Tex source for notes on a talk about quantum teleportation

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\title{Notes for \emph{Using the peculiar properties of quantum entanglement to teleport quantum states}}
\author{Matthew Wetmore}
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\maketitle
\abstract{Erwin Schr\"odinger called quantum entanglement ``not one but rather the characteristic trait of quantum mechanics" when he introduced the term in 1935. The phenomenon, which was used by Albert Einstein and others as an argument against the framework of quantum mechanics that allowed it to arise, has long served as a reminder that classical lines of thought are often useless when considering quantum systems. Yet despite its perplexing behaviour, quantum entanglement has stood the tests of time and experiment, and its peculiar characteristics have served as a useful tool in the fields of quantum information theory (QIT) and quantum computation (QC). 

In this talk, I will outline the history of entanglement, and describe the phenomenon mathematically. I will then present the groundbreaking 1993 approach to using entanglement to ``teleport" a quantum state from one place to another, which has applications in QIT and QC. The relevant quantum mechanics will be briefly introduced before their application, so no technical understanding of QM is assumed.}

\tableofcontents

\section{Quantum Mechanics}

\subsection{Motivation}

For a period during the late nineteenth century, physicists thought they more or less completely understood the physical laws of the world around them. The culmination of centuries of experiment and theory, physics had come to understand complex classical phenomenon. Many believed that all of the great discoveries had been made, and the rest of the work would be to refine our understanding.

Experiments in the 1880s and 1890s, however, hinted that the current understanding of physics was not complete. Behaviour such as the photoelectric effect was still unexplained, and if current theories were taken to extremes their behaviour was... strange. Planck's 1900 theory that energy was discretized - came in discrete ``quanta" - introduced the beginning of quantum mechanics (as many people learn in chemistry classes). As physicists began to use this new theory to explore and explain the behaviour of the very small (atoms and their ilk), a new revolution in physics was kicked off. 

 

\section{A brief history of entanglement}

\subsection{The EPR paper}

The framework of quantum mechanics had proven to accurately describe a lot of previously-baffling behaviour. However, its inherently probabilistic nature bothered some people - as Einstein famously remarked, ``God doesn't play with dice" (the actual quote may differ in wording slightly). In the 1935 paper ``Can Quantum-Mechanical Description of Physical Reality Be Considered Complete?", Einstein, along with Boris Podolsky and Nathan Rosen, raised an apparent paradox of sorts - often referred to as the EPR paradox. They considered what was soon after labeled "entanglement" between two particles, showing that at any distance, the results of certain measurements on both particles would correlate - that is, a measurement of one particle would directly influence the results of a subsequent measurement of the other particle.

Such behaviour is always astounding when it's first heard - how could this be? Could it lead to information somehow travelling faster than light (violating relativity)? Thus it was a paradox, and one that caused some unrest. While other explanations were proposed, it turns out that quantum mechanics gives the only explanation for the kind of correlation between particles we're talking about. Even though the EPR paper raised the topic of entanglement critically, it ended up having profound influence in future quantum theory. This notion of entanglement was surely something to think about.

\subsection{What is entanglement mathematically?}

It's actually not so bad once you look at it mathematically. What we're going to look at is the following state, which is an example of something called a \textbf{Bell state}:

$$ \ket{\Psi^+_{12}} = \frac{1}{\sqrt{2}}(\ket{\uparrow_1}\ket{\downarrow_2} + \ket{\downarrow_1}\ket{\uparrow_2}) $$

The subscript on the left hand side indicates that we're dealing with a state of 2 particles - namely, particle 1 and particle 2 (so creative!). We see from the arrows that we're dealing with states of spin up and spin down, so we know we'll be dealing with spin-$\frac{1}{2}$ particles.

So what does this equation mean? We'll bring in the famous Alice and Bob, who have served in so many mathematical explanations before. Alice has particle 1 in her possession (just don't ask where and how fast), and Bob has particle 2 (without loss of generality). This equation tells us that if Alice makes a measurement on her particle, she will get a state of $\ket{\uparrow}$ or $\ket{\downarrow}$ - with equal, 50\% chance of either outcome. However, if Bob makes a subsequent measurement in the same basis of his particle, his state would be dependent on what Alice got. Whichever term in the above equation contains Alice's outcome also contains Bob's outcome. If she gets up, he gets down, and vice versa. 

It's worth noting that to Bob, his measurement also appears random. It's only after Alice and Bob tell each other their results that they realize the results are correlated (or, in this case, anticorrelated, as the results are opposites). There is no way to separate this equation into two disjoint terms, one for particle 1 and the other for particle 2. This is the mathematical consequence of entanglement - the state equation isn't separable.

To get another look at it, here are the rest of the Bell states - the four maximally entangled states possible for a system of two spin-$\frac{1}{2}$ particles:

\begin{align*}
\ket{\Psi^-_{12}} = \frac{1}{\sqrt{2}}(\ket{\uparrow_1}\ket{\downarrow_2} - \ket{\downarrow_1}\ket{\uparrow_2})\\
\ket{\Phi^+_{12}} = \frac{1}{\sqrt{2}}(\ket{\uparrow_1}\ket{\uparrow_2} + \ket{\downarrow_1}\ket{\downarrow_2})\\
\ket{\Phi^-_{12}} = \frac{1}{\sqrt{2}}(\ket{\uparrow_1}\ket{\uparrow_2} - \ket{\downarrow_1}\ket{\downarrow_2})
\end{align*}

\section{Using entanglement for quantum teleportation}

Alright, here we are at the meat of the talk: let's learn how to teleport.

\subsection{Wait a second, what does teleportation really mean?}

When we talk about quantum teleportation, we are talking about taking the quantum state $\ket{\phi}$ of one particle, and giving it to a new particle. I don't mean that we copy it - we can't actually do that unfortunately because of the rather descriptive \textbf{no-cloning theorem}. Instead, the original particle is given a new state, one that is no longer $\ket{\phi}$. But luckily the second particle preserves $\ket{\phi}$ for us now.

In order to motivate this rather peculiar action, let's consider the restrictions present in quantum mechanics when it comes to moving states around. As we just noted, a quantum state cannot be cloned - we can't make a copy without destroying the original. This naturally means we can't broadcast the state since that would require many copies of it. Hmm. Perhaps we can measure our state sufficiently to rebuild it somewhere else - this is known as \emph{classical} teleportation (note the distinction). Unfortunately, we run into another theorem, the \textbf{no-teleportation theorem}, which basically says that there is no reliable way to measure a quantum state completely with a single measurement.

So I guess the only thing we can do now is just physically give the state to someone else - but that's very prone  to failure. And now we're in a bind. This Gordian knot was cut in 1993 by Charles Bennett, Giles Brassard, and McGill's very own Claude Cr\'epeau\footnote{Unfortunately the paper was published before Cr\'epeau came to McGill, where he is now a professor of computer science.}, among others, when they applied entanglement to the problem, introducing \emph{quantum} teleportation.

\subsection{Going through the steps}

Let's set up the situation here. We'll invite Alice and Bob back in. Alice has a particle, which we will call particle 1, in the state $\ket{\phi_1}$, and she wants to give the state to Bob, via quantum teleportation. To do this, we start with an entangled pair, one of the Bell states from before. The Bell state

$$
\ket{\Psi^-_{23}} = \frac{1}{\sqrt{2}}(\ket{\uparrow_2}\ket{\downarrow_3} - \ket{\downarrow_2}\ket{\uparrow_3})
$$
is a \textbf{singlet} state, meaning the total spin of the system vanishes. Actually, it's commonly referred to as an EPR singlet, due to the paper we mentioned before. Note the 2 and 3 in the subscript - we will call the paired particles 2 and 3. We will give particle 2 to Alice and particle 3 to Bob. The total state of the system at this point, considering the three particles, is $\ket{\phi_1}\ket{\Psi^-_{23}}$ - a pure product state, meaning no sort of entanglement exists between particle 1 and the EPR pair. 

Let's consider the subsystem consisting of particles Alice has; that is, particles 1 and 2. Currently the two particles are not correlated - however, we can couple the two particles with a measurement. Earlier we mentioned that a state can have different representations if we make a change of basis, and we can make measurements in such a basis. In order to couple her two particles, Alice performs a measurement on the system in the Bell operator basis - that is, the basis consisting of $\ket{\Psi^\pm_{12}}$ and $\ket{\Phi^\pm_{12}}$ (note the subscripts). These four states form a complete set of orthonormal basis vectors for the system of particles 1 and 2. Note we can represent any possible outcome from the standard, spin-up-spin-down basis with Bell states as a result:

\begin{equation}
\ket{\uparrow_1}\ket{\uparrow_2} = \frac{1}{\sqrt{2}}(\ket{\Phi^+_{12}} + \ket{\Phi^-_{12}}), \quad\ket{\downarrow_1}\ket{\uparrow_2} = \frac{1}{\sqrt{2}}(\ket{\Psi^+_{12}} - \ket{\Psi^-_{12}})
\label{bell-basis}
\end{equation}
and so on.\footnote{Note the similarity to finding the real and imaginary parts of a complex number. This is a result of the fact that $\ket{\Phi^+_{12}}$ is in a way the conjugate of $\ket{\Phi^-_{12}}$ (similarly for $\ket{\Psi^\pm_{12}}$).}

In the standard basis, we can write the state $\ket{\phi_1}$ as follows:

$$ \ket{\phi_1} = \alpha\ket{\uparrow_1} + \beta\ket{\downarrow_1} $$
where $\alpha$ and $\beta$ are complex numbers such that $|\alpha|^2 + |\beta|^2 = 1$. Using this, we can rewrite our product state of the three particles as 

\begin{align*}
\ket{\phi_1}\ket{\Psi^-_{23}} &= (\alpha\ket{\uparrow_1} + \beta\ket{\downarrow_1})\frac{1}{\sqrt{2}}(\ket{\uparrow_2}\ket{\downarrow_3} - \ket{\downarrow_2}\ket{\uparrow_3}) \\
&= \frac{\alpha}{\sqrt{2}}(\ket{\uparrow_1}\ket{\uparrow_2}\ket{\downarrow_3} - \ket{\uparrow_1}\ket{\downarrow_2}\ket{\uparrow_3}) + \frac{\beta}{\sqrt{2}}(\ket{\downarrow_1}\ket{\uparrow_2}\ket{\downarrow_3} - \ket{\downarrow_1}\ket{\downarrow_2}\ket{\uparrow_3})
\end{align*}

As we showed in Equation~\ref{bell-basis} we can rewrite direct product states of the form $\ket{\,\cdot\,_1}\ket{\,\cdot\,_2}$ in the Bell operator basis. Let's look at our total state again:

$$ \frac{\alpha}{\sqrt{2}}(\underbrace{\ket{\uparrow_1}\ket{\uparrow_2}}_{\frac{\ket{\Phi^+_{12}} + \ket{\Phi^-_{12}}}{\sqrt{2}}}\ket{\downarrow_3} - \underbrace{\ket{\uparrow_1}\ket{\downarrow_2}}_{\frac{\ket{\Psi^+_{12}} + \ket{\Psi^-_{12}}}{\sqrt{2}}}\ket{\uparrow_3}) + \frac{\beta}{\sqrt{2}}(\underbrace{\ket{\downarrow_1}\ket{\uparrow_2}}_{\frac{\ket{\Psi^+_{12}} - \ket{\Psi^-_{12}}}{\sqrt{2}}}\ket{\downarrow_3} - \underbrace{\ket{\downarrow_1}\ket{\downarrow_2}}_{\frac{\ket{\Phi^+_{12}} - \ket{\Phi^-_{12}}}{\sqrt{2}}}\ket{\uparrow_3}) $$

Substituting in the Bell operator basis representations we identify above, we end up with the following expression for the total state of the system, where particles 1 and 2 are in the Bell operator basis.

\begin{align*}
\frac{1}{2}\bigg[
&\ket{\Psi^-_{12}}(-\alpha\ket{\uparrow_3} - \beta\ket{\downarrow_3})\\ +
&\ket{\Psi^+_{12}}(-\alpha\ket{\uparrow_3} + \beta\ket{\downarrow_3})\\ + 
&\ket{\Phi^-_{12}}(\beta\ket{\uparrow_3} + \alpha\ket{\downarrow_3})\\ + 
&\ket{\Phi^+_{12}}(-\beta\ket{\uparrow_3} + \alpha\ket{\downarrow_3})
\bigg]
\end{align*}

In the above equation, first note that we have 4 terms, each representing an outcome of a measurement on the system of particles 1 and 2 in the Bell operator basis. By distributivity, each term has the scalar $\frac{1}{2} = \frac{1}{\sqrt{4}}$ in front, meaning each outcome has the equally likely probability of $1 / 4$ of occurring.   Examining the terms, we see that each one is a direct product of some Bell state with particles 1 and 2 (indicating they will be entangled after the measurement is made), and some state of particle 3 in the standard basis\footnote{Notice that the state of particles 1 and 2 is separable from the state of particle 3, so particles 2 and 3 are no longer entangled.}. What's so important about this term is that the amplitudes for the outcomes are some form of $\alpha$ and $\beta$, just like our initial state $\ket{\phi_1}$ of particle 1! Note that even though Alice is performing her measurement only on particles 1 and 2, it's affecting the outcome for a measurement of particle 3 as well, because of the original entanglement between particles 2 and 3.

This leads us to our final step. After Alice makes her measurement, she ends up with her two particles in an entangled Bell state - and there are 4 possible states they can end up in. As her result determines the resulting state of Bob's particle 3, Alice can communicate her result to Bob with a minimum of 2 classical bits (as $2^2 = 4$). Once Bob hears what Alice got, he can perform an operation on his particle, so that he ends up with $\ket{\phi_3} = \alpha\ket{\uparrow_3} + \beta\ket{\downarrow_3}$ - the same state as $\ket{\phi_1}$, just encoded in particle 3.

Let's look at what Bob needs to do in each outcome to recover $\ket{\phi_3}$. We'll go down the terms consecutively in the above equation. For the first outcome, particle 3 is in the state $-\alpha\ket{\uparrow_3} - \beta\ket{\downarrow_3} = -(\alpha\ket{\uparrow_3} + \beta\ket{\downarrow_3}) = e^{i\pi}\ket{\phi_3}$. The $e^{i\pi}$ represents an overall phase, which we can ignore. Thus we see that in the first outcome, Bob's particle is already in the state $\ket{\phi_3}$, and he doesn't need to do anything. If 

In order to look at the other 3 outcomes, we can represent the state ket $\ket{\phi_3}$ as a column vector, using the spin-up and spin-down states as a basis. Thus

$$ \ket{\phi_3} = \alpha\ket{\uparrow_3} + \beta\ket{\downarrow_3} = \colvec{2}{\alpha}{\beta} $$
and so for our second term, the state of particle 3 after the measurement is

$$
-\alpha\ket{\uparrow_3} + \beta\ket{\downarrow_3} =
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}\ket{\phi_3}
$$
Similarly we can represent the state of particle 3 in the remaining terms as

$$ \beta\ket{\uparrow_3} + \alpha\ket{\downarrow_3} =
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}\ket{\phi_3} $$
and 

$$ -\beta\ket{\uparrow_3} + \alpha\ket{\downarrow_3} =
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}\ket{\phi_3} $$

So for each term, we need some unitary operation to apply to the state to cancel the matrix and recover $\ket{\phi_3}.$\footnote{At least to some overall phase factor. Remember, overall phase does not affect the behaviour of the system.} It turns out that for each term, we can actually just apply the operation represented by the matrix applied to $\ket{\phi_3}$. So for the state of particle 3 in the last outcome above, we can apply the unitary operation $\op Y$ to the state, where 

$$ \op Y = \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix} $$
obtaining

CHANGE THIS STUFF

$$ \op Y\big[-\beta\ket{\uparrow_3} + \alpha\ket{\downarrow_3}\big] =
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}^2\ket{\phi_3} =
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}\ket{\phi_3} = -\ket{\phi_3} $$
where once again we can ignore the $-1$, which is an overall phase and as such can be ignored. The operator $\op Y$ corresponds to a 180$^\circ$ around the $\op Y$ axis. Since

$$ \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}^2 = \begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}^2 = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$$
it follows that if we define

$$ \op X = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} \quad\text{and}\quad \op Z = \begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix} $$
then 

$$ \op X\big[\beta\ket{\uparrow_3} + \alpha\ket{\downarrow_3}\big] = \ket{\phi_3}\quad\text{and}\quad\op Z\big[-\alpha\ket{\uparrow_3} + \beta\ket{\downarrow_3}\big] = \ket{\phi_3} $$

So for each outcome of Alice's measurement, we have a unitary transformation (namely and 180$^\circ$ rotation about the axis in the operator's name) to apply to Bob's particle 3 to obtain $\ket{\phi_3}$, the same state that Alice's particle 1 initially held.

And now we know how to teleport a quantum state!

\subsection{TL;DR}

\textbf{Objective} Alice wants to send the quantum state $\ket{\phi_1}$, currently on particle 1, to Bob.
\textbf{Prerequisites} Alice and Bob need to share an entangled pair.

\begin{enumerate}

\item

Start with an EPR singlet of the form $\ket{\Psi^-_{23}} = \frac{1}{\sqrt{2}}(\ket{\uparrow_2}\ket{\downarrow_3} - \ket{\downarrow_2}\ket{\uparrow_3})$ - an entangled state between particles 2 and 3.

\item

Alice takes particle 2, Bob takes particle 3.

\item

Alice makes a measurement on her two particles (1 and 2) in the Bell operator basis. There are four equally-likely outcomes.

\item

Alice encodes her result in 2 bits and sends it to Bob (directly or via broadcast).

\item

Based on Alice's result, Bob applies a rotation of some sort to his particle 3, or simply does nothing.

\item Bob's particle 3 is now in the state $\ket{\phi_3}$ as desired.

\end{enumerate}
\textbf{Side effects} Particles 1 and 2 are now entangled in a Bell state.

\subsection{Taking this further}

Teleportation as we have constructed it is a linear operation that can be applied to a state $\ket{\phi}$. This means that it can work on entangled, mixed, or pure states. In fact, we can use it to swap entanglement between different particles! Say Alice's particle 1 is itself entangled to another particle, particle 0. Then after we make the teleportation, particle 0 is entangled with particle 3 in a singlet state.

We can also generalize the operation to transport states of more than 2 orthogonal states. We start with an entangled pair of $N$-state particles, in order to teleport a particle of $N > 2$ orthogonal states. This state is of the form 

$$ \sum_j \ket{j}\ket{j} / \sqrt{N} \quad\text{for $j = 0, 1, \ldots, N -1$} $$
and we note the similarity to the Bell state $\ket{\Phi^+_{23}}$. She then, as before, makes a measurement on the subsystem of particles 1 and 2, in the basis with basis vectors

$$ \ket{\psi_{nm}} = \sum_j e^{2\pi i j n / N}\ket{j}\ket{(j+m) \!\!\!\mod N} / \sqrt{N} $$
where $n, m$ can be varied from $0$ to $N - 1$. Note we can generate the Bell states for $N = 2$ using this formula, such that

$$ \ket{\psi_{00}} = \ket{\Phi^+} \quad \ket{\psi_{01}} = \ket{\Psi^+} \quad \ket{\psi_{10}} = \ket{\Phi^-} \quad \ket{\psi_{11}} = \ket{\Psi^-} $$
So far this approach is a generalization of our approach for spin-$\frac{1}{2}$ particles.

The result of this measurement is some $\ket{\psi_{nm}}$, and when Bob is told what the result is he applies the unitary transformation given by

$$ U_{nm} = \sum_k e^{2\pi i k n / N}\ket{k}\bra{(k+m)\!\!\!\mod N} $$

This completes the teleportation.

\subsection{Can we do better?}

As we mentioned earlier, teleportation through purely classical or quantum means is not possible. This doesn't stop us from thinking about what could happen without the classical channel in this protocol. It's certainly limiting - so what if Bob guessed what result Alice got and operated on his particle before he got the message? This would, of course, be useless - there is no possible correlation between his guess and the result. If there was he could send a superluminal message - violating relativity. It's possible to prove that a minimum of 2 bits is required to complete the teleportation.

\subsection{So are we doing this thing or what?}

Yup. It took 5 years, but in 1998 the polarization state of a photon was teleported a short distance. Over the years, the distance of successful teleportations increased, reaching a peak of 143km (89 miles) between islands off of the coast of Africa. Distance is not the only metric of efficacy though. Researchers have also teleported quantum information between gas clouds at the Niels Bohr Institute, and teleported strongly entangled states as well, preserving the entanglement (which is important in quantum information applications).

Unfortunately it's not as easy as the theory might make it seem. For some perspective, consider this excerpt from the abstract of the paper announcing the 143km teleportation:

\begin{quote}``To achieve this, the experiment had to employ novel techniques such as a frequency-uncorrelated polarization-entangled photon pair source, ultra-low-noise single-photon detectors, and entanglement-assisted clock synchronization."
\end{quote}
As the technology allowing long-term storage and measurement of EPR pairs evolves, the barriers making quantum teleportation such an ordeal will hopefully be broken down.



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