# Question about the probability of failure of the bit flip code

For the bit flip superoperator is $$\mathcal{E}_{BF}(\rho) = (1-p)\rho + p X \rho X$$

where the first term refers to no bit flip and the second term refers to the bit flipping.

A single qubit pure state $$|\phi \rangle = \alpha|0 \rangle + \beta |1 \rangle$$ corresponds to the density operator $$\rho = |\phi \rangle \langle \phi|$$. So the bit flip superoperator becomes $$\sigma = (1-p)|\phi \rangle \langle \phi| + p X |\phi \rangle \langle \phi| X$$

A book I'm reading states that the probability of the failure is error due to noise: $$p_{err} = 1- \langle \phi|\sigma | \phi \rangle$$

I'm having trouble understanding this part. I am interpreting 'failure' as a bit flip with probability $$p$$ but this clearly is wrong. What is 'failure' in this regard?

The quantity $$\langle \phi | \sigma | \phi \rangle$$ is precisely the fidelity between $$| \phi \rangle$$ and $$\sigma$$; you can think of this as how "close" your state with the bit-flip channel is to the pure state. $$1 - \langle \phi | \sigma | \phi \rangle$$ is basically how much the bit flip makes you deviate from the pure state.

In other words, it's not the probability of the bit flip error occurring; it's how badly it mangles your quantum state.

For example: take the superposition $$| \phi \rangle = \frac{1}{\sqrt{2}} ( | 0 \rangle + | 1 \rangle)$$ Then $$\sigma = (1-p)|\phi \rangle \langle \phi| + p X |\phi \rangle \langle \phi| X = |\phi \rangle \langle \phi|$$ so $$p_{err} = 1- \langle \phi|\sigma | \phi \rangle = 1 - \langle \phi| \phi \rangle \langle \phi| \phi \rangle = 0$$ No matter the probability $$p$$ of the bit flip, the state will be unaffected.

• The next sentence in the book says that the goal is to maintain coherence, so we must correct the error and recover $| \phi \rangle$. So does this mean that if the bit flip channel changes the state, then the state has decohered? Is $p_{err}$ the probability of decoherence? Jun 26, 2022 at 1:13