91Ó°ÊÓ

Show that the following quantum circuit prepares the Bell state $\mathbf{\left|}\mathbf{}\mathbf{Ψ}\mathbf{â\S ½}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{}\mathbf{\right|}\mathbf{00}\mathbf{â\S ½}\mathbf{+}\frac{\mathbf{}\mathbf{}\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{}\mathbf{|}\mathbf{11}\mathbf{â\S ½}\mathbf{}$on input $\mathbf{|}\mathbf{00}\mathbf{âŸ\copyright }$: apply a Hadamard gate to the first qubit followed by a CNOT with the first qubit as the control and the second qubit as the target.

What does the circuit output on input 10 , 01 and 11 ? These are the rest of the Bell basis states.

What is the quantum Fourier transform modulo M of the uniform superposition $\frac{\mathbf{1}}{\sqrt{\mathbf{M}}}\underset{\mathbf{j}\mathbf{=}\mathbf{0}}{\overset{\mathbf{M}\mathbf{−}\mathbf{1}}{\mathbf{∑}}}\mathbf{ }\mathbf{|}\mathbf{j}\mathbf{âŸ\copyright }$?

What is the QFT modulo M of $\overline{)\mathbf{j}\mathbf{>}}$

$$\begin{gathered} \mathbf{Convocation}\mathbf{-}\mathbf{Multiplication}\mathbf{}\mathbf{.}\mathbf{}\mathbf{Suppose}\overline{)\mathbf{Î\pm }\mathbf{âŸ\copyright }\mathbf{=}\underset{\mathbf{j}}{\mathbf{∑}}{\mathbf{a}}_{\mathbf{j}}\overline{)\mathbf{j}\mathbf{âŸ\copyright }\mathbf{}\mathbf{by}\mathbf{}\mathbf{I}\mathbf{}\mathbf{to}\mathbf{}\mathbf{get}\mathbf{}\mathbf{the}\mathbf{}\mathbf{superposition}\mathbf{}\overline{)\mathbf{Î\pm }}}}\mathbf{\text{'}}\mathbf{âŸ\copyright }\mathbf{=}\underset{\mathbf{j}}{\mathbf{∑}}{\mathbf{Î\pm }}_{\mathbf{j}}\overline{)\mathbf{j}\mathbf{+}\mathbf{I}\mathbf{âŸ\copyright }\mathbf{.}} \\ \mathbf{If}\mathbf{}\mathbf{the}\mathbf{}\mathbf{QFT}\mathbf{}\mathbf{of}\mathbf{}\overline{)\mathbf{Î\pm }\mathbf{âŸ\copyright }}\mathbf{}\mathbf{is}\mathbf{}\overline{)\mathbf{β}\mathbf{âŸ\copyright }\mathbf{,}\mathbf{}\mathbf{show}\mathbf{}\mathbf{that}\mathbf{}\mathbf{the}\mathbf{}\mathbf{QFT}\mathbf{}\mathbf{OF}\mathbf{}\mathbf{Î\pm }\mathbf{\text{'}}}\mathbf{}\mathbf{IS}\mathbf{}\mathbf{β}\mathbf{\text{'}}\mathbf{,}\mathbf{}\mathbf{Where}\mathbf{}\mathbf{βÂá}\mathbf{\text{'}}\mathbf{=}{\mathbf{β}}_{\mathbf{j}}{\mathbf{Ó\neg }}^{\mathbf{ij}}\mathbf{.}\mathbf{}\mathbf{Conclude}\mathbf{}\mathbf{that}\mathbf{}\mathbf{if}\mathbf{}\overline{)\mathbf{Î\pm }\mathbf{\text{'}}\mathbf{âŸ\copyright }\mathbf{=}\underset{\mathbf{j}\mathbf{=}\mathbf{0}}{\overset{{\displaystyle \frac{\mathbf{M}}{\mathbf{K}\mathbf{-}\mathbf{1}}}}{\mathbf{∑}}}\sqrt{\frac{\mathbf{K}}{\mathbf{M}}}\overline{)\mathbf{jk}\mathbf{+}\mathbf{I}\mathbf{âŸ\copyright }\mathbf{,}}} \\ \mathbf{then}\mathbf{}\overline{)\mathbf{β}\mathbf{\text{'}}\mathbf{âŸ\copyright }\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{k}}}\underset{\mathbf{j}\mathbf{=}\mathbf{0}}{\overset{\mathbf{k}\mathbf{-}\mathbf{1}}{\mathbf{∑}}}}{\mathbf{Ó\neg }}^{\frac{\mathbf{IjM}}{\mathbf{K}}}\overline{)\frac{\mathbf{jM}}{\mathbf{k}}\mathbf{âŸ\copyright }}\mathbf{.} \end{gathered}$$

Show that if you apply the Hadamard gate to the inputs and outputs of a CNOT gate, the result is a CNOT gate with control and target qubits switched:

The CONTROLLED SWAP ( $\mathbf{C}\mathbf{-}\mathbf{SWAP}$) gate takes as input$\mathbf{3}$ qubits and swaps the second and third if and only if the first qubit is a $\mathbf{1}$.

1. Show that each of the$\mathbf{NOT}\mathbf{,}\mathbf{\text{ }}\mathbf{CNOT}\mathbf{\text{ }}\mathbf{and}\mathbf{\text{ }}\mathbf{C}\mathbf{-}\mathbf{SWAP}$ gates are their own inverses.
2. Show how to implement anrole="math" localid="1658207684748" $\mathbf{AND}$ gate using a$\mathbf{C}\mathbf{-}\mathbf{SWAP}$ gate, i.e., what inputs$\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{c}$ would you give to a$\mathbf{C}\mathbf{-}\mathbf{SWAP}$ gate so that one of the outputs is $\mathbf{a}\mathbf{âˆ\S }\mathbf{b}$?
3. How would you achieve fanout using just these three gates? That is, on input $\mathbf{a}$and$\mathbf{0}$ , output$\mathbf{a}$ and$\mathbf{a}$ .
4. Conclude therefore that for any classical circuit $\mathbf{C}$there is an equivalent quantum circuit$\mathbf{Q}$ using just NOT and C-SWAP gates in the following sense: if$\mathbf{C}$ outputs $\mathbf{Y}$on input $\mathbf{x}$, then $\mathbf{Q}$outputs$\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{âŸ\copyright }$ on input $\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{âŸ\copyright }$. (Here$\mathit{z}$ is some set of junk bits that are generated during this computation.)
5. Now show that that there is a quantum circuit ${\mathbf{Q}}^{-1}$ that outputs$\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{âŸ\copyright }$ on input$\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}âŸ\copyright$ .
6. Show that there is a quantum circuit$\mathbf{Q}\mathbf{\text{'}}$ made up of$\mathbf{NOT}\mathbf{,}\mathbf{\text{ }}\mathbf{CNOT}\mathbf{\text{ }}\mathbf{and}\mathbf{\text{ }}\mathbf{C}\mathbf{-}\mathbf{SWAP}$gates that outputs$\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{0}âŸ\copyright$ on input $\mathbf{|}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}âŸ\copyright$.

In this problem we will show that if N=pq is the product of two odd primes, and if x is chosen uniformly at random between 0 and N-1, such that $\mathbf{gcd}\left(x,N\right)\mathbf{=}\mathbf{1}$, then with probability at least role="math" localid="1658908286522" $\frac{\mathbf{3}}{\mathbf{8}}$, the order r of x mod N is even, and more over ${\mathbf{x}}^{\frac{\mathbf{r}}{\mathbf{2}}}$is a nontrivial square root of 1 mod N.

a) Let p be an odd prime and let x be a uniformly random number modulo p. Show that the order of x mod p is even with probability at least$\frac{\mathbf{1}}{\mathbf{2}}$ (Hint:Use Fermat’s little theorem (Section 1.3).)

b) Use the Chinese remainder theorem (Exercise 1.37) to show that with probability at least $\frac{\mathbf{3}}{\mathbf{4}}$, the order r of x mod N is even.

c) If r is even, prove that the probability that role="math" localid="1658908648251" ${\mathbf{x}}^{\frac{\mathbf{r}}{\mathbf{2}}}\mathbf{≡}\mathbf{Â\pm }\mathbf{1}$is at most$\frac{\mathbf{1}}{\mathbf{2}}$.