Quantum Information

Introduction

Quantum Cryptography

Classical: use binary bits.

Quantum: Encode information through superpositions, entaglement, and wavefunction collapse (measurement).

No-cloning Theorem

No guarantee to make an exact replica of an arbitrary quantum state.

Example - Two Level System

Why do we stop at a two level system for Quantum Computing. $\{\Psi_a,\Psi_b\}$.

Given an initial two particle state: $|\Psi_a\rangle_1|\Psi_s\rangle_2$ Let $|\Psi_s\rangle$ denote the starting state.

Let say we have $\mathcal{U}|\Psi_a\rangle_1|\Psi_s\rangle_2=\Psi_a\rangle_1|\Psi_a\rangle_2$.

Then, lets say that also does $\mathcal{U}|\Psi_b\rangle_1|\Psi_s\rangle_2=\Psi_b\rangle_1|\Psi_b\rangle_2$.

So, let $|\Psi\rangle_1=\frac{1}{\sqrt{2}}(|\Psi_a\rangle_1+|\Psi_b\rangle_1)$

Then, $U\frac{1}{\sqrt{2}}(|\Psi_a\rangle_1+|\Psi_b\rangle_1)|\Psi_s\rangle_2=\frac{1}{\sqrt{2}}\left[|\Psi_a\rangle_1|\Psi_a\rangle_2+|\Psi_b\rangle_1|\Psi_b\rangle_2\right]$ But we wanted, $\frac{1}{\sqrt{2}}(|\Psi_a\rangle_1+|\Psi_b\rangle_1)\frac{1}{\sqrt{2}}(|\Psi_a\rangle_2+|\Psi_b\rangle_2)$.

Example Encryption Scheme (Bennett and Brassard in 1984)

Using single photons, over 48km of Optical Fiber and 1.6km of free space (shown in 2000).

Alice: $|V\rangle=1$ and $|H\rangle = 0$.

Bob: $|V\rangle=1$ and $|H\rangle = 0$.

Assume Eve is inserted in the middle with another polarizer.

If Alice and Bob rotates their polarizers by 45 degrees, $\frac{1}{\sqrt{2}}(|H\rangle\pm|V\rangle)$, then Eve’s polarizer would make the transmittance be reduced.

Simple scheme would be that Alice and Bob exchange ahead of time which bits should be recorded and which are junk.

Are there Security Risks associated with different Implementations
Different implementations to change states:
- Electrons use magnetic field
- Photons use optical components
- Atoms use E+M waves

Aside

Shor’s Algorithm

Post Quantum Cryptography

Lattice, CRYSTALS-KYBER, CRYSTALS-DILITHIUM

Quantum Computing

Classical: input bits $\Rightarrow$ |Black box| $\Rightarrow$ output bits

QM: input qubits $\Rightarrow$ |Oracle (With an additional clock input for trigger operations)| $\Rightarrow$ output bits (measurements) (Note the Blackbox is called an oracle, and the input qubits are any QM two-level system)

Question: How is this clock any different from the clock in a classical black box.

$|\Psi\rangle = \frac{\alpha}{\sqrt{\alpha^2+\beta^2}} |0\rangle + \frac{\beta}{\sqrt{\alpha^2+\beta^2}}|1\rangle$

Hadamard gate: $U_H=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\\1 & -1\end{pmatrix}, U_z=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}, U_{NOT X}=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$.

Bloch Sphere Represenatation. $|0\rangle=|+\rangle$ is along $+z$ and $|1\rangle=|-\rangle$ is along $-z$. So, $|s\rangle = \cos\theta/2|+\rangle + e^{i\varphi}\sin\theta/2|-\rangle$.

2-qubit states. $|\psi=C_{0,0}|0\rangle_{control}|0\rangle_{target}+C_{0,1}|0\rangle_{control}|1\rangle_{target}+C_{1,0}|1\rangle_{control}|0\rangle_{target}+C_{1,1}|1\rangle_{control}|1\rangle_{target}=\begin{pmatrix}C_{0,0}\\C_{0,1}\\C_{1,0}\\C_{1,1}\end{pmatrix}$ $U_{CNOT}=\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix}$.

Quantum Teleportation

Transfer of a quantum state.

Bell states

Also called entangled states or EPR pairs.

$|\Phi^\pm\rangle = \frac{1}{\sqrt{2}}(|H\rangle_1|H\rangle_2\pm|V\rangle_1|V\rangle_2)$

$|\Psi^\pm\rangle = \frac{1}{\sqrt{2}}(|H\rangle_1|V\rangle_2\pm|V\rangle_1|H\rangle_2)$

Using a nonlinear crystal crystal so that: $\omega=\omega_1+\omega_2$ and $\vec{k}=\vec{k}₁+\vec{k}₂, the energy and momentum are conserved.

Suppose we have Alice and Bob. Alice recieves $|\Psi\rangle_1=C_H|H\rangle_1+C_V|V\rangle_1$. Goal: Bob needs to output same photon.

Exchange EPR pair, $|\Psi^-\rangle_{23}=\frac{1}{\sqrt{2}}(|H\rangle_2|V\rangle_3 - |V\rangle_2|H\rangle_3)$. Photon 2 goes to Alice and Photon 3 goes to Bob. Alice sees $|\Psi\rangle_{123}=|\Psi\rangle|\Psi^-\rangle$. We can show: $|\Psi\rangle_{123} = \frac{1}{2}[|\Phi^+\rangle_{12}(C_H|V\rangle_3-C_H|H\rangle_3)+|\Phi^-\rangle_{12}(C_H|V_3\rangle+C_V|H_3\rangle) + |\Psi^+\rangle_{12}(C_V|V\rangle_3-C_H|H\rangle_3)-|\Psi^-_{12}(C_H|H_3\rangle+C_V|V\rangle_3)]$ Alice then measures Bell state. Say she gets $|\varphi^-\rangle_{12}$ so she gets $|\Phi^-\rangle_{12}(C_H|V\rangle_3+C_V|H\rangle_3)$. She then communicates over classical phone and says that she got the result she got. So Bob knows the state of his third state is similar to Alice’s original state $|\Psi\rangle_1$. Then Bob applies the unitary transformation that swaps the vertical and horizontal polarizations to recover the original $|\Psi\rangle_1$ in $|\Psi\rangle_3$.

EPR

The product of EPR gave rose to multiple interpretations: Copenhagen (Spearheaded by Bohr), Hidden Variables (John Bell 1964, testable inequality, no hidden variables), Subjective Theories.

Subjective Theories

E.P. Wigner: Wavefunction collapse happens at human brain
Everett: Multiverse - Each measurement creates universes of all possible universes, but we only exist in one of them (Where does the energy required come from, is this equivalent to hidden variables)