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I'm trying to understand how Y-axis rotations are represented in ZX Calculus. In the paper, wikipedia, everywhere I look, it's as if there is no such thing as Y-axis rotations, only X and Z.

I understand that I can translate a Y-axis rotation to X and Z, for example (using qiskit):

circuit = QuantumCircuit(1)
circuit.ry(pi/2, 0)

is equivalent to

circuit = QuantumCircuit(1)
circuit.rx(pi/2, 0)
circuit.rz(pi/2, 0)
circuit.rx(-pi/2, 0)

and the ZX-diagram looks like this

enter image description here

But I don't understand, if ZX calculus doesn't allow Y-axis rotations, is it stated anywhere? What is the logic. Thanks.

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  • $\begingroup$ Welcome to QC exchange! $\endgroup$
    – R.W
    Aug 25 at 14:54

2 Answers 2

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The ZX calculus doesn't have a Y rotation nodes. All Y rotations must be decomposed into a series of X and Z rotations.

The reason for this asymmetry comes down to the fact that $Y \neq Y^T$ whereas $X=X^T$ and $Z=Z^T$. This is relevant because travelling around a Bell pair transposes gates. For all single qubit gates $U$, is is the case that $U_a(|0_a0_b\rangle + |1_a1_b\rangle) = U_b^T(|0_a0_b\rangle + |1_a1_b\rangle)$. This generalizes to multi-qubit gates. Note that this doesn't work if you use $U_b$ instead of $U_b^T$.

One of the very common things to do in a ZX graph is slide a node around a U turn. That's equivalent to moving a gate to the other side of a Bell pair. If Y nodes were allowed, this U turn would suddenly need to involve changing the node to be its transpose. You would lose the property that the layout of the graph was irrelevant to its function, or you would need a node that had an orientation which could be mirrored. When decomposing Y into X and Z rotations the need to transpose is naturally dealt with by the fact that the ordering of the decomposition gets flipped around as part of going around the U turn.

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First important fact about ZX-calculus: it is universal.

Theorem (informal, van de Wetering, 2020): Every complex matrix of size $2^n \times 2^m$ for some $n$ and $m$ can be represented in the ZX-calculus.

In particular, every map from qubits to qubits can be written as a ZX-diagram. Universality can be showed simply showing that a universal set of quantum gates have ZX-diagram representation.

Theorem (Coecke and Kissinger, 2018): Any $n$-qubit unitary can be constructed out of CNOT gates and phase gates.

For the second part, the $Y$-axis rotation can be written in terms of Z and X consecutive rotations together with a phase. Therefore a $Y$ gate can be written as,

enter image description here

as was written (van de Wetering, 2020), see the cheatsheet A.2.

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