MAWI's Grenoble Conference

25-29th March 2024

Warning

These notes were taken as the lectures were happening. I need to revise them for errors. All errors are mine, all credit goes to the lecturers.

Day 1

Lecture 1: Introduction to Python

Lecture 2: Theory of two mode Quantum Interferometers

We will consider the lossless / ideal case

We will work with the Mach-Zehnder interometer

The goal is to count the number of particles in the result ( screens above ), and based on that understand the test.

The overall goal is to remove the Measurement uncertainty $\Delta \theta$ , where $\theta$ represents the Operator in our Sample Beam (SB). We need a Reference Beam (RB) to compare the 2 modes of the beam.

Our system satisfies the Canonical Commutation Relations

$$\{\begin{array}{rl}[\hat{a},{\hat{a}}^{†}]& =1\\ [\hat{a},\hat{l}]& =0\\ [\hat{a},{\hat{a}}^{†}]& =0\\ [\hat{l},{\hat{l}}^{†}]& =1\end{array}$$

We have

$${\left(\begin{array}{r}\hat{a}\\ \hat{b}\end{array}\right)}_{out}=\left(\begin{array}{rl}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right){\left(\begin{array}{r}\hat{a}\\ \hat{b}\end{array}\right)}_{in}$$

where we have the relation $m_{ij}=|m_{i}j|e^{i{\theta }_{ij}}$ , which defines the angle ${\theta }_{ij}$.

We have

$$\begin{array}{rl}{N}_{Out}& ={\hat{a}{\hat{a}}^{†}+{\hat{l}}^{†}\hat{l}|}_{out}\\ & =(m_{11}^{\ast }{\hat{a}}^{†}+m_{12}^{\ast }{\hat{l}}^{†})(m_{11}\hat{a}+m_{12}\hat{l})(m_{21}^{\ast }{\hat{a}}^{†}+m_{22}^{\ast }{\hat{l}}^{†})(m_{21}\hat{a}+m_{12}\hat{l}\\ & ={\hat{a}}_{out}^{†}{\hat{a}}_{out}{\hat{l}}_{out}^{†}{\hat{l}}_{out}\\ & ={\hat{a}}^{†}\hat{a}+{\hat{l}}^{†}\hat{l}\end{array}$$

We require

$$\{\begin{array}{rll}|m_{11}{|}^2+|m_{21}{|}^2& =1& \\ |m_{22}{|}^2+|m_{12}{|}^2& =1& \\ m_{11}{m}_{12}^{\ast }+m_{21}{m}_{22}& =0& \\ |m_{11}{|}^2+|m_{12}{|}^2& =1& \text{This condition is derived from the commutation relations applied to aout, lout, lin and ain}\\ |m_{21}{|}^2+|m_{22}{|}^2& =1& \text{As above}\end{array}$$

The condition $|m_{11}{|}^2+|m_{21}{|}^2=1$ is solved as

$$\{\begin{array}{rl}{m}_{11}& =e^{i{\theta }_{11}}\cos (\alpha )\\ m_{21}& =e^{i{\theta }_{21}}\sin (\alpha )\end{array}$$

The condition $|m_{22}{|}^2+|m_{12}{|}^2=1$ is solved as

$$\{\begin{array}{rl}{m}_{22}& =e^{i{\theta }_{11}}\cos (\alpha )\\ m_{12}& =e^{i{\theta }_{21}}\sin (\alpha )\end{array}$$

Which results in the relation ${\theta }_{21}-{\theta }_{11}=\pi +{\theta }_{11}-{\theta }_{12}$.

So we can write $M$ as

$$M=e^{i{\theta }_0}\left(\begin{array}{rl}{e}^{-i{\theta }_t}\cos (\frac{\theta }{2})& e^{-i{\theta }_t}\sin (\frac{\theta }{2})\\ e^{i{\theta }_{\pi }}\sin (\frac{\theta }{2})& e^{i{\theta }_{\pi }}\cos (\frac{\theta }{2})\end{array}\right)$$

$M$ satisfies the properties

• $\det (M)=e^{-2i{\theta }_0}$
• $M\in SU(2)$, the Special Unitary Group of dimention $2$

Setting ${\theta }_0=0\Rightarrow \det (M)=1$, which allows us to see $M$ in this Intorfermeter as applying rotations in space, as we are now working in the Orthogonal Group $M\in SO(3)\subset SU(2)$.

• See the result by Schwinfer

We define the Angular Momentum Operator, which are Casimir invariant:

$$\{\begin{array}{rl}{\tau }_X& =\frac{1}{2}({\hat{a}}^{†}\hat{l}+{\hat{l}}^{†}a)\\ {\tau }_Y& =\frac{1}{2i}({\hat{a}}^{†}\hat{l}-{\hat{l}}^{†}a)\\ {\tau }_Z& =\frac{1}{2}({\hat{a}}^{†}\hat{a}-{\hat{l}}^{†}l)\end{array}$$

Which satisfies the commutation relations

$$\{\begin{array}{rl}[{\tau }_X,{\tau }_Y]& =i{\tau }_Z\\ [{\tau }_X,{\tau }_Z]& =-i{\tau }_Y\\ [{\tau }_Y,{\tau }_Z]& =i{\tau }_X\end{array}$$

and ${\tau }^2={\tau }_X^2+{\tau }_Y^2+{\tau }_Z^2=\frac{\hat{N}}{2}\left(\frac{\hat{N}}{2}+1\right)$.

We can write our operator $M$ as mapping

$$M:\{\begin{array}{r}{\hat{a}}_{out}={\theta }_a{a}_{in}\\ {\hat{b}}_{out}={\theta }_b{b}_{in}\end{array}$$

Our Phase Shift Operator $PS$ is given by

$$M=e^{i{\theta }_0}\left(\begin{array}{rl}{e}^{-i{\theta }_a}& 0\\ 0& e^{-i{\theta }_b}\end{array}\right)\stackrel{\text{Ignoring the global phase}}{=}\left(\begin{array}{rl}{e}^{-i({\theta }_a-{\theta }_b)}& 0\\ 0& e^{i({\theta }_a-{\theta }_b)}\end{array}\right)\stackrel{\theta :={\theta }_a-{\theta }_b}{=}\left(\begin{array}{rl}{e}^{-i\theta }& 0\\ 0& e^{i\theta }\end{array}\right)$$ $$\{\begin{array}{l}{{\tau }_X|}_{out}=(...)=\cos (\theta ){{\tau }_X|}_{in}-n\theta {{\tau }_Y|}_{in}\\ {{\tau }_Y|}_{out}={{\tau }_Y|}_{in}a\theta -n\theta {{\tau }_Y|}_{in}\\ {{\tau }_Z|}_{out}={{\tau }_Z|}_{in}\end{array}$$

Which is the rotation around $Z$ described by

$${\left(\begin{array}{r}{\hat{\tau }}_X\\ {\hat{\tau }}_Y\\ {\hat{\tau }}_Z\end{array}\right)}_{out}=\left(\begin{array}{rll}\cos (\theta )& -\sin (\theta )& 0\\ \sin (\theta )& \cos (\theta )& 0\\ 0& 0& 1\end{array}\right){\left(\begin{array}{r}{\hat{\tau }}_X\\ {\hat{\tau }}_Y\\ {\hat{\tau }}_Z\end{array}\right)}_{in}$$

We need the Baker-Campbell-Hausdorff formula as a way to expand ${{\tau }_X|}_{out}={\left(e^{i\theta {\hat{\tau }}_Z}{\hat{\tau }}_X{e}^{i\theta {\hat{\tau }}_Z}\right)}_{in}$. This equation describes the transition of the phase in the Heisenberg picture.

If we’re interested in the Expectation

$$\begin{array}{rl}{⟨{\tau }_X⟩}_{out}& ={⟨{\psi }_{in}|{\tau }_X|{\psi }_{in}⟩}_{out}\\ & =...\end{array}$$

Giving us the relations

$$\{\begin{array}{rll}PS& \Rightarrow e^{-i\theta {\tau }_Z}& \text{Z Rotation}\\ BS& \Rightarrow ???& \text{Mixes the inputs}\end{array}$$

We describe our Beam Splitter as

$$M_{BS}=\left(\begin{array}{rl}\cos (\frac{\theta }{2})& i\sin (\frac{\theta }{2})\\ i\sin (\frac{\theta }{2})& \cos (\frac{\theta }{2})\end{array}\right)$$

Which gives the matrix in 3D

$${\left(\begin{array}{r}{\hat{\tau }}_X\\ {\hat{\tau }}_Y\\ {\hat{\tau }}_Z\end{array}\right)}_{out}=\left(\begin{array}{rll}1& 0& 0\\ 0& \cos (\theta )& -\sin (\theta )\\ 0& \sin (\theta )& \cos (\theta )\end{array}\right){\left(\begin{array}{r}{\hat{\tau }}_X\\ {\hat{\tau }}_Y\\ {\hat{\tau }}_Z\end{array}\right)}_{in}$$

Which allows us to correct the initial splitters as

$$\{\begin{array}{rll}PS& \Rightarrow e^{-i\theta {\hat{\tau }}_Z}& \text{Z Rotation}\\ BS& \Rightarrow e^{-i\theta {\hat{\tau }}_X}& \text{X Rotation}\end{array}$$

The Mach-Zehnder interferometer aggregates these operations as $BS\circ PS\circ BS$, as per the image on top. Hence

$$\begin{array}{rll}MZ& =e^{-i\theta {\hat{\tau }}_X}{e}^{-i\theta {\hat{\tau }}_Z}{e}^{-i\theta {\hat{\tau }}_X}& \\ & =\exp \left(-i\theta [e^{-i\theta {\hat{\tau }}_X}{\hat{\tau }}_Z{e}^{-i\theta {\hat{\tau }}_X}]\right)& \text{Apply the BCH formula}\end{array}$$

Setup

We setup $N$ particles in $2$ modes, creating the $N+1$ dimensional space spanned by the basis $|m,N-m⟩$, for $m\in \{0,1,...,N\}$. We write $|\psi ⟩={\sum }_{m=0}^N{c}_m{|m⟩}_{\alpha }{|N-m⟩}_e$ and ${|\psi ⟩}_{out}=e^{-i\theta {\hat{\tau }}_Y}{|\psi ⟩}_{in}$.

For simplicity we write $|\mu ⟩$ instead of $|m,N-m⟩$, with $\mu :=\frac{m-(N-m)}{2}=$\frac{N}{2}-m$.Inthissettingthe$\tau_X$and$\tau_Y$ act as the raising operator and lowering operator, respectively.

We define the polarised state as ${|N⟩}_a{|0⟩}_b$, i.e. all particles are in the Ground state. This is an Eigenstate of ${\tau }_Z$ , with eigenvalue $\frac{N}{2}$.

• Our Phase Shift represents a rotation around $Z$ ( green above )
• Our Beam Splitter represents a rotation around $X$ ( red above )

Hence

$$\begin{array}{rll}{⟨{\tau }_Z⟩}_{out}& =⟨\mu =\frac{N}{2}|e^{i\theta {\hat{\tau }}_Y}{\tau }_Z{e}^{-i\theta {\hat{\tau }}_Y}|\mu =\frac{N}{2}⟩& \text{Using eiθτ^YτZe−iθτ^Y=τZcos(θ)+τXsin(θ) }\\ & =⟨\mu =\frac{N}{2}|{\tau }_Z|\mu =\frac{N}{2}⟩\cos (\theta )-⟨\mu =\frac{N}{2}|{\tau }_X|\mu =\frac{N}{2}⟩\sin (\theta )& \\ & =\frac{N}{2}\cos (\theta )-0=\frac{N}{2}\cos (\theta )& \end{array}$$

Hence by making the Measurement ${⟨{\tau }_Z⟩}_{out}$ we can recover information on $\theta$ through the relation ${⟨{\tau }_Z⟩}_{out}=\frac{N}{2}\cos (\theta )$.

• Apply the Maximum Likelihood Estimation with observations of ${\tau }_Z$ above, where we want the value of $\theta$.
  • Find the parameterized distribution for the errors of ${\tau }_Z$.

Letting $\bar{\theta }$ be our estimation of the true value of $\theta$, we have $\tau -⟨\tau ⟩=\frac{\partial }{\partial \theta }⟨{\tau }_Z⟩\left(\theta -\hat{\theta }\right)$.

By the Central Limit Theorem, the estimated error is approximately $\frac{\Delta {\tau }_Z}{zsqrtm}$.

Exercises: Introduction to Python

Day 2

Lecture 1: More Python

No notes taken

Lecture 2: Monte Carlo Methods

Setup

We build a well model with Hamiltonian $H$, and we want to build a high precision model of how it works

$$H\mapsto \text{"High precision"}$$

We want to use a $t$ to describe the transition rate matrix for our Monte Carlo process.

We describe our system as a set of spherical particles in a box

We defined

• Our partition function as $Z={\sum }_{C}W(C)$, where $C$ represents a particle configuration, and $W(C)=\{\begin{array}{cc}1,& \text{2 particles overlap}\\ 0,& \text{No particles overlap}\end{array}$
• The Potential $V$ is described as ${\sum }_{i<j}v(v_{ij})$
• $Z=\int e^{-\beta V(R)dR}$
• $⟨A(R)⟩=\frac{1}{Z}\int e^{-\beta V(R)}A(R)dR$

We can consider a few different cases:

• Classical fluids/solids
• Spin models, such as the Ising Model. ( $E=-y{\sum }_{ij}{S}_{Z,i}{S}_{Z,j}$ )
• Lattice fields

we can consider the system

$$\{\begin{array}{rl}p& =m\dot{x}\\ \dot{p}& =-\frac{\partial }{\partial x}V\end{array}\Leftarrow \text{Molecular Dynamics}$$

The Metropolis–Hastings algorithm consists of two sets:

1. Generate a new sample based on our known but incomplete information on how the system evolves
2. Reject/Approve the new sample based on it’s properties. The probability of acceptance is usually based on some invariants of the configuration that we want to preserve. This could be
   1. ensuring that the new energy is similar to the previous energy
   2. The momentum is maintained through time
   3. etc

Setup

• Data $(y,\sigma )$.
• Parameters $\theta \to y_{\theta }$. The goal is to find $\mathbb{P}[\theta |y]$ ( i.e. find the true underlying parameters based on our observations )

By Bayes theorem, $\mathbb{P}[\theta |y]=\mathbb{P}[y|\theta ]\mathbb{P}[\theta ]\mathbb{P}[y{]}^{-1}$. We assume that

• $\mathbb{P}[y|\theta ]$ is normally distributed
• $\mathbb{P}[y]={\mathbb{E}}_{\theta }[y]=\int \mathbb{P}[\theta ]\mathbb{P}[y|\theta ]d\theta$, where ${\mathbb{E}}_{\theta }[y]$ is the Expectation of $y$ given a certain distribution of $\theta$. This can be as simple as $d\theta =1_{\text{valid configuration}}$.

I’m not familiar with the concept of a partition function in statistical mechanics? Help me ❓

In quantum systems, we can have $\hat{H}=\hat{K}+\hat{V}=\frac{{\Delta }^2}{2m}+V(x)$, using the position basis. We always have that

$$\begin{array}{rl}{E}_0& \le \frac{⟨\psi |H|\psi ⟩}{⟨\psi |\psi ⟩}\\ & =\frac{\int {\psi }^2(R)H\psi (R)dR}{\int |\psi (R){|}^{2}dR}\end{array}$$

( What is a mean field ?? Help me ❓)

$$\{\begin{array}{rl}\mathbb{P}[R]& =\frac{\psi (R{)}^2}{\int \psi (R{)}^{2}dR}\\ E_u(R)& =\frac{H\psi (R)}{\psi (R)}\\ \psi (R)& \approx {\prod }_{i<j}\varphi (v_{ij})=e^{-{\sum }_{i<j}u(v_{ij})}\end{array}$$

With this

$$\begin{array}{rl}{E}_0& \le \underset{\theta }{\min }{E}_{\theta }\\ & =\underset{\theta }{\min }\frac{⟨{\psi }_{\theta }|H|{\psi }_{\theta }⟩}{⟨{\psi }_{\theta }|{\psi }_{\theta }⟩}\end{array}$$

with $$\begin{gathered} {\theta }_{n+1}={\theta }_n \\ -{ϵ}_n{\partial }_{\theta }{E}_{\theta } \end{gathered}$$ as our “evolution” of the algorithm. We want ${ϵ}_n\stackrel{n\to \infty }{\to }0$, but ${\sum }_n{ϵ}_n=\infty$ to ensure that we don’t get locally stuck. In order to iterate, we need to calculate ${\partial }_{\theta }{E}_{\theta }$. This is calculated as

$$\begin{array}{rll}{\partial }_{\theta }{E}_{\theta }& ={\partial }_{\theta }\frac{⟨{\psi }_{\theta }|H|{\psi }_{\theta }⟩}{⟨{\psi }_{\theta }|{\psi }_{\theta }⟩}& \text{By the chain rule}\\ & =\frac{⟨{\psi }_{\theta }|H|{\partial }_{\theta }{\psi }_{\theta }⟩}{⟨{\psi }_{\theta }|{\psi }_{\theta }⟩}-\frac{⟨{\psi }_{\theta }|H|{\psi }_{\theta }⟩}{⟨{\psi }_{\theta }|{\psi }_{\theta }⟩}\frac{⟨{\psi }_{\theta }|\partial {\psi }_{\theta }⟩}{⟨{\psi }_{\theta }|{\psi }_{\theta }⟩}& \text{Applying the Hamiltonian H to the left}\\ & ={\psi }_{\theta }^2(R)\left(E(R)-E_{\theta }\right)\frac{{\partial }_{\theta }{\psi }_{\theta }}{{\psi }_{\theta }}& \end{array}$$

Improvement by considering $|{\psi }_{\beta }⟩:=e^{-\beta H}|\psi ⟩$. We get that $\frac{⟨{\psi }_{\beta }|H|{\psi }_{\beta }⟩}{⟨{\psi }_{\beta }|{\psi }_{\beta }⟩}=(...)$ ( I didn’t get this )

$$\begin{array}{rll}(e^{-\frac{\beta H}{M}}{)}^M& =e^{-iH}{e}^{iH}(e^{-\frac{\beta H}{M}}{)}^M& \text{Multiplying by an identity}\\ & =\int e^{-iH}|R⟩⟨R|e^{iH}|R⟩⟨R|(e^{-\frac{\beta H}{M}}{)}^{M}dR& \end{array}$$

Lecture 3: Phase estimation numerics

We use $|{\psi }_{\theta }⟩$ as a, and Pauli Matrix

$$\begin{array}{rll}|{\psi }_{\theta }⟩& =e^{-i\frac{\theta }{2}{\sigma }_Y}& \text{By a Taylor expansion}\\ & ={\sum }_{n=0}^{\infty }\frac{1}{n!}{\left(-i\frac{\theta }{2}{\sigma }_Y\right)}^n& \text{Matching the coefficients with the taylor series of trignometric functions}\\ & =i\sin (\frac{\theta }{2}){\sigma }_Y-cos(\frac{\theta }{2})1& \end{array}$$

With this we get

$$\{\begin{array}{rl}\mathbb{P}[\uparrow |\theta ]& ={\cos }^2(\frac{\theta }{2})\\ \mathbb{P}[\downarrow |\theta ]& ={\sin }^2(\frac{\theta }{2})\end{array}$$

We make a single measurement to get $r$

$$\{\begin{array}{rl}\mathbb{P}[\uparrow |\theta ]& ={\cos }^2(\frac{\theta }{2})\\ \mathbb{P}[\downarrow |\theta ]& ={\sin }^2(\frac{\theta }{2})\end{array}$$

By measuring on ${\sigma }_Z$ we get the expectation

$$\begin{array}{rl}⟨{\sigma }_Z⟩& =1\mathbb{P}[\uparrow |\theta ]+(-1)\mathbb{P}[\downarrow |\theta ]\\ & ={\cos }^2(\frac{\theta }{2})-{\sin }^2(\frac{\theta }{2})\\ & =\cos (\theta )\\ & \Downarrow \\ \theta & =\arccos (⟨{\sigma }_Z⟩)\end{array}$$

Day 3

Lecture 1: Phase estimation numerics ( continued )

• No notes taken. I need to add the blackboard photos taken during class

Lecture 2: Monte Carlo Methods ( continued )

We want to calculate the integral

$$\begin{array}{rl}I& ={\int }_{-\infty }^{\infty }\tilde{f}(R)dR\\ & =\sum \sum {\sum }_{\Delta }\end{array}$$

Where

$$\{\begin{array}{rll}\bar{f}& =\frac{1}{P}{\sum }_{i}f(R_i)\to ???\approx \tau \sqrt{\frac{{\sigma }^2}{P}},& \tau \ge 1\\ {\sigma }^2& =\int (\tilde{f}(R)-I{)}^{2}dR& \\ p(\tilde{f})& \approx \exp (-\frac{?}{?})& \end{array}$$

We can reduce the Variance by rewriting

$$\begin{array}{rll}{\sigma }^2& ={\int }_{-\infty }^{\infty }\tilde{f}(R)p(R)dR& \text{Where p(r)≥0 and ∫p(R)dR=1}\end{array}$$

As a Markov Chain

We have a transition rate matrix $T$ gives us

$$\begin{array}{rll}\mathbb{P}[X_{t+1}=R]& ={\sum }_{R^{\prime }}T(R^{\prime },R)\mathbb{P}[X_t=R^{\prime }]& \text{Separating out the case R′=R}\\ & =T(R,R)\mathbb{P}[X_t=R]+{\sum }_{R^{\prime }\ne R}T(R^{\prime },R)\mathbb{P}[X_t=R^{\prime }]& \text{Subtracting P[Xt=R] from both sides}\\ & \Downarrow & \\ \mathbb{P}[X_{t+1}=R]-\mathbb{P}[X_t=R]& =-\mathbb{P}[X_t=R]+T(R,R)\mathbb{P}[X_t=R]+{\sum }_{R^{\prime }\ne R}T(R^{\prime },R)\mathbb{P}[X_t=R^{\prime }]& \text{In a stationary state the LHS equals zero}\\ & \Downarrow & \\ 0& =(T(R,R)-1)\mathbb{P}[X_t=R]+{\sum }_{R^{\prime }\ne R}T(R^{\prime },R)\mathbb{P}[X_t=R^{\prime }]& \\ & \Downarrow & \\ (1-T(R,R))\mathbb{P}[X_t=R]& ={\sum }_{R^{\prime }\ne R}T(R^{\prime },R)\mathbb{P}[X_t=R^{\prime }]& \\ & \Downarrow & \\ {\sum }_{R^{\prime }\ne R}T(R^{\prime },R)\mathbb{P}[X_t=R]& ={\sum }_{R^{\prime }\ne R}T(R^{\prime },R)\mathbb{P}[X_t=R^{\prime }]& \text{aka the detailed balanced equation}\end{array}$$

It is unclear why is detailed balance needed, with equality without the sums ( see https://en.wikipedia.org/wiki/Detailed_balance#Reversible_Markov_chains ). Help me ❓

Applying the Metropolis–Hastings algorithm

With the detail balanced equation above satisfied, we can apply the Metropolis–Hastings algorithm with

1. Acceptance rate given by $T(R,R^{\prime }):=\min \{1,\frac{{\pi }_R^{\prime }}{{\pi }_R}\}$ where ${\pi }_R$ being the $\mathbb{P}[X_{\infty }=R]$, with $X_{\infty }$ being the stationary distribution of the Markov Chain.
2. New state generated by $R\to R^{\prime }=R+J$, where we generate $J\sim \mathcal{N}(0,{\sigma }_2)$ ( i.e. normally distributed ).

Examples

In all the approaches below we apply Markov Chain Monte Carlo:

1. We initialise $R$
2. We take a step using the transition matrix $T$ and acceptance rate, a la Metropolis–Hastings algorithm
3. We take in an observable

Ising

$$\{\begin{array}{rll}E(s_1,s_2,...,s_N)& =-{\sum }_{⟨ij⟩}{s}_i{s}_j& \\ \Pi (S)& =e^{-\beta E(s)}& \text{The stationary distribution}\end{array}$$

VMC

aka Variational Monte Carlo

$$\{\begin{array}{rll}\psi (x)& =e^{-\alpha x^2}& \\ \Pi (X)& =|\psi (x){|}^2& \text{The stationary distribution}\\ T(R,R^{\prime })& \propto R+J& J\sim \exp (-\frac{J^2}{2{\sigma }^2})\\ A(R,R^{\prime })& =\frac{{\Pi }_{R^{\prime }}}{{\Pi }_R}<U& \text{Acceptance function, where U∼Uniform[0,1]}\end{array}$$

Day 4

Lecture 1: Learning the noise fingerprint of quantum devices

Presentation based on the paper Learning the noise fingerprint of quantum devices.

Overall idea:

• Run the circuit above in different quantum computers
  • This is done in groups of 1000 times, to get an observed probability distribution function.
• We consider the probability distributions as our $x$, used to predict the machine $y$ they ran on.
  • We use Support Vector Machines to make this prediction, which lies under the binary classification group under Machine Learning.

Lecture 2:

Added the information directly to the Grover’s Algorithm /

• Add a note on the quantum threshold theorem

$|{\psi }_3⟩=U_3{U}_2{U}_1|\psi ⟩$

Day 5

Presentation on Qiskit