Quantum Computing Past Paper

Lecture 1: Bits and qubits

Given a complex number $c = a + b i$, where $a, b \in \mathbb{R}$,

• its norm is defined as $|c| = c^* c = \sqrt{a^2 + b^2}$, where $c^*$ is the complex conjugate of $c$.
• its square is defined as $c^2 = (a + ib)^2 = a^2 - b^2 + 2ab i$.

Valid qubit state $| \psi \rangle = a |0 \rangle + b |1 \rangle$, where $|a|^2 + |b|^2 = 1$.

• y2010p9q11 (a)

Lecture 2: Linear algebra

Inner product $\langle \psi | \times | \phi \rangle = \langle \psi | \phi \rangle = \sum_{i=1}^n \psi_i^* \phi_i$. Outer product $| \psi \rangle \langle \phi | = \sum_{i=1}^n \sum_{j=1}^n \psi_i \phi_j^* |i \rangle \langle j|$.

Tensor / Kronecker product $| \psi \rangle \otimes | \phi \rangle = | \psi_1 \phi, \psi_2 \phi, ... ,\psi_n \phi \rangle$,

$$A \otimes B = \begin{bmatrix} a_{11}B & \cdots & a_{1n}B \\ \vdots & \ddots & \vdots \\ a_{m1}B & \cdots & a_{mn}B \end{bmatrix}.$$

Hadamard / Element-wise product $| \psi \rangle \circ | \phi \rangle = | \psi \rangle \odot | \phi \rangle = | \psi \phi \rangle = | \psi_1 \phi_1, \psi_1 \phi_2, ... ,\psi_n \phi_n \rangle$,

$$A \circ B = A \odot B = \begin{bmatrix} a_{11}b_{11} & \cdots & a_{1n}b_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1}b_{m1} & \cdots & a_{mn}b_{mn} \end{bmatrix}.$$

Eigenvalues $\lambda_i$ / (normalised) eigenvectors $|v_i \rangle$ $\boxed{U |v_i \rangle = \lambda_i |v_i \rangle}$, for unitary matrix $U$.

• y2024p9q11 (b), y2023p9q11 (d.i), y2020p9q14 (b)
  • for 4x4 matrix, use block matrix representation.
  • $U = T \otimes H$, $U | v_i \rangle = \lambda_i | v_i \rangle$, we can decompose it into two eigenvalue problems, for any matrix $T$ and $H$,
    • $T | v_t \rangle = \lambda_t | v_t \rangle$, $H | v_h \rangle = \lambda_h | v_h \rangle$
    • $(T \otimes H)(| v_t \rangle \otimes | v_h \rangle) = (T | v_t \rangle) \otimes (H | v_h \rangle) = (\lambda_t | v_t \rangle) \otimes (\lambda_h | v_h \rangle) = (\lambda_t \lambda_h)(| v_t \rangle \otimes | v_h \rangle)$.
    • Hence, $| v_i \rangle = | v_t \rangle \otimes | v_h \rangle$ and $\lambda_i = \lambda_t \lambda_h$.

For diagonalisable matrix, spectral decomposition $U=\sum_{i=1}^n \lambda_i |v_i \rangle \; \langle v_i |$.

• y2021p9q14 (a)

Unitary $\cap$ Hermitian: $A^2=I$ (self-inverse), e.g. $X, Y, Z, H$.

$\subseteq$ Unitary $A^{\dagger}A = I$ $\implies$ $A^{-1} = A^{\dagger}$ (unique inverse) $\lor$ Hermitian $A=A^{\dagger}$ (self-adjoint).

$\subseteq$ normal matrices $A^{\dagger}A = AA^{\dagger}$.

• y2020p9q14 (a)

Lecture 3: Quantum mechanics postulates

Superposition Hadamard gate $H$,

$$\begin{aligned} H | x \rangle & = \frac{1}{\sqrt{2}} (| 0 \rangle + e^{i \pi x} | 1 \rangle) \\ & = \frac{1}{\sqrt{2}} [| 0 \rangle + (-1)^{x} | 1 \rangle] \quad \text{, since } e^{i \pi x} = (-1)^x. \\ & = \frac{1}{\sqrt{2}} \sum_{z \in \{0, 1\}} (-1)^{x \cdot z} | z \rangle . \end{aligned}$$

Interference discerns some global property of the state, e.g. Hadamard gate $H$, QFT, QPE.

Entanglement non-separability via Hadamard-CNOT combination $\boxed{\text{CNOT} (H \otimes I)|00 \rangle= \frac{1}{\sqrt{2}} (|00 \rangle + |11 \rangle)}$, transferring from standard basis to Bell basis.

• y2022p9q11 (a.ii.)
  • single qubit unitary (local) applied to a 2-qubit state cannot change its (global) entanglement.

Three-qubit entanglement: collapse it to two-qubit via $I_2 \otimes \text{CNOT}$ gate.

• y2022p9q11 (b.i.)

Measurement in computational basis or (|+>, |->) basis

• y2022p8q11 (b)
• y2021p9q14 (a)
• y2010p9q11 (b)
• y2009p9q11 (a.iv.)
  • measurement operator

Measuring entangled states: sum of the squared amplitudes of the states in the basis.

• y2015p8q8 (a), y2012p8q11 (a.ii.)
• y2023p9q11 (b, c)

Lecture 4: Quantum mechanics concepts

Perfectly distinguish pair of orthogonal states.

• (1) either transfer qubits to computational basis,
• (2) or measure in their basis.
• y2022p8q11 (c,d)

Helstrom-Holevo bound $p \leq \frac{1+\sin\theta}{2}$, where $|\langle \psi_a | \psi_b \rangle|= \cos\theta$. Hence, $\sin\theta = \sqrt{1-|\langle \psi_a | \psi_b \rangle|^2}$.

Proof: $|\langle \psi_a| v_a \rangle|^2 = cos^2(\frac{\frac{\pi}{2}-\theta}{2})=\frac{1+cos(\frac{\pi}{2}-\theta)}{2}=\frac{1+sin\theta}{2},$ where the chosen measurement basis $v_i$ spread equally each side of the states $\psi_i$ to be distinguished.

• y2012p8q11 (c)
  • disprove: measure non-orthogonal states with certainty.

No-signalling principle: impossible to use entanglement or any quantum operation to infer whether a distant party has measured their qubit. After measurement, the entanglement is collapsed, thus not possible to transmit information faster than light.

• y2022p9q11 (b.iii.)

No-cloning principle: impossible to copy an unknown quantum state. $\nexists U. U (| \psi \rangle |0 \rangle) = | \psi \rangle | \psi \rangle$.

• y2023p9q11 (a)
  • (ii.) design copy qubits via standard quantum gates.
• y2012p8q11 (b), y2008p9q4 (a)

No-deleting principle: impossible to delete one of the unknown quantum state copies. $\nexists \tilde{U}. \tilde{U} (| \psi \rangle |\psi \rangle) = |\psi \rangle | 0 \rangle$.

Lecture 5: Quantum circuits

Universal gate set: $\{H, T, CNOT\}$. Pauli gates $X=HZH$, $Y=iXZ=SXSZ$.

Proof for $Z=HXH$ (L8. quantum search)

• either by matrix multiplication.
• or geometric interpretation ($X/Z$: rotate 180 degree about x/z-axis, $H$: swap x and z axis).

Rotation $R_k = \text{diag}(1, e^{i \frac{2\pi}{2^{k}}})$, $R_k^{\dagger} = \text{diag}(1, e^{-i \frac{2\pi}{2^{k}}})$. $R_0 = I$, $R_1 = Z$, $R_2 = S$, $R_3 = T$, ... .

$R_z (\theta) = \text{diag}(e^{-i \frac{\theta}{2}}, e^{i \frac{\theta}{2}})$, ignoring the global phase.

$$\begin{aligned} T & = \text{diag}(1, e^{i \frac{\pi}{4}}) &= R_3 & = R_z(\frac{\pi}{4}) = e^{i \frac{\pi}{8}} \text{diag}(e^{- i \frac{\pi}{8}}, e^{i \frac{\pi}{8}}) . \\ S & = T^2 = \text{diag}(1, e^{i \frac{\pi}{2}}=i) & = R_2 & = R_z(\frac{\pi}{2}) = e^{i \frac{\pi}{4}} \text{diag}(e^{- i \frac{\pi}{4}}, e^{i \frac{\pi}{4}}) . \\ Z & = S^2 = \text{diag}(1, e^{i \pi}=-1) & = R_1 & = R_z(\pi) = e^{i \frac{\pi}{2}} \text{diag}(e^{- i \frac{\pi}{2}}, e^{i \frac{\pi}{2}}) . \\ I & = Z^2 = \text{diag}(1, 1) & = R_0 & = R_z(0) . \end{aligned}$$

[$T, S$ are not self-invertible and $Z$ is self-inverse].

Toffoli gate

• y2012p8q11 (a)
  • matrix form.
• y2021p8q14 (e)
  • decomposition.

SWAP can be decomposed into 3 CNOTs.

• y2009p9q11 (a)
• y2013p8q11
• y2024p9q11 (c)

Design via standard quantum gates

• y2023p9q11 (a.ii)
• y2015p8q8 (b)
  • two qubit QFT circuit
• y2022p9q11 (a.i.)
  • two qubit entanglement
    • $Z=HXH$ (L8. quantum search). Hence, $CNOT=CX=(I \otimes H) \times CZ \times (I \otimes H)$, by self-inverse of $X,Z$.
    • or $(U_1 \otimes U_2) \times \text{CZ} \times (H \otimes H)|00 \rangle = (U_1 \otimes U_2) \times \frac{1}{2} \left( |00\rangle + |01\rangle + |10\rangle - |11\rangle \right)$, but hard to decode.
    • or use outer product matrix form, with (input, output) pair.

Lecture 6: Quantum information

Teleportation, send a qubit via two bits.

• y2018p8q11, y2008p9q4 (b)
• y2022p9q11 (b)
  • three-qubit teleportation.

Super_dense coding, send two bits via one qubit.

• sender: $|00 \rangle$ $\rightarrow_{superposition}^{H \otimes I + \text{CNOT}}$ Bell state $\rightarrow_{\text{two bits}}^{\{I,X,Z,XZ\}}$ four Bell states.
• receiver: four Bell states $\rightarrow_{interference}^{\text{CNOT} + H \otimes I}$ two bits.

Past paper

• y2022p9q11 (a.iii-iv.)
  • self-inverse and sender preparing entangled state.
• y2020p8q14

QKD: BB84

Eve's (intercept, measure, resend) attack.

Lecture 7: Deutsch-Jozsa algorithm

$\boxed{f: \{0, 1\}^n \rightarrow \{0, 1\}}$, which is either constant or balanced.

• Prepare state: $| \psi \rangle |-\rangle$.
  • where the uniform superposition $| \psi \rangle = | + \rangle^{\otimes n} = \frac{1}{\sqrt{2^n}} \sum_{x \in \{0, 1\}^n} | x \rangle$ of all $N=2^n$ states.
• Unitary operation $U_f$, a phase operator on the state $| x \rangle$,
  • $U_f | x \rangle | y \rangle = | x \rangle | y \oplus f(x) \rangle$, where $y \in \{0, 1\}$.
  • $U_f | x \rangle | - \rangle = (-1)^{f(x)} | x \rangle | - \rangle$.
• Interference $H^{\otimes n}$ and measure the first $n$ qubits in $| 0 \rangle^{\otimes n}$ basis.

$$\boxed{ H^{\otimes n} | x \rangle = \frac{1}{\sqrt{2^n}} \sum_{z \in \{0, 1\}^n} (-1)^{x \cdot z} | z \rangle }$$

Proof: as $| x \rangle = | x_1 ... x_n \rangle$, where $x_i \in \{0, 1\}$ and

$$\begin{aligned} H |x_i \rangle & = \frac{1}{\sqrt{2}} (|0 \rangle + (-1)^{x_i} |1\rangle ) \\ & = \frac{1}{\sqrt{2}} (|z_1=0 \rangle + (-1)^{x_i} |z_j=1\rangle) \\ & = \frac{1}{\sqrt{2}} ((-1)^{x_i \times 0}|z_1=0 \rangle + (-1)^{x_i \times 1} |z_2=1\rangle) \\ & = \frac{1}{\sqrt{2}} ((-1)^{x_i \times z_1}|z_1=0 \rangle + (-1)^{x_i \times z_2} |z_2=1\rangle) \\ & = \frac{1}{\sqrt{2}} \sum_{z_j \in \{0, 1\}} (-1)^{x_i \times z_j} |z_j \rangle \end{aligned}$$

$H^{\otimes n} | x_1 ... x_n \rangle = \otimes_i ( H |x_i \rangle)$, and the power of the function is $\sum_{i} x_i \times z_i = x \cdot z$, we are done.

Past paper

• y2016p8q10 (a, b), y2008p9q4 (c)
• y2022p8q11 (a), y2016p8q10 (c)
  • Toffoli gates as oracle $U_f$.

Lecture 8: Grover's search

• Quadratic speedup over unstructured classical search, from $O(N)$ to $O(\sqrt{N})$.
• $M$ is the number of solutions (marked states $f(x)=1$) to the search problem.
• Prepare state: $| \psi \rangle |-\rangle$.
  • where the uniform superposition $| \psi \rangle = | + \rangle^{\otimes n} = \frac{1}{\sqrt{2^n}} \sum_{x \in \{0, 1\}^n} | x \rangle$ of all $N=2^n$ states.
• With the target state $| x_t \rangle = \frac{1}{\sqrt{M}} \sum_{x \text{ s.t. } f(x) = 1} | x \rangle$,

$$\begin{aligned} | \psi \rangle & = \frac{1}{\sqrt{N}} [ \sum_{x \text{ s.t. } f(x) = 0} | x \rangle + \sum_{x \text{ s.t. } f(x) = 1} | x \rangle] \\ &= \frac{1}{\sqrt{N}} [ \sum_{x \text{ s.t. } f(x) = 0} | x \rangle + \sqrt{M} \frac{1}{\sqrt{M}} \sum_{x \text{ s.t. } f(x) = 1} | x \rangle] \\ &= \frac{1}{\sqrt{N}} [ \sum_{x \text{ s.t. } f(x) = 0} | x \rangle + \sqrt{M} | x_t \rangle ] \end{aligned}$$

• Each iteration $(W \otimes I) V$: rotate the state towards the target state $| x_t \rangle$ by $2 \theta$, where $\theta = \arcsin\frac{\sqrt{M}}{\sqrt{N}}$.

  • Oracle $V$ flips the sign of the target state $| x_t \rangle$, i.e. $V | x \rangle = (-1)^{\mathbb{I}(x=x_t)} | x \rangle$.

  • Diffusion operator $W = 2 | \psi \rangle \langle \psi | - I$, rotating $| \psi' \rangle =V | \psi \rangle$ around the axis $| \psi \rangle$.

    • reflected vector: $| \psi'' \rangle = 2 | \psi \rangle \langle \psi| | \psi' \rangle - | \psi' \rangle$, where the former is the projected vector of $| \psi' \rangle$ onto the axis $| \psi \rangle$.

    |a> = |psi> <psi|psi'>, projected vector
    |b> = 2|psi> <psi|psi'>,
    |psi''> = |b> - |psi'>,  by parallelogram.
            = 2 |psi> <psi|psi'> - |psi'>.
    
        /> |psi''>
       /  
      /  |a>   |b> 
     ---->|----->-----> |psi>
      \   |   /> |psi''>
       \  |  /
        \ | /  
         \> |psi'>
    

• After $n_{it}$ iterations, the angle between the final state and the target state is $(2n_{it}+1) \theta$.
  • $n_{it} = \frac{\frac{\pi}{2} - \theta}{2\theta} = \frac{\pi}{4\theta} \approx \frac{\pi}{4\sin\theta}$.
  • the final state is $\frac{1}{\sqrt{N}}[ \cos((2n_{it}+1)\theta) \sum_{x \text{ s.t. } f(x) = 0} | x \rangle + \sin((2n_{it}+1)\theta) | x_t \rangle]$.
  • the probability of measuring the target state is $\sin^2((2n_{it}+1)\theta)$.

Past papers

• y2023p8q11
• y2019p9q13 (a-d)
• y2007p8q12

Lecture 9: QFT & QPE

Quantum Fourier Transform (QFT)

QFT transforms a sequence of $N$ complex numbers $\{x\}$ into another $\{y\}$ of the same length,

$|x \rangle$ $\rightarrow$ $|y \rangle$: $\sum_{j=0}^{N-1} x_j |j \rangle$ $\rightarrow$ $\sum_{k=0}^{N-1} y_k |k \rangle$, where $\boxed{y_k = \frac{1}{\sqrt{N}} \sum_{j=0}^{N-1} w^{jk} x_j}$ and $w = e^{i \frac{2\pi}{N}}$.

• The normalization term is $\frac{1}{N}$ and exponential term is negated in DFT.
  • here, we use $\frac{1}{\sqrt{N}}$ to satisfy the unitary condition, where the dimension of Hilbert space for $n$ qubits is $N=2^n$.
• The time series coefficients $x_j$ are transformed into the frequency domain coefficients $y_k$.
  • DFT is the change of basis operator that converts from euclidean basis to the Fourier basis.
  • each $y_k$ corresponds to how much of the sinusoid with frequency $f=\frac{k}{N}$ [cycles per sample] is present in the signal.
  • $w^{jk} = e^{i \frac{2\pi}{N} jk} = \cos(\frac{2\pi k}{N} j) + i \sin(\frac{2\pi k}{N} j)$, forming an orthogonal basis over the space of $N$ complex vectors.
  • note that $w^{N} = e^{i 2\pi} = 1$.

Alternatively, we can express the QFT as a matrix transformation $\mathbf{M}$, where $N$ is the dimension of the Hilbert space.

The DFT is thus $y = \mathbf{M} x$, which in the matrix form is expressed as,

$$\begin{bmatrix} y_0 \\ \dots \\ y_k \\ \dots \\ y_{N-1} \end{bmatrix} = \frac{1}{\sqrt{N}} \begin{bmatrix} 1 & 1 & 1 & \dots & 1 \\ 1 & \omega & \omega^2 & \dots & \omega^{N-1} \\ 1 & \omega^2 & \omega^4 & \dots & \omega^{2(N-1)} \\ 1 & \dots & \dots & \dots & \dots \\ 1 & \omega^{N-1} & \omega^{2(N-1)} & \dots & \omega^{(N-1)(N-1)} \end{bmatrix} \cdot \begin{bmatrix} x_0 \\ \dots \\ x_j \\ \dots \\ x_{N-1} \end{bmatrix}, \text{where }\omega = e^{i \frac{2\pi}{N}}.$$

The matrix $\mathbf{M}$ can be expressed as a sum of outer products of the basis states $|k \rangle$, and $\langle j|$,

$$\mathbf{M} = \frac{1}{\sqrt{N}} \sum_{j=0}^{N-1} \sum_{k=0}^{N-1} \omega^{jk} |k \rangle \langle j|.$$

where the outer product maps the state from $|j \rangle$ to $|k \rangle$,

$${\mathbf{M}} |j \rangle = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} \omega^{jk} |k \rangle \langle j | j \rangle, \\ |j \rangle \rightarrow^{\mathbf{M}} \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} \omega^{jk} |k \rangle. \\$$

inverse QFT (iQFT)

$|y \rangle$ $\rightarrow$ $|x \rangle$: $\sum_{k=0}^{N-1} y_k |k \rangle \rightarrow \sum_{j=0}^{N-1} x_j |j \rangle$, where $\boxed{x_j = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} w^{-jk} y_k}$ and $w^{-jk} = e^{-i \frac{2\pi}{N} jk }$.

• The exponential term is negated from the QFT.

Quantum Phase Estimation (QPE)

If given the eigenvector $|u \rangle$ of $U$ and eigenvalue $e^{i 2\pi \phi}$ with phase $\phi \in [0,1)$, we have $U |u \rangle = e^{i 2\pi \phi}|u \rangle$, we can estimate the phase $\phi$ via QPE with $t$ bits of precision.

• preparation
  • $1^{st}$ register: $H ^ {\otimes t} |0 \rangle^{\otimes t}=\frac{1}{\sqrt{2^t}}\sum_{x\in \{0,1\}^t}|x \rangle$ (superposition)
  • $2^{nd}$ register: the (superposition of) given eigenvector(s) $|u \rangle$ with eigenvalue $e^{i 2 \pi \phi}$,
    • e.g. $|0 \rangle = a |u_1 \rangle + b |u_2 \rangle$, where $|u_1 \rangle$ and $|u_2 \rangle$ are the eigenvectors of $U$ with eigenvalues $e^{i 2 \pi \phi_1}$ and $e^{i 2 \pi \phi_2}$.
• oracle $U^j$ on the $1^{st}$ register (Entanglement)
  • $\frac{1}{\sqrt{2}} (|0 \rangle + |1 \rangle) \rightarrow \frac{1}{\sqrt{2}} (|0 \rangle + (e^{i 2 \pi \phi })^j |1 \rangle)$
  • $\frac{1}{\sqrt{2^t}} \sum_{x=0}^{2^t-1}|x \rangle \rightarrow \frac{1}{\sqrt{2^t}} \sum_{j=0}^{2^t-1} (e^{i 2 \pi \phi })^j |j \rangle$
  • $2^{nd}$ register: respective $|u \rangle$ with eigenvalue $e^{i 2 \pi \phi}$ and phase $\phi$.
• iQFT (Interference)
• measurement
  • $1^{st}$ register: t bits approximation of $| \tilde{\phi} \rangle$
  • $2^{nd}$ register: $|u \rangle$ with phase $\phi$.
    • $| u_1 \rangle$ with probability $|a|^2$ and $| u_2 \rangle$ with probability $|b|^2$.

Past papers: QFT

• y2015p8q8 (b)
  • design two qubit QFT circuit
• y2009p9q11 (b)
  • (i) QFT in the matrix and outer product form $D$.
  • (ii) the sum of geometric series for $\sum_l$ (case split on $q=1$ and $q \neq 1$).
  • (iii) relationship between QFT and iQFT is $D = D^{2} D^{-1}$,
  • The sum $\sum_{i=1}^{M-1} | M-1 \rangle \langle i |$ maps the state $|i \rangle$ to $|M-1 \rangle$, shifting by one the nonzero basis labels.

iQFT / QPE

• y2024p9q11 (a, c-d)
• y2023p9q11 (d.ii)
• y2021p9q14 (c)
• y2020p9q14 (c-e)

Lecture 10: QFT & QPE: factoring

order finding: for coprime $x$ and $N$, find $x^r \equiv 1 \mod N$, where $r$ is the least positive integer.

$U |r \rangle = |(x \cdot r) \mod N \rangle$ $\implies$ For eigenstates $s\in [0, r-1],$ we have eigenvectors $|u_s \rangle = \frac{1}{\sqrt{r}} \sum_{j=0}^{r-1} e^{-i 2\pi \frac{s}{r}}j |x^j \mod N \rangle$ with phase $\phi = \frac{s}{r}$.

Use QPE, $2^{nd}$ register prepared with equal superposition of unknown eigenvectors $\frac{1}{\sqrt{r}} \sum_{j=0}^{r-1} |u_j \rangle = |1 \rangle$ (shallow-depth quantum circuit $X$).

factoring: for composite integer $N$, $N=p \cdot q$, where $p$ and $q$ are prime numbers.

Shor's algorithm

• y2024p8q11 (a, b)

Lecture 11: QFT & QPE: quantum chemistry

Trotter formula: $U = e^{-i(H_1+H_2)t} = U_1 U_2 = e^{-iH_1 t} e^{-iH_2t} + O(t^2)$, where $U_1$ and $U_2$ don't commute.

• y2021p9q14 (b)

Projective measurement with (normalized) eigenvectors

Ground state energy estimation $| e_0 \rangle$ of a $H$ with eigenvalue $\lambda_0 = E_0$.

Use QPE, $2^{nd}$ register should be prepared as close to the eigenvector such that it's sufficiently dominated by the ground state $| e_0 \rangle$

• L15. adiabatic state preparation.

Lecture 12: Quantum automata and complexity

• y2019p9q13 (e)

Trick: try different combinations of matrix multiplication for state transition.

• y2024p8q11 (c)
• y2017p8q10 (c,d)
• y2007p9q12

Lecture 13: Quantum error correction

• y2021p8q14 (a,c)

New addition of Practical QEC (Google implementation).

Lecture 14: Fault tolerance

bit-flip, phase-flip, Shor code, Steane code

Fault tolerance threshold $p_{th} = \frac{1}{c}$, for suppressed error rate $p = c p_e^2 + O(p_e^3)$. Per-gate error rate $\frac{(cp_e)^{2^k}}{c}$ after $k$ concatenation.

• y2021p8q14 (b,c)

Lecture 15: Adiabatic quantum computing

The adiabatic quantum theory, as a universal computing model that is polynomially equivalent to gate-based.

Quantum annealing, D-Wave.

QAOA

Lecture 16: Near-term case studies

Superconduct, trapped-ion, silicon, optical.

• y2024p9q11