Week 1: QUANTUM BASICS, SPIN & QUBITS | ECE 802-730

A conceptual introduction to Quantum Mechanical Systems, their properties and qubits. Research focused on Chapters 1 – 3 of Quantum Computing for Everyone by Chris Bernhardt. The weekly content here is representative of assignments and exercises related to my coursework for ECE 802 – 730 | Quantum Sensor and System Engineering at Michigan State University.

Superposition

Quantum superposition is the linear and simultaneous combination of qubit states. An example of quantum superposition is the pattern of quantum particles shot through two narrow slits. Traditional logic would dictate that the particle would travel through one slit or the other. A quantum particle can traverse both paths simultaneously when in superposition. Later measurements of the particle leave evidence of wave collisions from the two paths validating that there was activity present at both potential paths through the slits. Another example would be to assume a quantum particle is known to be in superposition. That particle is then measured and then has the 50% probability of being in either a basis state of | 0 > or | 1 >. If | 0 > is observed, it will always result in | 0 >. Likewise, if | 1 > is observed, it will always result in | 1 >. A consequence of this basic quantum-mechanical property is Wave-Particle Duality, which allows for atoms, electrons, molecules and photons to appear in either wave or particle states. Tout Est Quantique designed a great concept video illustrating this phenomenon.

Entanglement

The inability to define the quantum state of two or more particles independently from each other. An example of quantum entanglement can be represented by two entangled electrons. If one electron is found to be in the spin up state then the other electron will be in the spin down state. These two particles will always be in some combination of balanced polarity with one another until entanglement is broken.

Quantum Key Distribution (QKD)

Similar to Cryptographic Key Distribution, QKD secures communication by providing keys to users on a network for securing private or sensitive information. The key differentiator is that QKD leverages quantum-mechanics secure keys and their distribution channels. The main quantum-mechanical dependencies involved are: Entanglement (mentioned above), the No-Cloning Theorem and Hiesenberg’s Uncertainty Principle.

The No-Cloning Theorem states it’s impossible to create a duplicate copy of an arbitrary unknown quantum state. Attempting to clone such a state irreversibly collapses the quantum system into some eigenstate. The theorem proves that no such unitary operator could clone all possible states of the time-independent Hamiltonian. The first quantum cryptography protocol, BB84, relies explicitly on the No-Cloning Theorem; where information gain can only occur when signals between two non-orthogonal states are disturbed and in the presence of a classical public channel.

How Quantum Key Distribution (BB84 Protocol) Works

In order to send a private key to a recipient in BB84, two strings ( $a$ and $b$ ) must be encoded as a tensor product of $n$ qubits.

$\displaystyle|\psi\rangle = \bigotimes_{i=1}^{n} |\psi\rangle_{{a_i}{b_i}},$ where $a_i$ and $b_i$ are the $i$ -th bits of $a$ and $b$

$a_i$ $b_i$ provides an index for the following four qubit states:
$|\psi\rangle_{00} = |0\rangle,$
$|\psi\rangle_{01} = |+\rangle = \frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}} |1\rangle,$
$|\psi\rangle_{10} = |1\rangle,$
$|\psi\rangle_{11} = |+\rangle = \frac{1}{\sqrt{2}} |0\rangle - \frac{1}{\sqrt{2}} |1\rangle,$

The qubit states are now mutually non-orthogonal, such that $b_i$ decides which basis to encode $a_i$ with. Therefore, it’s impossible to distinguish them with any sort of certainty without knowing what $b$ is.

Consider the following scenario. A sender transmits $|\psi\rangle$ over a publicly authenticated quantum channel $\varepsilon$ to a receiver. The receiver obtains a state $\varepsilon(\rho) = \varepsilon(|\psi\rangle \langle\psi |)$ , where $\varepsilon$ represents both channel noise and any possible eavesdropping that may be occurring on the channel. Both sender and receiver have their own unique states at this point. However, since the sender only knows $b$ , it’s impossible for the receiver and/or the eavesdropper to distinguish the current qubit states. The eavesdropper also cannot obtain copies of the qubits sent without making measurements. Doing so risks disturbing the qubits with a probability of $\frac{1}{2}$ , due to the No-Cloning Theorem discussed earlier.

After generating a random string of bits $b^{\prime}$ (the same length as $b$ ), the receiver can publicly announce reception of the transmission and measure the qubits of the senders message: $a^{\prime}$ . The sender can now safely announce across the public channel $b$ (the basis in which the qubits were prepared). Sender and receiver can now both communicate over the channel which $b_i$ and $b^{\prime}_{i}$ are not equal. Doing so, they both now discard the bits in $a$ and $a^{\prime}$ where $b$ and $b^{\prime}$ are not equal.

The sender now discloses which randomly chosen bits from $\frac{k}{2}$ are to be discarded from the remaining $k$ bits measured from the basis. When this check passes, the sender & receiver can use information reconciliation and privacy amplification to create a set of shared secret keys. Otherwise, the exercise is repeated, ran and checked over the public channel to see if the sample of $\frac{k}{2}$ bits reconcile.

Exercise 1

Show that
$|\leftarrow\rangle \; and \;|\rightarrow\rangle \;$
are an orthonormal basis.

From Chapter 3 of Quantum Computing for Everyone.

If the following vectors:

$\; |\leftarrow\rangle \;=\begin{bmatrix}\frac{1}{\sqrt{2}} \\\frac{1}{\sqrt{2}}\end{bmatrix}$

and

$\; |\rightarrow\rangle \;=\begin{bmatrix}\frac{1}{\sqrt{2}} \\-\frac{1}{\sqrt{2}}\end{bmatrix}$

then

$\langle\rightarrow\; | \rightarrow\rangle\;=\begin{bmatrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}} \\-\frac{1}{\sqrt{2}}\end{bmatrix}\;=\frac{1}{2}\; + \frac{1}{2}\; = 1$

$\langle\leftarrow\; | \leftarrow\rangle\;= 1$

$\langle\leftarrow\; | \rightarrow\rangle\;=\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}} \\-\frac{1}{\sqrt{2}}\end{bmatrix}\;=\frac{1}{2}\; - \frac{1}{2}\; = 0$

$\langle\rightarrow\; | \leftarrow\rangle\;= 0$

$\therefore\; |\leftarrow\rangle \; and \;|\rightarrow\rangle \;$ is an orthonormal basis.

Exercise 2

Consider putting a particle with spin through two measurements. The first measures spin of ‘up’ or $|\uparrow\rangle$ and the second measurement is done at an and of 45°. What is the probability of the particle measured at 45° and -135° after the second measurement?

From Chapter 3 of Quantum Computing for Everyone.

The first measurement yields: $\begin{bmatrix}\;1 \\ \;0\end{bmatrix}$ which is the standard basis

From Figure 3.3 of Quantum Computing for Everyone, we can see that the new basis from the second measurement at 45° is:

$\begin{bmatrix}cos\frac{\theta}{2} & sin\frac{\theta}{2} \\-sin\frac{\theta}{2} & cos\frac{\theta}{2} \end{bmatrix}\implies\begin{bmatrix}.9238 & .3826 \\-.3826 & .9328 \end{bmatrix}= A$

The probability in the new basis is $A^T |\uparrow\rangle$

$A^T\begin{bmatrix}\;1 \\ \;0\end{bmatrix} = \begin{bmatrix}.9238 & .3826 \\-.3826 & .9328 \end{bmatrix}\begin{bmatrix}\;1 \\ \;0\end{bmatrix} = .9328 |\nearrow^{45^\circ}\rangle - .3826|\swarrow^{-135^\circ}\rangle$

At 45°

$P = (.9328)^2 \approx 85\%$

And at -135°

$P = (-.3826)^2 \approx 15\%$