# Store/load superposition state of a qubit into/from classical register

I know this question looks ridiculous.

But I simply want to know if possible to store/load a superposition state into classical register from qubit. Specifically two bit of classical register.

Consider 1-qubit $$q$$ and 2-bit $$c$$:

• If $$q=|0\rangle$$, then $$c=00$$.
• If $$q=|1\rangle$$, then $$c=01$$.
• If $$q=|+\rangle$$, then $$c=10$$.
• If $$q=|-\rangle$$, then $$c=11$$.

Basically left classical bit (MSB) will set to 1 if in superposition state, otherwise 0.

Right classical bit (LSB) was about binary state in qubit. But depends on context too, if in superposition state (MSB=1) then it represent previous value before became superposition state.

So is it possible to just using one qubit in this problem? I don't mind using many logic gates in single qubit.

What I mean with store and load is basically copy from/to classical register. Store with measurement, Load with using conditional X gate in classical bit, here is what I mean.

As any pure quantum state of $$n$$-qubits can be described by a vector from vector space $$\mathbb{C}^{2^n}$$, you can of course store it classically. However, with increasing number of qubits, a memory requiremens increase exponentially. That is the reason why quantum computers are built, they cannot be simulated classically with reasonable (i.e. polynomial) computing resources.