A Christmas story about quantum teleportation

The first example is part of a larger research project with the aim of developing a 20-hour course for high school students on quantum computing where quantum teleportation is a sub-topic in the course [18]. As part of the pilot study, the materials were evaluated with a small group of students between 16 and 17 years old. The researchers set out to create resources on quantum teleportation and to find ways to make students aware of the revolution in quantum technologies.

Design of the classroom materials on teleportation followed the model of education reconstruction (MER) [19], which consists of three main parts. First, the researchers catalogued existing teleportation experiments to find the one most suitable for use in the classroom. Second, student conceptions (naive or otherwise) in learning in general quantum physics were taken into account so that the content was made accessible to secondary students. The authors relied on their combined experience in the teaching of quantum physics. Third, the materials were then tested in the classroom with a small group of students. Through this process, the researchers [18] developed a resource that followed the quantum teleportation experiment in Vienna from 2004 where photons were teleported over a distance of 600 meters across the river Danube [20].

In the second example, the researchers present a quantum curriculum transformation framework (QCTF), which can be used to create new materials for quantum technologies courses [21]. In the paper, the researchers focus on examples pertaining to quantum teleportation. The materials in the paper are certainly suitable for higher education, but they could also be implemented with the same group of students from the previous study considered [18].

Central to the design of their resources are a number of components. First, the researchers used the European Competence Framework for Quantum Technologies, which is a tool designed by the EU's Quantum Flagship, as the basis for the materials to be created. Second, they identify three main skills to be targeted with the materials: namely theory and analytics, computation and simulation, and experiment and real world. The authors note that quantum teleportation, which encapsulates key concepts such as superposition and entanglement, can be used to develop all or some of these skills [21]. Thereafter, the authors present descriptive (text), symbolic (mathematical) and graphical (pictures and diagrams) visual representations of the quantum teleportation process for the three main skills outlined above. The final key aspect relates to the teaching approach for the materials. The authors argue that there are 'infinitely possible ways' to teach quantum teleportation in the classroom when the requirements are for learners to understand the theory of quantum teleportation and recognise applications in the real world. Suggested teaching approaches include the open exploration of the materials by the student to guided learning where students are asked to focus on specific details [21].

In this study, we align with the approach of Satanassi et al [18], where the design process is inspired by two of the steps from the MER. First, rather than using an existing quantum teleportation example, we devised a new explanation for quantum teleportation with a connection to popular culture without too many technical details.

Second, in creating the content and the worksheet, we took into account some naive conceptions of students in two different ways. We looked at a selection of the recent research on quantum physics to gain an appreciation for the concepts that may confuse students [4, 5, 10, 17, 22].

Finally in this section, we reflect on the importance of enthusing students about the concept of quantum teleportation, and quantum physics in general. By means of example, in a recent PhD thesis, one researcher reflected on their own experiences in the high school classroom when teaching quantum physics and noted that they struggled to enthuse physics teaching about quantum physics [22]. Their experiences formed the basis of their research in which they explored the use of freely-available digital resources to teach high school students about quantum physics.

In one part of their study involving three high school physics teachers, the researcher noted that all three teachers struggled to enthuse their students [5]. This then led to a compromised educational experience for teachers and students alike. In fact, two teachers and a number of the students in their respective classes all highlighted the need for practical and technological examples and applications of quantum physics [5]. While a range of digital materials consisting of animations and interactive simulations [23, 24] are available to the educator, many of these rely on traditional contexts similar to those seen in textbooks.

The outcomes of the aforementioned study motivate in part the content and worksheet presented here. In addition, one author has experience in the design and evaluation [25–27] of lesson materials based on popular culture (in this case for teaching the electromagnetic spectrum), which has proven to be effective at engaging with students.

In summary, the content presented here is designed to enthuse educators and students alike in three ways. First, it presents a concept from quantum physics in an unconventional manner, using the gift-bringer Santa Claus. Second, it utilises a fictional character that would be recognisable in educational settings in many countries. Third, the example is based upon a practical implementation of the principles of quantum teleportation, albeit an unusual practical example, and does not focus solely on mathematical notation, as in previous classroom implementations [18].

Let us imagine that Alice needs to tell Bob something very important, but they are not in the same room. A protocol describes the steps needed by Alice and Bob to successfully transfer information. A simple example of a protocol is a phone call between Alice and Bob. If Alice wants to speak to Bob on the phone, she must dial Bob's phone number, and then press the call button. On Bob's side, he has to accept the call so that Alice can speak to him and share the information. The protocol relies on both parties having certain resources (a telephone each and Alice needs Bob's number) and performing well-defined actions (Bob must accept Alice's call before hearing what Alice has to say).

In quantum teleportation, the protocol is similar to a phone call in the sense that information is transferred, but it is quantum information that is transferred rather than spoken information. In the classical computers that you will find in almost every electronic device on the planet from smartphones to flying drones, information is encoded in bits as 1s and 0s, the basic units of classical information in computers. The bits 1 and can represent everything from a traffic light being red or green (two states) to a coin showing head or tails. In real devices, these 1s and 0s can be, for example, realised by the orientation of tiny magnets in a hard drive device where the magnets can be orientated in one of two directions.

On the other hand, quantum information is information stored in qubits. Short for quantum bit, a qubit is the fundamental unit of information in quantum computing. Like the bit in a classical computer, a qubit also has two states, and while these states are not 1 and 0, there are similarities. Rather than using 1 and 0, the qubit states are $\vert 1\rangle$ and $\vert 0\rangle$. The strange notation $\vert \cdot\rangle$ indicates that qubits are different from classical bits. The key difference between classical and qubits is that qubits can be in a quantum superposition.

So, what is the difference then between bits in classical computers and qubits in quantum computers? An easy way to realise that qubits can represent more information than classical bits is by visualising the qubits in a coordinate system. In figure 1(a), the two quantum bits are used as the axes of a coordinate system. Classical bits 0 and 1 are always located exactly on these axes: there are only two choices 0 and 1. However, there is a lot of white space left between the two axes, where classical bits can never be.

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This same is not true for qubits. The marvels of quantum physics means that a qubit can be in any state between the two axis. This is known as quantum superposition. We can actually understand the word superposition quite literally. The states $\vert 0\rangle$ and $\vert 1\rangle$ form the axis of the coordinate system. Instead of the x- and y-direction, the directions of the coordinate system are called $\vert 0\rangle$ and $\vert 1\rangle$. These vectors actually represent the same vectors as in everyday vector calculus

$$\begin{align} \begin{aligned} \vert 0\rangle& = \left(\begin{array}{c}1\\0\end{array}\right)\ \text{unit vector in}\ x\ \text{direction,} \\ \vert 1\rangle& = \left(\begin{array}{c}0\\1\end{array}\right)\ \text{unit vector in}\ y\ \text{direction.} \end{aligned} \end{align} \tag{ 1 }$$

By super-imposing these two directions with certain coefficients or weights α and β, we can form an arbitrary quantum state, which we write as

$$\begin{align} \vert \psi\rangle = \alpha \vert 0\rangle + \beta \vert 1\rangle. \end{align} \tag{ 2 }$$

Here, $\vert \psi\rangle$ represents the superposition state of the qubit. Any direction in the two-dimensional plane (cf figure 1(a)) can be understood as a quantum state. Since only the direction encodes a quantum state, the length of the vector does not play a role. This means that we can limit ourselves to vectors of length one. In the illustration, this is illustrated by the black arc connecting the $\vert 0\rangle$ and the $\vert 1\rangle$ axis. If we translate the condition to the coefficients of the state, we notice that they must fulfill $|\alpha|^2+|\beta|^2 = 1$.

To be fair, the coefficients α and β could be taken to be complex numbers, so instead of a circle in the plane, the actual mathematical model of the qubit is a sphere, but that is a technical issue we do not need to consider here. Therefore, we use a simplified model where α and β are real numbers and therefore can be interpreted as the x- and y- coordinates of a circle, respectively.

Although we can prepare an arbitrary quantum state, observing which quantum state we have is more difficult. Performing a measurement on a qubit means that this qubit will return to $\vert 0\rangle$ or $\vert 1\rangle$. Why is that so? In fact, it is important to note that we have to display the result of that measurement on a classical display (there are no such things as displays for quantum information). Thus, we must either get 0 or 1, corresponding to $\vert 0\rangle$ and $\vert 1\rangle$, respectively. If we look at figure 1, we know two notable points already: The coordinate axes corresponding to the classical bits 0 and 1.

So, what happens if the qubit is neither exactly $\vert 0\rangle$ nor $\vert 1\rangle$? How does it decide which one to become after the measurement? The answer is related to α and β. If we measure a quantum state, the length of α and β are related to the probabilities of measuring the state in 0 and 1. In technical terms, the probabilities are given by the squared projections of the vector to the axis $\vert 0\rangle$ and $\vert 1\rangle$.

Intuitively, the projection process can be imagined as shadows cast by the vectors along the two axes as shown in figure 1(b). If we think of the quantum state as a stick, the projections on the axes $\vert 0\rangle$ and $\vert 1\rangle$ are the shadows cast by the green and the blue lamp. The green lamp casts a shadow onto the $\vert 1\rangle$ axis. The length of that shadow is equivalent to the amount of the stick that points in the $\vert 1\rangle$ direction. In other words, it is the part of the quantum state in the $\vert 1\rangle$ direction in equation (2). However, we cannot observe the coefficients α and β directly, unless we have many copies of the quantum state. If you want to measure them, you have to prepare and measure the quantum state many times. The number of times that either 0 or 1 is measured can be used to estimate the values for α and β.

Importantly, a measurement changes the quantum state. Remember, it is not really a stick after all. After measurement, the state will be pointing exactly in the direction that was measured (and its length will be one again). The quantum superposition is lost, and the qubit points either in the $\vert 0\rangle$ or $\vert 1\rangle$ direction, according to the measurement result. The information that we get as the result of a measurement is classical again: it points exactly along one axis. In quantum physics, this is called the collapse of a quantum state.

The collapse of the quantum states leads to an important feature of quantum information. A general quantum state can never be copied or cloned, it can only be transferred [28–30]. This is in stark contrast to our experience with classical bits. Every film that you stream is copied from a server and sent to you. This is not possible with quantum states, because trying to learn the exact values of α and β destroys the quantum state itself.

But what do qubits look like in the real world? In analogy to the variety of systems where we can store classical information (traffic lights, hard drives, paper, etc) there is a whole plethora of platforms that can encode quantum information. For example, qubits in quantum devices can be represented by the spin of an electron (an electron has two spin values—spin up and spin down) or the polarisation of photons where the two states are vertical polarisation or horizontal polarisation, just to name two. Here, we do not focus on the technology used to store the qubits, and instead focus on how qubits relate to quantum teleportation.

In the macro-world that we live in, we can experience entanglement on a daily basis. Just think of entangled shoelaces, ropes, or even long hair. However, in the quantum world of qubits, quantum entanglement is something different. It is a special kind of interrelationship that two or more quantum systems (qubits in our example) have with each other.

Let us go back to the example of Alice and Bob from the description of a protocol. Imagine that Alice and Bob have one qubit each, q_A and q_B , respectively. Let us check which quantum states we can realise with two qubits. Using the intuition from the one-qubit case, we can list all directions the qubit could point to. In the one-qubit case, these are $\vert 0\rangle$ and $\vert 1\rangle$, which is just a list of all classical values a bit can take (0 and 1). Let us list the classical states two bits can take, and there are four values in total—00, 01, 10, and 11. Thus, two qubits can, in general, take all states of the form $\vert \psi\rangle = \alpha_1\vert 0_A,0_B\rangle+\alpha_2\vert 0_A,1_B\rangle+\alpha_3\vert 1_A,0_B\rangle+\alpha_4\vert 1_A,1_B\rangle$, where the α_i 's determine the superposition of the four possible states. As in the one-qubit case, we just enter the numbers in fancy brackets to show that we are not considering classical bits anymore. The subscripts A and B in the formula are inserted to remind us that one qubit belongs to Alice and one qubit belongs to Bob. The mathematical details of this notation are explained in appendix B.

Some states with two qubits are special, since we can learn something about one qubit by measuring the other qubit. These states are said to be entangled. One special case is the maximally entangled state

$$\begin{equation} \vert \phi\rangle = \frac{1}{\sqrt{2}}\left(\vert 0_A,0_B\rangle+\vert 1_A,1_B\rangle\right). \end{equation} \tag{ 3 }$$

This state is a superposition state of the two-qubit states $\vert 0_A,0_B\rangle$ and $\vert 1_A,1_B\rangle$. As in the one qubit case, we can only measure the outcomes that are in the superposition. The state will always collapse to either the classical state 00 or 11. The probabilities are determined by the coefficients in front of the qubits.

Let us consider what happens if we measure the maximally entangled state. If Alice measures her state, she obtains either 0 or 1, both with an equal 50% probability, since the coefficients in front of both qubits are the same. We cannot predict which result she will get, as it is probabilistic after all. Once she has measured the status of her qubit, Alice will know Bob's result! It has to be the same as her own result. First, if Alice measures 0, then the state collapses to $\vert 0_A,0_B\rangle$ and, as a result, Bob will also measure 0. Second, if Alice measures 1, then the basis state collapses to $\vert 1_A,1_B\rangle$ and, as a result, Bob will also measure 1.

However, not all two-qubit states are entangled. It is important to distinguish between quantum correlation and classical correlation. The simple fact that Alice and Bob measure the same outcome is not surprising. This effect can be easily obtained with classical objects. For instance, imagine you can pick either red or blue socks in the morning. Since you like symmetry, you always choose both socks to have the same color. Thus, if someone sees only one of your socks, that person can be certain that the other sock has the same color. So far, this is the same as Alice measuring her qubit and knowing what Bob will get. Nonetheless, the randomness of the outcome in the case of qubits is a critical part of the story. Alice's outcome is only fixed once she measures it, and Bob will still have the same outcome no matter if Alice measures 0 or 1. This is the beauty and puzzle of quantum correlations.

We note that we said nothing about the distance between Alice and Bob so far. They could even be on the opposite sides of the galaxy and the argument would still work. The fact that Alice and Bob get the same result (either both of them get 0 or both of them get 1) does not require that they are close to each other. Would it then be possible to communicate instantaneously across large distances with some sort of quantum telephone? Unfortunately, no. This would violate one of the key postulates of special relativity: no information transfer can happen faster than the speed of light.

Unfortunately, we cannot build a quantum telephone that operates in this way. In the case above, Alice knows Bob's measurement, but Bob does not know Alice's measurement. Bob would have to check his qubit himself, and because the qubits are maximally entangled, he would then know the value of Alice's qubit. If Alice wanted to tell Bob about her measurement directly, she would have to somehow send that information to Bob in the form of a standard transmission such as a text file or short audio message. This latter communication is limited by the speed of light.

To use entanglement between qubits for quantum teleportation, we have to find a method to distribute entanglement between Alice and Bob. As the aim is to transmit quantum information from Alice to Bob over a distance, it would be counterproductive if they had to meet in person to exchange entangled qubits.

The solution to distributing entanglement can come in the form of entangled photons. As photons travel at the speed of light, this makes them perfectly suited to be 'entanglement messengers'. There are a variety of quantum optics experimental approaches available that can produce entanglement between two photons. We can send them to Alice and Bob to have them share an entangled pair. Once the entangled photons arrive at Alice and Bob, they must store the quantum state in some quantum device since photons cannot be stopped and always travel at the speed of light. Such a storage process has been demonstrated, but is still under active development [31–33].

So far, we have covered the fundamentals of quantum information and entanglement. Here, we show how all of this comes together to facilitate quantum teleportation.

For the protocol to be successful, three qubits are required. Alice is assigned two qubits $\vert \psi_{A_1}\rangle$ and $\vert \psi_{A_2}\rangle$, while one qubit $\vert \psi_B\rangle$ is intended for Bob's side of the communication. Our goal is to transfer the quantum information from Alice's qubit $\vert \psi_{A_1}\rangle$ to Bob's qubit $\vert \psi_B\rangle$.

Alice's second qubit A₂ is maximally entangled with Bob's qubit B leading to

$$\begin{align} \vert \phi_{A_2B}\rangle = \frac{1}{\sqrt{2}}\left( \vert 0_{A_2}0_B\rangle + \vert 1_{A_2}1_B\rangle\right). \end{align} \tag{ 4 }$$

Note that this is the same as equation (3), just with adapted indices. Although the notation seems a bit cumbersome, we use the indices here to highlight who holds which qubit. Now, the state of the whole system comprises qubit $\vert \psi_{A_1}\rangle$ and the entangled pair $\vert \phi_{A_2B}\rangle$, which is distributed across Alice and Bob, and can be written as is $\vert \psi_{A_1} \phi_{A_2B}\rangle$. The overall goal is to transfer the state of Alice's first qubit $\vert \psi_{A_1}\rangle$ to Bob's qubit $\vert \psi_B\rangle$. The initial setup is illustrated in step 1 of the quantum teleportation protocol shown in figure 2.

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The second step is to share the entanglement of Alice's second qubit (denoted A₂) and Bob's qubit with Alice's first qubit A₁ as illustrated in step 2 in figure 2. To achieve this, Alice performs two operations on her side (cf appendix B for further details). This manipulation on Alice's side leads to a special structure of the overall state of the system whereby Bob's qubit is in one of four states. Each of these states only slightly differs from the target state $\vert \psi_{A_1}\rangle$. Through the application of some correction operations, Bob could obtain the target state $\vert \psi_{A_1}\rangle$, but he does not know which operation he should perform as of yet.

In step 3, Alice obtains the information necessary for Bob's correction. Alice measures her two qubits and the superposition between the qubits of Alice and Bob collapses. As a result, Alice's pair of qubits $\vert \psi_{A_1}\rangle$ and $\vert \psi_{A_2}\rangle$ are then in one of four output pairs, which are 00, 01, 10, or 11. Now, Alice calls Bob and tells him the result of her measurement.

Knowing the result on Alice's side, Bob can correct his qubit (step 4). After the correction, Bob's qubit $\vert \psi_B\rangle$ then contains the state that was originally stored in Alice's qubit $\vert \psi_{A_1}\rangle$.

It is Christmas, and this year, Santa Claus will be delivering one type of present to every child around the world—a book.

During the protocol, the books represent the qubits while the content of the books represent the quantum information stored in the books. Importantly, a book is never teleported, it is always an individual book's contents that is transferred. The protocol involving Alice and Bob uses qubits which can be described with two basis states. These basis states are represented by a blank page ($\vert 0\rangle$) or a written page ($\vert 1\rangle$). A book contains much more information than just one qubit or one bit. For the sake of the metaphor, we will pretend that the book's contents can be described by just one qubit.

Unfortunately, hiding presents in all homes across the world is a time-consuming and arduous task, which essentially takes the whole year and cannot be completed in a single night. Santa likes to plan ahead, and he would love to distribute the books well ahead of time. However, he will only receive the children's wishlist at the start of December. Santa and his elves, therefore, resort to quantum teleportation to deliver the books on time for Christmas Eve.

To teleport the contents in the book to every child anywhere around the world, three books are required and they correspond to the three qubits needed for quantum teleportation. Two books $\vert \psi_{A_1}\rangle$ and $\vert \psi_{A_2}\rangle$ are based at the North Pole (A stands for Arctic Circle), while the third book $\vert \psi_B\rangle$ is one that needs to be placed at the home of the child prior to Christmas Eve. Here, we choose a child in Barcelona, hence the name $\vert \psi_B\rangle$. Long before Christmas, all books are at one of Santa's workshops close to the North Pole.

First, before any present book is delivered to a home, Santa and the elves use a special Christmas entanglement machine to entangle the present book $\vert \psi_B\rangle$ with the workshop book $\vert \psi_{A_2}\rangle$. The books are placed side-by-side in the machine, and the machine then evaluates if the content on any given page in both books will be either blank or full of text. Note, the entanglement machine cannot tell what the actual content is and nobody is allowed to open the books afterwards until the protocol is completed.

There are two key features here: the question that creates entanglement and the inability of the machine to tell the precise contents. The machine asks the question whether a certain page in both books is either blank or has text. Here, the two states of the page represent the two states of a qubit: $\vert 0\rangle$ means blank and $\vert 1\rangle$ stands for text. Since the question allows only two outcomes, both containing text or both blank, we ensure that the two-book system (two-qubit system) is in a state of the form $\vert BB\rangle+\vert TT\rangle$, where T stands for text and B stands for blank. This is the same structure as the maximally entangled state in equation (3). The two books are now playing the role of the maximally entangled state in the quantum teleportation protocol.

After the entangling procedure, nobody is allowed to open the book and the machine is not allowed to 'read' the text on a given page. This would correspond to measuring and the superposition created in the step would vanish. We would be left with purely classical and random information, and not with a quantum state.

The elves' job is not done after entangling the books. One half of the entangled book-pair has to find its way to the children's homes. Santa's elves are expert at hiding the present books in homes, and do so in a top-secret manner. This arduous task is completed over the course of the year so that the elves have time to visit all homes around the world. They have a lot of time since the actual present, which is the contents of the book stored in $\vert \psi_{A_1}\rangle$, will only be teleported on Christmas Eve.

Of course, it can happen that a child finds the present book at home, which would ruin their Christmas present from Santa Claus. This is not an issue though because if a child were to open the present book without proper completion of the quantum teleportation process, the entangled state would be broken. As a result, the child would see gibberish on the pages and it would make no sense.

The idea of opening the book early corresponds to measuring the quantum state. The entanglement will be destroyed and the child is just left with random information that is unintelligible and does not make for a nice present for the child.

The next stage of the process begins just before Santa Claus leaves his workshop on Christmas Eve.

First, for each child, the elves entangle the present book $\vert \psi_{A_1}\rangle$ with entangled book at the workshop $\vert \psi_{A_2}\rangle$ using the special Christmas entanglement machine. As before, the books are placed side-by-side in the entanglement machine, and the machine assesses whether on any given page from both books, there will be either a page full of text or a blank page in the two books. Then the elves add that information about the entanglement to a ledger for Santa Claus known as the Ledger of Good Nature to keep a record of the entanglement process. Whether a child receives a book on Christmas or not is influenced by their level of good nature over the course of the year.

The entanglement operation between the two books at the Arctic Circle workshop corresponds to the entanglement operation on Alice's side in the second step of the protocol (cf figure 2). In this Santa Claus example, the second step and the third step of the protocol are combined. The elves have to record the outcome of the measurement (step 3) because Santa needs the information to correct the book at the child's home in the final step of the process. If Santa decides not to bring a gift to the child, he can do so at the last minute, simply by not correcting the book on the child's side.

As a result of the creation of entanglement between $\vert \psi_{A_1}\rangle$ and $\vert \psi_{A_2}\rangle$, the elves have also created entanglement between $\vert \psi_{A_2}\rangle$ and $\vert \psi_B\rangle$ from Part 1 of the process.

The reason that book $\vert \psi_{A_2}\rangle$ is used in this process, and not $\vert \psi_{A_1}\rangle$ and $\vert \psi_B\rangle$ right away, is that the books stored in $\vert \psi_{A_1}\rangle$ are not assigned to children until very close to Christmas. This is because the wishlists from children only start to arrive from the start of December approximately. Using the first entanglement pair allows the elves and Santa to prepare an entangled book well in advance of Christmas, and then once the book $\vert \psi_{A_1}\rangle$ is selected for the child, it is entangled with $\vert \psi_{A_2}\rangle$, and as a result with $\vert \psi_B\rangle$.

The pair of entangled books is a tool to enable quantum teleportation later at a flexible and more opportune moment. For instance, imagine bringing a fork and a plate to a buffet dinner, but now knowing what you are going to eat at the buffet dinner. All you know for certain is that you will use the fork and plate to eat some food, but you do not know yet what food you will be eating. It is the same process with regards to the use of the first pair of entangled books—the elves and Santa Claus know they are going to need them, but they do not know yet what they will send with them.

On Christmas Eve, Santa Claus travels around the world to deliver the presents to the children. In effect, he travels around the world to complete the quantum teleportation process.

During the journey, Santa Claus checks the Ledger of Good Nature to see whether each child will receive a present or not. If a child has been well behaved during the year, then the child will get the book that they have asked for. However, if the ledger indicates that the child has misbehaved too much, then the child will not get the book that they asked for. Instead, they will get random, gibberish text in their book, because Santa does not correct the information in the book.

This part of the process is all about Santa Claus initiating the final step of the quantum teleportation process, the correction step. The result of the ensuing measurement was recorded in the ledger. Recall that $\vert \psi_{A_1}\rangle$ and $\vert \psi_B\rangle$ are linked to each other by the process.

With this in mind, Santa Claus must complete the final step of the quantum teleportation by travelling around the world. When Santa Claus arrives at the home of each child, he quickly locates the hidden book $\vert \psi_B\rangle$, and corrects its contents according to the information in his ledger. He sets up his present scanner with the right information from the Ledger of Good Nature, which adapts the book's contents to ensure that the present book reads the same as the originally intended book stored in $\vert \psi_{A_1}\rangle$.

On Christmas morning, when the child opens the book $\vert \psi_B\rangle$, they can open the book without worrying about seeing gibberish as a result of breaking the quantum teleportation process. Of course, if the child has been deemed to have misbehaved over the year, they will receive a book filled with gibberish text anyway!

We have designed a worksheet similar in form to those worksheets that one author has used to support lessons on the electromagnetic spectrum using the superhero genre [25–27], which were well received by students as demonstrated by results from a post-test survey of students. We envision that the worksheet presented here can also have a similar impact on the teaching of quantum teleportation in the classroom, which we intend to evaluate as part of a future study. The goal of a future study would be to assess the content and worksheet suitability for communicating on quantum teleportation and use student input on the worksheet to revise the worksheet where required.

The worksheet has one primary goal, which relates to the student learning objective, and that is to give students the opportunity to demonstrate their understanding of quantum teleportation. This is achieved in two ways. First, the students are asked to define the key attributes of quantum teleportation such as a qubit, superposition, and the key steps in quantum teleportation. Second, with the final question, the students are invited to critically assess the quantum teleportation process and to identify the steps that require the most attention in the Santa Claus example. In terms of the revised Bloom's taxonomy [36], this question engages the higher levels of cognitive complexity such as analyse, evaluate, and create. As a secondary goal, we aspire to excite students about the real possibilities of quantum teleportation.

The worksheet questions are intended to address naive conceptions that students have in relation to quantum physics. Below we list such naive conceptions and the questions from the worksheet in appendix C that are intended to help address them:

• Students struggle to transit from the deterministic world of their common experiences to the probabilistic world of quantum physics [17]—Question 2;
• Students have difficulties with interpreting and comprehending counter-intuitive quantum physics concepts such as qubits and superposition [12, 17]—Questions 1 and 2;
• Students struggle with the language of quantum concepts [17]—Questions 4 and 5;

The only question not categorised here is question 3 from the worksheet, which simply asks the students to give the number of qubits needed for quantum teleportation. This question acts as the bridge between the first two questions where the students are asked to expand on two quantum concepts, and the final two questions where the students are required to give details related to a quantum teleportation process.

The quantum teleportation worksheet (cf appendix C) is split into two sections. First, a motivation for the quantum teleportation worksheet based on the Christmas Eve adventures of Santa Claus is presented as well as the delivery expectations on Christmas Eve.

Second, the students are then invited to put their own stamp on the quantum teleportation process. They are invited to select a present type that they would like to distribute on Christmas Eve. Thereafter, the students can confine the delivering process to a particular city, country, or continent. The only invariable part of the worksheet is the inclusion of Santa Claus.

Finally, the students are posed a series of questions in relation to the key steps of the quantum teleportation process, as well as open questions to discuss with each other in relation to the process.

This worksheet can be used to encourage students to immediately apply their new knowledge related to quantum teleportation. Additionally, the questions encourage critically thinking about the concept. The combination of the worksheet and the lesson is intended to present a clear application of quantum teleportation, albeit using an unconventional and unlikely situation.

Such an unusual approach towards covering quantum concepts in the classroom could be adapted for other learning objectives in quantum physics. In a recent study exploring both teacher and student experiences of quantum physics lessons, students noted that the quantum physics topics are difficult, while teachers found it difficult to identify clear applications, which can lead students to question its relevance [5]. Furthermore, students have difficulty relating concepts from quantum physics to the macro-world in which they live [4]. Therefore, it is imperative that teachers are challenged to adopt new ways of enthusing students in their learning of quantum physics. An unconventional approach based on Santa Claus as outlined here with regards to quantum teleportation would certainly garner student attention given that it is based upon a popular culture icon. We envisage that a similar lesson plan and worksheet could be formulated for other concepts from quantum physics, although this is outside the scope of the current work.

The goal of quantum teleportation is to transfer the quantum state $\vert \psi\rangle = \alpha\vert 0\rangle+\beta\vert 1\rangle$ from Alice to Bob. Expressed as a vector, this equation reads

$$\begin{align}\begin{aligned}\vert\psi\rangle& =\alpha\vert 0\rangle+\beta\vert 1\rangle \\&= \alpha\left(\begin{array}{c}1\\0\end{array}\right)+\beta\left(\begin{array}{c}0\\1\end{array}\right) = \left(\begin{array}{c}\alpha\\\beta\end{array}\right).\end{aligned}\end{align} \tag{ B1 }$$

Here, we used the definitions of $\vert 0\rangle$ and $\vert 1\rangle$ in equation (1). In general, the coefficients can be complex, $\alpha,\beta\in\mathbb{C}$. For the sake of simplicity, we will assume for the rest of the example that the coefficients are real $\alpha,\beta\in\mathbb{R}$. The same computation works for complex coefficients.

As described in section 3.4, Alice and Bob have a resource to perform the quantum teleportation. An entangled state $\vert \phi\rangle$, where one of the qubits is located at Alice's side and the other is located at Bob's. In the main text, we wrote the state as $\vert \phi\rangle = \frac{1}{\sqrt{2}}(\vert 0_A0_B\rangle+\vert 1_A1_B\rangle)$. This state consists of two qubits: one on Alice's side and one on Bob's side. While a single qubit has two possible states, $\vert 0\rangle$ and $\vert 1\rangle$, this system as four possible states: $\vert 00\rangle$, $\vert 10\rangle$, $\vert 01\rangle$ and $\vert 11\rangle$. The state $\vert \phi\rangle$ can also be written as a vector

$$\begin{align} \vert \phi\rangle & = \frac{1}{\sqrt{2}}\left(\vert 0_A0_B\rangle+\vert 1_A1_B\rangle\right)\nonumber\\& = \frac{1}{\sqrt{2}}\left(\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)+\left(\begin{array}{c}0\\0\\0\\1\end{array}\right)\right) = \left(\begin{array}{c}1/\sqrt{2}\\0\\0\\1/\sqrt{2}\end{array}\right). \end{align} \tag{ B2 }$$

Instead of listing the basis states of a system, there is a more structured approach to combining quantum systems. By taking the tensor product, denoted by ⊗, the vectors grow automatically to the appropriate size. The tensor product takes two matrices of size m × n and r × k and returns a matrix of size mr × nk. A vector is interpreted as a n × 1 matrix and the same rules apply. We can directly apply the tensor product to two one-qubit vectors and see that we obtain the same vectors as we had before in equation (B2)

$$\begin{align} \vert 01\rangle & = \vert 0\rangle\otimes\vert 1\rangle = \left(\begin{array}{c}1\\0\end{array}\right)\otimes\left(\begin{array}{c}0\\1\end{array}\right)\nonumber\\& = \left(\begin{array}{c}1\left(\begin{array}{c}0\\1\end{array}\right)\\0\left(\begin{array}{c}0\\1\end{array}\right)\end{array}\right) = \left(\begin{array}{c}0\\1\\0\\0\end{array}\right) \end{align} \tag{ B3 }$$

$$\begin{align} \vert 00\rangle& = \vert 0\rangle\otimes\vert 0\rangle = \left(\begin{array}{c}1\\0\end{array}\right)\otimes\left(\begin{array}{c}1\\0\end{array}\right)\nonumber\\& = \left(\begin{array}{c}1\left(\begin{array}{c}1\\0\end{array}\right)\\0\left(\begin{array}{c}1\\0\end{array}\right)\end{array}\right) = \left(\begin{array}{c}1\\0\\0\\0\end{array}\right). \end{align} \tag{ B4 }$$

The notation of multiple numbers in the same ket, as in $\vert 01\rangle$, implicitly expresses a tensor product: $\vert 01\rangle = \vert 0\rangle\otimes\vert 1\rangle$.

Using the same procedure, we notice that a three-qubit system must have eight-dimensional vectors. When we first combine two single qubit systems, we obtain a four-dimensional object like $\vert 00\rangle$. Combining it with another two-dimensional qubit, we obtain an eight-dimensional state like $\vert 000\rangle$.

For the protocol we do not need only qubits, but also operators acting on the qubits. Operators are manipulations that change the state of the qubit. We start with an example on one qubit. The operation of flipping a qubit from $\vert 0\rangle$ to $\vert 1\rangle$ is performed by the X operator. You can think of the X operator as the logical $\mathrm{NOT}$. In matrix language, we express the operator as

$$\begin{equation} X = \left(\begin{matrix}0&1\\1&0\end{matrix}\right). \end{equation} \tag{ B5 }$$

We can check directly that it acts in the proper way

$$\begin{align} X\vert 0\rangle& = \left(\begin{matrix}0&1\\1&0\end{matrix}\right)\left(\begin{matrix}1\\0\end{matrix}\right) = \left(\begin{matrix}0\\1\end{matrix}\right);\nonumber\\ X\vert 1\rangle& = \left(\begin{matrix}0&1\\1&0\end{matrix}\right)\left(\begin{matrix}0\\1\end{matrix}\right) = \left(\begin{matrix}1\\0\end{matrix}\right). \end{align} \tag{ B6 }$$

The state $\vert 0\rangle$ is transformed into $\vert 1\rangle$, and vice versa.

The quantum teleportation protocol needs only two more operators. The first one is the Z operator and it leaves the computational basis states $\vert 0\rangle$ and $\vert 1\rangle$ invariant. It only adds a minus sign to $\vert 1\rangle$. Similarly, we apply the Z gate as

$$\begin{align}\begin{aligned}Z & =\left(\begin{array}{cc}1 & \\ & -1\end{array}\right)\\ Z\vert 0\rangle & =\left(\begin{array}{cc}1& \\ & -1\end{array}\right)\left(\begin{array}{c}1 \\ 0\end{array}\right)=\left(\begin{array}{c}1\\0\end{array}\right)\\ Z\vert 1\rangle & =\left(\begin{array}{cc}1 & \\ & -1\end{array}\right)\left(\begin{array}{c}0 \\ 1\end{array}\right)=-\left(\begin{array}{c}0\\1\end{array}\right).\end{aligned}\end{align} \tag{ B7 }$$

In addition to the X and the Z gate, we need the so-called Hadamard gate. It helps us to generate a superposition from the states $\vert 0\rangle$ and $\vert 1\rangle$. We can write it as

$$\begin{equation} H = \frac {1}{\sqrt {2}}\left(\begin{matrix}1&1\\1&-1\end{matrix}\right). \end{equation} \tag{ B8 }$$

When it acts on $\vert 0\rangle$ and $\vert 1\rangle$, we obtain

$$\begin{align} H\vert 0\rangle& = \frac{1}{\sqrt{2}}\left(\begin{matrix}1&1\\1&-1\end{matrix}\right)\left(\begin{matrix}1\\0\end{matrix}\right) = \frac{1}{\sqrt{2}}\left(\begin{matrix}1\\1\end{matrix}\right) \nonumber\\& = \frac{1}{\sqrt{2}}\left(\vert 0\rangle+\vert 1\rangle\right), \end{align} \tag{ B9 }$$

$$\begin{align} H\vert 1\rangle& = \frac{1}{\sqrt{2}}\left(\begin{matrix}1&1\\1&-1\end{matrix}\right)\left(\begin{matrix}0\\1\end{matrix}\right) = \frac{1}{\sqrt{2}}\left(\begin{matrix}1\\-1\end{matrix}\right) \nonumber\\& = \frac{1}{\sqrt{2}}\left(\vert 0\rangle-\vert 1\rangle\right). \end{align} \tag{ B10 }$$

Acting on a single qubit alone will not be enough as figure 2 already suggests with the bubbles wrapping around two qubits. We need one two qubit gate for the protocol: the so-called controlled-not $\mathrm{cNOT}$ operation. Since it acts on two qubits, it must be a $4\times 4$ matrix instead of a $2\times 2$ matrix: two-qubit states have four dimensions. The idea of the $\mathrm{cNOT}$ is to flip the second qubit depending on the first qubit (called control qubit). If the first qubit is in state $\vert 1\rangle$, the second qubit will be flipped. The matrix is given by

$$\begin{align} \mathrm{cNOT}& = \left(\begin{matrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{matrix}\right). \end{align} \tag{ B11 }$$

The action of the $\mathrm{cNOT}$ is illustrated by

$$\begin{align} \mathrm{cNOT}\vert 00\rangle& = \left(\begin{matrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{matrix}\right)\left(\begin{matrix}1\\0\\0\\0\end{matrix}\right) = \left(\begin{matrix}1\\0\\0\\0\end{matrix}\right) = \vert 00\rangle;\nonumber\\[4pt] \mathrm{cNOT}\vert 10\rangle& = \left(\begin{matrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{matrix}\right)\left(\begin{matrix}0\\0\\1\\0\end{matrix}\right) = \left(\begin{matrix}0\\0\\0\\1\end{matrix}\right) = \vert 11\rangle. \end{align} \tag{ B12 }$$

Note that the state of the first qubit always stays unchanged and the second qubit flips depending on the state of the first qubit.

Equipped with these tools, we are ready to write down the state of the full system and perform a quantum teleportation. We write the state of the system including all three qubits as

$$\begin{align}\begin{aligned}\vert S\rangle& = \vert \psi\rangle\otimes\vert \phi\rangle \\&= \left(\alpha\vert 0\rangle+\beta\vert 1\rangle\right)\otimes\frac{1}{\sqrt{2}}\left(\vert 00\rangle+\vert 11\rangle\right) \\&= \frac{1}{\sqrt{2}}\left(\alpha\vert 000\rangle+\alpha\vert 011\rangle+\beta\vert 100\rangle+\beta\vert 111\rangle\right).\end{aligned}\end{align} \tag{ B13 }$$

Here, we used the distributive property of the tensor product: we can directly compute

$$\begin{align} \left(\vert 0\rangle+\vert 1\rangle\right)\otimes\vert 1\rangle& = \vert 0\rangle\otimes\vert 1\rangle+\vert 1\rangle\otimes\vert 1\rangle\nonumber\\& = \vert 01\rangle+\vert 11\rangle \end{align} \tag{ B14 }$$

without writing the vector notation. The numbers in the ket expressions like $\vert 011\rangle$ follow the convention $\vert A_1 A_2 B\rangle$. The first qubit A₁ on Alice's side holds the information that we would like to teleport. Alice's second qubit A₂ is part of the Bell pair and the third qubit B is Bob's qubit, also part of the entangled pair $A_2 B$.

The first step of the protocol is an entanglement operation between Alice's two qubits. This operation is performed with a $\mathrm{cNOT}$ operation on both of Alice's qubits. Note that all operation of the protocol are local. That means that there are not two-qubit operations that act on a qubit of Alice and a qubit of Bob at the same time. We obtain the state after the first operation $\vert S_1\rangle$ by applying the $\mathrm{cNOT}$ gate

$$\begin{align}\begin{aligned}\vert S_1\rangle& = \left(\mathrm{cNOT}_{A_1A_2}\otimes \unicode{x1D7D9}_B\right)\vert S\rangle\\& = \left(\mathrm{cNOT}_{A_1A_2}\otimes \unicode{x1D7D9}_B\right)\frac{1}{\sqrt{2}}\left(\alpha\vert 000\rangle+\alpha\vert 011\rangle\right.\nonumber\\& \left. \quad +\beta\vert 100\rangle+\beta\vert 111\rangle\right)\\& = \frac{1}{\sqrt{2}}\left(\alpha\vert 000\rangle\!+\!\alpha\vert 011\rangle\!+\!\beta\vert 110\rangle\!+\!\beta\vert 101\rangle\!\right)\!.\end{aligned}\end{align} \tag{ B15 }$$

Before we write the same calculation in terms of matrices, let us have a look at the operator $\mathrm{cNOT}_{A_1A_2}\otimes \unicode{x1D7D9}$. As introduced above, $\mathrm{cNOT}$ acts on 2 qubits, but $\vert S\rangle$ is a three qubit state. The $\mathrm{cNOT}$ acts on the first two (Alice's) qubits, thus we have to define what happens on with the third qubit. since we do not want any action on the third qubit, we build the tensor product with the identity of size 2, written as $\unicode{x1D7D9}$.

Now, we will do the same computation in equation (B15) in terms of matrices. This computation adds nothing new to the derivation, it is just a different of writing it. First, we have to write down the matrix of the operation $\left(\mathrm{cNOT}_{A_1A_2}\otimes \unicode{x1D7D9}_B\right)$

$$\begin{align} \mathrm{cNOT}_{A_1A_2}\otimes \unicode{x1D7D9}_B& = \left(\begin{matrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{matrix}\right) \otimes \left(\begin{matrix}1&\\ &1\end{matrix}\right)\nonumber\\& = \left(\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\end{matrix}\right). \end{align} \tag{ B16 }$$

After writing the state $\vert S\rangle$ (cf equation (B13)) as

$$\begin{align}\begin{aligned}\vert S\rangle& = \vert \psi\rangle\otimes\vert \phi\rangle \\&= \left(\alpha\vert 0\rangle+\beta\vert 1\rangle\right)\otimes\frac{1}{\sqrt{2}}\left(\vert 00\rangle+\vert 11\rangle\right) \\& =\left(\begin{matrix}\alpha\\\beta\end{matrix}\right)\otimes \left(\begin{matrix}\frac{1}{\sqrt{2}}\\0\\0\\\frac{1}{\sqrt{2}}\end{matrix}\right)\nonumber\\& = \frac{1}{\sqrt{2}}\left(\begin{matrix}\alpha & 0 & 0 & \alpha & \beta & 0 & 0 & \beta\end{matrix}\right)^T\end{aligned}\end{align} \tag{ B17 }$$

we can transform the operation in equation

$$\begin{align}\begin{aligned}\vert S_1\rangle& = \left(\mathrm{cNOT}_{A_1A_2}\otimes \unicode{x1D7D9}_B\right)\vert S\rangle\nonumber\\& = \frac{1}{\sqrt{2}}\left(\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\end{matrix}\right)\left(\begin{matrix}\alpha \\ 0 \\ 0 \\ \alpha \\ \beta \\ 0 \\ 0 \\ \beta\end{matrix}\right) \\& = \frac{1}{\sqrt{2}} \left(\begin{matrix}\alpha & 0 & 0 & \alpha & 0 & \beta & \beta &0 \end{matrix}\right)^T.\end{aligned}\end{align} \tag{ B18 }$$

Here, we wrote the last vector in its transposed form for convenience. At this point we will continue with the algorithm in the bra-ket notation and show the equivalence with matrix calculus again after the next step. To see the direct correspondence between the last line of equations (B13) and (B17), we give the full basis for three qubits in appendix B.7.

The second operation of the quantum teleportation protocol is a Hadamard gate acting on the first qubit A₁. Note that this is still the first step as described in the main text

$$\begin{align}\begin{aligned}\vert S_2\rangle& = \left(H_{A_1}\otimes \unicode{x1D7D9}\otimes\unicode{x1D7D9}\right)\vert S_1\rangle\nonumber\\& = \frac{1}{\sqrt{2}}H_1\left(\alpha\vert 000\rangle+\alpha\vert 011\rangle+\beta\vert 110\rangle+\beta\vert 101\rangle\right)\\&=\frac{1}{2}\left(\alpha\left[\vert 000\rangle+\vert 100\rangle+\vert 011\rangle+\vert 111\rangle\right]\right.\nonumber\\&\left.\quad +\beta\left[\vert 010\rangle+\vert 001\rangle-\vert 110\rangle-\vert 101\rangle\right]\right).\end{aligned}\end{align} \tag{ B19 }$$

We can perform the equivalent operation by first writing the Hadamard gate in the three qubit basis as

$$\begin{align} H_{A_1}\otimes\unicode{x1D7D9}\otimes\unicode{x1D7D9} = \left(\begin{matrix} \frac{1}{\sqrt{2}} & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & -\frac{1}{\sqrt{2}}\end{matrix}\right). \end{align} \tag{ B20 }$$

The computation in equation (B19) is equivalent to

$$\begin{align}\begin{aligned}\vert S_2\rangle& = \left(H_{A_1}\otimes \unicode{x1D7D9}\otimes\unicode{x1D7D9}\right)\vert S_1\rangle \\& = \frac{1}{\sqrt{2}}\left(\begin{matrix}\frac{1}{\sqrt{2}} & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0\\ 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & -\frac{1}{\sqrt{2}}\end{matrix}\right) \left(\begin{matrix}\alpha \\ 0 \\ 0 \\ \alpha \\ 0 \\ \beta \\ \beta \\0\end{matrix}\right)\\& = \frac{1}{2} \left(\begin{matrix}\alpha & \beta & \beta & \alpha & \alpha & -\beta & -\beta & \alpha \end{matrix}\right)^T.\end{aligned}\end{align} \tag{ B21 }$$

Starting from equation, we can use the distributive property of the tensor product in the opposite direction and write equivalently

$$\begin{align} \begin{aligned} \vert S_2\rangle& = \frac{1}{2}\left(\vert 00\rangle\otimes\left(\alpha\vert 0\rangle+\beta\vert 1\rangle\right)\right)\\ & \quad+\frac{1}{2}\left(\vert 01\rangle\otimes\left(\alpha\vert 1\rangle+\beta\vert 0\rangle\right)\right)\\ & \quad+\frac{1}{2}\left(\vert 10\rangle\otimes\left(\alpha\vert 0\rangle-\beta\vert 1\rangle\right)\right)\\ & \quad+\frac{1}{2}\left(\vert 11\rangle\otimes\left(\alpha\vert 1\rangle-\beta\vert 0\rangle\right)\right). \end{aligned} \end{align} \tag{ B22 }$$

On the first glance this might look like a mathematical trick, but it actually has a physical interpretation. The coefficients α and β are now prefactors of the third qubit. This is Bob's qubit! That means that the information about the original state on Alice's side $\vert \psi\rangle = \alpha\vert 0\rangle+\beta\vert 1\rangle$ is now on Bob's side. The same step is much harder to see in matrix notation, but the decomposition is equally possible in the vector–matrix picture.

One problem remains: the prefactors are not precisely correct in all cases. Sometimes, they are in front of the wrong qubit ($\alpha\vert 1\rangle$ instead of $\alpha\vert 0\rangle$) and sometimes the signs do not match ($-\beta$ instead of β). We can fix this by allowing Alice to measure her qubits in the computational basis. By measuring, she obtains the bit values of her states, i.e. she reads out 01 for the state $\vert 01\rangle$. With this additional information, the problem is solved! The superposition in equation (B22) collapses and only one part of it is realised. Since Alice knows her measurement, e.g. 01, she knows that Bob's qubit must be in the state $\frac{1}{\sqrt{2}}\left(\alpha\vert 1\rangle+\beta\vert 0\rangle\right)$.

Alice submits her bit string to Bob and he knows now the operation that he has to perform to obtain the proper state $\vert \psi\rangle$. If Alice measures, for example, 01, he knows that the signs of the coefficients are correct, but they are in front of the wrong states. By applying an X gate to this state, he will obtain the correct state

$$\begin{align} X\left(\beta\vert 0\rangle+\alpha\vert 1\rangle\right) = \alpha\vert 0\rangle+\beta\vert 1\rangle. \end{align} \tag{ B23 }$$

All four possible outcomes and the corresponding operations are listed in table 1.

Table 1. Operations on Bob's side to adapt the state according to Alice's measurement.

Bob's state	Alice's measuremnt	Bob's operation
$\alpha\vert 0\rangle+\beta\vert 1\rangle$	00	$\unicode{x1D7D9}$
$\alpha\vert 1\rangle+\beta\vert 0\rangle$	01	X
$\alpha\vert 0\rangle-\beta\vert 1\rangle$	10	Z
$\alpha\vert 1\rangle-\beta\vert 0\rangle$	11	ZX

The measurement on Alice's side is not only essential for the protocol, it is also important to avoid a contradiction with the no-cloning theorem [28, 29]. The no-cloning theorem is a so-called no-go theorem, a proof that shows the impossibility of something. In this case, the no-cloning theorem states that it is impossible to copy an arbitrary quantum state. Since we do not know anything about the quantum state $\vert \psi\rangle$ of Alice, i.e. α and β are unkown, we should not be allowed to copy it. And indeed, we are not copying it. Alice destroys her quantum state by measuring her two qubits. Thus, the state is transferred (or teleported) and not copied.

The second remark is related to special relativity. One of the postulates of special relativity is that no information can be transferred faster than light. If we could transfer the state from Alice to Bob without Alice telling Bob about her measurements, we would have indeed transferred information faster than light. However, as Bob will not know whether he has the correct state or a modified one until Alice tells him her measurements. Alice will communicate her measurement results to Bob via a classical communication channel, e.g. telephone, email or any other means of non-quantum communication. This communication is limited by the speed of light. In the end, no information is transmitted faster than light and quantum teleportation does not contradict special relativity.

As a reference and to ease the translation between bra-ket notation and matrix-vector notation, we provide here the all computational basis states for three qubits

$$\begin{align} \begin{aligned} \vert 000\rangle& = \left(\begin{matrix}1&0&0&0&0&0&0&0\end{matrix}\right)^T\\ \vert 001\rangle& = \left(\begin{matrix}0&1&0&0&0&0&0&0\end{matrix}\right)^T\\ \vert 010\rangle& = \left(\begin{matrix}0&0&1&0&0&0&0&0\end{matrix}\right)^T\\ \vert 011\rangle& = \left(\begin{matrix}0&0&0&1&0&0&0&0\end{matrix}\right)^T\\ \vert 100\rangle& = \left(\begin{matrix}0&0&0&0&1&0&0&0\end{matrix}\right)^T\\ \vert 101\rangle& = \left(\begin{matrix}0&0&0&0&0&1&0&0\end{matrix}\right)^T\\ \vert 110\rangle& = \left(\begin{matrix}0&0&0&0&0&0&1&0\end{matrix}\right)^T\\ \vert 111\rangle& = \left(\begin{matrix}0&0&0&0&0&0&0&1\end{matrix}\right)^T. \end{aligned} \end{align} \tag{ B24 }$$