# Quantum state where phase information is unknown

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?

• 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? – 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).