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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – roshoka
    Jun 26, 2022 at 1:13

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