I'm trying to obtain a more intuitive understanding of the notion of quantum coherence and how to mathematically represent it. I know that coherence has to do with the interaction of phases between the basis states, so, if I suppose that I have a qubit in some state $|\psi\rangle = a|0\rangle + b|1\rangle,$ where $a$ and $b$ are unknown values but $p_{1} = |a|^2$ and $p_{2} = |b|^2$ are known values.

Would I be correct in saying that, in this scenario, I have a state that exhibits decoherence? Would the density matrix for this state be diagonal?

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    $\begingroup$ Welcome to Quantum Computing SE! Without actually answering the question, it's maybe worth mentioning first that decoherence is a process which a state undergoes, so I assume that what you're asking is if this is a state which has already decohered? $\endgroup$
    – Mithrandir24601
    Jun 11 '19 at 23:02

The state you've given, $|\psi\rangle=a|0\rangle+b|1\rangle$ is a pure state. It has not been decohered. Decoherence is a process which turns pure states into mixed states. We usually write these in terms of density matrices. A pure state can be written in this form: $$ |\psi\rangle\langle\psi|=\left(\begin{array}{cc} |a|^2 & ab^\star \\ a^\star b & |b|^2 \end{array}\right). $$ It is very much not diagonal.

Decoherence (I'm going to use a particular form called dephasing) can alter this matrix, reducing the off-diagonal elements to $$ \rho=\left(\begin{array}{cc} |a|^2 & \gamma ab^\star \\ \gamma a^\star b & |b|^2 \end{array}\right) $$ for $0\leq \gamma\leq 1$. So, when we talk about coherence, people often mean comparing the product of the off-diagonal elements to the product of the diagonal elements (or some equivalent procedure). $$ \gamma^2=\frac{\langle0|\rho|1\rangle\langle1|\rho|0\rangle}{\langle0|\rho|0\rangle\langle1|\rho|1\rangle} $$ If the state is pure, and therefore has not been decohered, the two values are equal. The larger the difference, the more decoherence has occurred (because the coherence has been reduced).


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