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4

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 $n$ ...

3

This depends on the context in which you're using the operators. You're talking about multiplying them, so I guess you're thinking of, for example, unitaries (and other circuit model elements). In this case, not only are terms $A\otimes I$ local operators, but $A\otimes B$ is also a local operator. For example, on a quantum circuit, two Hadamard gates ...

1

There are two inequivalent definitions of "local operator" used in quantum information theory. The first definition is used in the context of communication over a classical channel (e.g. LOCC). In this context, you have a fixed partition of the complete Hilbert space into a tensor product of $k$ different subsystems, and the subsystems are assumed to be so ...

1

By taking into account this representation of the CNOT gate: $$CNOT = | 0 \rangle \langle 0 | \otimes I + | 1 \rangle \langle 1 | \otimes X$$ We can write: $$CNOT(1, 3) = | 0 \rangle \langle 0 | \otimes I \otimes I + | 1 \rangle \langle 1 | \otimes I \otimes X$$ $$CNOT(1, 2) = | 0 \rangle \langle 0 | \otimes I \otimes I + | 1 \rangle \langle 1 | \otimes ... 2 Let H_1,H_2 be two Hilbert spaces and for unit vectors u_1,v_1 \in H_1 and u_2,v_2 \in H_2 we have$$ u_1 \otimes u_2 = v_1 \otimes v_2 \in H_1 \otimes H_2. $$Then it must be$$ v_1 = \lambda u_1, ~~ v_2 = \frac{1}{\lambda} u_2, $$for some \lambda \in \mathbb{C}, |\lambda|=1. To see why consider a scalar product$$ 1 = (u_1\otimes u_2, u_1\...

3

Let's say I know that $$U_1|\Psi_1\rangle\otimes U_2|\Psi_2\rangle=e^{i\theta}|\Psi_1\rangle\otimes|\Psi_2\rangle.$$ Now, let's imagine that $U_1|\Psi_1\rangle=|\phi_1\rangle$ and $U_2|\Psi_2\rangle=|\phi_2\rangle$. So, $$|\phi_1\rangle\otimes |\phi_2\rangle=e^{i\theta}|\Psi_1\rangle\otimes|\Psi_2\rangle.$$ Now, I would just read off your desired ...

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