Module 6 Solved Problems

Problem:
Derive the matrix for $R_Y(\theta)$.

Solution:
$$\eqb{ R_Y(\theta) & \eql & e^{\frac{-i\theta}{2}Y} \\ & \eql & I\cos\frac{\theta}{2} - iY\sin\frac{\theta}{2} \\ & \eql & \mat{\cos\frac{\theta}{2} & 0 \\ 0 & \cos\frac{\theta}{2}} - i\sin\frac{\theta}{2} \mat{0 & -i\\ i & 0} \\ & \eql & \mat{\cos\frac{\theta}{2} & 0 \\ 0 & \cos\frac{\theta}{2}} + \mat{0 & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & 0} \\ & \eql & \mat{\cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2} } }$$

Problem:
Show that $$\sqrt{X} \eql \frac{1}{2} \mat{1+i & 1-i\\ 1-i & 1+i}$$

Solution:
Using the approach shown for powers of Pauli matrices, $$\eqb{ X^{\frac{1}{2}} & \eql & e^{i \frac{\pi}{4} } e^{-i \frac{\pi}{4} X} \\ & \eql & e^{i \frac{\pi}{4} } e^{-i \frac{\pi}{4} X} \\ & \eql & e^{i \frac{\pi}{4} } \left( I\cos\frac{\pi}{4} - iX \sin\frac{\pi}{4} \right) \\ & \eql & e^{i \frac{\pi}{4} } \left( \mat{ \isqt{1} & 0 \\ 0 & \isqt{1} } - i \mat{ 0 & \isqt{1} \\ \isqt{1} & 0 } \right) \\ & \eql & \parenl{ \isqt{1} + \isqt{i} } \left( \mat{ \isqt{1} & -\isqt{i} \\ -\isqt{i} & \isqt{1} } \right) \\ & \eql & \frac{1}{2} \parenl{ 1 + i } \left( \mat{ 1 & -i \\ -i & 1 } \right) \\ & \eql & \frac{1}{2} \mat{1+i & 1-i\\ 1-i & 1+i} }$$

Problem:
For a 1-qubit operator such that $A^2=I$ show that $\left( e^{i\theta A} \right)^\dagger = e^{-i\theta A^\dagger}$ and use this to show that $R_X(\theta)$ is not Hermitian.

Solution:
$$\eqb{ \left( e^{i\theta A} \right)^\dagger & \eql & \parenl{ I\cos\theta }^\dagger + \parenl{ iA\sin\theta }^\dagger \\ & \eql & I\cos\theta + (i)^* \sin\theta A^\dagger \\ & \eql & I\cos\theta - i \sin\theta A^\dagger \\ & \eql & e^{-i\theta A^\dagger} }$$ Thus, $$\eqb{ R_X(\theta)^\dagger & \eql & \left( e^{i\theta X} \right)^\dagger \\ & \eql & e^{-i\theta X^\dagger} \\ & \eql & e^{-i\theta X} & \neq & e^{i\theta X} }$$

Problem:
Show that $H = K(-\frac{\pi}{4})\, X \, Y^{\frac{1}{2}}$

Solution:
$$X \, Y^{\frac{1}{2}} \eql \mat{0 & 1\\ 1 & 0} e^{\frac{\pi}{4}} \mat{1 & -1\\ 1 & 1} \eql e^{\frac{\pi}{4}} \mat{1 & 1\\ 1 & -1} \eql e^{\frac{\pi}{4}} H$$ Hence, a correcting global phase of $K(-\frac{\pi}{4})$ results in $H$.

Problem:
Recall that the $P(\theta)$ gate was used in setting a qubit's state to a desired state $\alpha\kt{0}+\beta\kt{1}$. Show how the $R_Z(\theta)$ gate instead of the $P(\theta)$ gate can be used for the same purpose of setting qubit

Solution:
Recall that we converted to polar form and wrote: $$\alpha\kt{0}+\beta\kt{1} \eql r_1\kt{0} + r_2 e^{i\theta_2-\theta_1}\kt{1}$$ where $$\eqb{ \alpha & \eql & r_1 e^{i\theta_1} \\ \beta & \eql & r_2 e^{i\theta_2} \\ }$$ And then we showed that $$R_Y(2\cos^{-1}\alpha) \kt{0} \eql r_1\kt{0} + r_2 \kt{1}$$

Next, let's recall $$R_Z(2\phi) \eql \mat{ e^{-i\phi} & 0\\ 0 & e^{i\phi} }$$ and apply $$\eqb{ R_Z(2\phi) R_Y(2\cos^{-1}\alpha) \kt{0} & \eql & R_Z(2\phi) \parenl{ r_1\kt{0} + r_2 \kt{1}} \\ & \eql & \mat{ e^{-i\phi} & 0\\ 0 & e^{i\phi} } \mat{r_1\\ r_2}\\ & \eql & e^{-i\phi} r_1 \kt{0} + e^{i\phi} r_2 \kt{1} \\ & \eql & e^{-i\phi} \parenl{ r_1 \kt{0} + e^{i2\phi} r_2 \kt{1} } \\ }$$ By choosing $2\phi = \theta_2-\theta_1$, and ignoring the global phase, we have $$R_Z(2\phi) R_Y(2\cos^{-1}\alpha) \kt{0} \eql \alpha\kt{0} + \beta\kt{1}$$ as desired.

Problem:
Show that

Solution:
Let's start by computing $H\otimes H$: $$H\otimes H \eql \frac{1}{2} \mat{1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1}$$ Then $$\cnot (H\otimes H) \eql \frac{1}{2} \mat{1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0} \mat{1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1} \eql \mat{1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & -1 & -1 & 1\\ 1 & 1 & -1 & -1}$$ And finally, $$(H\otimes H) \cnot (H\otimes H) \eql \frac{1}{4} \mat{1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1} \mat{1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1\\ 1 & -1 & -1 & 1\\ 1 & 1 & -1 & -1} \eql \frac{1}{4} \mat{4 & 0 & 0 & 0\\ 0 & 0 & 0 & 4\\ 0 & 0 & 4 & 0\\ 0 & 4 & 0 & 0}$$ Which gives the desired result $$\bar{C}_{\scriptsize NOT} \eql \mat{1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0}$$

Problem:
Recall the example from the section on applying SWAP to move qubits and now suppose we measure the first qubit in standard basis:

How does this impact the second qubit? The third qubit?

Solution:
Recall that the circuit, just before measurement, produces the 3-qubit state $$\ksi \eql \isqts{1} \parenl{ \kt{0}\kt{\phi}\kt{0} + \kt{1}\kt{\phi}\kt{1} }$$ There are two projectors here: $$\eqb{ P_0 & \eql & \otr{0}{0} \otimes I \otimes I \\ P_1 & \eql & \otr{1}{1} \otimes I \otimes I \\ }$$ Applying each gives $$\eqb{ P_0 \isqts{1} \parenl{ \kt{0}\kt{\phi}\kt{0} + \kt{1}\kt{\phi}\kt{1} } & \eql & \isqts{1} \kt{0}\kt{\phi}\kt{0} \\ P_1 \isqts{1} \parenl{ \kt{0}\kt{\phi}\kt{0} + \kt{1}\kt{\phi}\kt{1} } & \eql & \isqts{1} \kt{1}\kt{\phi}\kt{1} \\ }$$ From which we get $$\eqb{ \mbox{Outcome } \kt{0}\kt{\phi}\kt{0} & \mbox{ with probability } \smf{1}{2} \\ \mbox{Outcome } \kt{1}\kt{\phi}\kt{1} & \mbox{ with probability } \smf{1}{2} \\ }$$ As we can see, the third qubit is affected while the second is not.

Problem:
Show that three $\cnot$'s can be combined to create SWAP as follows: $$\mbox{SWAP} \eql \cnot \bar{C}_{\scriptsize NOT} \cnot$$ That is,

Solution:
Here, we will exploit both the compactness of binary-variable notation and known properties of the $\oplus$ binary operator.

Let $x,y$ be two binary variables, i.e., $x,y\in\{0,1\}$. Then, for any values of $x,y$, the 2-qubit vector $\kt{x}\kt{y}$ is a standard-basis vector. Let's apply the circuit on the right to $\kt{x}\kt{y}$: $$\eqb{ \parenl{\cnot \bar{C}_{\scriptsize NOT} \cnot} \: \kt{x}\kt{y} & \eql & \parenl{\cnot \bar{C}_{\scriptsize NOT} } \cnot \kt{x}\kt{y} & \mbx{First: the leftmost gate}\\ & \eql & \parenl{\cnot \bar{C}_{\scriptsize NOT} } \kt{x}\kt{x \oplus y} & \mbx{Apply \(\cnot\)} \\ & \eql & \parenl{\cnot } \bar{C}_{\scriptsize NOT} \kt{x}\kt{x \oplus y} & \mbx{Next gate}\\ & \eql & \parenl{\cnot } \kt{(x \oplus y) \oplus x}\kt{x \oplus y} & \mbx{Apply reverse \(\cnot\)}\\ & \eql & \parenl{\cnot } \kt{x \oplus (x \oplus y) }\kt{x \oplus y} & \mbx{Properties of XOR}\\ & \eql & \cnot \kt{y}\kt{x \oplus y} & \mbx{Properties of XOR}\\ & \eql & \kt{y}\kt{y\oplus (x \oplus y)} & \mbx{Apply third gate}\\ & \eql & \kt{y}\kt{(x \oplus y) \oplus y} & \mbx{Properties of XOR}\\ & \eql & \kt{y}\kt{x}\\ }$$ It is common to shorten the steps in the following way: $$\eqb{ \kt{x}\kt{y} & \xrightarrow{\cnot} & \kt{x}\kt{x \oplus y} \\ & \xrightarrow{\bar{C}_{\scriptsize NOT}} & \kt{y}\kt{x \oplus y} \\ & \xrightarrow{\cnot} & \kt{y}\kt{x} }$$ (But this obscures some of the binary operator simplifications)

Thus, if $$U \eql \parenl{\cnot \bar{C}_{\scriptsize NOT} \cnot}$$ then we've shown that $$U\kt{x}\kt{y} \eql \kt{y}\kt{x}$$ for the standard-basis vectors. Does this mean it'll work for any tensored 2-qubit vectors? That is, is it true that $$U \kt{\psi}\kt{\phi} \eql \kt{\phi}\kt{\psi}?$$ Here, we'll exploit linearity to show that this is true. Let $$\eqb{ \ksi & \eql & \alpha_1 \kt{0} + \alpha_2 \kt{1} \\ \khi & \eql & \beta_1 \kt{0} + \beta_2 \kt{1} \\ }$$ Then, $$\eqb{ \parenl{ \alpha_1 \kt{0} + \alpha_2 \kt{1} } \otimes \parenl{ \beta_1 \kt{0} + \beta_2 \kt{1} } & \xrightarrow{\text{expand}}& \alpha_1\beta_1 \kt{00} + \alpha_1\beta_2 \kt{01} \alpha_2\beta_1 \kt{10} + \alpha_2\beta_2 \kt{11} \\ & \xrightarrow{U} & \beta_1\alpha_1 \kt{00} + \beta_1\alpha_2 \kt{01} \beta_2\alpha_1 \kt{10} + \beta_2\alpha_2 \kt{11} \\ & \xrightarrow{\text{factor}}& \parenl{ \beta_1 \kt{0} + \beta_2 \kt{1} } \otimes \parenl{ \alpha_1 \kt{0} + \alpha_2 \kt{1} } \\ & \eql & \khi\ksi }$$ Note that we applied $U$ to each term (linearity) in going from the second to the third step.