This might be a straightforward problem for you guys; it would be helpful if you can explain it in simple language.

I have $n$-qubits given as $$\frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta_{1}}|1\rangle \right) \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta_{2}}|1\rangle \right) \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta_{3}}|1\rangle \right) \otimes \cdots \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+e^{\iota\theta_{n}}|1\rangle \right)$$ where $\theta_i$'s are same except for only a few $\theta$s which are diffferent from rest.

How to apply an error correction on these states to correct those tiny fractions of $\theta$ to get all qubits in the same state?

For simplest case if I have $$\frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta}|1\rangle \right) \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\theta}|1\rangle \right) \otimes \cdots \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+ e^{\iota\beta}|1\rangle \right) \otimes \cdots \otimes \frac{1}{\sqrt{2}} \left(|0\rangle+e^{\iota\theta}|1\rangle \right)$$ all same but the $i$th $\theta$ is different, then how to correct that qubit?

Any help or suggestions would be greatly appreciated!


2 Answers 2


I believe what you try to do is impossible, because it violates the no-cloning theorem. Note that what you want to do is not the standard way to do quantum error correction, because you have $n-1$ copies of a state, not an encoded state. If you have some more knowledge about the state, you might want to look into stuff like magic state distillation, e.g. here.

One more comment: If $n$ is large you might be able to determine $\theta$, then prepare $n$ new copies of the state.


I'm not sure how possible what you described is. I think the best you can possibly do is to project into the $n$ qubit symmetric space by applying

$$\text{SymmetricProjector}(n) = \sum_{k=0}^{n} \left|{n \atop k}\right\rangle\left\langle{n \atop k}\right|$$

where the "n choose k state" is the superposition of all n-bit numbers with k bits set:

$$\left|{n \atop k}\right\rangle = \frac{1}{\sqrt{n \choose k}} \sum_{x \in \left\{{n \atop k}\right\}} |x\rangle$$

$$\left\{{n \atop k}\right\} = \left\{x \in \mathbb{Z} | 0 \leq x < 2^n \land \text{HammingWeight}(x) = k \right\}$$

But that's more of an error detection scheme than an error correction scheme (because the projection can fail), and honestly it won't work particularly well.

A similar process that is possible is magic state distillation. The key differences that make it work are that the target state is known, and the number of output states is less than the number of input states.


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