Problem Set 9. Introduction to Quantum Computing¶
Problem 1 Antipodal points on the Bloch sphere¶
Discarding the global phase, we can write any qubit wave vector in the form
$$|{\psi}\rangle = \cos (\theta/2) |{0}\rangle + \sin(\theta/2) e^{i\phi} |1 \rangle$$with angles $\theta \in [0,\pi]$ and $\phi \in [0, 2\pi)$. The position on the Bloch sphere is
$$(x,y,z) = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) .$$Show that any two antipodal points on the Bloch sphere correspond to orthogonal wave vectors. In other words if $|\phi_1\rangle$ and $|\phi_2\rangle$ are antipodal on the Bloch sphere, then $\langle{\phi_1}|{\phi_2}\rangle = 0$.
Hint: For $(x,y,z)$ a point on the sphere, $(-x,-y,-z)$ is its antipodal point.
Problem 2. A Hadamard transformation on a Bell pair state¶
Starting with a wave vector that is a Bell pair for a two qubit state, $$|{\psi}\rangle_{Bell} = \frac{1}{\sqrt{2}} (|{00}\rangle + |{11}\rangle) $$ perform a Hadamard transformation on one of the qubits and then measure one of the qubits in the $|0\rangle$, $|1\rangle$ basis.
Show that the probability of measuring a 0 or 1 (corresponding to the $|0\rangle$ or $|1\rangle$ states) during the measurement does not depend on which bit is measured or on which bit you operated on with the Hadamard operation.
On a single qubit with basis $$|0\rangle = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} \qquad |1\rangle = \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix}, $$ the Hadamard operator $${\bf H} = \frac{1}{\sqrt{2}} \begin{pmatrix}1 & 1 \\ 1 & -1 \\ \end{pmatrix} .$$
Note: When performing the Hadamard operation on the first qubit, you would actually be applying the operator ${\bf H} \otimes {\bf I}$ where $\bf I$ is the identity matrix.
Problem 3. On quantum circuits with controlled gates¶
Consider the basis \begin{align*} |00\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} \qquad |01\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} \qquad |10\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix} \qquad |11\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix} \end{align*} for a 2 qubit quantum system.
Find a matrix representation (in the above basis) for the controlled Hadamard gate $\Lambda(H,1,0)$ where the control bit is the second one (with index 1) and the bit that is operated on is the first one (with index 0). If the second bit is 1, then the Hadamard operator is applied to the first bit.
Problem 4. On creating a cat state with 2-bit quantum gates¶
Consider a 3 qubit system which begins in the state
$ |000\rangle$. By operating on this state with a series of quantum gates it is possible to
put the system in the state
$$\psi_{cat} = \frac{1}{\sqrt{2}} ( |000\rangle + |111\rangle ) .$$
This state is known as a cat state.
- Design a quantum circuit that constructs this cat state with a series of simple 1-bit and 2-bit quantum gates starting with $|000\rangle$.
(Hint: use a Hadamard and CNOTs).
- For $n$ qubits and starting with state $|000....\rangle$ design a quantum circuit that constructs the n-bit cat state $$\psi_{cat,n}=\frac{1}{\sqrt{2}} ( |0000... \rangle + |1111 ..\rangle) $$
Problem 5. Classical logic operations with 3-bit gates¶
Using 3-bit Toffoli gates and other two and single bit gates, construct a quantum circuit that can compute the OR logic gate.
In other words construct a quantum circuit that takes as inputs $|x\rangle \in |0\rangle, |1\rangle$ for one qubit and $|y\rangle \in |0\rangle, |1\rangle$ for another qubit and returns a qubit that contains $|x \vee y\rangle$ where $\vee$ is the OR logic operation.
Use as many extra qubits as you need. You can set the extra bits to whatever you want!
Hint:
(x OR y) is equivalent to (x+y+xy) = x XOR y XOR (x AND y)
(x OR y) is equivalent to NOT( (NOT x) AND (NOT y) )
Here addition is mod 2.
The CNOT lets you do an XOR. The Pauli-X lets you do a NOT. The Toffoli lets you do an AND or a NAND depending upon how you set the third bit.