# What does the notation $U(B,\beta) = \prod_{j =1}^n e^{-i \beta \sigma_j^x}$ mean in the context of QAOA?

In the article Quantum Observables for continuous control of the Quantum Approximate Optimization Algorithm via Reinforcement Learning, the following notation is used to describe an Unitary operation over $$n$$ qubits :

$$U(B,\beta) = \prod_{j =1}^n e^{-i \beta \sigma_j^x}$$

with $$\beta$$ a real parameter and $$\sigma^x_j$$ the pauli matrix $$\sigma^x$$ applied to the $$j^{th}$$ qubit. I first thought that this should be read as a matrix product, but since it's applied over $$n$$ qubits, I believe it should be a tensor product, otherwise there's a dimensionality issue. I wonder if using the classical matrix product $$\prod_{j =1}^n$$ notation instead of the $$\otimes_{j =1}^n$$ notation is conventionnal in Quantum computing litterature ? If it is, I would also be interested with some explanation on why it is conventionnal, as I find it confusing.

I think there are two ways that you could denote the same thing. The first is what is done here: $$\prod_{j =1}^n e^{-i \beta \sigma_j^x}$$ The second is $$\bigotimes_{j-1}^ne^{-i \beta \sigma^x},$$ which I imagine is what you're thinking of.
In the first expression, note the subscript on the Pauli matrix. This means that it's an operator over all $$n$$ qubits, $$\sigma^x_j=I^{\otimes(j-1)}\otimes \sigma^x\otimes I^{\otimes(n-j)}.$$ Hence the matrix product does make sense (and because all terms commute, we don't have to worry about ordering).
• Thanks a lot @DaftWullie, it makes sense. Could you please explain to me why all terms commute ? I was trying to answer it myself as the paper contains no notion of order, so I deduced it must commute but I'm unsure why. I found this property over matrix exponential, saying that if X and Y commutes then $e^X$ and $e^Y$ also commutes. $\sigma^x$ and $\sigma^x$ commutes , and multiplying them with a scalar (here $-i \beta$) preserves commutation, but what about $\sigma_j^x$ and, let's say, $\sigma_{j+1}^x$ ? – nathan raynal Apr 3 at 13:23
• Well, the easiest way to think about it in this specific case is that each term acts on different qubits. If I do a unitary on qubit $j$ followed by a unitary on qubit $j+1$, that should be the same as a unitary on qubit $j+1$ followed by a unitary on qubit $j$. Those two qubits could be light years apart (where time ordering of events might not even make sense)! – DaftWullie Apr 3 at 15:03