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).

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  • $\begingroup$ 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, $\sigma_{j+1}^x$ ? $\endgroup$ – nathan raynal Apr 3 at 13:23
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    $\begingroup$ 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)! $\endgroup$ – DaftWullie Apr 3 at 15:03

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