# Quantum Teleportation: Faster Than Light Transfer Of Information?

Even thought I work in Molecular Biology, I nominally sit in a computer science department. As a consequence, those of us in our group occasionally have the pretence (rightly or wrongly) of being computer scientists and debating questions more fundamental to that field …

The issue came up the other week as to whether quantum entanglement, whereby under quantum mechanics two particles are joined in such a way that certain actions upon one will instantly affect the other over great distances of space, would allow you to transmit information faster than light.

The short answer is no. I’m going to give a quick, qualitative answer before I jump into a bit of maths to make the reasoning more rigorous. As I said, entanglement is a means to join two particles, like photons or electrons, together so that they are ‘linked’. What do I mean by ‘linked’? You know how twins are supposed to be able to ‘know’ when something good/bad is happening to the other, even when they’re miles apart? It’s a bit like that, but not just an old wives tale.

Entanglement allows you to take particles in an undetermined state (in QM, a system can be in a variety of states and only settle down upon one when a measurement is made upon it) and smack them together in a way that, whilst preserving this fuzzy, undetermined state business, adds a further constraint that the result of a measurement of one particle completely determines the state of the other. So up until a measurement is made, who knows what state they’re in – they behave completely randomly. But as soon as one particle has a measurement performed upon it, the other instantly snaps to a well defined state that is completely correlated with its partner – no matter how far apart they are in space. Weird huh? Einstein, who never truely accepted QM as a complete theory of nature, dreamed the situation up as an example of how poor QM was at describing things. But thirty years later, independent experiments confirmed that the effect really does happen!

So if this change between the two particles is instant, why can’t we use it to transmit information by giving one particle of an entangled quibit (as these two particle systems are called) to Alice here on Earth and the other to a guy called Bob on a planet a million, billion miles away? Crudely put, it’s because Bob has no idea what the state should have looked like. Sure, he can measure his part of the system and reliably extract the collapsed state value, but this is no different to if they had made an observation upon the uncollapsed state before Alice performed a measurement. So no information gain occurs.

BUT, there is a way to transmit information via a different means – quantum teleportation. Again, this can not occur fast than light. This time round, Alice takes her half of the quibit and entangles it further with another quantum system that she wishes to send Bob. This affects Bob’s particle, so that it contains some information about this new system. Alice then performs a particular experiment, such that Bob’s particle collapses a bit, but not totally. Alice then sends Bob two pieces of information about her result, which allows Bob to manufacture an identical re-creation of Alice’s quantum system. Encoded in this ‘teleported’ system could be a small amount of information. Alice’s two pieces of additional information need to travel along a ‘classical’ route, which is of course slower than light speed (particularly if it’s by second class post). However, as these two bits of information don’t contain the description of the system, this technique of transmitting information is completely secure against eavesdroppers. It is this fact that it is impossible for a malicious third party to access the information being sent by Alice to Bob that makes this phenomenum so interesting and important; although it turns out that no extra information can actually be sent over the channel than you would be able to send classically, you can send it SECURELY.

For the sake of completeness, I’ve included a little record of the treatment of the effect in mathematical physics. Notice that this is a much smaller section than my qualitative explanation above, but I promise you it has no further details. We use the maths as a notation for scaffolding our thinking …

We take two non-interacting quantum systems of two states $$A$$ and $$B$$:

\begin{aligned} |\psi\rangle_A = \sum_{i}\;\; c^A_{i}|i\rangle_A \\ |\psi\rangle_B = \sum_{j}\;\; c^B_{j}|i\rangle_B \end{aligned}

And we ‘mix’  or ‘entangle’ them, such that we can describe the system by:

$|\psi\rangle_{AB} = \sum_{i,j}\; c_{ij} |i\rangle_A \otimes |j\rangle_B$

Where it is not possible to seperate $$c_{ij}$$ as $$c_{ij} \neq c^A_ic^B_j$$.

There are four possible maximally entangled states that we could construct:

$|\Phi^+\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_{B} + |1\rangle_A \otimes |1\rangle_{B})$

$|\Phi^-\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_{B} - |1\rangle_A \otimes |1\rangle_{B})$

$|\Psi^+\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |1\rangle_{B} + |1\rangle_A \otimes |0\rangle_{B})$

$|\Psi^-\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |1\rangle_{B} - |1\rangle_A \otimes |0\rangle_{B})$

We shall use $$|\Phi^+\rangle_{AB}$$ and hand one half of the quibit to Alice, the other to Bob. The principle is the same with any of the others. Say that Alice has a further two state system $$C$$, which she wishes to teleport to Bob:

$|\psi\rangle_C = \alpha |0\rangle_C + \beta|1\rangle_C$

The total system wavefunction is:

$|\Phi^+\rangle_{AB} \otimes |\psi\rangle_C = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_B + |1\rangle_A \otimes |1\rangle_B)\otimes (\alpha |0\rangle_C + \beta|1\rangle_C)$

If we note that:

$|0\rangle \otimes |0\rangle = \frac{1}{\sqrt{2}} (|\Phi^+\rangle + |\Phi^-\rangle)$

$|0\rangle \otimes |1\rangle = \frac{1}{\sqrt{2}} (|\Psi^+\rangle + |\Psi^-\rangle)$

$|1\rangle \otimes |0\rangle = \frac{1}{\sqrt{2}} (|\Psi^+\rangle - |\Psi^-\rangle)$

$|1\rangle \otimes |1\rangle = \frac{1}{\sqrt{2}} (|\Phi^+\rangle - |\Phi^-\rangle)$

We can rewrite the wavefunction of the system as follows (note that this is completely equivalent to the above description of the the three particle system):

$\frac{1}{2} (\ |\Phi^+\rangle_{AC} \otimes (\alpha |0\rangle_B + \beta|1\rangle_B)\ +\ |\Phi^-\rangle_{AC} \otimes (\alpha |0\rangle_B - \beta|1\rangle_B)\ +\ |\Psi^+\rangle_{AC} \otimes (\beta |0\rangle_B + \alpha|1\rangle_B)\ + |\Psi^-\rangle_{AC} \otimes (\beta |0\rangle_B - \alpha|1\rangle_B)\ )$

Alice now performs an observation on her $$AC$$ system, collapsing the total system to one of the following four scenarios:

$|\Phi^+\rangle_{AC} \otimes (\alpha |0\rangle_B + \beta|1\rangle_B)$

$|\Phi^-\rangle_{AC} \otimes (\alpha |0\rangle_B - \beta|1\rangle_B)$

$|\Psi^+\rangle_{AC} \otimes (\beta |0\rangle_B + \alpha|1\rangle_B)$

$|\Psi^-\rangle_{AC} \otimes (\beta |0\rangle_B - \alpha|1\rangle_B)$

Notice that $$B$$ looks remarkably similar to $$C$$. All Alice need do now it transmit which of these scenarios has occurred, which she knows from information gleaned from her measurements of $$A$$ and $$C$$, to Bob (things like the system angular momentum or the projection of said angular momentum on a given axis). This constitutes two bits (or yes/no questions) of information, which must be sent via a classical means. Upon receiving these two bits, Bob performs an operation upon his system to completely recreate system $$C$$, only at his end. These operations vary from rotating about $$90^{\circ}$$ to turning it upside down or even doing nothing – the exact details aren’t that important. All that matters is that at the end, Bob has a system:

$|\psi_{B}\rangle = \alpha|0\rangle_B) + \beta |1\rangle_B$

Which is a recreation of Alice’s quantum system $$C$$.