$ and $(0,0,\pm 1)$ for $Z$.

(4.2) Let $x\in \mathbb{R}$ and $A$ be a matrix that satisfies $A^2=I$. Show that

$$\exp \left(iAx\right)=\cos \left(x\right)I+i\sin \left(x\right)A$$

From the power series definition of $e^z$, we have that

$$\begin{gathered} \exp \left(iAx\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{(iAx{)}^n}{n!} \\ =\sum _{n=0}^{\mathrm{\infty }}\frac{(iAx{)}^{2n}}{(2n)!}+\sum _{n=0}^{\mathrm{\infty }}\frac{(iAx{)}^{2n+1}}{(2n+1)!} \\ \sum _{n=0}^{\mathrm{\infty }}\frac{(iAx{)}^{2n}}{(2n)!}=\sum _{n=0}^{\mathrm{\infty }}\frac{i^{2n}{A}^{2n}{x}^{2n}}{(2n)!} \\ =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{2n}I{x}^{2n}}{(2n)!} \\ =I\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{2n}{x}^{2n}}{(2n)!}=\cos \left(x\right)I \\ \sum _{n=0}^{\mathrm{\infty }}\frac{(iAx{)}^{2n+1}}{(2n+1)!}=\sum _{n=0}^{\mathrm{\infty }}\frac{i^{2n+1}{A}^{2n+1}{x}^{2n+1}}{(2n+1)!} \\ =\sum _{n=0}^{\mathrm{\infty }}\frac{i(-1{)}^{2n+1}A{x}^{2n+1}}{(2n+1)!} \\ =iA\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{2n+1}{x}^{2n+1}}{(2n+1)!}=i\sin \left(x\right)A \end{gathered}$$

$X,Y,Z$ give rise to three useful classes of unitary matrices when they are exponentiated, the rotation operators about $\hat{x}$, $\hat{y}$, and $\hat{z}$,

$$\begin{gathered} R_x(\theta )\equiv e^{-i\theta X/2} \\ {R}_y(\theta )\equiv e^{-i\theta Y/2} \\ {R}_z(\theta )\equiv e^{-i\theta Z/2} \end{gathered}$$

We can use exercise 4.2 to write the above equations more conveniently.

(4.3) Show that, up to a global phase, the $\pi /8$ gate satisfies $T=R_z(\pi /4)$.

Note that

$$T=\left[\begin{array}{cc}1& 0\\ 0& \exp \left(i\pi /4\right)\end{array}\right]=\exp \left(i\pi /8\right)\left[\begin{array}{cc}\exp (-i\pi /8)& 0\\ 0& \exp \left(i\pi /8\right)\end{array}\right]$$

Now, using the definition of $R_z$,

$$\begin{gathered} e^{-iZ\frac{\pi }{8}}=\cos (-\pi /8)I+i\sin (-\pi /8)Z \\ =\cos \left(\pi /8\right)I-i\sin \left(\pi /8\right)Z \\ =\left[\begin{array}{cc}\cos \left(\pi /8\right)-i\sin \left(\pi /8\right)& 0\\ 0& \cos \left(\pi /8\right)+i\sin \left(\pi /8\right)\end{array}\right] \\ =\left[\begin{array}{cc}\exp (-i\pi /8)& 0\\ 0& \exp \left(i\pi /8\right)\end{array}\right] \end{gathered}$$

(4.4) Express the Hadamard gate $H$ as a product of $R_x$ and $R_z$ rotations and $e^{i\varphi }$ for some $\varphi$.

We’ve discussed a procedure for expressing $H$ as a product of rotations on the Bloch sphere, by considering its actions on a basis of the Bloch sphere. We showed that $R_x(-\pi /2)R_z(-\pi /2)R_x(-\pi /2)$ suffices. We can verify this result a second way by considering the respective rotation matrices.

We know that $H=\frac{1}{\sqrt{2}}(X+Z)$. Furthermore,

$$\begin{gathered} R_x(-\pi /2)=\left[\begin{array}{cc}\cos \left(\pi /4\right)& i\sin \left(\pi /4\right)\\ i\sin \left(\pi /4\right)& \cos \left(\pi /4\right)\end{array}\right] \\ =\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& i\\ i& 1\end{array}\right]=\frac{1}{\sqrt{2}}(I+iX) \\ {R}_z(-\pi /2)=\left[\begin{array}{cc}\cos \left(\pi /4\right)+i\sin \left(\pi /4\right)& 0\\ 0& \cos \left(\pi /4\right)-i\sin \left(\pi /4\right)\end{array}\right] \\ =\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1+i& 0\\ 0& 1-i\end{array}\right]=\frac{1}{\sqrt{2}}(I+iZ) \end{gathered}$$

We’ll use that

$$\begin{gathered} XZ=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right] \\ =\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]=iY \\ ZX=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]=-iY \\ ⟹XZ+ZX=0 \\ ⟹XZX+Z{X}^2=0 \\ ⟹XZX=-Z \end{gathered}$$

Note that the above is simply showing that the anti-commutator of $X$ and $Z$, $X,Z=0$. This holds for any pair of distinct Pauli matrices (Exercise 2.41).

Hence,

$$\begin{gathered} \frac{1}{2\sqrt{2}}(I+iX)(I+iZ)(I+iX)=\frac{1}{2\sqrt{2}}[I+iX+iZ+i^{2}ZX+iX+i^2{X}^2+i^{2}XZ+i^{3}XZX] \\ =\frac{1}{2\sqrt{2}}[I+iX+iZ-ZX+iX-I-XZ+i^{3}XZX] \\ =\frac{1}{2\sqrt{2}}[i(X+Z)+iX-iXZX] \\ =\frac{1}{2\sqrt{2}}[i(X+Z)+iX+iZ] \\ =\frac{1}{\sqrt{2}}[i(X+Z)] \end{gathered}$$

which gives the Hadamard transform with phase $e^{i0}$.

(4.5) Prove that $(\hat{n}\cdot \hat{\sigma }{)}^2=I$, and use this to verify the following equation

$$R_n(\theta )\equiv \exp (-i\theta n\cdot \sigma /2)=\cos \left(\theta /2\right)I-i\sin \left(\theta /2\right)(n_{x}X+n_{y}Y+n_{z}Z)$$

Evidently, $\hat{n}\cdot \hat{\sigma }=(n_{x}X+n_{y}Y+n_{z}Z)$ so, recalling that distinct Pauli matrices anti-commute,

$$\begin{gathered} (n_{x}X+n_{y}Y+n_{z}Z{)}^2=n_x^2{X}^2+n_x{n}_{y}XY+n_x{n}_{z}XZ+n_x{n}_{y}YX+n_y^2{Y}^2+n_y{n}_{z}YZ+n_x{n}_{z}ZX+n_y{n}_{z}ZY+n_z{Z}^2 \\ =(n_x^2+n_y^2+n_z^2)I+n_x{n}_z(XZ+ZX)+n_y{n}_z(YZ+ZY)+n_x{n}_y(XY+YX) \\ =(n_x^2+n_y^2+n_z^2)I=I \end{gathered}$$

because $\hat{n}$ is a unit vector.

Therefore, using Exercise 4.2 (Nielsen & Chuang), if we let $A=\hat{n}\cdot \hat{\sigma }$, then the result follows directly.

(4.7) Show that $XYX=-Y$ and use this to prove that $X{R}_y(\theta )X=R_y(-\theta )$.

From above, we have that distinct Pauli matrices anti-commute. Furthermore, the Pauli matrices are hermitian and unitary $⟹$ ${\sigma }_i^2=0,i\in x,y,z$. Hence,

$$\begin{gathered} XY+YX=0 \\ XYX+Y{X}^2=0 \\ XYX+Y=0 \\ XYX=-Y \end{gathered}$$

So,

$$\begin{gathered} X{R}_y(\theta )X=X[\cos \left(\theta /2\right)I-i\sin \left(\theta /2\right)Y]X \\ =\cos \left(\theta /2\right)X^2-i\sin \left(\theta /2\right)XYX \\ =\cos \left(\theta /2\right)I+i\sin \left(\theta /2\right)Y \\ =\cos (-\theta /2)I-i\sin (-\theta /2)Y \\ =R_y(-\theta ) \end{gathered}$$

using that $\cos (-x)=\cos \left(x\right),\sin (-x)=-\sin \left(x\right)$.

(4.12) Give $A,B,C$, and $\alpha$ for the Hadamard gate.

Using Lemma 4.12 (Nielsen & Chuang) above we can solve, assuming $\gamma =\pi /2$, $$\begin{gathered} H=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]=\left[\begin{array}{cc}{e}^{i(\alpha -\beta /2-\delta /2)}& -e^{i(\alpha -\beta /2+\delta /2)}\\ e^{i(\alpha +\beta /2-\delta /2)}& e^{i(\alpha +\beta /2+\delta /2)}\end{array}\right] \\ \alpha -\beta /2-\delta /2=0 \\ \alpha -\beta /2+\delta /2=\pi \\ \alpha +\beta /2-\delta /2=0 \\ (\alpha +\beta /2+\delta /2)=\pi \\ ⟹\alpha =\pi /2,\beta =0,\delta =\pi \end{gathered}$$

So, the proof of Corollary 4.2 in Nielsen & Chuang tells us to set

$$\begin{gathered} A=R_z(\beta )R_y(\gamma /2) \\ =R_z(0)R_y(\pi /4) \\ =\left[\begin{array}{cc}\cos \left(\pi /8\right)& -\sin \left(\pi /8\right)\\ \sin \left(\pi /8\right)& \cos \left(\pi /8\right)\end{array}\right] \\ B=R_y(-\gamma /2)R_z(-(\delta +\beta )/2) \\ =R_y(-\pi /4)R_z(-\pi /2) \\ =\left[\begin{array}{cc}\cos \left(\pi /8\right)& \sin \left(\pi /8\right)\\ -\sin \left(\pi /8\right)& \cos \left(\pi /8\right)\end{array}\right]\left[\begin{array}{cc}{e}^{i\pi /4}& 0\\ 0& e^{-i\pi /4}\end{array}\right] \\ C=R_z((\delta -\beta )/2) \\ =R_z(\pi /2) \\ =\left[\begin{array}{cc}{e}^{i\pi /4}& 0\\ 0& e^{-i\pi /4}\end{array}\right] \end{gathered}$$

and $\alpha$ remains set $\alpha =\pi /2$.

(4.16)

For the first circuit, we consider action on the computational basis.

$$|x_1⟩|x_2⟩\to |x_1⟩H|x_2⟩=(I\otimes H)|x_1⟩|x_2⟩$$

Now, given that $H=\left[\begin{array}{ccc}1& 11& -1\end{array}\right]$ w.r.t the computation basis, then

$$(I\otimes H)=\left[\begin{array}{cccc}1& 1& 0& 0\\ 1& -1& 0& 0\\ 0& 0& 1& 1\\ 0& 0& 1& -1\end{array}\right]$$

Similarly, for the second circuit we have

$$(H\otimes I)=\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 0& -1& 0\\ 0& 1& 0& -1\end{array}\right]$$

(4.17)

Construct a CNOT gate from one controlled-$Z$ gate, that is, the gate whose action in the computational basis is specified by the unitary matrix

$$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right]$$

and two Hadamard gates, specifying the control and target qubits.

Recall that, in terms of the computational basis, the action of the CNOT is given by $|c⟩|t⟩\to |c⟩|t\oplus c⟩$ and that $Z=\left[\begin{array}{ccc}1& 00& -1\end{array}\right]$.

We construct our algorithm by first making the observation that $H|+⟩=|0⟩$ and $H|-⟩=|1⟩$. Hence, beginning with state $|c⟩|t⟩$ we can initially apply $H$ to $|t⟩$. Now, using the control-$Z$ gate with $|c⟩$ as the control and $H|t⟩$ as the target, we have two cases:

(1) If $|c=1⟩$, then the second qubit will swap either from $|+⟩$ to $|-⟩$ or vis versa. Therefore, we can apply another Hadamard to the second qubit and have $|t\oplus c⟩$ at the second qubit, as expected. The first qubit is unaltered, as expected.

(2) If $|c=1⟩$, then the second qubit will remain unchanged. Hence, if we apply another Hadamard to the second qubit, then $|t⟩$ is recovered since $H^2=I$. So, we have the expected behavior.

In summary, we have the circuit, beginning with state $|c⟩|t⟩$:

(1) Apply $H$ to the second qubit

(2) Controlled-$Z$ with the first qubit as the control and second as the target

(3) Apply $H$ to the second qubit.

From the next exercise, we’ll see that it didn’t actually matter whether we used the first or second qubit as control/target:

(4.18)

We simply prove the statement for the computational basis.

(1) $|0⟩|0⟩$: Both circuits give the identity transform since they are conditioned on a qubit which is $|0⟩$, in either case.

(2) $|1⟩|0⟩$: The first circuit is conditioned on $|1⟩$, so it applies $Z$ to $|0⟩$ which gives $|0⟩$. Hence, we have $|1⟩|0⟩$. The second circuit is conditioned on $|0⟩$, so we have the identity transform which gives $|1⟩|0⟩$, similarly.

(3) $|0⟩|1⟩$: By symmetry, we have the same outcome as in (2).

(4) $|1⟩|1⟩$: The first circuit is conditioned on the first $|1⟩$, so it applies $Z$ to the second qubit which gives $-|1⟩|1⟩$. Similarly, the second circuit gives $-|1⟩|1⟩$.

(4.19) The CNOT gate is a simple permutation whose action on a density matrix $\rho$ is to rearrange the elements in the matrix. Write out this action explicitly in the computational basis.

$$\begin{gathered} |00⟩⟨00|=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right] \\ |01⟩⟨01|=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right] \\ |10⟩⟨10|=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right] \\ |11⟩⟨11|=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right] \end{gathered}$$

Now,

$$\begin{gathered} C_1(X)|00⟩=|00⟩ \\ {C}_1(X)|01⟩=|01⟩ \\ {C}_1(X)|10⟩=|11⟩ \\ {C}_1(X)|11⟩=|10⟩ \end{gathered}$$

Hence, the permutation matrix acting on the computational basis as

$$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right]$$

satisfies this permutation.

(4.20)

Consider action on $|c⟩|t⟩$ by the circuit on the LHS. The action of this circuit is given by $(H\otimes H)C^1(X)|c⟩|t⟩(H\otimes H)$ using $c$ as the control and $t$ as the target for the controlled operation. So, in Exercise 4.17, we showed that we can decompose $C^1(X)$ as $H{C}^1(Z)H$ using the same control and target as used for $C^1(X)$ originally, and with the $H$ transforms acting on the target qubit. Hence, we can rewrite action by the LHS circuit as $(H\otimes H)(I\otimes H)C^1(Z)|c⟩|t⟩(I\otimes H)(H\otimes H)=(H\otimes I)C^1(Z)|c⟩|t⟩(H\otimes I)$.

Similarly, for the circuit on the RHS, action on $|c⟩|t⟩$ is given by $C^1(X)|c⟩|t⟩$ where in this case $t$ is the control and $c$ is the target. Hence, using the same result, we can rewrite this as $(H\otimes I)C^1(Z)|c⟩|t⟩(H\otimes I)$ with $t$ as control and $c$ as target. Finally, using Exercise 4.18, we can swap which qubits we regard as control/target in a controlled-$Z$ operation. Hence, we have the action $(H\otimes I)C^1(Z)|c⟩|t⟩(H\otimes I)$ with $c$ as control and $t$ as target, as in the LHS.

Now, using that $H^2=I$, we note that the identity given by the circuit is equivalent to $C^1(X)(H\otimes H)|c⟩|t⟩=C^1(X)|t⟩|c⟩(H\otimes H)$ (applying $H\otimes H$ to the end of both circuits). Hence, this directly gives the effect of CNOT on the basis $|\pm ⟩$.