# Beam Me Up, Qubit!

**Prepare for Transport**

It's been a while, but if you have followed this website throughout the past couple of weeks, then you have (hopefully) learned that

- Scientists have not teleported "objects" into space,
- What a quantum wave function is and how we use them,
- What quantum entanglement looks like, and
- A little mathematical notation!

Now, the time has finally come for us to put this knowledge to use and figure out how to transmit a qubit from one point to another. Joining us in this adventure are the two heroes of our tale: a brilliant scientist named Alice and her faithful assistant, Bob. To begin, Alice and Bob both have qubits with a \(50\%–50\%\) chance of either being a \(\lvert 0 \rangle\) or a \(\lvert 1 \rangle\):\begin{equation} \label{080117ABqb} \begin{aligned}

\lvert \psi \rangle_A &= \frac{1}{\sqrt{2}} \left( \lvert 0 \rangle_A + \lvert 1 \rangle_A \right),

\\[2ex]

\lvert \psi \rangle_B &= \frac{1}{\sqrt{2}} \left( \lvert 0 \rangle_B + \lvert 1 \rangle_A \right).

\end{aligned} \end{equation}Their goal is to create a "quantum comm-link" that will allow them to send quantum information in the form of a third qubit from one person to another, and the first step in this procedure is for Alice and Bob to entangle their qubits together. Earlier, we saw with our quantum playing cards that Alice and Bob's qubits automatically exist together in a non-entangled state that is a combination of two two different entangled pairs:\begin{equation} \label{080117ABwf} \lvert \psi \rangle _A\otimes \lvert \psi \rangle _B = \frac{1}{\sqrt{2}} \left( \lvert \Psi \rangle_{AB} + \lvert \Phi \rangle_{AB} \right). \end{equation}At this point, it therefore becomes necessary to spend a little more time with these entangled wave functions. In the last post, I introduced two of them; but in reality, there are four:\begin{equation} \begin{aligned}

\lvert \Psi^+ \rangle_{AB} &= \frac{1}{\sqrt{2}} \left( \lvert 0 \rangle_A \otimes \lvert 1 \rangle_B + \lvert 1 \rangle_A \otimes \lvert 0 \rangle_B \right),

\\[2ex]

\lvert \Psi^- \rangle_{AB} &= \frac{1}{\sqrt{2}} \left( \lvert 0 \rangle_A \otimes \lvert 1 \rangle_B - \lvert 1 \rangle_A \otimes \lvert 0 \rangle_B \right),

\\[2ex]

\lvert \Phi^+ \rangle_{AB} &= \frac{1}{\sqrt{2}} \left( \lvert 0 \rangle_A \otimes \lvert 0 \rangle_B + \lvert 1 \rangle_A \otimes \lvert 1 \rangle_B \right),

\\[2ex]

\lvert \Phi^- \rangle_{AB} &= \frac{1}{\sqrt{2}} \left( \lvert 0 \rangle_A \otimes \lvert 0 \rangle_B - \lvert 1 \rangle_A \otimes \lvert 1 \rangle_B \right).

\end{aligned} \label{080117bell} \end{equation}

Meet the Bell States, named after physicist John Bell. They describe the four maximally-entangled possibilities for two qubits. The two states labeled with a capital Greek "Psi," \(\Psi\), represent states in which the qubits always have opposite values, while those labeled with the capital Greek "Phi," \(\Phi\), are states in which the qubits always have the same value. The "\(+\)" and "\(-\)" superscripts denote whether the wave function is symmetric or anti-symmetric, respectively, which is not particularly important for our current project. Just know that there are four different Bell states, and they are listed above.

When Alice and Bob entangle their qubits, they are collapsing their shared wave function, \(\lvert \psi \rangle _A\otimes \lvert \psi \rangle _B\), into one of those four states, and this is doable by sending their qubits through some kind of filter. For example, say Alice and Bob represent their qubits with photons, and that the two different photon polarization states represent \(\lvert 0 \rangle\) and \(\lvert 1 \rangle\). Sending these photons through a polarizing beam splitter divides them into two groups: those with polarization \(\lvert 0 \rangle\) and those with polarization \(\lvert 1 \rangle\). However, we don't know who's photon wound up in which state, and so the system as a whole is described by\begin{equation}

\lvert \Psi^\pm \rangle_{AB} = \frac{1}{\sqrt{2}} \left( \lvert 0 \rangle_A \otimes \lvert 1 \rangle_B \pm \lvert 1 \rangle_A \otimes \lvert 0 \rangle_B \right). \end{equation}

(For the record, this is essentially how the Chinese researchers entangled their qubits!)

For the sake of this example, let's say that Alice and Bob's qubits wind up in the symmetric \(\lvert \Psi^+ \rangle\) state, and are therefore now described by the entangled wave function\begin{equation}

\lvert \Psi^+ \rangle_{AB} = \frac{1}{\sqrt{2}} \left( \lvert 0 \rangle_A \otimes \lvert 1 \rangle_B + \lvert 1 \rangle_A \otimes \lvert 0 \rangle_B \right). \end{equation}Again, what this wave function says is that it is *absolutely certain* that we will find Alice and Bob's qubits in different states then they are measured, but there is a \(50\%–50\%\) uncertainty of whose qubit will be in which state. It is also worth pointing out what this wave function *does not say*, such as *anything* about the locations of these two qubits! That's a fun property of entanglement: the two qubits can be ** anywhere in the Universe**, and still be entangled! Best of all, that is not merely some strange feature of quantum mechanics—

*it's just the way things are.*

For example, consider our entangled **non**-quantum playing cards; the setup where one is **definitely** a diamond and the other is **definitely** a spade. We turn them face down, we shuffle them up, and then we take them to opposite ends of the observable universe. I turn mine face-up and discover that it's a spade. Which card do you have? A diamond, right? You *know* this is true, despite the fact that our cards are 93 billion light-years apart! That is because entanglement is *really* a statement of logic, and logic (hopefully) works anywhere:

- One card is a spade,
- The other is a diamond.
- I have a spade,
- So you must have a diamond.

\(\therefore \text{QED}\)

*Location irrelevant!*

Of course, you cannot *confirm* that I have a spade until I send that information to you, and the fastest possible way to do that is via an electromagnetic wave. This wave could be a radio transmission, a laser pulse, or any other form of light, but no matter what form it takes, it is going to travel *at the speed of light*. In fact, the speed of light, \(c\), is really the **speed limit of information**. No information—not even quantum information—can travel through the Universe faster than \(c\), and this fact will become apparent as we continue our discussion of quantum teleportation.

So, let's get back to it! When we left off, Alice and Bob had entangled their qubits in the \(\lvert \Psi^+ \rangle_{AB}\) state. Now that entanglement has been achieved, let us say that Bob takes his qubit off to space! (*Lucky him...*) Since the locations of the qubits are irrelevant to their entanglement, *they are still entangled, and still described by* \(\lvert \Psi^+ \rangle_{AB}\). Meanwhile, back down on Earth, Alice meets up with her colleague, Carol, who is in possession of a *third* qubit,\begin{equation} \lvert \psi \rangle_C = \alpha \vert 0 \rangle_C + \beta \vert 1 \rangle_C. \label{080117psic} \end{equation}

(Remember, the parameters \(\alpha\) and \(\beta\) can be *anything*, so long as \(\lvert \alpha \rvert^2 + \lvert \beta \rvert^2 = 1. \))

*Just by merely existing* (no physical manipulation has occurred, yet), Alice's, Bob's, and Carol's qubits are described by the combined wave function\begin{equation} \begin{aligned}

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

\\[2ex]

&= \frac{1}{\sqrt{2}} \left( \alpha \lvert 0 \rangle_A \otimes \lvert 1 \rangle_B \otimes \lvert 0 \rangle_C + \beta \lvert 0 \rangle_A \otimes \lvert 1 \rangle_B \otimes \lvert 1 \rangle_C

\\[2ex]

\quad \quad \quad \; \; + \alpha \lvert 1 \rangle_A \otimes \lvert 0 \rangle_B \otimes \lvert 0 \rangle_C + \beta \lvert 1 \rangle_A \otimes \lvert 0 \rangle_B \otimes \lvert 1 \rangle_C \right)

\\[2ex]

&= \frac{1}{\sqrt{2}} \left[ \alpha \left( \lvert 0 \rangle_A \otimes \lvert 0 \rangle_C \right) \otimes \lvert 1 \rangle_B + \beta \left( \lvert 0 \rangle_A \otimes \lvert 1 \rangle_C \right) \otimes \lvert 1 \rangle_B

\\[2ex]

\quad \quad \quad \; \; + \alpha \left( \lvert 1 \rangle_A \otimes \lvert 0 \rangle_C \right) \otimes \lvert 0 \rangle_B + \beta \left( \lvert 1 \rangle_A \otimes \lvert 1 \rangle_C \right) \otimes \lvert 0 \rangle_B \right].

\end{aligned} \label{080117vfABC} \end{equation}

That may look a little messy, but what I have done is expand out the tensor product \(\lvert \Psi^+ \rangle_{AB} \otimes \lvert \psi \rangle_C\) and then group together the tensor products of Alice's and Carol's qubits. As we have already seen with other tensor products of non-entangled quantum states, the factors in parentheses in the last lines can be rewritten in terms of the the Bell states:\begin{equation} \begin{aligned}

\lvert 0 \rangle_A \otimes \lvert 1 \rangle_C &= \frac{1}{\sqrt{2}} \left( \lvert \Psi^+ \rangle_{AC} + \lvert \Psi^- \rangle_{AC} \right),

\\[2ex]

\lvert 1 \rangle_A \otimes \lvert 0 \rangle_C &= \frac{1}{\sqrt{2}} \left( \lvert \Psi^+ \rangle_{AC} - \lvert \Psi^- \rangle_{AC} \right),

\\[2ex]

\lvert 0 \rangle_A \otimes \lvert 0 \rangle_C &= \frac{1}{\sqrt{2}} \left( \lvert \Phi^+ \rangle_{AC} + \lvert \Phi^- \rangle_{AC} \right),

\\[2ex]

\lvert 1 \rangle_A \otimes \lvert 1 \rangle_C &= \frac{1}{\sqrt{2}} \left( \lvert \Phi^+ \rangle_{AC} - \lvert \Phi^- \rangle_{AC} \right).

\end{aligned} \label{080117bb} \end{equation}

(*Pro-tip: It is a fantastic exercise for the reader to verify these equations using *\eqref{080117bell}

*. Just replace the Bell states in*\eqref{080117bb}

*with thier expressions in*\eqref{080117bell}

*, switching all of the subscript*\(B\)

*'s to*\(C\)

*'s, and simplify!*)

The next step is therefore to substitute \eqref{080117bb} into \eqref{080117vfABC}, expand it out, and then collect all of the terms into multiples of the four Bell states:\begin{equation} \begin{aligned}

\lvert \Psi^+ \rangle_{AB} \otimes \lvert \psi \rangle_C

&= \frac{1}{2} \left[ \alpha \left( \lvert \Phi^+ \rangle_{AC} + \lvert \Phi^- \rangle_{AC} \right) \otimes \lvert 1 \rangle_B + \beta \left( \lvert \Psi^+ \rangle_{AC} + \lvert \Psi^- \rangle_{AC} \right) \otimes \lvert 1 \rangle_B

\\[2ex]

\quad\quad \, + \alpha \left( \lvert \Psi^+ \rangle_{AC} - \lvert \Psi^- \rangle_{AC} \right) \otimes \lvert 0 \rangle_B + \beta \left( \lvert \Phi^+ \rangle_{AC} - \lvert \Phi^- \rangle_{AC} \right) \otimes \lvert 0 \rangle_B \right]

\\[2ex]

&= \frac{1}{2} \left( \lvert \Phi^+ \rangle_{AC} \otimes \alpha \lvert 1 \rangle_B + \lvert \Phi^- \rangle_{AC} \otimes \alpha \lvert 1 \rangle_B

\\[2ex]

\quad \; \; \, + \lvert \Psi^+ \rangle_{AC} \otimes \beta \lvert 1 \rangle_B + \lvert \Psi^- \rangle_{AC} \otimes \beta \lvert 1 \rangle_B

\\[2ex]

\quad \; \; \; + \lvert \Psi^+ \rangle_{AC} \otimes \alpha \lvert 0 \rangle_B - \lvert \Psi^- \rangle_{AC} \otimes \alpha \lvert 0 \rangle_B

\\[2ex]

\quad \; \; \; \; \; + \lvert \Phi^+ \rangle_{AC} \otimes \beta \lvert 0 \rangle_B - \lvert \Phi^- \rangle_{AC} \otimes \beta \lvert 0 \rangle_B \right)

\\[2ex]

&= \frac{1}{2} \left[ \lvert \Phi^+ \rangle_{AC} \otimes \left( \beta \lvert 0 \rangle_B + \alpha \lvert 1 \rangle_B \right) + \lvert \Psi^+ \rangle_{AC} \otimes \left( \alpha \lvert 0 \rangle_B + \beta \lvert 1 \rangle_B \right)

\\[2ex]

\quad \quad \; \; + \lvert \Psi^- \rangle_{AC} \otimes \left( \beta \lvert 1 \rangle_B - \alpha \lvert 0 \rangle_B \right) + \lvert \Phi^- \rangle_{AC} \otimes \left( \alpha \lvert 1 \rangle_B - \beta \lvert 0 \rangle_B \right) \right].

\end{aligned} \label{080217cb} \end{equation}

Alright, if you were able to follow along with \eqref{080117vfABC}–\eqref{080217cb}, then ya done good! If you were not able to follow, try going back and looking at it again more carefully. The symbols might be a bit obtuse, but the math is basic high school algebra—junior high school material in some places! If you are still lost, just know this: Alice and Bob have entangled qubits, and Carol has a *separate* qubit whose state she would like to send from Alice's location to Bob's. \(\lvert \Psi^+ \rangle_{AB} \otimes \lvert \psi \rangle_C\) describes this system by saying that Alice and Bob's qubits are entangled and Carol's **just ****exists somewhere in the same universe.** (Note once again that the wave function says *nothing specific* about the locations of any of the qubits, meaning they could be *anywhere!*) Then, the only thing we do in equations \eqref{080117vfABC}–\eqref{080217cb} is rewrite \(\lvert \Psi^+ \rangle_{AB} \otimes \lvert \psi \rangle_C\) in a more useful form. Physically, **nothing has happened yet**, and the last lines of \eqref{080217cb} *still* describe the state in which Alice and Bob's qubits are entangled and Carol's exists separately somewhere within the same universe.

However, the last lines of equation \eqref{080217cb} also tell us something new. Take a close look, and try to use your knowledge of quantum wave functions to determine the predictions this expression makes about measurements of our system. It has four terms and an overall factor of \(1/2\), meaning there is a \((1/2)^2 = 1/4 = 25\%\) chance of obtaining one of the following four results when the appropriate measurement is made:\begin{equation} \begin{array}

\lvert \lvert \Psi^+ \rangle_{AC} \otimes \left( \alpha \lvert 0 \rangle_B + \beta \lvert 1 \rangle_B \right),

\\[2ex]

\lvert \Psi^- \rangle_{AC} \otimes \left( \beta \lvert 1 \rangle_B - \alpha \lvert 0 \rangle_B \right),

\\[2ex]

\lvert \Phi^+ \rangle_{AC} \otimes \left( \beta \lvert 0 \rangle_B + \alpha \lvert 1 \rangle_B \right),

\\[2ex]

\lvert \Phi^- \rangle_{AC} \otimes \left( \alpha \lvert 1 \rangle_B - \beta \lvert 0 \rangle_B \right).

\end{array} \label{080217res} \end{equation}

What exactly is this "appropriate measurement," one might ask? It is one that entangles Alice's and Carol's qubits into a Bell state, of course, according to table \eqref{080217res}. When this process is done, Bob's qubit *looses* its entanglement with Alice's, and it acquires a state that is *very similar to, if not identical to* the original wave function of Carol's qubit!

For example, Alice and Carol could subject their qubits to the same process that Alice and Bob first used to entangle their qubits. Then, Alice and Carol's qubits would wind up in the \(\lvert \Psi^+ \rangle_{AC}\) state, and Bob's qubit would be in the \( \alpha \lvert 0 \rangle_B + \beta \lvert 1 \rangle_B \) state. But wait a minute, that's the *original* state of Carol's qubit, equation \eqref{080117psic}. That means **quantum information that was initially held by Carol is now in the hands of Bob!**

But what if Alice and Carol's qubits become entangled in a different Bell state—after all, all four are equally possible. In that case, Bob's qubit collapses into a different superposition of \(\lvert 0 \rangle\) and \(\lvert 1 \rangle\); i.e., whichever one is in a tensor product with Alice's and Carol's measured Bell state. Alas, it seems that our quest to transfer Carol's qubit is foiled. But not so fast! Notice how all of the \(B\) states in table \eqref{080217res} are simply reshufflings of the others, with a possible sign change. In quantum mechanics, one achieves such shufflings of a wave function through a unitary operation, which by definition are always undoable via another unitary operation. An example of this is a rotation: say you perform the unitary operation of rotating an object *clockwise*. You can then return this object to its original state by merely performing the unitary operation of rotating it *counterclockwise*. Thus, any states that are unitary transformations of each other can be transformed into each other through unitary transformations. (If you find that statement confusing, go back to the clockwise-counterclockwise rotation example above.)

So, regardless of which state Bob's qubit winds up in, he can easily transform it into the one matching equation \eqref{080117psic} by applying the right unitary operation—he just needs to know which operation that is. Unfortunately, he has no way of knowing this, *but Alice and Carol do!* Remember, they are the ones who performed the measurement that collapsed the whole system's wave function, and they, therefore, know within which Bell state their qubits have become entangled. Furthermore, because they are both smart people who can do the same calculations that we did in equations \eqref{080117vfABC}–\eqref{080217cb}, they also know which state Bob's qubit is in: the one that's in a tensor product with their measured Bell state. Therefore, they can tell Bob which unitary operation he needs to perform to recover Carol's original state!

However, to get this information to Bob, they have to send him a signal, and as we have already discussed, ** that information cannot travel faster than light.** For example, if Bob is in space, then Alice and Carol might transmit their message to him via radio. Radio waves are a form of electromagnetic radiation, i.e.,

*light*, and therefore they travel

*at the speed of light*. Or Alice and Carol could shine a laser at Bob, the color of which tells him which unitary operation he must perform. Laser beams are beams of light, and therefore

*they travel at the speed of light*. This even applies if Bob is not in space; say he is standing right next to Alice and Carol and can therefore see the result of their measurement. Well,

*seeing*things occurs when light comes from those things and enters one's eyes, and it should go without saying that this happens

**at the speed of light!**What this means is that even though it *seems* that Bob's qubit's wave function changes at the same instant as Alice's and Carol's, he cannot actually decode the quantum state they are trying to send him, and thus obtain its information, faster than the speed of light. *That* is why I said earlier that the "speed of light" is really **the speed limit of information.** It turns out that **NO ****information, not even quantum information, can travel from one point in the universe to another faster.**

Furthermore, there is another significant result to all of this, one that makes data transfer via quantum teleportation *very interesting* to spy agencies around the world: the fact that it is *physically impossible* to intercept and decode this message. Radio transmissions can be tuned into and deciphered, notes can be intercepted and read, computers are monitorable, but entangled qubits? Short of punching Bob and stealing his, there is no way to eavesdrop on that! Even if someone intercepts the signal from Alice and Carol telling Bob which unitary operation he must perform to recover the state they are sending him, the only information this would-be spy gains is *knowledge of which unitary operation Bob must perform to recover the state Alice and Carol are sending him*. The spy **cannot** obtain the actual wave function unless she steals it directly from Bob!

Therefore, you can think of quantum teleportation as *more* than just a way of transmitting quantum information from one point to another, which is an absolute necessity if one intends to build a quantum network. It is also true that quantum teleportation is the **best virtual private network in the Universe. ***That* is what makes this technology so interesting! It is **NOT** going to lead to Star Trek transporters, but it has potential to revolutionize the Information Age, which has already transformed human civilization. This research lays the foundation for a future *Quantum Information Age*, one in which all of your qubits are safe.

*That* is why you should get excited. That and the fact that you now understand quantum teleportation, and you can share it with all of your friends! Feel free to share these articles, too, for that helps me out a lot. And please stay tuned for more! We have finally reached the peak of quantum teleportation, but I feel that there is one more blog post on this topic in order—one that explains what the Chinese researchers actually did. Now that you know how quantum teleportation works, you should be able to follow their method, and perhaps even devise an experiment of your very own. When you do, be sure to share it with me, for I would love to see what you are capable of!

Thanks, Y'all, and I'll see you next time,

Aaron