Categories: Physics, Quantum information, Quantum mechanics.

Quantum entanglement

Consider a composite quantum system which consists of two subsystems AA and BB, respectively with basis states ∣an⟩\Ket{a_n} and ∣bn⟩\Ket{b_n}. All accessible states of the sytem ∣Ψ⟩\Ket{\Psi} lie in the tensor product of the subsystems’ Hilbert spaces HA\mathbb{H}_A and HB\mathbb{H}_B:

∣Ψ⟩∈HA⊗HB\begin{aligned} \Ket{\Psi} \in \mathbb{H}_A \otimes \mathbb{H}_B \end{aligned}

A subset of these states can be written as the tensor product (i.e. Kronecker product in a basis) of a state ∣α⟩\Ket{\alpha} in AA and a state ∣β⟩\Ket{\beta} in BB, often abbreviated as ∣α⟩∣β⟩\Ket{\alpha} \Ket{\beta}:

∣Ψ⟩=∣α⟩∣β⟩=∣α⟩⊗∣β⟩\begin{aligned} \Ket{\Psi} = \Ket{\alpha} \Ket{\beta} = \Ket{\alpha} \otimes \Ket{\beta} \end{aligned}

The states that can be written in this way are called separable, and states that cannot are called entangled. Therefore, we are dealing with quantum entanglement if the state of subsystem AA cannot be fully described independently of the state of subsystem BB, and vice versa.

To detect and quantify entanglement, we can use the density operator ρ^\hat{\rho}. For a pure ensemble in a given (possibly entangled) state ∣Ψ⟩\Ket{\Psi}, ρ^\hat{\rho} is given by:

ρ^=∣Ψ⟩⟨Ψ∣\begin{aligned} \hat{\rho} = \Ket{\Psi} \Bra{\Psi} \end{aligned}

From this, we would like to extract the corresponding state of subsystem AA. For that purpose, we define the reduced density operator ρ^A\hat{\rho}_A of subsystem AA as follows:

ρ^A=TrB(ρ^)=∑m⟨bm∣(ρ^)∣bm⟩\begin{aligned} \boxed{ \hat{\rho}_A = \Tr_B(\hat{\rho}) = \sum_m \Bra{b_m} \Big( \hat{\rho} \Big) \Ket{b_m} } \end{aligned}

Where TrB(ρ^)\Tr_B(\hat{\rho}) is called the partial trace of ρ^\hat{\rho}, which basically eliminates subsystem BB from ρ^\hat{\rho}. For a pure composite state ∣Ψ⟩\Ket{\Psi}, the resulting ρ^A\hat{\rho}_A describes a pure state in AA if ∣Ψ⟩\Ket{\Psi} is separable, else, if ∣Ψ⟩\Ket{\Psi} is entangled, it describes a mixed state in AA. In the former case we simply find:

∣Ψ⟩=∣α⟩⊗∣β⟩  ⟹  ρ^A=∣α⟩⟨α∣\begin{aligned} \boxed{ \Ket{\Psi} = \Ket{\alpha} \otimes \Ket{\beta} \quad \implies \quad \hat{\rho}_A = \Ket{\alpha} \Bra{\alpha} } \end{aligned}

We call ∣Ψ⟩\Ket{\Psi} maximally entangled if its reduced density operators are maximally mixed, where NN is the dimension of HA\mathbb{H}_A and I^\hat{I} is the identity matrix:

ρ^A=1NI^\begin{aligned} \hat{\rho}_A = \frac{1}{N} \hat{I} \end{aligned}

Suppose that we are given an entangled pure state ∣Ψ⟩≠∣α⟩⊗∣β⟩\Ket{\Psi} \neq \Ket{\alpha} \otimes \Ket{\beta}. Then the partial traces ρ^A\hat{\rho}_A and ρ^B\hat{\rho}_B of ρ^=∣Ψ⟩⟨Ψ∣\hat{\rho} = \Ket{\Psi} \Bra{\Psi} are mixed states with the same probabilities pnp_n (assuming HA\mathbb{H}_A and HB\mathbb{H}_B have the same dimensions, which is usually the case):

ρ^A=TrB(ρ^)=∑npn∣an⟩⟨an∣ρ^B=TrA(ρ^)=∑npn∣bn⟩⟨bn∣\begin{aligned} \hat{\rho}_A = \Tr_B(\hat{\rho}) = \sum_n p_n \Ket{a_n} \Bra{a_n} \qquad \quad \hat{\rho}_B = \Tr_A(\hat{\rho}) = \sum_n p_n \Ket{b_n} \Bra{b_n} \end{aligned}

There exists an orthonormal choice of the subsystem basis states ∣an⟩\Ket{a_n} and ∣bn⟩\Ket{b_n}, such that ∣Ψ⟩\Ket{\Psi} can be written as follows, where pnp_n are the probabilities in the reduced density operators:

∣Ψ⟩=∑npn(∣an⟩⊗∣bn⟩)\begin{aligned} \Ket{\Psi} = \sum_n \sqrt{p_n} \Big( \Ket{a_n} \otimes \Ket{b_n} \Big) \end{aligned}

This is the Schmidt decomposition, and the Schmidt number is the number of nonzero terms in the summation, which can be used to determine if the state ∣Ψ⟩\Ket{\Psi} is entangled (greater than one) or separable (equal to one).

By looking at the Schmidt decomposition, we can notice that, if O^A\hat{O}_A and O^B\hat{O}_B are the subsystem observables with basis eigenstates ∣an⟩\Ket{a_n} and ∣bn⟩\Ket{b_n}, then measurement results of these operators will be perfectly correlated across AA and BB. This is a general property of entangled systems, but beware: correlation does not imply entanglement!

But what if the composite system is in a mixed state ρ^\hat{\rho}? The state is separable if and only if:

ρ^=∑mpm(ρ^A⊗ρ^B)\begin{aligned} \boxed{ \hat{\rho} = \sum_m p_m \Big( \hat{\rho}_A \otimes \hat{\rho}_B \Big) } \end{aligned}

Where pmp_m are probabilities, and ρ^A\hat{\rho}_A and ρ^B\hat{\rho}_B can be any subsystem states. In reality, it is very hard to determine, using this criterium, whether an arbitrary given ρ^\hat{\rho} is separable or not.

As a final side note, the expectation value of an obervable O^A\hat{O}_A acting only on AA is given by:

⟨O^A⟩=Tr ⁣(ρ^O^A)=TrA ⁣(TrB(ρ^O^A))=TrA ⁣(TrB(ρ^)O^A))=TrA ⁣(ρ^AO^A)\begin{aligned} \expval{\hat{O}_A} = \Tr\!\big(\hat{\rho} \hat{O}_A\big) = \Tr_A\!\big(\Tr_B(\hat{\rho} \hat{O}_A)\big) = \Tr_A\!\big(\Tr_B(\hat{\rho}) \hat{O}_A)\big) = \Tr_A\!\big(\hat{\rho}_A \hat{O}_A\big) \end{aligned}

References

  1. J.B. Brask, Quantum information: lecture notes, 2021, unpublished.