$\rho_{SE}(0)=\rho_S(0)\otimes\rho_E(0)$: No coupling or no entanglement?


We know that the entangled states cannot be expressed like product state, e.g. $|\omega\rangle = |\psi\rangle\otimes|\phi\rangle$. In the density matrix describing the correlations between system $S$ and environment $E$ we sometimes assume that there are no correlations between $E$ and $S$ at $t=0$: $\rho_{SE}(0)=\rho_S(0)\otimes\rho_E(0)$. I'm wondering what 'correlation' means here? Is this equation implying there's 'no coupling' or 'no entanglement' between $S$ and $E$? (I'm still a bit confused with the difference)

Also in a quantum circuit, can we say that the qubits are coupled as long as there's some gate(s) acting on each qubit, but they are entangled only if some states in the final simulation result cannot be decomposed like the product state?


Posted 2021-01-31T00:14:49.663

Reputation: 988



Coupling is a dynamic concept that characterizes the evolution of a composite system. It means that the evolution involves an interaction between subsystems. In a quantum circuit, coupling corresponds to multi-qubit gates.

Entanglement is a static concept that characterizes the state of a system. It is related to coupling in that it arises due to coupled evolution.

Correlation is an overloaded term. In statistics and probability, it usually refers to a linear dependence between two random variables, but in the broadest sense it denotes non-linear dependence also. In quantum physics, the term also encompasses entanglement, because like classical correlation entanglement gives rise to dependence between observables.

Quantum states can exhibit either type of correlation. For example, two qubits $A$ and $B$ in the joint state

$$ \rho_{AB} = \frac{1}{2}\begin{pmatrix} 0 & & & \\ & 1 & & \\ & & 1 & \\ & & & 0 \end{pmatrix} $$

are correlated, because the knowledge that the qubit $A$ is in the state $|0\rangle$ reveals that qubit $B$ is in the state $|1\rangle$. However, the correlation is classical since the state describes a probability distribution over product states

$$\rho_{AB} = \frac{|01\rangle\langle 01| + |10\rangle\langle 10|}{2},$$

not a superposition. It contains no entanglement, in contrast to a state such as $\beta_{AB} = (|01\rangle + |10\rangle)/\sqrt{2}$ which is a superposition in the joint Hilbert space of the qubits.

Product states like $\rho_S \otimes \rho_E$ contain neither classical correlations (like those in $\rho_{AB}$) nor entanglement (like that in $\beta_{AB}$). Consequently, outcome distributions of any pair of measurements on $S$ and $E$ are independent. Therefore, people often say that these states have no correlations.

Adam Zalcman

Posted 2021-01-31T00:14:49.663

Reputation: 4 307

Many thanks for the answer, that's really clear and helpful:) Could the density matrix sometimes describe superposition? Or it only tells us the probability distribution? – Zhengrong – 2021-01-31T05:35:44.047

1You're welcome! Yes, it could. Any pure quantum state $|\psi\rangle$ has a corresponding density matrix $\rho = |\psi\rangle\langle\psi|$, so if $|\psi\rangle$ is entangled, $\rho$ is entangled. – Adam Zalcman – 2021-01-31T07:10:15.950

Thanks, does coupled evolution always produce entangled states? – Zhengrong – 2021-02-06T04:54:54.417

1No. For example, CNOT turns a Bell state into a product state. – Adam Zalcman – 2021-02-07T21:05:03.770