# The Life of Psi

Philosophical Musings on the Foundations of Physics and Chemistry

# Beam me up, Scotty! How to build a quantum teleportation device

Update 11/06: I'm happy to announce that my post just appeared on the blog of the KU Leuven. Check it out!

The content of this post is based on a three minute talk I gave last week during the FameLab Benelux finals at Science Center NEMO in Amsterdam. So in case you wouldn't feel like reading, just poor yourself another cup of coffee and watch my three minute pitch which nicely summarizes this post (see the video below).

Imagine a world without traffic jams or airport security lines. A world where you could have a morning cup of (Arabica) coffee in Paris, an afternoon walk in the Brazilian rainforest, and a dinner on top of an Egyptian pyramid. Now wouldn't that be cool? Well, all you’d need is a teleportation device such that, at the flick of a switch, you could beam yourself across space to whatever destination you'd like, without physically crossing the space in between.

## Classical teleportation

The typical beam-me-up-Scotty-kind of teleportation works pretty much like a cross between a fax machine and a 3D printer. Your entire body would first be scanned in some futuristic MRI device, and the details about your atomic configuration would then be send to a distant location, where it would be used to construct an exact replica of yourself.

No more traffic jams or airport security lines. In the future, we might go to a teleport terminal where we would step into a transporter to beam ourselves to whatever destination we'd like. Source: Zeilinger, A. "Quantum Teleportation." Scientific American (April 2000): 50-59.

If you don’t want to arrive at your destination with a leg sticking out of your head or your organs inside-out (remember the inside-out baboon in The Fly?), you would of course want to scan yourself accurately enough. But that is where the problem lies. You see, there’s this little thing in physics, called the Heisenberg uncertainty principle, which says that the more accurately you try to scan something, the more you have to disturb it, and the more you will change it, making it fundamentally impossible to extract the information of an object without destroying it entirely.

## Quantum teleportation

So for years, scientists believed teleportation was just a science fiction pipe dream. But in 1993, a bunch of smart-ass physicists found a way to circumvent Heisenberg’s principle by using something called entanglement.

Entanglement is one of those quirky concepts from quantum mechanics which even Einstein called "spooky". If you entangle two particles, they will share a magical but invisible bond, such that if one particle gets affected, the other one will feel it too, no matter how far apart they are. "It's like two people playing dice, each in a different galaxy, and always getting the same result, even though the result is determined by pure chance," said Rupert Ursin. (If you'd like to know more about this mysterious quantum link, don't hesitate to read my previous post on Bertlmann's Socks and the Nature of Reality.)

Artistic depiction of two entangled atoms. There is no visible connection: no forces, no pulleys, no telephone wires, no nothing. And yet, somehow the atoms seem to communicate with one another, allowing scientists to teleport information from one place to another.

The key to teleportation is to use this entangled pair as a communication channel to transmit the information you want to teleport from one place to another. The exact teleportation protocol is a little tricky, so bear with me for a moment. Let's imagine you wanted to teleport a particle (let's call it A) from the Earth to the Moon. First of all, you would need to generate a pair of entangled particles B and C. One of those (B), you would keep here on Earth, while the other one (C) would be send to the Moon.

Now, the aim of quantum teleportation is to change the identity of the particle C on the Moon and to turn it into an exact replica of A. In order to do this, we somehow need to extract all the information about A, and teleport that information to the Moon, so that C can use it as a kind of blueprint to turn itself into A. The problem however is that we cannot simply scan A to extract its information (remember Heisenberg?).

$$Earth:A\,-\,B\,--------------------\,C:Moon\notag$$

So instead, we entangle A with the other particle B here on Earth. That way, some of A's information gets shared with B, and since B is entangled with C on the Moon, the information is passed on further to C, enabling C to transform itself into a copy of A. (I've added a green box at the bottom of this post with a more detailed explanation of the quantum teleportation protocol if you'd like more mathematical detail.)

There is a price to pay however: the original particle A will be destroyed in the process as it loses all its information and thus also its identity. (The same also happens to B.) In summary, you would see the original particle A disappear here on Earth, and reappear an instant later on the Moon. Congratulations! You successfully teleported your first particle.

## Practical teleportation

I realize the whole teleportation procedure might sound rather magical, but two years ago, a team of scientists set foot on the Canary Islands, and managed to teleport a particle of light from La Palma to Tenerife over a distance of 143 kilometers! A few months earlier, a group of Chinese scientists performed a similar feat by teleporting a photon over a distance of 97 km across a lake in China. Soon, scientists might be zapping photons via satellites through interstellar space over hundreds (if not thousands) of kilometers.

In April 2012, a team of international researchers from Austria, Canada, Germany and Norway, led by the physicist and quantum wizard Anton Zeilinger, managed to teleport a photon from La Palma to Tenerife over a record breaking distance of 143 kilometers.

Admitted, at this point teleportation is still very much restricted to tiny particles. Photons and electrons, in particular, are now routinely being teleported across record smashing distances. But great advances are being made. Physicists recently teleported the properties of one atom to another one. And last year, researchers at the Niels Bohr Institute managed to teleport the information between two gas clouds!

In the not-too-distant future, scientists might be teleporting small molecules, like a water (H2O) molecule. Further on down the road, teleportation might be extended to larger (organic) molecules, such as DNA fragments or small proteins. And in a hundred years from now, scientists might be teleporting viruses, or even bacteria.

## Human teleportation

Now I know what you are thinking at this point. If we can already teleport entire gas clouds of atoms, will we ever be able to teleport a human being? Will we, in other words, turn science fiction into science fact?

Unfortunately, we're not quite there yet. And that's probably the understatement of the year. All in all, there are at least three problems that should be overcome before we can allow Scotty to beam us up:

1. The practical problem. First of all, our body consists of more than a quadrillion atoms, each of which should get entangled, digitized and teleported. But all that information would fill billions and billions of gigabytes of data, which would take more than 350.000 times the age of the universe to transmit! So basically, teleportation would take forever!

A teleportation accident.

2. The safety problem. Then, there's the problem that entanglement is a very fragile bond, which can easily break due to the phenomenon of environmental decoherence. Basically, even the slightest disturbance (such as a light ray falling in, or a slight temperature fluctuation) can shatter the entanglement, breaking the communication between the points of departure and arrival. Now, when we are dealing with more than 1028 atoms, keeping all the entanglement intact becomes extremely difficult. And if anything would go wrong, you might well lose half your body in transit, or be zapped right inside a mountain.

3. The philosophical problem. Finally, there’s the problem that teleportation does not actually involve the transmission of matter but only the transmission of the information encoded in that matter. To put it bluntly, as the information about you would be sucked out of your body, you would get pulverized to a heap of atoms, and the person walking out at the other end of the pipeline would be nothing more than an exact Doppelganger. It's as if your soul would have been re-injected into another body and the original you would have "died" in the process. Here's Sheldon, from the Big Bang Theory, explaining it:

But let’s not despair, because human teleportation is theoretically possible. That is, human teleportation does not violate any of the fundamental laws of nature, and the above mentioned problems are in the end nothing more than technological problems. So who knows who Scotty might be beaming up, say ten generations from now! But for the moment, it's probably quicker and safer to buy another plane ticket . . .

#### Articles

[Green boxes, like the one you are reading now, provide more technical information. Whereas the aim is to keep everything to a theoretical minimum, these boxes might assume a little more background in quantum mechanics. So if you are the kind of person whose head starts spinning the moment you see too many mathematical equations, feel free to skip these boxes entirely.]

#### Teleportation protocol

Step 1: Setting the stage. Imagine that Charlie, who lives here on Earth, would like to teleport an unknown qubit $\left|\psi\right\rangle$ to Bob, who lives on the Moon. We can expand Charlie's qubit as follows:

$$\left|\psi\right\rangle_C=\alpha\left|0\right\rangle_C+\beta\left|1\right\rangle_C,\label{O}$$

with $\left|\alpha\right|^2+\left|\beta\right|^2=1$ as usual. The subscript $C$ will be used to distinguish this qubit from the other two qubits, labelled $A$ and $B$, which we will introduce in a moment.

Step 2: Creating a maximally entangled pair. In order to teleport her qubit, Charlie will need the help of her good friend Alice, who lives on Earth too. Alice starts by generating a (maximally) entangled pair of qubits. In general, this could be any of the following four Bell states:

\left|\Phi^+\right\rangle_{AB}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_A\left|0\right\rangle_B+
\left|1\right\rangle_A\left|1\right\rangle_B\right);\label{A}

$$\left|\Phi^-\right\rangle_{AB}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_A\left|0\right\rangle_B-\left|1\right\rangle_A\left|1\right\rangle_B\right);\label{B}$$

\left|\Psi^+\right\rangle_{AB}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_A\left|1\right\rangle_B+
\left|1\right\rangle_A\left|0\right\rangle_B\right);\label{C}

\left|\Psi^-\right\rangle_{AB}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_A\left|1\right\rangle_B-
\left|1\right\rangle_A\left|0\right\rangle_B\right).\label{D}

Let us assume for what will follow that Alice created the first Bell state $\left|\Phi^+\right\rangle_{AB}$.

Step 3: Distributing the entangled pair between Alice and Bob. At this point, Alice needs to share her entangled pair with Bob. Typically, Alice will keep one of the qubits, labelled $A$, here on Earth, while sending the other one, labelled $B$, to Bob on the Moon via a laser beam. The entangled pair $AB$ will then function as a communication channel through which the (quantum) information about $C$ can be send.

Step 4: Introducing the third qubit. In the next step, Charlie hands over her qubit $C$ to Alice. Notice that Alice has now two particles: $C$, the one Charlie is asking her to teleport, and $A$, one of the two entangled particles. Bob, on the other hand, has only one particle, $B$. The complete state of these three particles can be written as:

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

Working out the brackets yields the following state:

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

While the above equation may seem daunting at first sight, I hope the notation involved is intuitively clear to you.

Step 5: Changing Alice's measurement basis. So far, so good. Alice now needs to perform a Bell state measurement on the two qubits ($A$ and $C$) in her possession. In order to simplify matters, we will switch to a different measurement basis for Alice. Instead of working in the computational basis, as we have done so far, it will be easier to perform Alice's measurement in a Bell basis, much like the one given in equations \eqref{A}—\eqref{D}:

\left|\Phi^+\right\rangle_{AC}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_A\left|0\right\rangle_C+
\left|1\right\rangle_A\left|1\right\rangle_C\right);\label{E}

$$\left|\Phi^-\right\rangle_{AC}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_A\left|0\right\rangle_C-\left|1\right\rangle_A\left|1\right\rangle_C\right);\label{F}$$

\left|\Psi^+\right\rangle_{AC}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_A\left|1\right\rangle_C+
\left|1\right\rangle_A\left|0\right\rangle_C\right);\label{G}

\left|\Psi^-\right\rangle_{AC}=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle_A\left|1\right\rangle_C-
\left|1\right\rangle_A\left|0\right\rangle_C\right).\label{H}

(The only change I've made, as compared to equations \eqref{A}—\eqref{D}, is to substitute the subscripts $B$ for $C$.) We can then rewrite the four computational measurement states in terms of the four Bell states \eqref{E}—\eqref{H}:

$$\left|0\right\rangle_A\left|0\right\rangle_C=\frac{1}{\sqrt{2}}\left(\left|\Phi^+\right\rangle_{AC}+\left|\Phi^-\right\rangle_{AC}\right);\label{I}$$

\left|0\right\rangle_A\left|1\right\rangle_C=\frac{1}{\sqrt{2}}\left(\left|\Psi^+\right\rangle_{AC}+\left|\Psi^-
\right\rangle_{AC}\right);\label{J}

$$\left|1\right\rangle_A\left|0\right\rangle_C=\frac{1}{\sqrt{2}}\left(\left|\Psi^+\right\rangle_{AC}-\left|\Psi^-\right\rangle_{AC}\right);\label{K}$$

$$\left|1\right\rangle_A\left|1\right\rangle_C=\frac{1}{\sqrt{2}}\left(\left|\Phi^+\right\rangle_{AC}-\left|\Phi^-\right\rangle_{AC}\right).\label{L}$$

You can easily verify the above identities by inserting the expressions for $\left|\Phi^\pm\right\rangle_{AC}$ and $\left|\Psi^\pm\right\rangle_{AC}$ and working out. Using equations \eqref{I}—\eqref{L}, we can then rewrite the total particle state \eqref{M} as the following four-term superposition:

\left|\Phi^+\right\rangle_{AB}\otimes\left|\psi\right\rangle_C=\frac{1}{2}\left|\Phi^+\right\rangle_{AC}\otimes\left(\alpha\left|0\right\rangle_B+\beta\left|1\right\rangle_B\right)\\

Agreed, this equation looks monstrous!! But please remember that nothing has happened so far, except for a change of Alice's basis. No measurement has been performed, no collapse has occurred; the three particles $A$, $B$ and $C$ are still in the same total state \eqref{M}, but due to Alice having rotated her measurement apparatus to align with the Bell basis, we have now re-expressed this state \eqref{M} in the alternative, but perfectly equivalent, way \eqref{N}.

Step 6: Performing a Bell state measurement. Alice now performs a (local) Bell state measurement on the total state $\left|\Phi^+\right\rangle_{AB}\otimes\left|\psi\right\rangle_C$. This is the point at which the teleportation occurs. It follows from equation \eqref{N} that the three-particle state will collapse, with an equal probability of $\frac{1}{4}$, to any of the following four states:

$$\left|\Phi^+\right\rangle_{AC}\otimes\left(\alpha\left|0\right\rangle_B+\beta\left|1\right\rangle_B\right);\label{P}$$

$$\left|\Phi^-\right\rangle_{AC}\otimes\left(\alpha\left|0\right\rangle_B-\beta\left|1\right\rangle_B\right);\label{Q}$$

$$\left|\Psi^+\right\rangle_{AC}\otimes\left(\beta\left|0\right\rangle_B+\alpha\left|1\right\rangle_B\right);\label{R}$$

$$\left|\Psi^-\right\rangle_{AC}\otimes\left(\beta\left|0\right\rangle_B-\alpha\left|1\right\rangle_B\right).\label{S}$$

As a result of Alice's measurement, the particles $A$ and $C$ have become entangled in one of the four Bell states $\left|\Phi^+\right\rangle_{AC}$, $\left|\Phi^-\right\rangle_{AC}$, $\left|\Psi^+\right\rangle_{AC}$ or $\left|\Psi^-\right\rangle_{AC}$. But Alice's measurement has also affected Bob's particle. The entanglement that was originally shared between Alice and Bob has been broken, and Bob's particle $B$ is now in one of the four superpositions given above.

But here comes the amazing thing. Take a closer look at Bob's qubits, and notice how similar they are to the original qubit $\left|\psi\right\rangle_C=\alpha\left|0\right\rangle_C+\beta\left|1\right\rangle_C$ in equation \eqref{O} that Charlie wanted to teleport!

Step 7: Communicating classical information to Bob. I should of course emphasize that whereas Bob's qubits are very similar to Charlie's qubit, they are not yet quite identical. That is, after Alice's measurement, Bob's qubit will be in one of the four states $\alpha\left|0\right\rangle_B+\beta\left|1\right\rangle_B$, $\alpha\left|0\right\rangle_B-\beta\left|1\right\rangle_B$, $\beta\left|0\right\rangle_B+\alpha\left|1\right\rangle_B$ or $\beta\left|0\right\rangle_B-\alpha\left|1\right\rangle_B$. The problem is that Bob doesn't know which one.

Alice is therefore forced to communicate which of the four possible outcomes she obtained upon measurement by sending Bob two bits of classical information via a classical channel (e.g. via radio waves or light signals). So depending on whether she obtained the state $\left|\Phi^+\right\rangle_{AC}$, $\left|\Phi^-\right\rangle_{AC}$, $\left|\Psi^+\right\rangle_{AC}$ or $\left|\Psi^-\right\rangle_{AC}$, she will send Bob the message $00$, $01$, $10$ or $11$. Bob can then use this information to transform the state of his qubit $B$ into the original state \eqref{O}.

But before we move to this final step of the teleportation protocol, let me repeat once again that whereas the collapse in step 6 occurs instantaneously, the teleportation speed is nevertheless limited to the speed of light $c$ because of Alice's sending two classical bits of information in this step.

Step 8: Transforming Bob's qubit. Imagine Alice obtained the state $\left|\Phi^+\right\rangle_{AC}$ in step 6, in which case Bob's particle $B$ is already in the desired state $\alpha\left|0\right\rangle+\beta\left|1\right\rangle$ that Charlie wanted to teleport [see equation \eqref{P}]. Alice sends Bob the message $00$, and Bob thereby knows he doesn't have to do anything anymore. Notice that this amounts to the trivial unitary operation where one acts on $B$ with the identity operator:

$$\hat{I}\left(\alpha\left|0\right\rangle+\beta\left|1\right\rangle\right)=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]\left[\begin{array}{c}\alpha\\ \beta\end{array}\right]=\left[\begin{array}{c}\alpha\\ \beta\end{array}\right]=\alpha\left|0\right\rangle+\beta\left|1\right\rangle.$$

But now imagine Alice obtained one of the other three states [equations \eqref{Q}—\eqref{S}]. In these cases, Bob still needs to transform his qubit into the state $\alpha\left|0\right\rangle+\beta\left|1\right\rangle$ via a unitary operation. He can achieve this via the use of quantum logic gates (similar to the logic gates AND, OR, NOT and NAND in classical computing):

I. If Alice sends the message $01$, Bob knows his particle is in the state $\alpha\left|0\right\rangle-\beta\left|1\right\rangle$ and that he needs to change the minus sign into a plus sign. This can be done with a Z gate:

$$\hat{Z}\left(\alpha\left|0\right\rangle-\beta\left|1\right\rangle\right)=\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]\left[\begin{array}{c}\alpha\\ -\beta\end{array}\right]=\left[\begin{array}{c}\alpha\\ \beta\end{array}\right]=\alpha\left|0\right\rangle+\beta\left|1\right\rangle.$$

II. If Alice sends the message $10$, Bob knows his particle is in the state $\beta\left|0\right\rangle+\alpha\left|1\right\rangle$ and that he needs to swap the order of the two states in the superposition. This is done with an X gate:

$$\hat{X}\left(\beta\left|0\right\rangle+\alpha\left|1\right\rangle\right)=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}\beta\\ \alpha\end{array}\right]=\left[\begin{array}{c}\alpha\\ \beta\end{array}\right]=\alpha\left|0\right\rangle+\beta\left|1\right\rangle.$$

III. Finally, if Alice sends the message $11$, Bob knows his particle is in the state $\beta\left|0\right\rangle-\alpha\left|1\right\rangle$ and that he needs to change both the minus sign and the order of the two states in the superposition. This is done by first applying the Z gate, and then the X gate:

$$\hat{X}\hat{Z}\left(\beta\left|0\right\rangle-\alpha\left|1\right\rangle\right)=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]\left[\begin{array}{c}\beta\\ -\alpha\end{array}\right]=\left[\begin{array}{cc}0&-1\\1&0\end{array}\right]\left[\begin{array}{c}\beta\\ -\alpha\end{array}\right]=\left[\begin{array}{c}\alpha\\ \beta\end{array}\right]=\alpha\left|0\right\rangle+\beta\left|1\right\rangle.$$

So no matter which outcome Alice obtained, after performing the correct procedure based on Alice's two bit message, Bob will always obtain the desired state $\alpha\left|0\right\rangle+\beta\left|1\right\rangle$. That is, Charlie has successfully teleported the state of her qubit $C$, here on Earth, to Bob's qubit on the Moon, thanks to Alice's help. Teleportation has been achieved, using one pair of entangled particles $A$ and $B$ and two bits of classical information!!

#### Pieter Thyssen

Whereas his left brain was trained as a theoretical scientist, his right brain prefers the piano. At work, Pieter builds time machines (on paper) and loves to dabble in the history and philosophy of science. He often gets stuck in another dimension, contemplating time travel and parallel universes, or thinking about ways to save Schrödinger's cat (maybe). He explores the world on foot, and takes life one cup of (Arabica) coffee at a time. Follow him on Twitter @PieterThyssen or at thelifeofpsi.com. You can reach Pieter via email at pieterthyssen@gmail.com.

## 4 comments for “Beam me up, Scotty! How to build a quantum teleportation device”

1. V. K. Robertson
June 12, 2015 at 4:51 pm

What if you were to flash freeze or place in suspended motion the original person/atom/molocule and then attempt to transport? I'm not a physicist ( hope i spelled that correctly) so this may sound unintelligent. Just want to see some super strives in my time.

2. Abe
December 13, 2015 at 9:07 pm

Nice, Thanks a lot.