Can Interaction with the environment cause $|0\rangle$ to evolve into a superposition like $|+\rangle$?

We say if a qubit is in state $$|0\rangle$$, and it is possible if it interacts with the environment it can result in a bit flip to a state $$|1\rangle$$, so I was wondering why can't it go to a superposition state instead? Is it possible? If not then why is it so?

• It should be certainly possible by application of e.g. Hadamard gate. Application of gate is, in certain sense, also interaction with an environment. Mar 30 at 15:06

If the qubit is in connection with an environment, most likely it will develop quantum correlations (entanglement) and will become a mixed state. If the state is pure, the most general state is a superposition of states for the qubit $$|0\rangle,|1\rangle$$ and those of the environment $$\{|e\rangle\}$$: $$\begin{eqnarray} |\Psi\rangle=\sum_{e}\left(\psi_{0e}|0,e\rangle + \psi_{1e}|1,e\rangle \right). \end{eqnarray}$$ This is a pure state because the combined system qubit+environment is considered a closed quantum system itself.

To look at the qubit alone, we need to trace out the environment degrees of freedom and look at the reduced density matrix of the qubit: $$\begin{eqnarray} \rho_q=\text{Tr}_e |\Psi\rangle\langle\Psi| \end{eqnarray}$$ This will yield in general a mixed state, unless the environment and the qubit are completely uncorrelated. The precise state of the qubit after interacting with the environment depends on the set of Kraus operators that describe such interaction with a bath. If the associated map for an interaction operation is a completely positive trace preserving one, then: $$\begin{eqnarray} \mathcal{E}(\rho_q)=\sum_i K_i\rho_q K_i^\dagger, \end{eqnarray}$$ where $$K_i$$ are the Kraus operators describing the operation and satisfying $$\sum_i K_i^\dagger K_i=I.$$ For a bit-flip noise model, there are $$K_0=I$$ and $$K_1=\sigma_x$$ as Kraus operators, depending on some probability $$1-p$$ the state of the qubit will remain invariant under the map, giving: $$\begin{eqnarray} \mathcal{E}(\rho_q)=(1-p)\rho_q + p\sigma_x\rho_q \sigma_x. \end{eqnarray}$$ This corresponds to a mixed state in general. It does not make much sense to talk about superposition state in an open system given that generally the state will have purity $$\text{Tr}(\rho_q^2)<1$$ due to interaction with an environment.

• This might be a little off topic but can you tell me when we say tracing out the environment, so does it give us the density matrix of the principal system? Mar 31 at 17:00
• If you trace out the degrees of freedom of the environment, it gives the reduced density matrix of the principal system. However, quite generally this will correspond to a mixed state if the environment and system interacted in some way. The way to know whether the stage is mixed is to look at the purity of the state, which is always less than 1 for mixed states. Mar 31 at 17:59
• got it thanks so much Mar 31 at 18:25

It entirely depends on what "environment" and "interaction" you're talking about. As an extreme example, your "environment" could be a qubit in the state $$|+\rangle$$, and the interaction a SWAP gate, and then the interaction with the environment will send whatever initial state you had into $$|+\rangle$$. As a quantum channel, you model this as the mapping $$\Phi(\rho)\equiv\operatorname{tr}(\rho)|+\rangle\!\langle +|$$.

If you have error correction in mind, a more relevant example might be a channel that flips after a rotation, as suggested in the comments. That is, the channel $$\Phi(\rho) = \frac12(H\rho H + XH\rho HX)$$.

Generally speaking, any environment interaction that can be modelled by a quantum channel is in principle possible (tautologically so).