Decoherence in quantum systems always produces $\vert0\rangle$

Question

I was recently asked two questions concerning error in quantum computing:

1. Is it possible for quantum computers to exhibit behavior similar to flip errors in classical computers where a state $\vert0\rangle$ becomes $\vert1\rangle$ due to this error?
2. Is it possible for a quantum computer to decohere to $\vert1\rangle$ or some probability distribution between $\vert0\rangle$ and $\vert1\rangle$ based on the state the system is currently in.

My own thoughts are that the first is possible depending on where in the Bloch sphere your qubits are and the operations you're trying to do. That's why we have quantum error-correction code.

The second question was more tricky. My explanation is that, if I have a density matrix representing the pure state $\begin{bmatrix}1 & 0 & 0 & 1\end{bmatrix} / \sqrt 2$: $$\begin{bmatrix}0.5 & 0 & 0 & 0.5 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0.5& 0& 0& 0.5\end{bmatrix}$$

then I can see the probabilities of all given outcomes for the system. However, if I introduce some noise, I might get something like:

$$\begin{bmatrix}0.3 & 0 & 0 & 0.1 \\ 0 & 0.2 & 0 & 0 \\ 0 & 0 & 0.2 & 0 \\ 0.1& 0& 0& 0.3\end{bmatrix}$$

If I try to measure the state corresponding to this density matrix, I get the result 0.08, which is very close to 0. What I glean from this experiment is that, as the system interacts with the environment, information is lost, returning the system to the ground state, which is 0.

Is this interpretation correct? If not, can somebody provide more detail into what's going on?

Answer 1

Is it possible for quantum computers to exhibit behavior similar to flip errors in classical computers where a state |0⟩ becomes |1⟩ due to this error?

Yes. It would be a very abrupt error if you're talking about errors on physical qubits. Usually, we'd think of an error as being a little bit of an X rotation (for example). However, the effect of performing an error correction step, and measuring the syndrome, divides that into two cases, "no error" and "bit flip", the latter occurring with small probability. Hence, in particular when you're talking about the effects of error correction/fault tolerance on an encoded qubit, the ultimate errors will generally be of this Pauli form.

Is it possible for a quantum computer to decohere to |1⟩ or some probability distribution between |0⟩ and |1⟩ based on the state the system is currently in.

That depends very much on your noise model. In the context of error correction, people often describe the errors as the action of Pauli operators (i.e. the depolarising channel on each qubit). The only stationary state of this channel is the maximally mixed state (NB Not $|0\rangle$), so the ultimate state after a long time is independent of the initial state. Perhaps you're thinking about the specific noise model that is relaxation/amplitude damping? In that case, anything in the excited state has a tendency to relax to the ground state, and so the only stationary state is $|0\rangle$.

However, I can easily give you noise models where the final state depends on the initial state. Dephasing noise (effectively, Pauli Z errors on each qubit) acting on several qubit is one such example. To see why, note that $Z$ never changes the number of qubits in the $|1\rangle$ state, so the final state must depend on how many $|1\rangle$s there were in the initial state. (Ultimately, you get a mixed state which is maximally mixed in each 'excitation subspace', but the weight on that subspace is determined by the initial state.)