You're reading from Quantum Machine Learning and Optimisation in Finance On the Road to Quantum Advantage

Product type Paperback

Published in Oct 2022

Publisher Packt

ISBN-13 9781801813570

Length 442 pages

Edition 1st Edition

Concepts

Machine Learning

Authors (2):

Alexei Kondratyev

Jacquier Antoine

View More author details

Table of Contents (4) Chapters

Preface

1. Chapter 1: The Principles of Quantum Mechanics FREE CHAPTER

2. Part I: Analog Quantum Computing – Quantum Annealing

3. Part II: Gate Model Quantum Computing

1.3 Pure and Mixed States

There are situations where the state of a quantum mechanical system cannot be described with the help of a state vector. Here, we look at such situations and provide a mathematical tool for describing them.

1.3.1 Density matrix

Let us start with the state of a combined two-component physical system given by (1.2.5). Let ( |i⟩ )_i=1,...,N and ( |j⟩ )_j=1,...,M denote, respectively, the standard orthonormal bases of the Hilbert spaces of systems A and B:

N M ∑ ∑ |ψA ⟩ = αi |i⟩, |ψB ⟩ = βj |j⟩, i=1 j=1

where (α_i)_i=1,...,N and (β_j)_j=1,...,M are some probability amplitudes. The states that allow the state vector representation (1.3.1) are called pure states. In this case, the state of the combined system is

∑N M∑ |ψ ⟩ = |ψA ⟩⊗ |ψB ⟩ = αiβj |i⟩⊗ |j⟩. i=1 j=1

However, in general, the state of the combined system would look like

∑N ∑M |ψ⟩ = γij |i⟩⊗ |j⟩, i=1 j=1

where γ_ij are probability amplitudes that may not necessarily be factorised as the product of probability amplitudes (α_i)_i=1,...,N and (β_j)_j=1,...,M. If γ_ij cannot be factorised as α_iβ_j, then the component systems A and B are entangled and their states cannot be represented by the state vectors (1.3.1). Such states of systems A and B are called mixed states.

The more general setup is that of an ensemble of states of the form {p_k, |ψk ⟩ }_k=1,…,N, where each |ψi⟩ is a quantum state whose wavefunction is known with certainty (although this does not necessarily provide full knowledge of the measurement statistics), and each p_k is the associated probability (not amplitude) in [0,1]. In order to define properly pure and mixed states, introduce the density operator as follows:

Definition 7. A density operator ρ is a positive semidefinite Hermitian operator with unit trace and takes the form

∑N ρ := pk |ψk⟩⟨ψk|, k=1

where ∑ _k=1^Np_k = 1 and ⟨ψ |ψ ⟩ k l equals 1 if k = l and zero otherwise.

Mathematically, such a density operator ρ corresponds to a density matrix (ρ_kl)_k,l=1,…,N such that

N ρ = ρ†, Tr(ρ) ≡ ∑ ρ = 1, ρ ≥ 0, for all k = 1,...,N. kk kk k=1

1.3.2 Pure state

A pure state is one that can be represented by a state vector

N∑ |ψ⟩ = αi |i⟩, i=1

where (α_i)_i=1,...,N are probability amplitudes in ℂ such that ∑ _i=1^N|α_i|² = 1. In the ensemble setup above, this means that there exists k^∗∈{1,…,N} such that p_k^∗ = 1 and hence |ψ⟩ = |ψk∗⟩ and therefore ρ = |ψ ⟩ ⟨ψ|. The density matrix also allows us to compute expectations of the form (1.2.4):

Lemma 3. Let ρ be the density matrix associated to the pure state (1.3.2) and let 𝒜 be an observable (Hermitian operator), then

⟨𝒜⟩ := ⟨ψ |𝒜 |ψ ⟩ = Tr (ρ𝒜 ).

Proof. The lemma follows from the immediate computation

⟨ψ\|𝒜	= ⟨ψ\|𝒜∑ _i=1^Nα _i
	= ∑ _i=1^Nα _i ⟨ψ\|𝒜
	= ∑ _i=1^N⟨ψ\|𝒜
	= ∑ _i=1^N ⟨i\|ρ𝒜 = Tr(ρ𝒜).

□

With the state |ψ ⟩ given by (1.3.2), we obtain

N∑ ∑N ⟨𝒜 ⟩ = αiα ∗j ⟨j|𝒜 |i⟩. i=1 j=1

At the same time we have

∑N ∑N ⟨𝒜 ⟩ = Tr(ρ𝒜) = ρij ⟨j|𝒜 |i⟩. i=1j=1

Comparison of (1.3.2) and (1.3.2) yields the following expression for the density matrix of a pure state:

∑N ∑N ρij = αiα∗, ρ = αiα∗ |i⟩⟨j| = |ψ ⟩⟨ψ |. j i=1 j=1 j

Example: An example of a pure state is the Hadamard state

⌊ ⌋ -1- -1- 1 |+ ⟩ = √2-(|0⟩+ |1⟩) = √2-⌈ ⌉, 1

with corresponding density matrix

⌊ ⌋ ρ = |+ ⟩⟨+ | = 1-⌈1 1⌉ . 2 1 1

1.3.3 Mixed state

A mixed state is one that cannot be represented by a single pure state vector, and is therefore represented as a statistical distribution of pure states in the form of an ensemble of quantum states {p_k, |ψk ⟩ }_k=1,…,N, where ∑ _k=1^Np_k = 1 and p_k ∈ [0,1] for each k. The density of a mixed state therefore reads

N ∑ ρ = pk |ψk ⟩⟨ψk|. k=1

Similarly to Lemma 3, we can write expectations of observables with respect to mixed states using the density matrix:

Lemma 4. Let ρ be the density matrix associated to the mixed state (1.3.3) and let 𝒜 be an observable (Hermitian operator), then

∑N Tr(ρ𝒜 ) = pk⟨ψk|𝒜 |ψk⟩ . k=1

Proof. The lemma follows from the immediate computation

Tr(ρ𝒜)	= ∑ _i=1^N ⟨i\|ρ𝒜
	= ∑ _i=1^N ⟨i\|𝒜
	= ∑ _k=1^Np _k
	= ∑ _k=1^Np _k ⟨ψ_k\|𝒜.

□

Let us see now how the density matrix formalism can help us describe the state of a combined system. Consider an entangled state of two systems, A and B, given by (1.3.1), and a Hermitian operator 𝒜 that only acts within the Hilbert space of system A. What would be the expectation value of 𝒜 in this state? Starting with (1.2.4), we obtain

N∑ ∑M ∑N M∑ ∗ ⟨𝒜 ⟩ = γijγ kl⟨k|𝒜 |i⟩⟨l|j⟩. i=1 j=1k=1 l=1

Since only terms with l = j survive in (1.3.3) due to the orthogonality of the basis states, we have

( ) ∑N ∑N ∑M ⟨𝒜 ⟩ = ( γijγ∗kj) ⟨k|𝒜 |i⟩. i=1 k=1 j=1

Thus, the density matrix that describes the mixed state of system A is

M ∑ ∗ ρik = γijγkj. j=1

Note that in the case where the probability amplitudes γ_ij can be factorised as the product of probability amplitudes (α_i)_i=1,...,N and (β_j)_j=1,...,M, we obtain

M∑ ∗ ∗ ∗∑M 2 ∗ ρik = αiβjαkβj = αiα k |βj| = αiαk, j=1 j=1

which describes a pure state.

A simple criterion to distinguish a pure state from a mixed state is the following:

Lemma 5. Let ρ be a density matrix. The inequality Tr(ρ²) ≤ 1 always holds and Tr(ρ²) = 1 if and only if ρ corresponds to a pure state.

Proof. Consider an ensemble of pure states {p_i, |ψi⟩ }_i=1,…,N, with density matrix given by (1.3.3). Therefore

Tr(ρ²)	= Tr
	= Tr
	= Tr = ∑ _i=1^Np _i²Tr⟨ψ_i\| = ∑ _i=1^Np _i² = ∑ _i=1^Np _i²,

which is smaller than 1 since the p_i are probabilities in [0,1] summing up to 1. Assume now that Tr(ρ²) equals one, then so does ∑ _i=1^Np_i². If p_i ∈ (0,1) for all i = 1,…,N, then

∑N 2 ∑N 1 = pi < pi = 1, i=1 i=1

which is a contradiction, and therefore there exists i^∗∈{1,…,N} such that p_i^∗ = 1, so that ρ = |ψi∗⟩ ⟨ψ_i^∗| is a pure state. Conversely, if ρ = |ψi⟩ ⟨ψ_i| for some i ∈{1,…,N} represents a pure state, then

Tr(ρ2) = Tr(|ψ ⟩⟨ψ | |ψ ⟩⟨ψ |) = Tr(|ψ ⟩⟨ψ |) = ⟨ψ |ψ ⟩ = 1. i i i i i i i i

□

Example: An example of a mixed state is a statistical ensemble of states |0⟩ and |1⟩ . If a physical system is prepared to be either in state |0⟩ or state |1⟩ with equal probability, it can be described by the mixed state

⌊ ⌋ 1- 1- 1-⌈1 0⌉ ρ = 2 |0⟩⟨0|+ 2 |1⟩⟨1| = 2 0 1 .

Note that this is different from the density matrix of the pure state

|ψ ⟩ = 1√--(|0⟩ + |1⟩), 2

which reads

⌊ ⌋ ρ = |ψ⟩⟨ψ | = 1(|0⟩+ |1⟩)(⟨0|+ ⟨1|) = 1(|0⟩⟨0|+|1⟩⟨0|+ |0⟩⟨1|+ |1⟩⟨1|) = 1⌈1 1⌉. ψ 2 2 2 1 1

Unlike pure quantum states, mixed quantum states cannot be described by a single state vector. However, the pure states and the mixed states can be described by the density matrix.