OPhO 2022, Теория — Задача 5. Quantum Computing

Introduction

Physicists use braket notation to describe vectors in quantum systems. When using a vector $\vec v$ to describe a quantum state, the ket, written as $|v\rangle$ can be used. Both notations below are equivalent:

The bra, on the other hand, is the conjugate transpose of the ket $\langle v|=(|v\rangle)^\dagger$. Given two vectors $|v\rangle$ and $|w\rangle$, the braket $\langle v|w\rangle=|v\rangle\cdot|w\rangle$ is the inner product of both vectors. This notation will be used throughout this problem.

In any digital device, information is communicated via 0s and 1s, or binary code. The simplest units of this information are called bits. Similar to a bit, the qubit can be represented as a linear combination of two orthogonal states: quantum-0 and quantum-1, which are typically $|0\rangle$ and $|1\rangle$. Here,

Typically, we write a single qubit state as

where $a,b\in\mathbb C$, and $\langle\Psi\rangle\Psi=1$.

a. (1 pt.) A qubit is prepared in the state $a|0\rangle+b|1\rangle$.

(i) What is the probability of measuring the qubit in the state $|0\rangle$?

(ii) What is the probability of measuring the qubit in the state $|-\rangle=|0\rangle-|1\rangle$?

Hint: If you are still confused about measurement (it’s tricky!), check out this qiskit article. You can ignore all the parts with code, we’ll save those for the computer science students writing OCSO.

A quantum gate performs an unitary operator on a quantum state. Applying an operator (sometimes known as a gate) to a qubit state can be represented in the diagram below.

where $\hat U$ is a local unitary since it only acts on a single qubit. There are five important gates:

Here, $I,X,Y,Z$ form the four Pauli matrices and $H$ is known as the Hadamard gate, which we will use later on when we talk about entanglement.

For example, if $|\Psi\rangle=0.6|0\rangle+0.8|1\rangle$ and apply the gate $X=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, we end up with

b. (1 pt.) A qubit is prepared in the state $|\Psi\rangle=a|0\rangle+b|1\rangle$. What is the probability of measuring the qubit $\hat U|\Psi\rangle$ in the state $|0\rangle$? Express your answer in terms of $a,b$, and properties of the unitary $\hat U$.

The heart of quantum information lies in what we can do with more than a single qubit. If one qubit has two dimensions ($|0\rangle$ and $|1\rangle$), then a two-qubit system can be represented in four dimensions. For a two qubit system, the state can be written as $a_0|00\rangle+a_1|01\rangle+a_2|10\rangle+a_3|11\rangle$, where $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ can be seen as the “basis vectors.” If we have two independent qubits, i.e. $|\Psi_1\rangle=a|0\rangle+b|1\rangle$ and $|\Psi_2\rangle=c|0\rangle+d|1\rangle$, we can represent their combined state using the tensor product, i.e.

Here, we can see that

Note that not every two qubit state can be written as a tensor product. When this occurs, we say that they are entangled. We can immediately determine if a state is entangled by calculating its concurrence, defined by

If $C=0$, then the two qubits are separate and the system is separable. If $C=1$, the system is maximally entangled, such as

Physically, this means that a measurement of one qubit directly leads to a “collapse” of the other qubit (this is the classic example shown in popular science media). Note that $0\le C\le1$.

We can change the concurrence using a control operation. For example,

performs the CNOT gate. The unitary $X$ is applied to $|b\rangle$ if $|a\rangle=1$, otherwise nothing is done. That is, we have:

The CNOT gate is an example of a global unitary, since it acts on more than one qubit. Global unitaries for 2 qubit systems can be written as a $4\times4$ matrix. For example, we can write

where $|00\rangle,\ldots,|11\rangle$ form the 4 standard basis vectors. We can combine local and global unitaries to create entangled states. For example, consider the following circuit:

The initial state is $|\Psi_{\mathrm{in}}\rangle=|00\rangle=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}$. After applying the Hadamard gate $H$, the state becomes

After applying the CNOT gate, the state becomes:

Note that we can avoid matrix multiplication in this last step by seeing what CNOT does on each term of $|\Psi_{\mathrm{middle}}\rangle$. CNOT will not have an effect on $\frac12|00\rangle$ since the first qubit is $|0\rangle$. CNOT will have an effect on $\frac12|10\rangle$ since the first qubit is a $|1\rangle$, so it’ll flip the second qubit to a $|1\rangle$, giving us the map $\frac12|10\rangle\mapsto\frac12|11\rangle$.

c. (1 pt.) Construct a quantum circuit where the input state is $|00\rangle$ and the output state is $\frac i{\sqrt2}(|0\rangle-|1\rangle)$ using only $X,Y,Z,H,CNOT$ gates.

Quantum Teleportation

Quantum teleportation is the transfer of the quantum state of one qubit to another (not the actual physical qubit) using a shared entangled resource and two classical bits of information. It is performed using the following circuit.

The gate measures the qubit (returns either a 0 or a 1) and the wider wire represents that information that flows through this wire is a classical bit.

d. (1 pt.) Verify that the above circuit does teleport the qubit from the top branch to the bottom branch by looking at the specific case of $\alpha=\beta=\frac1{\sqrt2}$

e. (3 pts.) After the first operation is performed on branch $C$, the branch is brought a very far distance from the other two branches. By doing so, it appears we can create faster-than-light communication during the teleportation process, which is impossible! Explain why there is no contradiction. Justify rigorously.

We can analyze this by performing matrix multiplication, but using a circuit-based approach is much cleaner. To do so, we need to use the Griffiths-Niu Theorem.

f. (2 pts.) The following circuits, according to the Griffiths-Niu Theorem, are equivalent:

Prove the Griffiths-Niu Theorem.

Using this theorem, we can redraw our circuit as:

g. (1 pts.) For a control-Z gate, it doesn’t matter which branch is the control. In other words,

Prove this relationship.

Using the above problem, we can flip the control-Z gate. Then using the identity $Z=HXH$, we can reduce it further:

Since $H^2=I$, we can simplify the top part. Furthermore, we can introduce another CNOT between the first and the second branch.

We were allowed to introduce this CNOT gate since $XH|0\rangle=H|0\rangle$. This actually makes it easier using the following problem:

h. (2 pts.) Prove that the below two circuits are equivalent.

Using this substitution, we end up with:

We can now introduce another CNOT gate, which doesn’t do anything since $C$ will always be $|0\rangle$.

Three alternating CNOT gates is equivalent to the SWAP gate, so we can write:

where we clearly see a swapping that occurs between the top and bottom branch!

Building Quantum Computers

According to theoretical physcist David P. Divencenzo, there are five necessary (but not necessarily sufficient) criteria to build a quantum computer:

• A well-characterized qubit.
• The ability to initialize qubits.
• Long and relevant decoherence times.
• A “universal set” of quantum gates.
• The ability to measure qubits.

In this section, we will focus on how we can create qubits and how we can create a universal set of quantum gates. Consider two energy levels $E_1,E_0$ as the qubit states $|1\rangle,|0\rangle$ respectively. Assume that

Also assume that the qubit state is time varying, in the form of:

i. (14 pts.) Using the above setup, show how we can implement the quantum gates $X,Y,Z$. Hint: The Schrodinger Equation tells us

where $\hat{\mathcal H}=\begin{pmatrix}E_0&0\\0&E_1\end{pmatrix}$.