
Quantum Machine Learning and Optimisation in Finance
By :

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.
Let us start with the state of a combined two-component physical system given by (1.2.5). Let ()i=1,...,N and (
)j=1,...,M denote, respectively, the standard orthonormal bases of the Hilbert spaces of systems A and B:
![]() |
where (αi)i=1,...,N and (βj)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
![]() |
However, in general, the state of the combined system would look like
![]() |
where γij are probability amplitudes that may not necessarily be factorised as the product of probability amplitudes (αi)i=1,...,N and (βj)j=1,...,M. If γij cannot be factorised as αiβj, 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 {pk,}k=1,…,N, where each
is a quantum state whose wavefunction is known with certainty (although this does not necessarily provide full knowledge of the measurement statistics), and each pk 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
![]() |
where ∑ k=1Npk = 1 and equals 1 if k = l and zero otherwise.
Mathematically, such a density operator ρ corresponds to a density matrix (ρkl)k,l=1,…,N such that
A pure state is one that can be represented by a state vector
![]() |
where (αi)i=1,...,N are probability amplitudes in ℂ such that ∑ i=1N|αi|2 = 1. In the ensemble setup above, this means that there exists k∗∈{1,…,N} such that pk∗ = 1 and hence =
and therefore ρ =
⟨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
Proof. The lemma follows from the immediate computation
⟨ψ|𝒜![]() |
= ⟨ψ|𝒜∑ i=1Nα i![]() |
= ∑ i=1Nα i ⟨ψ|𝒜![]() |
|
= ∑ i=1N![]() ![]() |
|
= ∑ i=1N ⟨i|ρ𝒜![]() |
With the state given by (1.3.2), we obtain
![]() |
At the same time we have
![]() |
Comparison of (1.3.2) and (1.3.2) yields the following expression for the density matrix of a pure state:
Example: An example of a pure state is the Hadamard state
with corresponding density matrix
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 {pk,}k=1,…,N, where ∑ k=1Npk = 1 and pk ∈ [0,1] for each k. The density of a mixed state therefore reads
![]() |
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
Proof. The lemma follows from the immediate computation
Tr(ρ𝒜) | = ∑ i=1N ⟨i|ρ𝒜![]() |
= ∑ i=1N ⟨i|![]() ![]() |
|
= ∑ k=1Np k![]() |
|
= ∑ k=1Np k ⟨ψk|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
![]() |
Since only terms with l = j survive in (1.3.3) due to the orthogonality of the basis states, we have
![]() |
Thus, the density matrix that describes the mixed state of system A is
Note that in the case where the probability amplitudes γij can be factorised as the product of probability amplitudes (αi)i=1,...,N and (βj)j=1,...,M, we obtain
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 {pi,}i=1,…,N, with density matrix given by (1.3.3). Therefore
Tr(ρ2) | = Tr![]() |
= Tr![]() |
|
= Tr![]() ![]() ![]() ![]() ![]() |
which is smaller than 1 since the pi are probabilities in [0,1] summing up to 1. Assume now that Tr(ρ2) equals one, then so does ∑ i=1Npi2. If pi ∈ (0,1) for all i = 1,…,N, then
which is a contradiction, and therefore there exists i∗∈{1,…,N} such that pi∗ = 1, so that ρ = ⟨ψi∗| is a pure state. Conversely, if ρ =
⟨ψi| for some i ∈{1,…,N} represents a pure state, then
Example: An example of a mixed state is a statistical ensemble of states and
. If a physical system is prepared to be either in state
or state
with equal probability, it can be described by the mixed state
![]() |
Note that this is different from the density matrix of the pure state
which reads
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.
Change the font size
Change margin width
Change background colour