Chapter 1: Quantum States

Chapter 1 Quantum States

•

Preskill, John. "Quantum computing and the entanglement frontier." arXiv preprint arXiv:1203.5813 (2012).
•

Ballentine, Leslie E. Quantum mechanics: a modern development. 1998.
•

Preskill, John. "Lecture notes for Physics 219: Quantum computation." California Institute of Technology 3 (2004): 70. (Chapter 4 on entanglement)

1.1 Introduction

The field of quantum information science has its roots in the careful reasoning and deep thought that went into understanding the foundational concepts of quantum theory. Quantum mechanics is a theory which was invented initially to do a number of “useful” things – to describe the radiation emitted from thermal bodies, the structure of atoms, their spectra, scattering probabilities for subatomic particles, and the properties of matter and radiation. However, already in the 1920s, it raised significant philosphical questions about the nature of reality and required a reworking of basic conceptual frameworks that had been used for centuries to understand the world around us. Thought experiments such like Schrödinger’s cat, the work of Einstein, Podolsky, and Rosen on entanglement in 1935, Bohr’s complementarity principle, and Heisenberg’s microscope thought experiment were developed initially to either create a philosophical scaffolding, or to form a basis for criticizing the nascent theory. These concepts now form the basis of fields in quantum computing, sensing, and communications. It was only in the 1960s and 1970s that some of the questions raised by these early pioneers began to be addressed in a rigorous manner, soon giving rise to the field of quantum communication, information, and computing. The field has come full circle. Today we try to use these philosophical advances in our understanding of nature, to again do useful things – to build quantum sensors to detect fields with unprecedented precision and develop enormously capable computers. Before diving into these new advances however, it is useful and important to remind ourselves of some of the the truly weird properties of quantum theory. We will also review a few practical things are important for describing real phenomena.

1.2 What is a quantum state?

Classical probability theory provides a framework to talk about and make calculations of the likelihood or relative frequency of events. It help us predict outcomes based on what we know about a system. At the heart of this framework is the probability distribution function, which gives us the probabilities of different outcomes of measurements on a given system. For instance, consider the probability distribution function:

p(x_{1},\dots,x_{N})=\text{Pr}[X_{1}=x_{1},\cdots,X_{N}=x_{N}],

(1.1)

This function encodes the likelihood that, upon measurement, the random variables $X_{1},\cdots,X_{N}$ yield the results $x_{1},\cdots,x_{N}$ . We can use this to predict potential outcomes of measurements. However, it’s essential to remember that while the probability distribution offers probabilities for various outcomes, it doesn’t usually precisely predict a specific outcome — for instance, it doesn’t indicate if $X_{1}$ will measure as $1.5$ or $2.2$ – it only offers the likelihood for either result.

Example 1.2.1.

Heads or tails (Bernoulli trials)

We define an experiment as such: we have an unbiased coin that we toss into the air, it lands, and we check whether it is heads (H) or tails ( $\text{T})$ . We do this experiment $N$ times, insuring that the trials are independent, resulting in a vector $(x_{1},x_{2},\cdots,x_{N})$ where each $x_{k}$ is either H or T. The probability distribution

\displaystyle p(x_{1},x_{2},\cdots,x_{N})

fully characterizes the result of any measurement. Convince yourself that $p(x_{1},x_{2},\cdots,x_{N})=2^{-N}$ .

In quantum mechanics, the theory’s mathematical formulation enables us to calculate the probability distribution functions $p(x_{1},\dots,x_{N})$ for any measurement.

The first element of this mathematical formulation is the Hilbert space ( $\mathcal{H}$ ), a complex vector space. The main objects we use in the theory are either vectors in this Hilbert space or operators acting on the Hilbert space. The observables ( $\hat{O}$ ) which are operators acting on the Hilbert space, can correspond to variables in classical physics (energy, momentum, position) or represent the question we are asking of the system (e.g., “what is your spin?”, “where are you?”, “Is the voltage 1.5 V?”, etc.). The state ( $\hat{\rho}$ ) is another operator which encodes all of our knowledge of the state of the system. Pure states, a category among them, are also indicated by a state vector $|\psi\rangle$ with the corresponding state or density matrix being $\hat{\rho}=|\psi\rangle\langle\psi|$ . All states, including pure states, can be represented via the density matrix. Remarkably, even given as complete knowledge as possible in quantum mechanics, the exact result of a specific measurement may still elude prediction.

This is already a significant departure from the classical, mechanistic understanding of the world. For example, while knowing a coin’s $50\%$ probability for heads doesn’t predict a specific flip’s result, theoretically, with sufficient understanding and modeling of the coin’s physics within a classical theory, one could predict the outcome of a single coin toss. This is perhaps best captured in Laplace’s famous claim “Give me the positions and velocities of all the particles in the universe, and I will predict the future”. Randomness in classical theories only arises from our limited knowledge or incomplete information of initial conditions and interactions. In contrast, quantum mechanics introduces a different kind of randomness that persists despite complete knowledge of the state and dynamics.

1.2.1 Combining states and observables

We combine states and observables to make predictions in quantum theory. These predictions are in the form of probabilities or probability distribution functions. The way the two operators, observable and state, are combined is through the trace operation.

First we introduce the trace operation:

Definition 1.2.2 (Trace).

The trace of a matrix $A$ is the sum of its diagonal elements. Mathematically, for an $n\times n$ matrix $A$ , the trace is given by:

\text{Tr}(A)=\sum_{i=1}^{n}A_{ii}

In the context of quantum mechanics, the trace of an operator $\hat{A}$ in a Hilbert space with an orthonormal basis $\{|\phi_{i}\rangle\}$ is given by:

\text{Tr}[\hat{A}]=\sum_{i}\langle\phi_{i}|\hat{A}|\phi_{i}\rangle,

where the sum runs over all basis vectors in the Hilbert space.

For operators with continuous spectrum,

\text{Tr}[\hat{A}]=\int\mathrm{d}x\langle x|\hat{A}|x\rangle.

Theorem 1.2.3 (Cyclic property of trace).

For any two operators $\hat{A}$ and $\hat{B}$ :

\text{Tr}[\hat{A}\hat{B}]=\text{Tr}[\hat{B}\hat{A}]

This property implies that the trace remains invariant under cyclic permutations of the matrices inside the trace.

Proof 1.2.4.

Using the orthonormal basis $\{|\phi_{i}\rangle\}$ , we can express the trace of the product of two operators as:

\text{Tr}[\hat{A}\hat{B}]=\sum_{i}\langle\phi_{i}|\hat{A}\hat{B}|\phi_{i}\rangle

Expanding the identity operator the same basis orthonormal basis, we have:

\hat{1}=\sum_{j}|\phi_{j}\rangle\langle\phi_{j}|

Inserting this into our expression for the trace, we get:

\text{Tr}[\hat{A}\hat{B}]=\sum_{i}\sum_{j}\langle\phi_{i}|\hat{A}|\phi_{j}% \rangle\langle\phi_{j}|\hat{B}|\phi_{i}\rangle

Since $\langle\phi_{i}|\hat{A}|\phi_{j}\rangle$ and $\langle\phi_{j}|\hat{B}|\phi_{i}\rangle$ are just numbers, we rearrange them to prove the cyclic property of the trace:

\text{Tr}[\hat{A}\hat{B}]=\sum_{i}\sum_{j}\langle\phi_{j}|\hat{B}|\phi_{i}% \rangle\langle\phi_{i}|\hat{A}|\phi_{j}\rangle=\sum_{j}\langle\phi_{j}|\hat{B}% \hat{A}|\phi_{j}\rangle=\text{Tr}[\hat{B}\hat{A}].

Quantum mechanics provides a framework for calculating the probabilities of different measurement outcomes and expected values of observables. Let’s consider an observable $\hat{O}$ . All observables are represented by Hermitian operators. By the spectral theorem, we can express any Hermitian operator as a sum of projectors:

\hat{O}=\sum_{k}o_{k}|o_{k}\rangle\langle o_{k}|

where $o_{k}$ are real eigenvalues (possible measurement outcomes) and $|o_{k}\rangle$ are the corresponding eigenstates forming an orthonormal basis.

For each eigenvalue $o_{k}$ , the corresponding projection operator

\hat{E}_{k}\equiv|o_{k}\rangle\langle o_{k}|

allows us to calculate the probability of a particular measurement outcome $o_{k}$ when the system is in state $\hat{\rho}$ :

p(\hat{O}=o_{k})=\text{Tr}[\hat{\rho}\hat{E}_{k}]=\langle o_{k}|\hat{\rho}|o_{% k}\rangle.

The expected value of the observable $\hat{O}$ is then given by the weighted average of all possible outcomes:

	$\displaystyle\langle\hat{O}\rangle$	$\displaystyle=\sum_{k}o_{k}p(\hat{O}=o_{k})$
		$\displaystyle=\sum_{k}o_{k}\langle o_{k}\|\hat{\rho}\|o_{k}\rangle$
		$\displaystyle=\text{Tr}[\hat{\rho}\sum_{k}o_{k}\|o_{k}\rangle\langle o_{k}\|]$
		$\displaystyle=\text{Tr}[\hat{O}\hat{\rho}]$

This last expression provides a general formula for calculating the expected value of any observable $\hat{O}$ when the system is in state $\hat{\rho}$ .

Example 1.2.5.

The position operator

Given a particle in one dimension, we consider the operator $\hat{X}$ representing its position. Since the outcome of measuring the position of the particle is a real number $x\in\mathbb{R}[-\infty,\infty]$ , we associate with each of these possible positions a distinct and orthogonal eigenvector $|x\rangle$ , so

\hat{X}=\int_{-\infty}^{\infty}x|x\rangle\langle x|\leavevmode\nobreak\ % \mathrm{d}x.

Exercise 1.2.6 (Projection Operators as Observables).

Consider a projection operator $\hat{P}=|\psi\rangle\langle\psi|$ , where $|\psi\rangle$ is a normalized state vector.

1.

Show that $\hat{P}$ is Hermitian.
2.

Find the eigenvalues and eigenvectors of $\hat{P}$ .
3.

If we consider $\hat{P}$ as an observable, what are the possible measurement outcomes and their corresponding probabilities when measuring a system in state $|\phi\rangle$ ?
4.

Given an arbitrary state $|\phi\rangle=\alpha|\psi\rangle+\beta|\psi_{\perp}\rangle$ , where $|\psi_{\perp}\rangle$ is orthogonal to $|\psi\rangle$ and $|\alpha|^{2}+|\beta|^{2}=1$ , calculate the expectation value $\langle\hat{P}\rangle$ .
5.

Express the variance of this observable, $\text{Var}(\hat{P})=\langle\hat{P}^{2}\rangle-\langle\hat{P}\rangle^{2}$ , in terms of $|\alpha|^{2}$ .

We can summarize these properties of the density matrix and their physical interpretation:

1.
Unit trace: $\text{Tr}[\hat{\rho}]=1$
- •
  
  This is a statement about probabilities. If we have an orthonormal basis $\{|\alpha_{k}\rangle\},$ then $p_{k}=\langle\alpha_{k}|\hat{\rho}|\alpha_{k}\rangle$ is the probability of being in state $k$ . The sum $\sum_{k}p_{k}=1$ , is exactly the trace. (probabilities add up to 1)
2.
The density matrix is Hermitian and positive.
- •
  
  $\hat{\rho}^{\dagger}=\hat{\rho}$ . (probabilities are real)
- •
  
  $\langle\psi|\hat{\rho}|\psi\rangle\geq 0$ for all vectors $|\psi\rangle$ . (probabilities are positive)

Some (but not all) density matrices can be expressed as $\hat{\rho}=|\psi\rangle\langle\psi|$ . The vector $|\psi\rangle$ is called a state vector and $\hat{\rho}$ is a pure state. A state that is not pure is called a mixed state. For example, we may not know whether a system is in a pure state $|\psi_{1}\rangle$ or $|\psi_{2}\rangle$ , but can assign probabilities $p_{1}$ and $p_{2}$ for these two different possibilities. The state of system would then be the linear combination $\hat{\rho}=p_{1}|\psi_{1}\rangle\langle\psi_{1}|+p_{2}|\psi_{2}\rangle\langle% \psi_{2}|$ .

1.

Conservation of trace over unitary evolution: $\hat{\rho}(t)=\hat{U}(t)\hat{\rho}_{0}\hat{U}^{\dagger}(t)$ , then using cyclic property of trace we have that $\text{Tr}[\hat{\rho}(t)]=\text{Tr}[\hat{U}(t)\hat{\rho}_{0}\hat{U}^{\dagger}(t% )]=\text{Tr}[\hat{U}^{\dagger}(t)\hat{U}(t)\hat{\rho}_{0}]=\text{Tr}[\hat{\rho% }_{0}]$ . (probabilities add up to 1 for all time)
2.
Purity: We define the purity $P=\text{Tr}[\hat{\rho}^{2}].$ $P=1$ if and only if $\hat{\rho}$ is a pure state ( $\hat{\rho}=|\psi\rangle\langle\psi|$ ). proof:
- •
  
  $\hat{\rho}=|\psi\rangle\langle\psi|$ implies $\text{Tr}[\hat{\rho}^{2}]=1$ trivially.
- •
  
  Assume $\text{Tr}[\hat{\rho}^{2}]=1$ . Since $\text{Tr}[\hat{\rho}]=1$ and all eigenvalues $p_{k}$ of $\hat{\rho}$ are greater than $0$ and less than 1 (property 1 and 2), for both $\sum_{k}p_{k}=1$ and $\sum_{k}p_{k}^{2}=1$ , we require that $p_{k^{\prime}}=1$ for some $k^{\prime}$ and zero otherwise, so $\hat{\rho}$ is a pure state.
3.
Convex combinations of density matrices are also density matrices.
- •
  
  If $\{\hat{\rho}_{s}\}$ are density matrices, then $\hat{\rho}=\sum_{s}p_{s}\hat{\rho}_{s}$ is also a valid density matrix if $p_{s}\geq 0$ and $\sum_{s}p_{s}=1$ .

Exercise 1.2.7 (States prepared using different procedures).

1.
You’re handed two boxes (1) and (2), (1) emits photons with polarization $|H\rangle$ or $|V\rangle$ randomly and with equal probability. (2) emits photons with polarization $|+\rangle$ or $|-\rangle$ (where $|\pm\rangle=(|H\rangle\pm|V\rangle)/\sqrt{2}$ ) randomly and with equal probability.
1. (a)
  
  What are the density matrices $\hat{\rho}_{1}$ and $\hat{\rho}_{2}$ representing the state of the photon in the basis with vector representation $|H\rangle=\left(\begin{array}[]{c}1\\ 0\end{array}\right)$ and $|V\rangle=\left(\begin{array}[]{c}0\\ 1\end{array}\right)$ .
2. (b)
  
  Can you think of an experiment to distinguish between the two boxes?
2.

(Tricky) Someone has made a box that emits photons with polarization $|H\rangle$ with $p_{H}=0.99$ and $|V\rangle$ with $p_{V}=0.01$ . Design a box that emits photons in state $|\psi\rangle_{1}$ or $|\psi\rangle_{2}$ randomly and with equal probability, and that is indistinguishable from the original box. What are $|x\rangle$ and $|y\rangle$ , expressed in the original basis?

1.2.2 Measurements

We will study measurements in quantum mechanics in significantly greater detail in later chapters. For now we will just consider projective measurements and how this formalism allows us to use the state and an observable to find the probability distribution for measurement result.

Definition 1.2.8 (Projective Measurements).

Projective Measurements

A measuring apparatus gives us a measurement result $s$ , and changes the state of the system. In a projective measurement, we characterize the measurement with set of projection operators $\{\hat{E}_{s}\}$ , with the following properties: (1) $\hat{E}_{s}=\hat{E}_{s}^{\dagger}$ , (2) $\sum_{s}\hat{E}_{s}=1$ , (3) $\hat{E}_{s}\hat{E}_{s^{\prime}}=\hat{E}_{s}\delta_{ss^{\prime}}$ .

Measurement is the processes by which a state vector $|\psi\rangle$ is transformed:

\begin{array}[]{ccc}|\psi\rangle&\underbrace{\longrightarrow}_{measurement}&% \begin{cases}|\psi_{s}\rangle=\frac{\hat{E}_{s}|\psi\rangle}{\sqrt{\langle\psi% |\hat{E}_{s}|\psi\rangle}}\end{cases}\end{array}

with the probability of obtaining measurement result $s$ leading to a final state $|\psi_{s}\rangle$ given by $p_{s}=\langle\psi|\hat{E}_{s}|\psi\rangle.$ Extending this definition to density matrices, we have:

\begin{array}[]{ccc}\hat{\rho}&\underbrace{\longrightarrow}_{measurement}&% \begin{cases}\hat{\rho}_{s}=\frac{\hat{E}_{s}\hat{\rho}\hat{E}_{s}^{\dagger}}{% p_{s}}\end{cases}\end{array}

with probability $p_{s}=\text{Tr}[\hat{E}_{s}\hat{\rho}].$

Example 1.2.9.

Measuring position

Consider the harmonic oscillator with Hamiltonian

\displaystyle\hat{H}

\displaystyle=

\displaystyle\frac{\hat{p}^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2}.

The position operator has a real continuous spectrum of eigenvalues $x\in\mathbb{R}[-\infty,\infty]$ . We assume we have a detector that tells us whether the particle is inside a bin $(x_{s},x_{s}+\Delta x]\subset\mathbb{R}[-\infty,\infty]$ for $x_{s}=s\Delta x$ , for every integer $s$ . The associated measurement operator is $\hat{E}_{s}=\int_{x_{s}}^{x_{s}+\Delta x}dx^{\prime}|x^{\prime}\rangle\langle x% ^{\prime}|$ . Verify that the set of operators ${\{\hat{E}_{s}\}}$ form a valid set of operators defining a projective measurement. Experimentally, such a system would be realized were we have a detector that takes a state, and gives us a number $s$ , corresponding to the position of the operator. The probability of measuring $s$ signifying that the particle has a position of $x_{s}<x\leq x_{s}+\Delta x$ is given by

\displaystyle p_{s}

\displaystyle=

\displaystyle\text{Tr}[\hat{E}_{s}\hat{\rho}].

1.3 Composite Systems and Entanglement

Let’s say we have two systems, system $A$ and system $B$ . We have a way of describing these two systems, including their Hilbert spaces, operators, and other relevant quantum mechanical properties. We should also have a way of describing both systems at once as a single larger composite or bipartite system.

For example, if the first system is in state $|\alpha_{i}\rangle$ and the second one is in state $|\beta_{j}\rangle$ , then the joint state of the composite system is $|\alpha_{i}\rangle\otimes|\beta_{j}\rangle$ , which we usually write more compactly as just $|\alpha_{i}\rangle|\beta_{j}\rangle$ . This new vector is obtained by performing the tensor product operation. It’s a vector that lives in a larger tensor product Hilbert space, which combines the Hilbert spaces of the individual systems. If the states $|\alpha_{i}\rangle$ and $|\beta_{j}\rangle$ form orthonormal bases for the $N_{A}$ - and $N_{B}$ -dimensional Hilbert spaces of system $A$ and $B$ , then $|\alpha_{i}\rangle|\beta_{j}\rangle$ forms an orthonormal basis for the the new tensor product space $\mathcal{H}_{\mathrm{tot}}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ , which consequently is of dimension $N_{A}\times N_{B}$ .

Since in quantum mechanics, any linear combination of these joint states is also a valid state, pure states of the bipartite will in general take the form:

|\Psi\rangle=\sum_{i,j}c_{ij}|\alpha_{i}\rangle|\beta_{j}\rangle\in\mathcal{H}% _{\mathrm{tot}}

where $c_{ij}$ are complex coefficients.

1.3.1 Density Matrix as the Partial Trace of a Pure State

Consider a composite system, composed of two subsystems $A$ and $B$ , with a state

|\Psi\rangle=\sum_{ij}c_{ij}|\alpha_{i}\rangle|\beta_{j}\rangle

(1.2)

Here, $\{|\alpha_{i}\rangle\}$ and $\{|\beta_{i}\rangle\}$ represent orthonormal bases for systems $A$ and $B$ respectively. Generally, observables or operators in this joint system influence both subsystems. However, suppose we only have experimental access to observables that act on system $A$ . This situation might arise when system $B$ is physically distant from $A$ , or in scenarios where one system’s properties are manipulated using another. For example we may be using an optical field described by system $A$ to make measurements on or modify the mechanical position or spin described by $B$ . In such cases, we deal with operators of the form $\hat{O}_{A}\otimes\hat{1}_{B}$ – operators that solely influence system $A$ and whose expectation values are independent of system $B$ ’s state. The expected value of these observables can be written as:

$\displaystyle\langle\Psi\|\hat{O}_{A}\otimes\hat{1}_{B}\|\Psi\rangle$	$\displaystyle=$	$\displaystyle\sum_{ij}\sum_{i^{\prime}j^{\prime}}c_{ij}^{\ast}c_{i^{\prime}j^{% \prime}}\langle\alpha_{i}\|\hat{O}_{A}\|\alpha_{i^{\prime}}\rangle\langle\beta_{% j}\|\beta_{j^{\prime}}\rangle$	(1.3)
	$\displaystyle=$	$\displaystyle\sum_{ij}\sum_{i^{\prime}}c_{ij}^{\ast}c_{i^{\prime}j}\langle% \alpha_{i}\|\hat{O}_{A}\|\alpha_{i^{\prime}}\rangle$
	$\displaystyle=$	$\displaystyle\sum_{ii^{\prime}}\left(\sum_{j}c_{ij}^{\ast}c_{i^{\prime}j}% \right)\langle\alpha_{i}\|\hat{O}_{A}\|\alpha_{i^{\prime}}\rangle$

Assuming no knowledge about system $B$ , let’s say we aim to determine the expected value of a local operator $\hat{O}_{A}$ , given the state of system $A$ . We represent the system’s state through a density matrix, $\hat{\rho}_{A}$ . The observable’s expectation can be calculated as:

\displaystyle\text{Tr}[\hat{\rho}_{A}\hat{O}_{A}]=\sum_{i^{\prime}}\langle% \alpha_{i^{\prime}}|\hat{\rho}_{A}\hat{O}_{A}|\alpha_{i^{\prime}}\rangle=\sum_% {ii^{\prime}}\langle\alpha_{i^{\prime}}|\hat{\rho}_{A}|\alpha_{i}\rangle% \langle\alpha_{i}|\hat{O}_{A}|\alpha_{i^{\prime}}\rangle

(1.4)

On comparing equations (1.3) and (1.4), we observe that as long as we restrict ourselves to observables acting only on system $A$ , the measurement outcome is adequately represented if we assume system $A$ is in state $\hat{\rho}_{A}=\sum_{ii^{\prime}}\sum_{k}c_{ik}^{\ast}c_{i^{\prime}k}|\alpha_{% i^{\prime}}\rangle\langle\alpha_{i}|$ .

The composite system’s state is given by

	$\displaystyle\hat{\rho}$	$\displaystyle=$	$\displaystyle\|\Psi\rangle\langle\Psi\|$
		$\displaystyle=$	$\displaystyle\sum_{ij}\sum_{i^{\prime}j^{\prime}}c_{ij}^{\ast}c_{i^{\prime}j^{% \prime}}\|\alpha_{i^{\prime}}\rangle\|\beta_{j^{\prime}}\rangle\langle\alpha_{i}% \|\langle\beta_{j}\|$

We find that the operation $\sum_{k}\langle\beta_{k}|\hat{\rho}|\beta_{k}\rangle=\sum_{ii^{\prime}}\sum_{k% }c_{ik}^{\ast}c_{i^{\prime}k}|\alpha_{i^{\prime}}\rangle\langle\alpha_{i}|$ provides us with $\hat{\rho}_{A}$ . This operation, $\sum_{k}\langle\beta_{k}|\hat{\rho}|\beta_{k}\rangle$ , is known as a partial trace and is denoted as $\text{Tr}_{B}[\hat{\rho}]$ . Thus, the state of the subsystem can be represented as

\hat{\rho}_{A}=\sum_{j}\langle\beta_{j}|\hat{\rho}|\beta_{j}\rangle\equiv\text% {Tr}_{B}[\hat{\rho}].

(1.5)

Let’s make things more concrete with a specific quantum system: a two-qubit system.

Exercise 1.3.1 (Two-Qubit System and Joint Measurement).

Consider a two-qubit system, $Q_{1}$ and $Q_{2}$ , which is prepared in the Bell state given by

|\Psi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)

(1.6)

Here, $|00\rangle$ and $|11\rangle$ are basis states representing both qubits being in state 0 and both qubits being in state 1, respectively.

1. Compute the density matrix $\hat{\rho}_{Q_{1}}$ for the first qubit $Q_{1}$ .
2. Suppose we want to measure the $z$ -component of the spin (also known as the Pauli- $Z$ operator) of the first qubit, represented by the observable $\hat{Z}$ . The Pauli- $Z$ operator can be represented in the computational basis as:

\hat{Z}=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}

(1.7)

Calculate the expected value of this measurement.
3. Now, consider a joint measurement on both qubits using the Pauli- $Z Z$ operator, defined as the tensor product of the Pauli- $Z$ operators acting on the two subsystems. Calculate the expected value of this joint measurement.

1.3.2 Entanglement

Entanglement is one of the most radical notions in quantum mechanics.

Definition 1.3.2 (Entangled States).

An entangled pure state is a state of a system that cannot be written as a product state:

\displaystyle|\Psi\rangle

\displaystyle\neq

\displaystyle|\alpha\rangle|\beta\rangle.

Similarly, an entangled mixed state is a state that cannot be expressed as a statistical (convex) mixture of product states

\displaystyle\hat{\rho}

\displaystyle\neq

\displaystyle\sum_{k}p_{k}\hat{\rho}_{A,k}\otimes\hat{\rho}_{B,k}.

Entangled states have counter-intuitive properties. Why is entanglement such a radical notion? For an entangled state, even if the state $|\Psi\rangle$ of the joint system is known perfectly, the reduced density matrix $\hat{\rho}_{A}$ describing a subsystem will be mixed. Loosely speaking, even if we know perfectly the state of the two systems taken together, we may still know very little about the state of each subsystem taken separately.

1.3.3 Bell’s Theorem

Background

Quantum mechanics is a statistical theory and has an irreducible randomness. Repeated measurements on identical quantum states can give different outcomes. Quantum theory only gives us the relevant probability distributions. This raises the question: does quantum mechanics emerge from some deeper theory with additional "hidden variables" that determine the outcomes of experiments? Is there some more fundamental theory from which quantum mechanics emerges, in analogy to how statistical physics emerges from an underlying deterministic set of classical theories? Theories postulating an underlying theory (with “hidden variables”) have been proposed to provide a more fundamental explanation.

Exercise 1.3.3 (EPR State).

In this exercise we go over some states and their properties which are useful for understanding EPR’s argument and Bell’s inequality.

1.

Consider two entangled particles 1 and 2 in the EPR state where their relative position and total momentum are precisely correlated:

$\hat{X}_{1}+\hat{X}_{2}=0$ and $\hat{P}_{1}-\hat{P}_{2}=0$
1. (a)
  
  Show that $[\hat{X}_{1}+\hat{X}_{2},\hat{P}_{1}-\hat{P}_{2}]=0$ , confirming that this is a valid quantum state.
2. (b)
  
  If particle 1’s position is measured to be $x$ , what can you conclude about particle 2’s position? What about their momenta if particle 1’s momentum is measured instead?
2.

Write the (unnormalized) wavefunction for the EPR state. (Hint: A state with the particle 1 at position $x_{1}$ and particle 2 at position $x_{2}$ is given by $|\Psi\rangle=|\hat{X}_{1}=x_{1}\rangle|\hat{X}_{2}=x_{2}\rangle$ .)

An influential argument challenging the completeness of quantum mechanics was presented by Einstein, Podolsky, and Rosen (EPR) in 1935. EPR argued that quantum mechanics faces a dilemma: either it is “incomplete” or it violates locality (the principle that distant objects cannot influence each other instantaneously). Here, “incomplete” means that quantum mechanics fails to account for all “elements of physical reality”. The argument considers two particles in an entangled state where their relative position and total momentum are precisely correlated, such that $\hat{X}_{1}+\hat{X}_{2}=0$ and $\hat{P}_{1}-\hat{P}_{2}=0$ . This is a state that’s allowed by quantum physics since the operators $\hat{X}_{1}+\hat{X}_{2}$ and $\hat{P}_{1}-\hat{P}_{2}$ commute and therefore their simultaneous eigenstate with eigenvalues $0$ for both is a valid state. It is now known as an EPR state. After the particles separate, the observer at one end has the choice of measuring either $\hat{X}_{1}$ , or $\hat{P}_{1}$ of particle 1. Let’s assume that they decide to measure $\hat{X}_{1}$ . Whatever the result of the measurement of $\hat{X}_{1}$ , we immediate know precisely what value of a $\hat{X}_{2}$ measurement would be if the second observer were to measure it. Assuming there are nonlocal effects allowed between the two particles, i.e., no so-called “Spooky Action at a Distance”, this means that position of the second particle, $\hat{X}_{2}$ , is fully determined and is thus an “element of physical reality.”¹¹ 1 EPR state that if we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. On the other hand, the observer at position 1 could just as well have decided to measure momentum instead, and so we can apply the same argument to $\hat{P}_{2}$ : it to has a certain value and is an element of reality. But this calls into question one of the fundamental precepts of quantum mechanics which states that noncommuting observable can not be assigned values simultaneously with certainty. It also contradicts the view that the quantum state provides a complete description of a physical system, as it cannot simultaneously represent definite values for both position and momentum.

In the wake of the EPR argument, physicists continued to grapple with the completeness of quantum mechanics. Some physcists, inspired by Einstein’s critique, tried to develop hidden variable theories that could provide a “complete,” deterministic description of quantum phenomena. These theories proposed that underlying the quantum world were hidden variables that, if known, would allow precise predictions of particle behavior. In 1932, John von Neumann claimed to prove that hidden variable theories were generally mathematically impossible, ostensibly closing the door on this line of inquiry. But then in 1952, David Bohm proposed a hidden variable theory that appeared to work, reproducing all the predictions of quantum mechanics, clearly showing that von Neumann’s proof was incorrect.²² 2 John von Neumann’s proof was “foolish” according to Bell’s later analysis. Bohm’s theory, while deterministic, was nonlocal, and allowed for instantaneous influences between distant particles. The natural question then became: is there “complete” and “local” hidden variable theory?

This is where John Bell’s work came in – he proved that if quantum mechanics is correct, then there can be no hidden variable theory that is both deterministic and local. Importantly, Bell’s work showed that the issue was not merely philosophical but experimentally testable. His theorem demonstrated that any hidden variable theory satisfying certain reasonable locality conditions must satisfy inequalities, now known as Bell inequalities, in its predictions for the outcomes of certain experiments. Quantum mechanics, on the other hand predicts violations of these inequalities. This insight transformed the landscape of quantum foundations. It shifted the debate from purely theoretical and philosophical grounds to the realm of experiment. Over the subsequent decades, a series of increasingly sophisticated experiments were conducted to test these predictions. While these experiments consistently confirmed the quantum mechanical predictions, by violating the Bell inequalities, they were subject to certain loopholes that could allow for alternative explanations. The two most significant were the "locality" loophole (the possibility that the detector settings could be communicated between the locations of the two particles) and the "detection" loophole (given inefficient measurements, there is always the possibility that the detected values were not a fair sample of all the values). It wasn’t until 2015 that experiments were finally conducted that closed both of these loopholes simultaneously, providing the most conclusive evidence to date against local hidden variable theories. Nearly sixty years after the original publication of Bell’s paper, the 2022 Nobel Prize in Physics was awarded to Alain Aspect, John Clauser, and Anton Zeilinger for their pioneering experiments in this field.

Proving Bell’s Theorem: CHSH Inequality

Refer to caption — Figure 1.1: Measurement directions for the CHSH inequality. The vectors $\mathbf{e}_{A,1}$ and $\mathbf{e}_{A,2}$ represent the measurement directions for system A. Vectors $\mathbf{e}_{B,1}$ and $\mathbf{e}_{B,2}$ represent those for system B. The vectors are arranged in the x-y plane with 45-degree separations.

We consider two separated systems with local observers, Alice ( $A$ ) and Bob ( $B$ ). At location $A$ , we have local observables $\hat{a}_{1}$ and $\hat{a}_{2}$ , while at location $B$ , we have $\hat{b}_{1}$ and $\hat{b}_{2}$ . Each of these observables yield a value of $\pm 1$ .

Let’s assume these observables have pre-assigned values $a_{1}$ , $a_{2}$ , $b_{1}$ , and $b_{2}$ . We collect the results and calculate $C=(a_{1}+a_{2})b_{1}+(a_{2}-a_{1})b_{2}$ . Note that $a_{1}+a_{2}$ or $a_{2}-a_{1}$ will be 0, and the other will be $\pm 2$ . Consequently, $C=\pm 2$ . If we repeat this measurement multiple times and average the result, we get $-2\leq\langle C\rangle\leq 2$ , or $|\langle C\rangle|\leq 2$ . It’s important to recognize that there’s an assumption of hidden variable theory present in this reasoning, since we assumed that that all of the variables have some definite values.

Now, let’s calculate the expected value of $C$ for a specific quantum mechanical setting. We assume that the observables are the spin along certain axes $\mathbf{e}_{j,k}$ where $j=A,B$ , and $k=1,2$ . Thus, $\hat{a}_{k}=\hat{\sigma}_{A}\cdot\mathbf{e}_{A,k}$ and $\hat{b}_{k}=\hat{\sigma}_{B}\cdot\mathbf{e}_{B,k}$ . We also assume that the two spins at locations $A$ and $B$ are in an entangled state $|\Psi^{-}\rangle=2^{-1/2}(|01\rangle-|10\rangle)$ .

Exercise 1.3.4.

Show that $\langle\Psi^{-}|\hat{\sigma}_{i}\hat{\sigma}_{j}|\Psi^{-}\rangle=-\delta_{ij}$ . Use this to demonstrate that $\langle\Psi^{-}|(\hat{\sigma}_{A}\cdot\mathbf{e}_{A})(\hat{\sigma}_{B}\cdot% \mathbf{e}_{B})|\Psi^{-}\rangle=-\mathbf{e}_{A}\cdot\mathbf{e}_{B}$ .

We choose the vectors for the observables $\hat{a}_{1}$ , $\hat{a}_{2}$ , $\hat{b}_{1}$ , $\hat{b}_{2}$ as follows (see Figure 1.1): $\mathbf{e}_{A,1}$ is along the x-axis, $\mathbf{e}_{B,1}$ is rotated by 45 degrees, $\mathbf{e}_{A,2}$ by 90 degrees, and $\mathbf{e}_{B,2}$ by 135 degrees about the z-axis. Evaluating $C$ , we find:

	$\displaystyle C$	$\displaystyle=\langle\hat{a}_{1}\hat{b}_{1}\rangle+\langle\hat{a}_{2}\hat{b}_{% 1}\rangle+\langle\hat{a}_{2}\hat{b}_{2}\rangle-\langle\hat{a}_{1}\hat{b}_{2}\rangle$
		$\displaystyle=-\mathbf{e}_{A,1}\cdot\mathbf{e}_{B,1}-\mathbf{e}_{A,2}\cdot% \mathbf{e}_{B,1}-\mathbf{e}_{A,2}\cdot\mathbf{e}_{B,2}+\mathbf{e}_{A,1}\cdot% \mathbf{e}_{B,2}$
		$\displaystyle=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{% 1}{\sqrt{2}}=2\sqrt{2}.$

This result violates the inequality $|\langle C\rangle|\leq 2$ derived under the hidden variable assumption, demonstrating Bell’s theorem.

1.3.4 Schmidt Decomposition

An entangled state $|\Psi\rangle$ can’t be expressed as $|\alpha\rangle|\beta\rangle$ . The closest we can come to this form is called the Schmidt decomposition. We can express

|\Psi\rangle=\sum_{k=1}^{\text{min}(N_{A},N_{B})}\sqrt{\lambda_{k}}|\alpha_{k}% \rangle|\beta_{k}\rangle,

with $\{|\alpha_{k}\rangle\}$ , and $\{|\beta_{k}\rangle\}$ with $\{0\leq\lambda_{k}\leq 1\}$ forming orthonormal sets of vectors in their respective Hilbert spaces. This decomposition is pretty remarkable. Remember that in general, the vector $|\Psi\rangle$ is expressed as $|\Psi\rangle=\sum_{ij}c_{ij}|\alpha_{i}\rangle|\beta_{j}\rangle$ and so $N_{A}\times N_{B}$ coefficients are required in an arbitrary basis. The Schmidt decomposition tells us that there is a basis within which only need $\min(N_{A},N_{B})$ coefficients. In fact the actual number of coefficients we need depends on how much entanglement there is. For separable states, the best basis is the one where $|\Psi\rangle$ is obviously a product state $|\alpha_{1}\rangle|\beta_{1}\rangle$ and so only single coefficient, $\lambda_{1}=1$ , is needed. The Schmidt decomposition effectively cuts through the chase and gives us a representation of the system that is commensurate to how much entanglement there really is in a state. This aspect of the Schmidt decomposition makes it the key component of numerical methods (like DMRG) and representation techniques (such as tensor networks) that have been developed to understand quantum correlated many-body states.

Theorem 1.3.5 (Schmidt Decomposition).

Schmidt Decomposition

Any state $|\Psi\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ can be expressed as

|\Psi\rangle=\sum_{k=1}^{\text{min}(N_{A},N_{B})}\sqrt{\lambda_{k}}|\alpha_{k}% \rangle|\beta_{k}\rangle,

with $\{|\alpha_{k}\rangle\}$ , and $\{|\beta_{k}\rangle\}$ with $\{0\leq\lambda_{k}\leq 1\}$ forming orthonormal sets of vectors in their respective Hilbert spaces.

Proof 1.3.6.

We start with the density matrix for the full system $\hat{\rho}=|\Psi\rangle\langle\Psi|$ . We take a partial trace over subsystem $B$ , so $\hat{\rho}_{A}=\text{Tr}_{B}[\hat{\rho}].$ Since this is a valid density matrix for the subsystem $A$ , it is also Hermitian, and so can be diagonalized. We call this basis where the density matrix is diagonal $\{|\alpha_{k}\rangle\}$ , and so

\displaystyle\hat{\rho}_{A}

\displaystyle=

\displaystyle\sum_{k}p_{k}|\alpha_{k}\rangle\langle\alpha_{k}|.

In this basis, we can express the original vector as $|\Psi\rangle=\sum_{jk}c_{jk}|\alpha_{k}\rangle|b_{j}\rangle$ for some basis $|b_{j}\rangle$ for system $B$ . Here we play with the order of summation a little:

	$\displaystyle\|\Psi\rangle$	$\displaystyle=$	$\displaystyle\sum_{jk}c_{jk}\|\alpha_{k}\rangle\|b_{j}\rangle$
		$\displaystyle=$	$\displaystyle\sum_{k}\|\alpha_{k}\rangle\left(\sum_{j}c_{jk}\|b_{j}\rangle\right)$

We label these vectors, $\sqrt{\lambda_{k}}|\beta_{k}\rangle$ , choosing $\lambda_{k}$ so that the $|\beta_{k}\rangle$ are normalized:

\displaystyle\sqrt{\lambda_{k}}|\beta_{k}\rangle

\displaystyle=

\displaystyle\sum_{j}c_{jk}|b_{j}\rangle.

We’ll show now that these vectors, $|\beta_{k}\rangle$ are also orthonormal, so that choosing the express the original wavefunction in this basis of the eigenvectors $\{|\alpha_{k}\rangle\}$ of $\hat{\rho}_{A}$ has also in sense “diagonalized” it in the basis for system $B$ . Notice that we can express the partial trace operation also as:

$\displaystyle\hat{\rho}_{A}$	$\displaystyle=$	$\displaystyle\text{Tr}_{B}\left[\sum_{kk^{\prime}}\sqrt{\lambda_{k}\lambda_{k^% {\prime}}}\|\alpha_{k}\rangle\|\beta_{k}\rangle\langle\alpha_{k^{\prime}}\|% \langle\beta_{k^{\prime}}\|\right]$
	$\displaystyle=$	$\displaystyle\sum_{kk^{\prime}}\text{Tr}_{B}\left[\sqrt{\lambda_{k}\lambda_{k^% {\prime}}}\|\beta_{k}\rangle\langle\beta_{k^{\prime}}\|\right]\|\alpha_{k}\rangle% \langle\alpha_{k^{\prime}}\|$
	$\displaystyle=$	$\displaystyle\sum_{kk^{\prime}}\sqrt{\lambda_{k}\lambda_{k^{\prime}}}\langle% \beta_{k^{\prime}}\|\beta_{k}\rangle\|\alpha_{k}\rangle\langle\alpha_{k^{\prime}% }\|.$

Comparing this expression to the initial $\hat{\rho}_{A}=\sum_{k}p_{k}|\alpha_{k}\rangle\langle\alpha_{k}|$ , and taking inner products with different $|\alpha_{k}\rangle$ on both sides, it is clear that $\langle\beta_{k^{\prime}}|\beta_{k}\rangle=\delta_{kk^{\prime}}$ , and that $\lambda_{k}=p_{k}$ . Since $0\leq p_{k}\leq 1$ , we’ve proven the proposition.

Exercise 1.3.7 (Entanglement of two qubits).

1.

Is the 2-qubit state $|\Psi\rangle=(|00\rangle+|01\rangle+|10\rangle+|11\rangle)/2$ separable? If yes express as a product.
2.

Is the 2-qubit state $|\Psi\rangle=(|00\rangle+|01\rangle+|10\rangle-|11\rangle)/2$ separable? If yes, express as a product.
3.

Is the $N$ -qubit state

$|\Psi\rangle=2^{-n/2}(|00\dots 00\rangle+|00\dots 01\rangle+|00\dots 10\rangle% +|00\dots 11\rangle+\cdots+|11\dots 11\rangle)$

separable?
4.
We start with a pure state that is separable, i.e. can be written as $|\Psi\rangle=|\alpha\rangle|\beta\rangle.$ Show that:
1. (a)
  
  The density matrix $\hat{\rho}_{A}$ is pure.
2. (b)
  
  The system now evolves according to a Hamiltonian that doesn’t have any interaction between $A$ and $B$ , i.e., $\hat{H}=\hat{H}_{A}+\hat{H}_{B}$ , where $\hat{H}_{A}$ and $\hat{H}_{B}$ generate the evolution of system $A$ and $B$ separately. Show that under this evolution, the state $|\Psi\rangle$ remains separable.
5.
Starting with a 2-qubit state (qubit A and B) $|\Psi\rangle=(|00\rangle+|01\rangle+|10\rangle+|11\rangle)/2$ , we perform a cPHASE gate, which results in a state $|\Psi\rangle=(|00\rangle+|01\rangle+|10\rangle+e^{i\phi}|11\rangle)/2$ .
1. (a)
  
  Calculate the reduced density matrix $\hat{\rho}_{A}$ in terms of $\phi$ .
2. (b)
  
  Calculate the purity of the reduced density matrix, $\text{Tr[}\hat{\rho}_{A}^{2}]$ .
3. (c)
  
  We can quantify entanglement using the purity of the partial trace. Let $\mathcal{E}=1-\text{Tr[}\hat{\rho}_{A}^{2}]$ . What is the entanglement in as a function of $\phi$ ?