I am reading about lattice surgery from this paper. I am interested in the merge operation which takes 2 qubits to 1 qubit. I want to understand the logical-level Kraus operation that the merge does.

For instance, the action of the rough merge between two states $\vert\psi\rangle = \alpha\vert 0\rangle + \beta\vert 1 \rangle$ and $\vert\phi\rangle$ is given by

$$|\psi\rangle \otimes |\phi\rangle\rightarrow \alpha|\phi\rangle+(-1)^M \beta X|\phi\rangle$$

where $M$ is the merge measurement result. However, this only tells me what happens to pure product states. What is this in terms of the Kraus operators of the channel that merges the two qubits?


2 Answers 2


CP maps for lattice surgery merge with post-selection

Post-selecting on $M_{ZZ}=0$, lattice surgery merge is described by the completely positive linear map with Kraus representation \begin{align} \mathcal{E}_0:L(\mathbb{C}^4)\to L(\mathbb{C}^2):\rho\mapsto\mathcal{E}_0(\rho)=K_0\rho K_0^\dagger\tag1 \end{align} where $K_0=|0\rangle\langle 00|+|1\rangle\langle 11|$. Note two implications of the form of this Kraus operator: it preserves coherence between $|00\rangle$ and $|11\rangle$ and it sends the pre-merge $XX$ observable to the post-merge $X$ observable$^1$. Similarly, post-selecting on $M_{ZZ}=1$, lattice surgery merge is described by the completely positive linear map with Kraus representation \begin{align} \mathcal{E}_1:L(\mathbb{C}^4)\to L(\mathbb{C}^2):\rho\mapsto\mathcal{E}_1(\rho)=K_1\rho K_1^\dagger\tag2 \end{align} where$^2$ $K_1=|0\rangle\langle 01|+|1\rangle\langle 10|$. Once again, quantum coherence between $|01\rangle$ and $|10\rangle$ is preserved and pre-merge $XX$ is sent to post-merge $X$.

CPTP maps for lattice surgery merge without post-selection

Note that the two maps above are not trace-preserving. This is expected, because they describe processes that generally occur with probability less than one due to post-selection. There are two ways we can obtain deterministic processes and corresponding CPTP maps. First, we can forget $M_{ZZ}$ which leads to \begin{align} \mathcal{E}_*:L(\mathbb{C}^4)\to L(\mathbb{C}^2):\rho\mapsto\mathcal{E}_*(\rho)=\mathcal{E}_0(\rho)+\mathcal{E}_1(\rho).\tag3 \end{align} Second, we can introduce a classical bit for recording the measurement outcome $M_{ZZ}$. We can model the bit as a completely decohered qubit which leads to \begin{align} \mathcal{E}_L:L(\mathbb{C}^4)\to L(\mathbb{C}^4):\rho\mapsto\mathcal{E}_L(\rho)=\mathcal{E}_0(\rho)\otimes|0\rangle\langle 0|+\mathcal{E}_1(\rho)\otimes|1\rangle\langle 1|\tag4 \end{align} where the qubit adjoined on the right records the measurement outcome $M_{ZZ}$. All of these maps preserve coherence within the equal-parity subspaces, but destroy coherence across the subspaces.

$^1$ Consider the action of $K_0$ on the shared eigenbasis of $XX$ and $ZZ$ and note that $K_0(|00\rangle\pm|11\rangle)=|\pm\rangle$.
$^2$ The choice of whether to send $01$ or $10$ to $0$ is arbitrary, but the two options yield different CP maps.


Lattice merge isn't faithfully captured by a CPTP map description (and therefore a Kraus representation). Tracking $M$, the measurement outcome, conditions what logical correction operation you will perform (or how you will track the stabilizers in software). You can write a Kraus description in the case where you are ignorant to $M$, but this isn't the same. Lattice merge is essentially a projection onto the $+ZZ$ or $-ZZ$ eigenspace of the two codepatches, but the logical dimensionality is halved in the sense that it take $|00\rangle \mapsto |0\rangle$.

In gist, in the abstraction hierarchy, lattice merge behavior is captured at the quantum instrument level, not as a channel.

  • $\begingroup$ How does one represent the logical operation that halves the dimensionality? It includes $\vert 00\rangle\rightarrow\vert 0\rangle$ but also has well-defined action on any two qubit state. $\endgroup$
    – qubit
    Commented May 13 at 17:17
  • $\begingroup$ A linear map doesn't have to be CPTP to admit a Kraus representation. Every CP map has Kraus representations. In particular, lattice surgery has Kraus representations. $\endgroup$ Commented May 14 at 2:55

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