###### 18.08.2016difficulty level - QQQ

## Quantum Teleportation Explained

*by Jérémy Ribeiro*

Have you ever dreamt about teleportation? You wonder if it is possible, or if we can use it to travel faster than light, or at least to communicate instantaneously. Then you are at the good place. Here I will explain what quantum teleportation is. Behind this very attractive name that reminds us of science fiction, a communication protocol is hidden which uses the mysterious quantum mechanics.

## Why do I talk about quantum teleportation ?

I wanted to write about this protocol because we hear a lot about it and a lot of information and explanations can be found about it. But sometimes those are partially wrong, or are a complete nonsense. For example we can read that quantum teleportation is an instantaneous transfer of information at a distance which respects special relativity… Well this is a contradiction.

These misconceptions of the protocol are not surprising since it relies on one of the most ununderstandable and less well understood phenomenon of quantum mechanics: the famous *entanglement*.

That’s why I will try to clarify what this notorious quantum teleportation actually is. For that I will have to introduce a little bit of quantum mechanics. Therefore there will be some mathematical expressions, but I’ll guide you through it to make you understand what is going on. It shouldn’t be too difficult since you should have already seen all the mathematical concepts in high school (vectors), and I think it is worth it and permits to really understand what quantum teleportation is.

*So what is that quantum teleportation ?*

The first thing to understand is that we only teleport a quantum *state* (we will call it $latex Phi$), and not a particle nor any other kind of matter. Only information is transmitted from one place to another. So somehow we will scan the physical system in the first place (which will destroy the state), send the information (by phone, internet or any other way of communication) to an other place and there reconstruct the state. But to do that there are some obstacles. The main obstacle the “scanning phase” is not trivial since we cannot (by the law of quantum mechanics) get all the

information on a state by measuring it, as I will explain later, but only partial information. To overcome that we will need to use a correlated state which will somehow compensate the lack of information.

But to make things simpler I will begin with a classical analogy to quantum teleportation (well, a non-quantum one…), where we will see how correlations can compensate a lack of information.

## ”Classical teleportation”

We can, in a certain way, think about a classical equivalent of the teleportation. We will deal here with our two favorite protagonists Alice and Bob. Let’s say that they each have a small electronic device with a screen and a big button in the middle (you can see the device in figure 1).

When they press the button, a number will appear on the screen. This number will be either a 0 or a 1. The value of this number is stored in the device, but neither Alice nor Bob know their value before they press the button, and there is no way to know it without pressing the button.

The goal for Alice will be to send the value stored in her device without pressing the button, which means Alice has to send the number without ever knowing it. Let’s call Alice’s device “a”.

*But why can’t she just press the button ? It would be simpler…*

Well… because… I… I’ve defined the rules of the game like this ! More precisely this rule is here in order to imitate one of the rules of quantum mechanics which is basically that we can only get partial information on a state when we measure it. But I’ll explain that in more detail later.

However, Alice will be allowed to perform some operations on the number stored in the device ‘a’. These could be addition, multiplication etc… (but these operations are done “modulo 2”; I’ll explain that later).

*Ok ok.. So how they do that ?*

So let’s divide that into four steps:

**First step**: Alice will have access to another device, let’s call it $latex A$, which is such that the value of $latex A$ is equal to the value of $latex B$ which is Bob’s device. That means that if Alice and

Bob press the button of respectively $latex A$ and $latex B$ the same number will be displayed on the screen, however neither Alice nor Bob know the value of $latex A$ or $latex B$. We say that the values of the

devices are correlated. This correlation will be useful in the next steps since Alice is not allowed to press the button on $latex a$.**Second step**: Alice now puts her two devices, namely $latex a$ and $latex A$, together. Let’s recall that Alice wants to send the value stored in $a$ to Bob. Here in the classical protocol,

this step is a bit trivial and maybe useless, but it will become

an interesting step in the quantum version.**Third step**: Alice plugs $latex a$ and $latex A$ together and makes her devices add their stored values modulo 2…

*Wow wow wow !!! What’s that thing, “modulo 2” ?.. you said that you’d explain this to us*

Oh sure, I made a box for you below called “Addition modulo 2”… So I was saying that Alice adds the value of $latex A$ to the value of $latex a$ (modulo $latex 2$) and stores it in a third device $latex tilde A$. So somehow we have $latex tilde A=A+a (text{mod} 2)$.

**Fourth step**: Alice presses the button of $latex tilde A$ (but not $latex a$, remember my weird rule !), reads the result and sends it to Bob. On his side Bob presses the button of his device and reads the value of $latex B$. Then, when he receives the value of $latex tilde A$ he just adds it modulo 2 to the value of $latex B$ which gives him $latex a$. That’s it !!

*Wait a minute, why do we get the value of $latex a$ when we add the value of $latex tilde A$ and $latex B$?*

Addition modulo 2 is a form of addition where we can only add $latex 0$s or $latex 1$s with the following rules: $latex 0+0=0 (text{mod} 2)$, $latex 0+1=1+0=1 (text{mod} 2)$ and $latex 1+1=0 (text{mod} 2)$.

Ok, let’s call the value of a device by the name of that device. We said that we have $latex tilde A= A+a (text{mod }2)$, so $latex B+ tilde A= B+A+a (text{mod }2)$, but we know that $latex A=B$ (they are correlated), so we have $latex B+ tilde A= B+B+a (text{mod }2)$. And as $latex B+B=0 (text{mod }2)$ (you can check that for all possible value of $latex B$, namely 0 and 1, with the rule of addition modulo $latex 2$ I gave you previously), we finally have $latex tilde A+B=B+B+a=a (text{mod }2)$.

Let’s take an example now, with $latex A=B=1$ and $latex a=0$. Then we have $latex tilde A= A+a = 1+0=1$ (mod 2), and $latex tilde A +B=1+1=0=a$ (mod 2).

**Important:** I want to add that in the third step, the information on the value of $latex a$ is not in the value of $latex tilde A$. In other words, knowing $latex tilde A$ doesn’t help us to guess $latex a$. In fact

the information on $latex a$ is in the correlation between $latex tilde A$ and $latex B$, that’s why Bob can recover $latex a$ when he knows $latex A$ and $latex B$ in the fourth step. As you can see here the word teleportation is an abuse, since no matter is teleported, and this is also the case for the quantum version of the protocol as we will see in the following section.

## The true quantum teleportation protocol

In the quantum version of the protocol one teleports a state, and to keep it simple I will only talk about two level states.

*Wowowowow !!! What are you talking about ?*

### A little bit of preliminary quantum mechanics

Don’t worry, I’am going to explain. A two levels system, is a quantum system such that when we measure it, we only get two possible outcomes. For instance we can take the spin of an electron, or of another particle of spin 1/2, or the polarization of a photon, or a two levels quantum dot or…

*Ok we got the point, you are muddling us with your examples…*

In short, this two levels systems are called qubits (for qu[antum] bits). So we here will see how to teleport a qubit, while we saw in the classical analogy how to teleport a bit.

Well, the thing is that a qubit is a little bit more complicated than a bit. A bit can only take two values: $latex 0$ or $latex 1$. A qubit can take an infinite amount of “values”, but for a fixed measurement, we can only measure two of its possible values.

*Here we go ! As soon as it’s quantum we don’t understand anything…*

No, don’t worry, I’ll explain. A quantum state (not necessarily a two level one), is unlike bits not represented by a number, but by a vector (of length $latex 1$). You know these things which looked like arrows

in high school math classes. Well here it is exactly (almost) the same. So for those who wondered: yes, vectors are useful! But don’t worry the goal here is not to do too much math.

So, a state is represented by a vector of length $latex 1$. We will choose two orthogonal ones, one will be called $latex |0rangle$ and the other $latex |1rangle$ (look at figure 3). These vectors form what we call a basis, and so we will call these vectors “basis vector(s)”. The set of possible states is a circle, where each point is a state.

Now for the measurements. When we measure the state, if the state is in $latex |0rangle$ then after the measurement the state stays in $latex |0rangle$ and we measure the value $latex 0$. The same for $latex |1rangle$, after the measurement nothing changes and we measure $latex 1$. But for all the other vectors, after the measurement they will be projected randomly on $latex |0rangle$ or $latex |1rangle$, and we get the value $latex 0$ or $latex 1$ depending on which basis vector the state is projected. When I say ‘randomly’ I mean that this happens with a probability that depends on the angle that the vector I ‘measure’ has with the basis vectors (more precisely it depends on the scalar product). For example in figure 3 the vector which is not one of the two basis vectors, has more chance to be projected onto $latex |0rangle$ than onto $latex |1rangle$ because the angle between the vector and $latex |0rangle$ is smaller than the one between the vector and $latex |1rangle$.

Do you remember in the classical protocol, the weird rule which said that Alice is not allowed to press the big button? Well, that’s because in quantum mechanics when we measure, the state is changed,

and then we don’t know in which state the qubit was before the measurement. So in the classical protocol the weird rule is there to imitate this quantum behaviour, and so we had to find a cleverer way to send the bit to Bob, as we will do in the quantum protocol.

## Quantum Teleportation Protocol

Well it is simple, just do like in the classical protocol ! That’s it,

you know everything!

*Hey !! We want explanations !! Else we would ask the Bogdanovs ^{[1]}*

Ok ok, I’ll explain. The goal of the protocol is to make Alice “send” a state $latex |Phirangle$ with only classical communication. We will proceed in four steps, like in the classical case.

**First step**: Alice and Bob share an entangled quantum state (one of the four Bell states). In figure 3 we can see the mathematical expression of this state.*And you were saying that we wouldn’t do math…*Indeed, we won’t. You don’t have to get the mathematical sense of these expressions, but just what they physically mean, which I will explain to you now. It won’t be difficult, don’t worry. First you can forget the constants like $latex frac{1}{2}$ or $latex sqrt{2}$, they are there only to assure that the vectors are length of 1. What we can see in these expressions is that we are adding terms of two or three qubits. This means that the state is a “superposition” of each term of the sum, and if we measure it, the state will be projected on one of these terms. For example in the first step Alice and Bob share a state which is a sum of two terms of two qubits: something like $latex |0rangle |0rangle + |1rangle|1rangle$. So this means that when they will measure it, the state will be projected either in $latex |0rangle |0rangle$ or in $latex |1rangle |1rangle$, that’s why they will both measure $latex 1$ or both measure $latex 0$, but never $latex 0$ for Alice and $latex 1$ for Bob, because the vector is never projected in $latex |0rangle |1rangle$. And it is because of these correlations that we say that the state is entangled. Remember that in the classical version of the protocol we had that the value of the device $latex A$ is equal to the value of the device $latex B$, this is somehow the classical analogy to the entanglement we have in the quantum version.

For those who want to know a bit more, an entangled state is also called non separable state, because this state cannot be written as a product of two qubits. An example of a “non separable” object would be the polynomial $latex xy+x^2y^2$. We cannot write it as a product of a polynomial depending only on $latex x$ with a polynomial depending only on $latex y$, that’s to say we cannot separate the $latex x$ from the $latex y$. It’s the same for a non separable state, we cannot separate what is on Alice’s side from what is on Bob’s side.

**Second step**: Alice makes the qubit she wants to teleport (the state $latex |Phirangle$) interact with her qubit which is entangled with Bob’s one. On the level of mathematical expression, we multiply the two qubit entangled state with the one qubit state $latex |Phirangle$. This makes a three qubit state. What is interesting is that this state can be rewritten in a form where the state $latex Phi$ is on Bob’s side! (Look in figure 4, the $latex |Phirangle$ has an index $latex B$). We are close to the goal. However the price to pay is that now there is an operator $latex U$ (this can be viewed as a transformation) in front of $latex |Phirangle$. This operator modifies $latex |Phirangle$, and will have to fix this problem. Moreover for each term of the expression the $latex U$ is different. But for now Bob cannot do so much since he does not know $latex U$.**Third step**: Alice will make a special measurement which is called a Bell measurement. This is a particular measurement because it measures two qubits at the same time, projecting them on one of the four Bell states (which are all maximally entangled states, i.e.~with perfect correlations). Alice will then measure a value $latex N$ ($latex N$ is $latex 1,2,3$ or $latex 4$), corresponding to the term on which the state will be projected. Then the sum disappears since the state has been projected onto one of the terms. We can see that the new state (after measurement) is now a product of a two qubit state on Alice side with a one qubit state on Bob side: the state is separable! We can now look what happens on Bob’s state independently of what happens on Alice’s state. But there still remains the problem of the $latex U$ which is still here. As there are four possible $latex U$s, Bob doesn’t know which operation to make in order to cancel the $latex U$ (such an operation is called the inverse of $latex U$).**Fourth step**: Therefore Alice will send $latex N$ to Bob. This $latex N$ tells to Bob which of the four possible $latex U$ operators he is dealing with and then he knows which transformation to make to cancel $latex U$ and then get $latex |Phirangle$. Ta-da ! The qubit is teleported !

### Important remarks

- Notice that each step of the teleportation has a classical analog.
- It is important to understand that in step 3, Bob doesn’t have any information about $latex Phirangle$, and this is precisely because of the fact that there are multiple possibilities for $latex U$. The fact that there are four possibilities and that he doesn’t know which one is the good one, hides all the information about $latex |Phirangle$. Also, at the end of the protocol, Alice doesn’t have any information about $latex |Phirangle$ anymore. The information is contained in the correlation between $latex N$ and the corresponding $latex U$. Therefore to get the information on $latex |Phirangle$ we need that the two pieces of information are in the same place, and that’s why Alice has to send $latex N$ to Bob. And she sends it through classical communication like fax, phone, e-mail etc. Then this $latex N$ travels at the speed of light (at most), so there is no instantaneous transfer of information, and hence teleportation respects relativity. (as it should!)
- Now to the question of whether or not it is possible to apply this quantum teleportation (seen as a very fast way of traveling
^{[2]}) to macroscopic objects (or more precisely to their state) is not clear. Indeed even if we admit that it is in principle possible by doing something very similar to what I described above (which is already still debated^{[3]}), then the complexity of a macroscopic object is so important that it is in practice not doable, or at least we are currently very very very far to be able to do so.

[1]: See the Bogdanov Affair

[2]: Basically at the speed of light.

[3] For example we don’t know at which size an object stops having a quantum behavior, or if we can force it to adopt such behavior.

Jérémy Ribeiro is a theorist at QuTech. He specialises in quantum cryptography: the use of quantum principles to design secure protocols for communication and other cryptographic tasks. In his free time (and sometimes during work hours) he enjoys practicing his Yo-yo skills and talking about open source software and undecidable problems.

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