New answers tagged tensor-product
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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