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Define $o = A \otimes B$. Compute the results of $o^{\otimes N}$. We have \begin{align} o^{\otimes N} &= (A \otimes B) \otimes (A \otimes B) \otimes ... \otimes (A \otimes B). \end{align} Can we groups the terms A and terms B and have \begin{align} o^{\otimes N} &= (A \otimes A \otimes ... \otimes A) \otimes (B \otimes B \otimes ... \otimes B)? \end{align}

I think this is wrong, but the eq. (2) in this question used this.

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Viewed rigorously, it is wrong. Consider a simple example $o = X\otimes Y, A = X, B = Y$. $o^3 = X\otimes Y\otimes X\otimes Y\otimes X\otimes Y \neq X\otimes X\otimes X\otimes Y\otimes Y\otimes Y$

It is often assumed without explicitly stating that when the operators are shuffled, the qubits are correspondingly permuted to reflect this reordering. When is this done? Often when dealing with algorithms that utilize entanglement, like QKD and teleportation that go from single bits/bell states and generalize to binary strings or sets of Bell pairs to represent binary strings, it is convenient to reorder and denote via bit strings.

$|\psi_{00}\rangle^{\otimes n} = \left(\frac{|00\rangle + |11\rangle}{\sqrt{2}}\right)^{\otimes n} = \left(\frac{|00\rangle + |11\rangle}{\sqrt{2}}\right)\otimes \left(\frac{|00\rangle + |11\rangle}{\sqrt{2}}\right)\otimes \cdots \otimes \left(\frac{|00\rangle + |11\rangle}{\sqrt{2}}\right)$

When two parties are sharing this entanglement, say Alexa and Bard, it is convenient to take the first qubits of each pair that Alexa holds to one side and the rest to the right and write as

$|\psi_{00}\rangle^{\otimes n} \equiv \frac{1}{2^{n/2}}\left(|\underbrace{0\cdots 0}_{n}\rangle|\underbrace{0\cdots 0}_{n}\rangle + |0\cdots 1\rangle|0\cdots 1\rangle + \dots |1\cdots 1\rangle|1\cdots 1\rangle\right)$

This way, it is easy to see if Alexa measures a bit string and obtains $\mathbf{b}$, it can be easily seen that Bard will also have $\mathbf{b}$ on his qubits.

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    $\begingroup$ This is correct and I don't think needs any additions, but I want to add one thing. When we have a tensor product, we are assuming there are different Hilbert spaces being acted upon. Those can have labels, so we might write $X_1\otimes Y_2\otimes X_3\otimes Y_4$, using subscripts to denote on which Hilbert space (or on which qubit) each operator acts. If we keep these labels, then the operator is equivalent to $X_1\otimes X_3\otimes Y_2\otimes Y_4$. These labels are often cumbersome and thus neglected, but it can help in these examples like saying Hilbert spaces 1 and 3 belong to Alexa $\endgroup$ Commented May 28, 2023 at 15:29
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    $\begingroup$ In my time, we used to call them Alice and Bob. Youth these days... $\endgroup$
    – Tristan Nemoz
    Commented May 28, 2023 at 21:49
  • $\begingroup$ About time we leave them alone and give them the privacy they've been trying to establish for years. $\endgroup$ Commented May 28, 2023 at 23:12

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