# Rearrangement of qubits in Quantum Teleportation Protocol

I have been reading Quantum Teleportation (Pg. 27) from Nielsen and Chuang and noticed that after the Hadamard operation, the state obtained was re-written by regrouping the terms to obtain the following states:

This state may be re-written in the following way, simply by regrouping terms:

\begin{align} |\phi_2\rangle=\frac{1}{2}\big[|00\rangle(\alpha|0\rangle+\beta|1\rangle)+|01\rangle(\alpha|1\rangle+\beta|0\rangle)\\ +|10\rangle(\alpha|0\rangle-\beta|1\rangle)+|11\rangle(\alpha|1-\beta|0\rangle)\big]\end{align}

but I am uncomfortable by this regrouping. Mathematical regrouping doesn't mean the qubits themselves get regrouped right? Moreover, how do we know that Alice is not measuring the unknown state (state to be teleported) after she measures her part of the EPR pair?

• It may be more clear which states belong to which system if you write the states with tensor products $\otimes$ rather than the shorthand grouping. The rewriting uses the fact that the tensor product is linear in each argument. But yes, a mathematical rewriting doesn't correspond to an actual transformation in the lab. If two states are equal then they both correspond to the same state in the lab. May 20, 2020 at 7:43

As @Rammus points out this is just shorthand for taking the tensor product, rather than algebraically multiplying out as you may first assume. If we take the first term of $$\phi_2$$ and expand it out more explicitly with subscripts $$A$$ for Alice and $$B$$ for Bob:

$$|0_A0_A\rangle(\alpha|0_B\rangle + \beta|1_B\rangle) = |0_A0_A\rangle \otimes (\alpha|0_B\rangle + \beta|1_B\rangle) = (|0_A0_A\rangle \otimes \alpha|0_B\rangle) + (|0_A0_A\rangle \otimes \beta|1_B\rangle) = \alpha|0_A0_A0_B\rangle + \beta|0_A0_A1_B\rangle$$.

If we start with the original $$|\phi_2\rangle$$ given by e.q. 1.31

$$|\phi_2\rangle = \frac{1}{2}[\alpha(|0_A\rangle + |1_A\rangle)(|0_A0_B\rangle + |1_A1_B\rangle)+ \beta(|0_A\rangle + |1_A\rangle)(|1_A0_B\rangle + |0_A1_B\rangle)]$$

if we expand this out keeping track of the subscripts and remembering to keep ordering from left to right (as the tensor product is noncommutive, this means $$A \otimes B \neq B \otimes A$$), you can see that we will get the same terms as expanding out the rewritten form with the same ordering (grouping) of qubits for both Alice and Bob.

The second part, Alice is measuring two qubits, one which is part of her EPR pair and the other that WAS $$|\phi\rangle$$ however during the teleportation protocol she acts on this state and mixes this with her part of the EPR pair so we no longer have the original pure state $$|\phi\rangle$$. We know she can't measure $$|\phi\rangle$$ otherwise this would violate the no-cloning theorem as Alice and Bob would then both have a quibit in state $$|\phi\rangle$$ at the end of teleportation!

Daftwullie gives a great explanation here Quantum teleportation: second classical bit for removing entanglement?