# Erasure errors in quantum error correction

Consider an $$[n, k, d]$$ classical code. This code can correct up to $$d-1$$ erasures. For example, if we have the code that maps $$0\rightarrow 00$$ and $$1\rightarrow 11$$, this code has distance $$2$$. Indeed, if we observe a symbol $$0X$$ where $$X$$ is an erased bit, then the decoder will output $$0$$.

In the quantum case, this seems like a problem. A typical encoding uses CNOT gates to entangle the logical qubit with redundant qubits. Let's assume a two-qubit quantum code although adding more qubits doesn't really help. We have that

$$\sqrt{\alpha}\vert 0\rangle + \sqrt{1-\alpha}\vert 1\rangle \rightarrow \sqrt{\alpha}\vert 00\rangle + \sqrt{1-\alpha}\vert 11\rangle$$

An erasure error should be thought of as partial tracing the second register. This leaves us with the state

$$\alpha\vert 0\rangle\langle 0\vert + (1-\alpha)\vert 1\rangle\langle 1\vert$$

This state no longer has coherence. If $$\alpha = \frac{1}{2}$$, we have gone from a $$\vert +\rangle$$ state to a maximally mixed state after the erasure error.

So how do quantum error correction protocols deal with this?

EDIT based on comment:

I see that the above code doesn't work, nor does any repetition code alone. The general principle is that an error on a physical qubit should be thought of as a physical channel $$\mathcal{E}$$ with Kraus operators $$\{E_i\}$$. In the case of qubits, one then writes the Kraus operators in the Pauli basis (see Nielsen and Chuang (10.14))

$$E_i=e_{i 0} I+e_{i 1} X+e_{i 2} Z+e_{i 3} XZ$$

For the partial trace, the Kraus operators are $$E_i=\langle i|$$ where the bra is on the register we are tracing out. It's not clear to me how the Pauli decomposition works since the dimension of the Kraus operator is $$2\times 1$$ while the linear combination of Pauli matrices gives us a $$2\times 2$$ matrix.

My question is still - how are erasures dealt with in QEC?

• The problem you're having is the distance of your example code: it is only distance 1 with respect to $Z$ errors. Thus, as you've found, it cannot cope with any erasure errors. Commented Dec 19, 2023 at 11:57
• Hey, thanks! That's a good point - I should not have used a repetition code. But even if I didn't, I'm not sure how the error correction formalism deals with erasures. I've updated the question to reflect this. Thanks! Commented Dec 19, 2023 at 13:34

You should think of error correction as the process of measuring syndromes (i.e. determining the $$\pm 1$$ values of stabilisers, which is the case I'll exclusively focus on) and then following a lookup table to see what correction to make.
• Thank you! Just to clarify one point: I should model an erasure error as a depolarizing channel (with probability $p = \frac{1}{4}$ for the case of no error, $X$ error, $Y$ error and $Z$ error) acting on that qubit and not a partial trace channel? This allows me to decompose the Kraus operators of that channel in the Pauli basis which I didn't know how to do for the partial trace channel. Is that right? Commented Dec 19, 2023 at 17:12
• Thanks, and sorry if this is a dumb question or I am missing something trivial but with the partial trace (or partial trace and replace with the 0 state) operation, it seems like the Kraus operators dimensions aren't $2\times 2$. How does one write that in the Pauli basis? My understanding is that one must first express the Kraus operators of the error-causing channel in the Pauli basis, then do the syndrome measurement which detects if there are Pauli errors. Commented Dec 20, 2023 at 12:09
• @user1936752 No that's not correct. If you followed that protocol assuming you don't know where the errors happen, it would be true. But the protocol really does deal with $2t$ erasure errors. Look at the smallest distance 2 quantum error correcting code, $[[4,1,2]]$. It has 3 stabilisers, so there are $2^3=8$ different syndromes. If you work it through, the errors $I$, $X$, $Y$ and $Z$ on a single (specified) qubit all have distinct syndromes, so all can be corrected, meaning you can recover from a single loss. Commented Apr 3 at 7:28