# Coherences in the density matrix

It is said that the off-diagonal elements of density matrix are "coherence". When a system interacts with its environment the off-diagonal elements decay and the final density matrix is the diagonal one, a statistical mixture. This process is called decoherence.

We know that every density matrix can be diagonalized in some basis.

What would decoherence be when the density matrix is diagonal in some basis?

• As Peter Shor says in this answer physics.stackexchange.com/questions/4284/… decoherence is actually basis dependent. – Antonio Ragagnin May 15 '14 at 11:51
• "When a system interacts with its environment the off-diagonal elements decay and the final density matrix is the diagonal one" ... this is only true if you write your system in the correct basis, which is determined by the interaction of the system and its environment. – Peter Shor May 15 '14 at 21:18
• @PeterShor. Could you please provide me with some scientific/technical articles which state and explain this basis dependence? The density operator is independent of basis. Even the Hamiltonian of the interaction is independent of basis. Actually, we can switch between every basis with a unitary transformation. So, to me it's unclear how decoherence is basis dependent. – MOON May 16 '14 at 20:44
• Decoherence is not basis dependent!!! I guess you didn't understand my earlier comment. This description of decoherence as the decay of the off-diagonal elements is only accurate if you write it in one certain basis. I've changed my answer to the linked question to say what I meant to say in the first place; namely, that this description of decoherence is basis-dependent. – Peter Shor May 16 '14 at 21:38

Regarding to the question of basis dependency of decoherence, I can perhaps give some examples to clear things up. For a pure qubit state $|0\rangle$, when complete decoherence happens, the basis where the maximum coherence loss is witnessable is in the maximum coherent state $$\rho= H | 0 \rangle \langle 0 | H^{\dagger} = \frac{1}{2} \left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right) \rightarrow \frac{1}{2} \left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right) ,$$ where $H$ is a unitary transform by a Hadamard matrix, $\rho$ is turned into the maximally mixed state. No matter what unitary operation you perform on it, coherence cannot be recovered.
However, if $|0\rangle$ and the environment interact in a way where partial decoherence happens and a full decoherence is only viewable from a non-maximal coherence basis $$\rho = U | 0 \rangle \langle 0 | U^{\dagger} = \frac{1}{3} \left( \begin{matrix} 1 & -\sqrt{2} \\ -\sqrt{2} & 2 \\ \end{matrix} \right) \rightarrow \frac{1}{3} \left( \begin{matrix} 1 & 0 \\ 0 & 2 \\ \end{matrix} \right),$$
where $$U= \left( \begin{matrix} \sqrt{\frac{1}{3}} & \sqrt{\frac{2}{3}} \\ -\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{3}} \\ \end{matrix} \right).$$ The sight from a Hadamard matrix point of view(where maximal coherence is exhibited) of the decohered state would be $$\frac{1}{6} \left( \begin{matrix} 3 & -1 \\ -1 & 3 \\ \end{matrix} \right),$$ which still has coherence.