Information and Entropy (3)

Principle of Maximum Entropy

Problem Setup

The probelm domain needs to be setup. We'll apply the general case to two examples, one a business model, and the other a model of physical system.

Berger's Burgers

The menu has been extended to include a gourmet low-fat tofu meal.

Magnetic Dipole Model

Consider a system containing one dipole and a two-side environment

Probabilities

We assume that each of the possible states $A_i$ has some probability of occupancy $p(A_i)$ where $i$ is an index running over the possible states.

\[ 1=\sum_{i}^{} p(A_i) \tag{9.1} \]

Probability, and all quantities that are based on probabilities, are subjective, or observer-dependent.

Entropy

\[ S=\sum_{i}^{} p(A_i)\log _{2}\left(\frac{1}{p(A_i)}\right) \] (9.2)

Information is measured in bits.

Constraints

We consider only one constraint here. If there is a quantity $G$ for which each of the states has a value $g(A_i)$ then we want to consider only those probability distributions for which the expected value is a known value $\widetilde{G}$ \[ \widetilde{G}=\sum_{i}^{} p(A_i)g(A_i) \tag{9.4} \]

Maximum Entropy, Single Constraint

Probability Formula

The probability distribution $p(A_i)$ we want has been derived by others. It is a function of the dual variable $\beta$: \[ p(A_i)=2^{-\alpha}2^{-\beta g(A_i)} \] (9.12)

which implies \[ \log _{2}\left(\frac{1}{p(A_i)}\right)=\alpha+\beta g(A_i) \] (9.13)

where \[ \alpha=\log _{2}\left(\sum_{i}^{} 2^{-\beta g(A_i)}\right) \] (9.14)

\[ S=\alpha+\beta G \tag{9.15} \] where $S$,$\alpha$, and $G$ are all functions of $\beta$.

Evaluating the Dual Variable

\[ 0=\sum_{i}^{} [g(A_i)-G(\beta)]2^{-\beta g(A_i)} \] (9.21)

If this equation is multiplied by $2^{\beta G(\beta)}$, the result is \[ 0=f(\beta) \] where the function $f(\beta)$ is \[ f(\beta)=\sum_{i}^{} [g(A_i)-G(\beta)]2^{-\beta[g(A_i)-G(\beta)]} \] (9.23)

It is not difficult to show that $f(\beta)$ is a monotonic function of $\beta$ since $G(\beta)$ is a monotonic function of $\beta$.

Examples

We only focus on the magnetic dipole example. \[ 1=p(U)+p(D) \] (9.33)

\[ \begin{aligned} \widetilde{E} &= e(U)p(U)+e(D)p(D) \\ &= m_{d}H[p(U)-p(D)] \\ \end{aligned} \] (9.34)

\[ S=p(U)\log _{2}\left(\frac{1}{p(A)}\right)+p(D)\log _{2}\left(\frac{1}{p(D)}\right) \] (9.35)

The entropy is the largest, for the energy $\widetilde{E}$ and magnetic field $H$, if

\[ p(U)=2^{-\alpha}2^{-\beta m_{d}H} \] (9.36)

\[ p(D)=2^{-\alpha}2^{\beta m_{d}H} \] (9.37)

where \[ \alpha=\log _{2}(2^{-\beta m_{d}H}+2^{\beta m_{d}H}) \] (9.38)

and an according $f(\beta)$.

Energy

Principle of Maximum Entrophy for Physical Systems

So far we have \[ 1=\sum_{i}^{} p_i \] \[ E=\sum_{i}^{} p_i E_i \] \[ S=k_B \sum_{i}^{} p_i \ln \left(\frac{1}{p_i}\right) \] \[ p_i=\mathrm{e}^{-\alpha} \mathrm{e}^{-\beta E_i} \] \[ \alpha=\ln \left(\sum_{i}^{} \mathrm{e}^{-\beta E_i} \right)=\frac{S}{k_B}-\beta E \]

Differential Forms

Suppose $E_i$ does not depend on an external parameter. Take differentiation. \[ 0=\sum_{i}^{} \mathrm{d}p_i \] \[ \mathrm{d}E=\sum_{i}^{} E_i \mathrm{d}p_i \] \[ \mathrm{d}S=k_B \beta \mathrm{d}E \] \[ \mathrm{d}\alpha=-E \mathrm{d}\beta \] \[ \mathrm{d}p_i=-p_i(E_i-E)\mathrm{d}\beta \] from which it is not difficult to show \[ \mathrm{d}E=-\left(\sum_{i}^{} p_i(E_i-E)^{2}\right)\mathrm{d}\beta \] \[ \mathrm{d}S=-k_B \beta\left(\sum_{i}^{} p_i(E_i-E)^{2}\right) \mathrm{d}\beta \] Note that the formula relating $\mathrm{d}E$ and $\mathrm{d}\beta$, implies that if $E$ goes up then $\beta$ goes down, and vice versa.

Differential Forms with External Parameters

Each $E_i$ could be written in the form $E_i(H)$. \[ E=\sum_{i}^{} p_i E_i(H) \tag{11.20} \] So we have \[ 0=\sum_{i}^{} \mathrm{d}p_i \] (11.21) \[ \mathrm{d}E=\sum_{i}^{} E_i(H)\mathrm{d}p_i+\sum_{i}^{} p_i \mathrm{d}E_i(H) \] (11.22) \[ \mathrm{d}S=k_B \beta \mathrm{d}E-k_B \beta \sum_{i}^{} p_i \mathrm{d}E_i(H) \] (11.23) \[ \mathrm{d}\alpha=-E\mathrm{d}\beta-\beta \sum_{i}^{} p_i \mathrm{d}E_i(H) \] (11.24) \[ \mathrm{d}p_i=-p_i(E(H)-E)\mathrm{d}\beta-p_i \beta \mathrm{d}E_i(H)+p_i \beta \sum_{j}^{} p_j \mathrm{d}E_j(H) \] (11.25) \[ \mathrm{d}E=-\left[\sum_{i}^{} p_i(E_i(H)-E)^{2}\right] \mathrm{d}\beta+\sum_{i}^{} p_i(1-\beta(E_i(H)-E))\mathrm{d}E_i(H) \] (11.26) \[ \mathrm{d}S=-k_B \beta\left[\sum_{i}^{} p_i(E_i(H)-E)^{2}\right] \mathrm{d}\beta-k_B \beta^{2}\sum_{i}^{} p_i(E_i(H)-E)\mathrm{d}E_i(H) \] (11.27)

System and Environment

Partition Model

Our model for the way the total combination is partitioned into the system and the environment. We will use index $i$ for the system and the index $j$ for the environment.

Whether system and environment are isolated or interacting does not affect the states or the physical properties associated with the states, although it may affect the probability of occupancy of the states.

Interaction Model

Consider two adjacent dipoles that exchange their orientations —— the one on the left ends up with the orientation that the one on the right started with, and vice versa. There are only a few cases.

First, if the two dipoles started with the same orientation, nothing would change. On the other hand,if the two dipoles started with different orientations, the effect would be that the pattern of orientationshas changed. This has happened even though the dipoles themselves have not moved. Since the energy associated with the two possible alignments is different, there has been a small change in the location of the energy, even though the total energy is unchanged.

Second, if both dipoles are in the system, or both are in the environment, then energy may have shifted position within the system or the environment, but has not moved between them.

Third, if the two dipoles started with different alignment, and they are located one on each side of the boundary between the system and the environment, then energy has flowed from the system to the environment or vice versa. This has happened not because the dipoles have moved, but because the orientations have moved.

Extensive and Intensive Quantities

The energies of the system state and of the environment state add up to form the energy of the corresponding total state: \[ E_{t,i,j}=E_{s,i}+E_{e,j} \tag{11.37} \]

The probability of occupancy of total state $k$ is the product of the two probabilities of the two associated states $i$ and $j$: \[ p_{t,i,j}=p_{s,i}p_{e,j} \]

The total energy is \[ E_t=\sum_{i,j}^{} E_{t,i,j}p_{t,i,j}=\sum_{j}^{} E_{e,j}p_{e,j}+\sum_{i}^{} E_{s,i}p_{s,i}=E_e+E_s \] (11.39)

This result holds whether the system and environment are isolated or interacting. Similarly, \[ S_t=\sum_{i,j}^{} p_{t,i,j}\ln \left(\frac{1}{p_{t,i,j}}\right)=\sum_{j}^{} p_{e,j}\ln \left(\frac{1}{p_{e,j}}\right)+\sum_{i}^{} p_{s,i} \ln \left(\frac{1}{p_{s,i}}\right)=S_{e}+S_{s} \] (11.40)

This kind of quantity with the property that its total value is the sum of the value for the two (or more) parts is known as an extensive quantity.

Not all quantities are extensive. In particular, $\alpha$ and $\beta$ are not. If the system and environment are isolated, then there is no reason why $\beta_s$ and $\beta_e$ would be related. However, if the system and environment are interacting, the distribution of energy between the system and the environment may not be known and therefore the Principle of Maximum Entropy can be applied only to the combination.

Quantities like $\beta$ that are the same throughout a system when analyzed as a whole are called intensive.

Equilibrium

After mixing the system and the environment, the total entrophy increases. The energies of the system and the environment have changed, and as a result the values of $\beta_s$ and $\beta_e$ have changed, in opposite directions. Their new values are the same (each is equal to $\beta_t$), and therefore this new value lies between the two old values.

Energy Flow, Work and Heat

Energy can be transferred by heat and work at the same time. Work is represented by changes in the energy of the individual states $\mathrm{d}E_i$, and heat by changes in the probabilities $p_i$. \[ \mathrm{d}E=\sum_{i}^{} E_i(H)\mathrm{d}p_i+\sum_{i}^{} p_i \mathrm{d}E_i(H) \] (11.52)

where the first term is heat and the second term is work.

Reversible Energy Flow

\[ \mathrm{d}S_{s}=k_{B} \beta_{s}\mathrm{d}E_s-k_{B}\beta_s\sum_{i}^{} p_{s,i}\mathrm{d}E_{s,i}(H)=k_{B}\beta_s\left[\mathrm{d}E_{s}-\sum_{i}^{} p_{s,i}\mathrm{d}E_{s,i}(H)\right]=k_{B}\beta_{s}\mathrm{d}q_{s} \] (11.55)

where $\mathrm{d}q_{s}$ stands for the heat that comes into the system due to the interaction mechanism.

This formula applies to the system and a similar formula applies to the environment: \[ \mathrm{d}S_{e}=k_{B}\beta_{e}\mathrm{d}q_{e} \tag{11.56} \] The two heats are the same except for sign \[ \mathrm{d}q_{s}=-\mathrm{d}q_{e} \tag{11.57} \]

and it therefore follows that the total entropy $S_{s}+S_{e}$ is unchanged (i.e., $\mathrm{d}S_{s}=-\mathrm{d}S_{e}$ ) if and only if the two values of $\beta$ for the system and environment are the same: \[ \beta_s=\beta_e \tag{11.58} \]

Reversible changes (with no changes in total entrophy) can involve work and heat and therefore changes in energy and entrophy for the system, but the system and the environment must have the same value of $\beta$. Otherwise, the changes are irreversible. Also, reversible changes result in no change to $\beta$.

The first-order change formulas given earlier can be written to account for reversible interactions with the environment by simply setting $\mathrm{d}\beta=0$ \[ 0=\sum_{i}^{} \mathrm{d}p_i \] (11.59) \[ \mathrm{d}E=\sum_{i}^{} E_i(H)\mathrm{d}p_i+\sum_{i}^{} p_i \mathrm{d}E_i(H) \] (11.60) \[ \mathrm{d}S=k_{B}\beta \mathrm{d}E-k_{B}\beta \sum_{i}^{} p_i \mathrm{d}E_i(H) \] (11.61) \[ \mathrm{d}\alpha=-\beta \sum_{i}^{} p_i \mathrm{d}E_i(H) \] (11.62) \[ \mathrm{d}p_i=-p_i \beta \mathrm{d}E_i(H)+p_i \beta \sum_{j}^{} p_j \mathrm{d}E_j(H) \] (11.63) \[ \mathrm{d}E=\sum_{i}^{} p_i(1-\beta(E_i(H)-E))\mathrm{d}E_i(H) \] (11.64) \[ \mathrm{d}S=-k_{B} \beta^{2}\sum_{i}^{} p_i(E_i(H)-E)\mathrm{d}E_i(H) \] (11.65)

Temperature

In this chapter we will interpret $\beta$ further, and will define its reciprocal as the temperature of the material.

Temperature Scales

Recall in Chapter 11 \[ \frac{\mathrm{d}E}{\mathrm{d}S}=\frac{1}{k_{B}\beta} \]

Define the "absolute temperature" as \[ T=\frac{1}{k_{B}\beta} \]

Heat Engine

The change in energy can be attributed to the effects of work $\mathrm{d}w$ and heat $\mathrm{d}q$ \[ \mathrm{d}w=\left(\frac{E}{H}\right) \mathrm{d}H \] \[ \mathrm{d}q=\sum_{i}^{} E_i(H) \mathrm{d}p_i =T \mathrm{d}S \]

Quantum Information

Quantum Information Storage

Model 1: Tiny Classical Bits

Like the magnetic dipole model, this model of the quantum bit behaves essentially like a classical bit except that its size may be very small and it mamy be able to be measured rapidly.

Model 2: Superposition of States (the Qubit)

\[ \psi=\alpha \psi_0+\beta \psi_1 \] When a measurement is made, the values of $\alpha$ and $\beta$ change so that one of them is $1$ and the other is $0$.

Model 3: Multiple Qubits with Entanglement

\[ \psi=\alpha_{00} \psi_{00}+\alpha_{01} \psi_{01}+\alpha_{10}\psi_{10}+\alpha_{11} \psi_{11} \]

This can be thought as two qubits, one corresponding to each of the two measurements. These qubits are not independent, but rather are entangled in some way. A measurement of, for example, the first qubit will return $0$ with probability $\lvert \alpha_{00} \rvert ^{2}+\lvert \alpha_{01} \rvert ^{2}$ and if it does the wave funtion collapses to only those stationary states that are consistent with this measured value \[ \psi=\frac{\alpha_{00}\psi_{00}+\alpha_{01} \psi_{01}}{\sqrt{\lvert \alpha_{00} \rvert ^{2}+\lvert \alpha_{01} \rvert ^{2}}} \]

Bracket Notation for Qubits

Kets, Bras, Brackets, and Operators

A ket is a column vector composed of complex numbers.

\[ |k\rangle = \begin{pmatrix} k_1 \\ k_2 \\ \end{pmatrix} = \vec k \]

The two kets $|0\rangle$ and $| 1 \rangle$ are used to represent the two logical states of qubits, and have a standard vector representation \[ |0 \rangle =\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} \quad |1 \rangle = \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} \]

The superposition of $|0 \rangle$ and $|1 \rangle$ can be written as \[ |\psi\rangle =\alpha |0\rangle+\beta|1 \rangle=\begin{pmatrix} \alpha \\ \beta \end{pmatrix} \]

Bras are the Hermitian conjugates of kets. \[ \langle \psi |=(|\psi\rangle)^{\dag}=(\alpha^{*} \ \beta^{*}) \] ($^{*}$) is the conventional notation for the conjugate of a complex number.

The dot product is the product of a bra by a ket. It is called bracket. \[ \langle q | k \rangle =\sum_{j}^{} q_j^{*}k_j \]

We have the very important property of kets.

\[ \langle \psi | \psi \rangle =\lvert \alpha \rvert ^{2}+\lvert \beta \rvert ^{2}=1 \]

The dot product can be used to compute the probability of a qubit of being in either one of the possible states $|0\rangle$ and $|1\rangle$. \[ \operatorname{Pr}(|0\rangle)=\lvert \langle 0 | \psi\rangle \rvert ^{2}=\lvert \alpha \rvert ^{2} \]

Operators are objects that transform one ket $|k\rangle$ into another ket $| q\rangle$. Operators are represented with hats: $\widehat{O}$. \[ \widehat{O} | k\rangle =\begin{pmatrix} o_{11} & o_{12} \\ o_{21} & o_{22} \\ \end{pmatrix} \begin{pmatrix} k_1 \\ k_2 \\ \end{pmatrix} = \begin{pmatrix} q_1 \\ q_2 \\ \end{pmatrix} = |q\rangle \]

Operators act on bras in a similar manner \[ \langle k| \hat{O}^{\dag}=\langle q | \]

All quantum operators must be unitary!

If we know the input and output kets, we have a easy way to construct an operator $\widehat{O}$ using the exterior product of a ket by a bra.

\[ \widehat{O} | k\rangle=(|q \rangle \langle k|)|k\rangle \]

Tensor Product —— Composite Systems

If we have two qubits $|\psi\rangle$ and $|\phi \rangle$, the system composed by these two qubits is represented by $|\psi\rangle \otimes |\phi\rangle$.

For $n$ qubits, \[ |q_1 \rangle \otimes |q_2 \rangle \otimes \cdots \otimes |q_n\rangle = \bigotimes_{j=1}^{n}|q_j\rangle \]

The dual of a tensor product of kets is the tensor product of the corresponding bras. \[ (|q_1q_2\rangle)^{\dag}=(|q_1\rangle \otimes |q_2 \rangle )^{\dag}=\langle q_1 | \otimes \langle q_2|=\langle q_1q_2| \]

No Cloning Theorem

Qubits cannot be cloned.

To show taht cloning is not possible, let us assume that an operator $\widehat{C}$ takes the information of one qubit $|\phi_1\rangle$ and copies it into another "blank" qubit, the result is a qubit $|\psi_1\rangle$ identical to $|\phi_1\rangle$, and the original $|\phi_1\rangle$ is unmodified. $\widehat{C}$ must be unitary. Thus we define $\widehat{C}$ \[ | Original \rangle \otimes | Blank \rangle \stackrel{\widehat{C}}{\longrightarrow} |Original \rangle \otimes | clone \rangle \]

We are now ready to clone two arbitrary qubits $|\phi_1\rangle$ and $| _2 $ separately. \[ \widehat{C} | \phi_1 \rangle | blank \rangle =| \phi_1 \rangle | \psi_1 \rangle \] \[ \widehat{C} | \phi_2 \rangle | blank \rangle =| \phi_2 \rangle | \psi_2 \rangle \] where it is understood that $| _1 =| _1 $ and $| \phi_2 \rangle =| \psi_2 \rangle$, and we have given them different names to distinguish original from copy.

Since the cloning machine is unitary, it preserves the dot products, so we can compare the dot product before and after cloning \[ \langle \phi_2 | \langle blank | | \phi_1 \rangle | blank \rangle = \langle \phi_2 | \langle \psi_2 | | \phi_1 \rangle | \psi_1 \rangle \] therefore \[ \langle \phi_2 | \phi_1 \rangle \langle blank | blank \rangle =\langle \phi_2 | \phi_1 \rangle \langle \psi_2 | \psi_1 \rangle \] The requirements that kets be normalized imposes that $\langle blank | blank \rangle =1$. the above equation can only be true in two cases: - $\langle \phi_2 | \phi_1 \rangle =0$, which means that $| _1 $ and $| _2 $ are orthogonal. This means that we can clone states chosen at random from a set of orthogonal states. And is equivalent to say that we can clone $| 0 $ and $| 1 $, which we already knew since we do that classically all the time. - $\langle \psi_2 | \psi_2 \rangle =1$, which means that $\psi_2=\psi_1$, that is, that clones obtained in each operation are identical. If the two originals were different, as we had assumed, what this result says is that the clone is independent from the original, which is quite a bizarre property for a clone!

Representation of Qubits

Qubits in the Bloch sphere

\[ | \psi \rangle =\mathrm{e}^{ia} \left( \cos \frac{\theta}{2}| 0 \rangle +\sin \frac{\theta}{2} \mathrm{e}^{i\varphi} | 1 \rangle \right) \]

If we ignore the global phase factor $^{ia} $, the two angles $\theta$ and $\varphi$ define a point in a unit sphere. This sphere is called the Bloch Sphere.

Qubits and symmetries

Pauli matrices: \[ \mathbb{I}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \quad \sigma_{x}= \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \quad \sigma_y= \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix} \quad \sigma_{z}= \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} \]

Action of Pauli matrices on an arbitrary qubit

An arbitrary superposition state \[ | \psi \rangle =\alpha| 0 \rangle +\beta | 1 \rangle =\cos \frac{\theta}{2} | 0 \rangle +\sin \frac{\theta}{2} \mathrm{e}^{i\varphi} | 1 \rangle \]

So \[ \sigma_x | \psi \rangle = \begin{pmatrix} \beta \\ \alpha \\ \end{pmatrix} \] (Rotation of $\pi$ about $x$ axis)

\[ \sigma_y | \psi \rangle =i \begin{pmatrix} -\beta \\ \alpha \\ \end{pmatrix} \] (Rotation of $\pi$ about $y$ axis)

\[ \sigma_z | \psi \rangle = \begin{pmatrix} \alpha \\ -\beta \\ \end{pmatrix} \] (Rotation of $\pi$ about $z$ axis)

Hence Pauli matrices are rotations of $\pi$ about each of the axis of the bloch sphere.

\[ \mathrm{e}^{i \sigma_x \theta/2} =\cos \frac{\theta}{2} \mathbb{I}+i \sin \frac{\theta}{2}\sigma_x \]

This result shows us how to do arbitrary rotations of an angle $\theta$ about the $x$ axis, the resulting operator is often called $R_x()=^{i_x /2} $. The cases of $R_y$ and $R_z$ are completely analogous.

Quantum Gates

Elementary Quantum Gates

Pauli $X$: $X=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\equiv \sigma_x$. It is equivalent to doing a $NOT$ or bit flip.
Pauli $Y$: $Y=\begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}\equiv \sigma_y$.
Pauli $Z$: $Z=\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}\equiv \sigma_z$. Changes the internal phase.
Hadamard: $H=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}$.
Phase: $S=\begin{pmatrix} 1 & 0 \\ 0 & i \\ \end{pmatrix}$.

Below we enumerates some of the properties of the elementary quantum gates.

\[ H=\frac{1}{\sqrt{2}}(X+Z) \quad HXH=Z \] \[ XYX=-Y \quad HYH=-Y \] \[ XZX=-Z \quad HZH=X \] \[ XR_{y}(\theta) X=R_y(-\theta) \quad XR_z(\theta)X=R_y(-\theta) \]

Two-qubit gates. Controlled Gates

Operators are unitary and square, so quantum gates will always have the same number of inputs and outputs. Another way to say it is that all quantum gates are naturally reversible.

The most important two qubit gates are the controlled gates. In a controlled gate the first input qubit is a control qubit. If is in the $| 1 $ state, it will trigger the gate that acts on the second qubit, otherwise, it will not trigger it and the second qubit will remain unaltered. Say, a control qubit $| \psi \rangle =\begin{pmatrix} \alpha \\ \beta \\ \end{pmatrix}$ with the second qubit $| $ and the gate $U$ results in $| +U | $.

There are two controlled gates that are very relevant to the algorithms we will describe later on, the $C-X$ also known as $C-NOT$ and the $C-Z$ also known as $C-Phase$.

Quantum Communication

Teleportation - Alice and Bob's story

Alice and Bob entangled a pair of qubits $| _{AB} $ when they first met, \[ | \phi_{AB} \rangle =\frac{1}{\sqrt{2}} (| 0_{A} \rangle \otimes | 0_{B} \rangle +| 1_{A} \rangle \otimes | 1_{B} \rangle ) \]

Life took each of them through separate paths. However, Alice decides to send Bob a letter in a qubit $| {L} =| 0{L} +| 1_{L} $.

Alice puts the qubit of the pair she once entangled with Bob in a composite system with $| _{L} $. The complete three-qubit system can be represented using tensor products \[ | \phi_{A}\psi_{L}\phi_{B} \rangle =\frac{1}{\sqrt{2}}\biggl( | 0_{A} \rangle \otimes (\alpha | 0_{L} \rangle +\beta | 1_{L}\rangle ) \otimes | 0_{B} \rangle + | 1_{A} \rangle \otimes (\alpha | 0_{L} \rangle +\beta | 1_{L} \rangle ) \otimes | 1_{B} \rangle \biggr) \]

In practice what the $C-NOT$ does is transfer the superposition to Alice's Qubit \[ \begin{aligned} &= \frac{1}{\sqrt{2}}\alpha\biggl( | 0_{A} \rangle \otimes | 0_{B} \rangle + | 1_{A} \rangle \otimes | 1_{B} \rangle \biggr) \otimes | 0_{L} \rangle \\ &+ \frac{1}{\sqrt{2}}\beta\biggl( | 1_{A} \rangle \otimes | 0_{B} \rangle +| 0_{A} \rangle \otimes | 1_{B} \rangle \biggr) \otimes | 1_{L} \rangle \end{aligned} \]

At this point Alice's and Bob's qubit have both the information of the superposition that was originally in the letter. The Hadamard gate produces a new superposition out of the letter as follows \[ \begin{aligned} &= \frac{1}{\sqrt{2}}\alpha\biggl( | 0_{A} \rangle \otimes | 0_{B} \rangle + | 1_{A} \rangle \otimes | 1_{B} \rangle \biggr) \otimes \frac{1}{\sqrt{2}}(| 0_{L} \rangle+| 1_{L} \rangle ) \\ &+ \frac{1}{\sqrt{2}}\beta\biggl( | 1_{A} \rangle \otimes | 0_{B} \rangle +| 0_{A} \rangle \otimes | 1_{B} \rangle \biggr) \otimes \frac{1}{\sqrt{2}}(| 0_{L} \rangle-| 1_{L} \rangle ) \end{aligned} \]

At this point the information about the superposition in the original letter is no longer in Alice's hands. To appreciate that it is so, we need to make some manipulations and reordering of the corss products. \[ \begin{aligned} &= \frac{1}{2}| 0_{A} \rangle \otimes | 0_{L} \rangle \otimes (\alpha | 0_{B} \rangle +\beta | 1_{B} \rangle ) \\ &+ \frac{1}{2}| 0_{A} \rangle \otimes | 1_{L} \rangle \otimes (\alpha | 0_{B} \rangle -\beta | 1_{B} \rangle ) \\ &+ \frac{1}{2}| 1_{A} \rangle \otimes | 0_{L} \rangle \otimes (\alpha | 1_{B} \rangle +\beta | 0_{B} \rangle ) \\ &+ \frac{1}{2}| 1_{A} \rangle \otimes | 1_{L} \rangle \otimes (\alpha | 1_{B} \rangle -\beta | 0_{B} \rangle ) \\ \end{aligned} \]

The next steps are the key to faultless teleportation - Alice measures her two qubits, she will obtain either of $| 0_{A}0_{L} $,$| 0_{A}1_{L} $, $| 1_{A}0_{L} $, or $| 1_{A}1_{L} $ with equal probability. - Upon Alice's measurement, Bob's qubit takes the value of one of the four possible superpositions. And so, the result of her measurement can help Bob unscramble his bit. - If she measured $| 0_{A}0_{L} $, she will tell Bob not to do anything to his qubit. If she measured $| 0_{A}1_{L} $, Bob will have to correct for the phase (that can be done with a $Z$ gate). If she measured $| 1_{A}0_{L} $, the states in the message have been flipped, and to unflip them Bob will have to use a bit-flip (a.k.a not, a.k.a $X$) gate. Finally if she measured $| 1_{A}1_{L} $, Bob will have to correct both for the phase and the bit flip.

Alice needs to communicate to bob 2 classical bits. The first bit of Alice informs about bit-flip errors and the second about phase errors.

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Information and Entropy (3)

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John Doe

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July 24, 2022

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