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Suppose I have a unitary operator acting on a state $|\psi\rangle$, such as:

$$\big (\sqrt A\sqrt B\sqrt C\sqrt D\big)|\psi\rangle.$$

Conventionally, $\sqrt D$ is performed first, then $\sqrt C$, then $\sqrt B$, and finally $\sqrt A$, i.e., the gates run in series, one right after the other, in reverse order as written.

However, suppose I know that each of $\sqrt A, \sqrt B,\sqrt C,$ and $\sqrt D$ commute with each other - indeed, $\sqrt C\sqrt D$ act on different qubits of $|\psi\rangle$ than $\sqrt A,\sqrt B$ do.

I'd prefer to have $\sqrt A$ and $\sqrt B$ run in series, reusing ancillae, and $\sqrt C$ and $\sqrt D$ run in series, also reusing ancillae, but otherwise $(\sqrt A\sqrt B)$ can run in parallel with $(\sqrt C\sqrt D)$, and can use different ancillae.

What is an intuitive notation for such parallel circuits? Of course, spelling it all out with a circuit diagram would be clear, but is there a succinct way to represent the above?

I'd propose:

$$\big(\sqrt A\sqrt B\big)\parallel\big(\sqrt C\sqrt D\big)|\psi\rangle,$$

but could this lead to confusion, or has anyone seen any comparable notation?

Mark Spinelli
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    Isn't it simply the tensor product notation that you are looking for? $A \otimes B$ means $A$ and $B$ are "ran in parallel on different qubits". – Marco Fellous-Asiani Feb 28 '22 at 17:40
  • Fair and good point, but, in general two operators could act on the same qubits of $|\psi\rangle$, but they still could commute. If $A$ and $B$ both act on qubits one and two of $|\psi\rangle$ but otherwise commute, then I can envision $A\parallel B$ as meaning $A$ acting on qubit one, $B$ acting on qubit two, $A$ acting on qubit one, and $B$ acting on qubit 2. – Mark Spinelli Feb 28 '22 at 17:49
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    @MarkS There is a notation for a symmetric tensor product: $A\odot B$. The issue is that this is also commonly used for the Hadamard product, but I think if you explicitly state what notation you are using, this would still be clear. – Tyberius Feb 28 '22 at 22:38
  • @MarkS "two operators could act on the same qubits of $|\psi\rangle$, but they still could commute" the order of application won't matter, but such operators will never run in parallel unless you can combine their action in some way. For example, $ZS = SZ$, but they can be compiled as a single $z$-rotation. – chrysaor4 Mar 01 '22 at 09:28
  • @chrysaor4, sure, that’s also reasonable. But in some examples, especially with large operators, you could “combine their action” by interlacing/dovetailing them- do part of $A$ then part of $B$ then part of $A$ again then part of $B$. – Mark Spinelli Mar 01 '22 at 13:41

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