# What is the main requirement for two arbitrary quantum states to be able to serve as qubit states?

Does it need to be orthogonal ? If now the state is not arbitrary but a coherent state $\lvert\alpha\rangle$, if we use $\lvert0\rangle=\lvert\alpha\rangle$ and $\lvert1\rangle=\lvert-\alpha\rangle$, what condition on the value of $\alpha$ does this requirement set for our qubit coding?

• There is no standard interpretation for the symbol used inside a ket, it is just some label. It does not even necessarily represent a number. (for example we often write $|\psi\rangle$ to mean "the state with wavefunction $\psi(x)$") Consequently asking "what are the conditions on $\alpha$" does not really make sense. Mar 8, 2018 at 13:59
• Thank you for your answer, but by $\alpha$ I meant coherent states, here used for continuous variables coding. I'll edit my question. Mar 8, 2018 at 14:18

Quantum computing with continuous variables is a full field in itself, with many different paradigms for how to actually do the information processing. Therefore, there is not necessarily a single straightforward answer to your question as this depends on what paradigm you're using, how you plan to physically implement this paradigm (i.e. what sort of errors are you expecting), and how you plan to do error-correction. For a reasonable review of part of this field, see Ref . Generally, however, you aren't limited by what values of $\alpha$ you choose to use so much as other factors such as how well you can entangle and how you do your error correction (which includes factors like sources of noise etc).

However, there is a proposal to use coherent states in exactly the way you mention as the computational qubits (Ref ). They restrict their values of $\alpha \ge 2,$ utilizing the fact that this makes the overlap of the two states roughly orthogonal as $\left|\langle\alpha|-\alpha\rangle\right|^2\le 1.1\times 10^{-7}$ for $|\alpha|\ge 2$. However, this is not a strict requirement (as pointed out in the paper) as one can modify their scheme to use $|0\rangle = |\alpha\rangle + |-\alpha\rangle$ and $|1\rangle = |\alpha\rangle - |-\alpha\rangle$ as an exactly orthogonal bases (with no limit on the values $\alpha$ can take).

### References

1. C Weedbrook, S Pirandola, R García-Patrón, NJ Cerf, TC Ralph, JH Shapiro, and S Lloyd, "Gaussian quantum information," Reviews of Modern Physics 84, 621–669 (2012). doi:10.1103/RevModPhys.84.621 arXiv:1110.3234
2. TC Ralph, A Gilchrist, GJ Milburn, WJ Munro, and S Glancy, "Quantum computation with optical coherent states," Physical Review A 68, (2003). doi:10.1103/PhysRevA.68.042319 arXiv:quant-ph/0306004

It depends on what you want to do with it. In some sense yes, orthogonality is enough to have a pair of states to act as an effective qubit.

If two states are not perfectly orthogonal, that can still be fine. This is the case for the example you mentioned: a state like $|\alpha\rangle+|-\alpha\rangle$, with $\alpha$ big enough, can still be considered as an effective qubit. This is because $\langle\alpha|-\alpha\rangle=e^{-2|\alpha|^2}$, so that the overlap is negligible. This kind of state is however usually referred to as a cat state more than as a "qubit".

It is to be noted that for many pairs of orthogonal states it is simply not of much use to think of them as a "qubit", because there is no (simple, known) way to use them as a qubit. By this I mean that it may be highly nontrivial to perform the equivalent of single-qubit operations for such systems, or even worse to perform many-qubit operations between sets of such "qubits".