# Why are Alice's states in teleportation superpositions, not “pure” tensor states?

I'm trying to understand the Wikipedia article on quantum teleportation. Since the two particles of Alice are not entangled, all base states $|i\rangle\otimes|j\rangle$ for $i,j\in\{0, 1\}$ of the tensor product are possible, so their common state is a superposition of these four states. The common state of the two entangled particles is a superposition of $|i\rangle\otimes|i\rangle$ for maximally positively correlated particles, because the correlation forbids anti-correlated states like $|0\rangle\otimes|1\rangle$.

Okay so far. Now Alice measures her two particles, and forces them both to collapse into one of the base states $|i\rangle$. Thus, I would expect the common state of her particles to be in one of the following states:

$$\tag{1} |0\rangle\otimes|0\rangle \\ |0\rangle\otimes|1\rangle \\ |1\rangle\otimes|0\rangle \\ |1\rangle\otimes|1\rangle \\$$

Instead, the Wikipedia article says they collapse into one of the Bell states:

$$\tag{2} \frac{1}{\sqrt{2}}\left(|0\rangle\otimes|0\rangle + |1\rangle\otimes|1\rangle\right) \\ \frac{1}{\sqrt{2}}\left(|0\rangle\otimes|0\rangle - |1\rangle\otimes|1\rangle\right) \\ \frac{1}{\sqrt{2}}\left(|0\rangle\otimes|1\rangle + |1\rangle\otimes|0\rangle\right) \\ \frac{1}{\sqrt{2}}\left(|0\rangle\otimes|1\rangle - |1\rangle\otimes|0\rangle\right) \\$$

Why do Alice's particles collapse into $(2)$ instead of into $(1)$?

Alice chooses to measure into a basis for the spin state of the pair Alice has so that the spin state of the third particle changes to have a state that has a relationship to the origin unshared state in a manner completely determined by the result of the pair measurement result.

In the basis you give, each of the four results have a well defined phase between the up and down of the particle.

So they collapse into (2) because Alice chooses to measure them into (2). The measurement that happens is the one you choose to do. When you choose the measurement you choose the device, then you connect that device (and not a different device) and then you let the device and the object interact. Then you read out the result.

Now Alice measures her two particles, and forces them both to collapse into one of the base states $|i\rangle$.

That is not what happens.

• Thanks but I don't get it. When I measure the energy of a particle in the box, the system collapses into one of the eigenstates of the Hamiltonian. I cannot choose the eigenstates by measuring the particle with a specific device. No matter which device I use, the eigenstates are determined by the operator that "is" the observable I'm measuring. So why can Alice "choose to measure in a basis"? – Bass Sep 10 '15 at 21:32
• Oh, are the eigenvalues degenerated, so the eigenspace has dimension $>1$? Is that the reason why the basis can be chosen arbitrarily? – Bass Sep 10 '15 at 21:35
• @BastianTreichler When you choose to measure energy you choose to measure with one device. When you choose to measure position you choose a different device. Similarly, you choose one device to measure (1) and you choose a different device to measure (2). – Timaeus Sep 11 '15 at 1:21
• @BastianTreichler Measurement (1) is simple, but even so, the device you use will depend on whether the two level state is the polarization of a photon, the spin 1/2 component of a (massive) spin 1/2 particle, etc. Measurement (2) is quite a bit harder, I think some of the first measurements could only do some of the four possibilities and you just had to fail half the time. Now they can do all four but it still depends on whether you have photons or electrons or something else. The wikipedia article says you can do it with lasers. – Timaeus Sep 11 '15 at 14:44
• @BastianTreichler sometimes there is an assumptions that you can measure on to any basis, but there is no proof. And there is no recipe from a basis that tells you experimentally how to measure it. Experimentalists are smart people and they have to work hard with their minds before they sit down to build something with their hands. – Timaeus Sep 11 '15 at 14:48

This is a misunderstanding: Alice does not measure the state in the basis $|ij\rangle$ and therefore the particles do not collapse to these states. She measures in the Bell basis, which is exactly basis (2) (and therefore the basis the states collaps to).

Note that if you pick a basis, this is arbitrary. You can always rotate the basis without anything happening that is physically relevant.

• You picked a local basis for the spins that were locally accessible. And this swapped the entanglement of the separated pair onto the local pair. There is no local way to just rotate away a wrong choice of basis for Alice to project onto into a thermodynamically accessible way. Measuring on the Bell basis swaps the entanglement onto the local pair, measuring on the other basis pushes the entanglement up the von Neumann chain. – Timaeus Sep 10 '15 at 15:25