# Quantum error correction using bit-flip code for the amplitude damping channel

I do not understand the error correction process that uses quantum codes for amplitude damping channel. I will take three bit-flip code for example.

The logical state of a three bit-flip code is $$|0\rangle_L=|000\rangle$$,$$|1\rangle_L=|111\rangle$$ with stabilizers $$Z_1Z_2, Z_2Z_3$$.

The amplitude damping channel on three qubits $$\mathcal{E}^{\otimes 3}$$ has Kraus operators $$\{E_0^{\otimes 3},E_0E_1^{\otimes 2},E_0E_1E_0,E_0^{\otimes 2}E_1,E_1^{\otimes 3},E_1E_0^{\otimes 2},E_1E_0E_1,E_1^{\otimes 2}E_0\}$$, where

$$\begin{matrix} E_0=\begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-r} \end{pmatrix}, E_1=\begin{pmatrix} 1 & \sqrt{r} \\ 0 & 0 \end{pmatrix}. \end{matrix}$$

Suppose I have a initial state $$|\psi\rangle=|111\rangle$$ and error $$E_1E_0^2$$ will transform $$|\psi\rangle$$ to $$\sqrt{r}(1-r)|011\rangle$$. By measuring the stabilizers $$Z_1Z_2, Z_2Z_3$$, I can find the first qubit is flipped. So I use $$X_1$$ to recover.

My question is that

1. the state could only be corrected back to $$\sqrt{r}(1-r)|111\rangle$$ after the recovery, can $$\sqrt{r}(1-r)$$ be recognized as the fidelity?

2. if I write the initial state in density matrix $$\rho=|111\rangle\langle 111|$$ and \begin{aligned} \mathcal{E}^{\otimes 3}(\rho)&=E_0^3 \rho (E_0^3)^\dagger+E_1E_0^2 \rho (E_1E_0^2)^\dagger+\cdots \\&= r^3|000\rangle\langle 000|+r(1-r)^2|011\rangle\langle 011|+\cdots. \end{aligned} How to perform projective measurements to obtain error syndromes and correct?

3. $$[5,1,3]$$ code can correct one arbitrary error but amplitude damping errors will occur on all $$5$$ qubits, does that mean $$[5,1,3]$$ code will have bad performance for amplitude damping errors?

Before starting, I should probably emphasise that, although useful for the practice of working through the maths of quantum error correction on a relatively simple case, amplitude damping combined with the repetition code is a really bad thing to be thinking about. This is because, if there's an error, and they you apply a syndrome measurement, so that you detect there's one bit that is different from the other two, then you know that your initial state was $$|111\rangle$$ and not $$|000\rangle$$ because under amplitude damping noise, $$|000\rangle$$ never changes. This means that your syndrome measurement, if it detects an error, constitutes a measurement on your encoded qubit. It destroys any of the coherence you were trying to protect!

1. Because of the issue described above, this question doesn't make the sort of sense you'd like it to. However, remember that there are measurements going on. When you measure, you get a probability of getting a given outcome, and the output state is renormalised. At this point, it is usually the probability of success that is evaluated (sum of the probabilities for measurement outcomes which are successful in correcting the error) rather than a fidelity.

2. How you perform the measurement has nothing to do with the state. Perhaps you feel unfamiliar with the formalism because it's applied to a density matrix rather than a pure state. Still, it's a sequence of applying unitaries ($$\rho\mapsto U\rho U^\dagger$$) and projections (probability $$p_i=\text{Tr}(P_i\rho)$$ and output state $$P_i\rho P_i/p_i$$).

3. It's not the issue that the errors could apply to all 5 qubits that's the problem. You should still think about these errors as an independent probability $$r$$ of the amplitude damping error acting on each qubit (if it's in the $$|1\rangle$$ state) and $$(1-r)$$ of not happening. So, there's a case of no error with probability $$(1-r)^5$$ and 5 cases of just one error happening, each of probability $$r(1-r)^4$$. All of these are cases that can be corrected by the 5-qubit code. So, the idea is that provided $$(1-r)^5+5r(1-r)^4>(1-r)$$, it is less likely that there's a error on the logically encoded qubit than there would have been on a single qubit with no encoding. That makes it a worthwhile procedure.

• Thank you so much. I had this question when I was reading this paper (journals.aps.org/pra/pdf/10.1103/PhysRevA.75.012338). In page $5$, figure.2 illustrates that given an initial state $\rho=1/2|0\rangle_{LL}\langle 0|+1/2|1\rangle_{LL}\langle 1|$ and the amplitude damping channel $\mathcal{E}^{\otimes 5}$, they can obtain the entanglement fidelity when using $[5,1,3]$ code to encode and projectors to recover (the blue line named QEC Recovery). I do not understand how this fidelity can be calculated if not knowing the way correcting amplitude damping errors with quantum codes? – Jacey Li Dec 13 '19 at 14:49
• If it were me, I’d take the initial density matrix before any error, $\rho$, and the one after error correction (a convex combination of the different outcomes following syndrome extraction and correction), $\rho_f$, and plug those into the standard fidelity formula, which is essentially $Tr(\rho\rho_f)$ assuming the input was pure. – DaftWullie Dec 13 '19 at 14:54

Firstly, I'm presuming that when you write $$E_0^3$$ it corresponds to $$E_0^{q_1} \otimes E_0^{q_2} \otimes E_0^{q_3}$$.

1. In question 2. you've written out how the Superoperator of the Kraus operators acts on the density matrix, In 1. we should also assume that this is the case, and the pre-factor of $$\sqrt{r}(1-r)$$ is a probability amplitude, so measuring with the stabilisers will collapse the state into $$|011\rangle$$ with probability $$\sqrt{r}(1-r)$$. So the state recovered after applying the corrective measurement has fidelity = 1.

2. To perform a projective measurement, we just need to write out the measurement operator, $$M = |\psi\rangle\langle\psi|$$ where $$|\psi\rangle\langle\psi|$$ is the measurement outcome, e.g. $$|\psi\rangle = |000\rangle$$ and the expectation value for this measurement is: $$\langle\rho\rangle = \textrm{Trace}(M\rho)$$. If the measurement outcome $$|000\rangle$$ is obtained, the state after measurement is $$\frac{M\rho M}{\textbf{Trace}(M\rho)}$$

3. All qubits will be subject to amplitude damping errors, and within a certain time can be expected to have relaxed into the ground state. This time is called $$T_1$$ time, and the trick is to do the stabiliser measurements before this time has elapsed. The stabiliser measurements are not performed at the end of the computation, but during it, and they project the logical qubits onto the `codespace' (like in Q1. where the qubits after measurement are in the $$|011\rangle$$ state).

• Thank you so much. I am still confused about the amplitude. If the initial quantum state is $|\psi\rangle=\alpha|00\rangle+\beta|11\rangle$. Then it could be changed to $|\psi\rangle'=\alpha'|10\rangle+\beta'|01\rangle$ by amplitude damping errors. I believe measuring stabilizers will obtain error syndromes that indicate the first qubit is flipped. So we can apply $X_1$ to $|\psi\rangle'$ and the quantum state becomes $|\psi\rangle'=\alpha'|00\rangle+\beta'|11\rangle$. But I found that the amplitudes are not corrected. – Jacey Li Dec 12 '19 at 10:20
• The fidelity $F(|\psi\rangle, |\psi\rangle')=\sqrt{\alpha^2+\alpha'^2+\beta^2\beta'^2+2\alpha\alpha'\beta\beta'}$ does not seem to be $1$. For the third question, if all qubits are subject to amplitude damping errors, then stabilizer measurements could not return the correct error syndromes since $[5,1,3]$ code only corrects one error. Then recovery operators conditioned on error syndromes cannot correct errors correctly. – Jacey Li Dec 12 '19 at 10:40