Why can't quantum teleportation be used to transport information?

Kaku Michio says in an interview that we've teleported photons, cesium atoms and beryllium atoms. Having watched a lot of Kaku as well as way too many astrophysics documentaries in general, I know he's exaggerating. From what I understand, some state of those particles were teleported.

Here's my question: what state exactly is being teleported, and why can't it be used to transfer classical information? How can we know experimentally that a teleportation occured, if the transported information can't be read classically?

Most of the news articles for the general public just throw around the word "state" without explaining the idea (or flat out tell us like Kaku that atoms were teleported), and Wikipedia uses math and terminology that's way over my head.

Is there a layman's explanation by any chance?

Answer

To elaborate on Misha's answer: here is a spiel on teleportation from quantum information theory. I'm going to do some actual computations, but they will be made as simple as possible for novice readers.

Beginning of the sketch

Let us simplify the discussion by limiting ourselves to the teleportation of a single bit's worth of information, e.g. a spin ½ particle. Let us denote spin-down by the bit-value 0, and spin-up by 1. A general (pure, i.e. non-randomized) state of the spin ½ particle is given by $|\sigma\rangle_s = \alpha_0 |0\rangle_s + \alpha_1 |1\rangle_s$: a superposition of the two spin-values, where $|\alpha_0|^2$ and $|\alpha_1|^2$ must sum to 1 and are otherwise unconstrained complex numbers. If you wanted to store a single bit, you could just use $\alpha_0 = 1$ or $\alpha_1 = 1$ to represent "0" or "1" respectively. (The subscript s just means that this indicates the state of a spin "s".)

For teleportation between two distant locations A and B (where A is in the same general area as the spin s), you need an entangled state as a resource, such as

$$|\beta_{00}\rangle_{A,B} = \frac{1}{\sqrt 2}\Bigl[\; |0\rangle_A|0\rangle_B + |1\rangle_A|1\rangle_B \Bigr]$$ which is a superposition of two "classical" states: one where both A and B are in the state 0, and one where both are in the state 1. This is a notation which represents a vector, but that's not important for this post.

• You should think of the symbols $|u\rangle_{\!\!X}$ as being like variables that you can multiply together in whichever order you like; each one describes the state of one part of the system; and combinations of these products can be used to describe superpositions. (This "multiplication" is actually a tensor product; but that's also not important for this post.)


• The factor of $1/\!\sqrt{2}\,$ is just for the sake of book-keeping, so that we can compute probabilities. We won't be doing a lot of probability calculations, but there's an interesting comparison coming below which makes it worthwhile to mention normalization. —— If you're happy with the notion of a vector, what this is doing is ensuring that the vector has norm 1; corresponding roughly to being in this entangled two-particle state with certainty. (Shorter vectors represent states which occur only with some probability: we will see an example later.) The terms $|0\rangle_A|0\rangle_B$ and $|1\rangle_A|1\rangle_B$ are also unit vectors, and are orthogonal to one another because they represent perfectly distinguishable classical states. Their sum has length $\sqrt 2$, so we divide by this to get another unit vector. States in quantum info are typically represented by unit vectors.) Another way of reading this is that if we measured this state to get bit values, we would have a probability of $(1/\!\sqrt 2)^2 = \frac12$ change of obtaining 00 for A and B, and the same chance of obtaining 11 for A and B.

Now, I'm going to make a sharp right turn, and talk for a little about cryptography, because the protocoal for teleportation bears a striking resemblance to the protocol for transmitting a message secretly with a one-time pad.

A digression into classical cryptography

A one-time pad is nothing more than two copies of the same sequence of random numbers, e.g. strings of random digits. When you want to encrypt a message which is n bits long, to send from Alice (A) to Bob (B), the two sequences of random bits must have length at least n. For example, if you want to encode a single bit of information, it is enough for A and B to share a single random bit. In a manner of speaking, the random bits in their position must be in a joint state of the form

$$\mathbf{Pads} = \frac{1}{2} \Bigl[ \Bigl(\langle 0\rangle_A \text{ and } \langle 0 \rangle_B\Bigr) + \Bigl(\langle 1 \rangle_A \text{ and } \langle 1 \rangle_B\Bigr) \Bigr]$$ where $\langle x \rangle_{\!A/B}\,$ just describes the scenario of Alice's/Bob's bit being x, the addition represents a sort of disjunction between the two (exclusive) events, and where the scaling of 1/2 represents the probabilities with which these two events occur. (I've written it this way deliberately; the mathematics for writing this as a classical probability distribution is almost identical to what one would use for writing quantum states such as $|\beta_{00}\rangle$.) If Alice has a single bit s ∈ {0,1} that she wants to send to Bob, she measures — yes, measures; you could also say compute if you like — the parity of s with her bit A, $$s' = s \oplus A = \begin{cases} 0,& \text{if $s = A$} \\ 1,& \text{if $s \ne A$} \end{cases}$$ where $\oplus$ is addition modulo two [$0 \oplus 0 = 1 \oplus 1 = 0$, and $0 \oplus 1 = 1 \oplus 0 = 1$] and then sends the result s' to Bob, who either flips his own bit B according to whether s' is zero or one: $$B' = B \oplus s'\;.$$ Exercise for the reader: show that B' = s, Alice's bit prior to encryption. Of course, before obtaining s', Bob has no way of guessing what the value of B' will be: no amount of staring at his copy of the original shared bit will reveal any operations Alice has done.

Back to teleportation!

How teleportation actually works (mathematically speaking)

In order to teleport $|\sigma\rangle_s$, Alice performs a joint measurement on s and A. There are four possible outcomes, so the most information Alice can obtain out of her measurement is two bits. The measurement she performs is in the basis

$$\begin{align*} |\beta_{00}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|0\rangle_A + |1\rangle_s|1\rangle_A \Bigr], \\ |\beta_{01}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|1\rangle_A + |1\rangle_s|0\rangle_A \Bigr], \\ |\beta_{10}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|0\rangle_A - |1\rangle_s|1\rangle_A \Bigr], \\ |\beta_{11}\rangle_{s,A} \;&=\; \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|1\rangle_A - |1\rangle_s|0\rangle_A \Bigr]. \end{align*}$$ There is a reason why we choose these states in particular; the high-level summary for why we choose these states is given in the final paragraph where I complete the analogy to the one-time pad. (This is not the only possible choice of basis, but it's the simplest.) She will obtain one of the states $$|\beta_{zx}\rangle_{s,A} = \tfrac{1}{\sqrt 2}\Bigl[\; |0\rangle_s|x\rangle_A + (-1)^z |1\rangle_s|1 \oplus x\rangle_A \Bigr]$$ as a result, which can be represented succinctly by the two bits x, z ∈ {0,1}. How do we compute the effect of this measurement? By applying the operator $$\langle \beta_{zx} |_{s,A} = \tfrac{1}{\sqrt 2}\Bigl[\; \langle 0|_s\langle x|_A + (-1)^z \langle 1|_s \langle 1 \oplus x|_A \Bigr]$$ to the input state $|\sigma\rangle_s |\beta_{00}\rangle_{A,B}$, distributing the $\langle \text{whatever}|_X$ symbols over addition like linear operators (which they are), expanding the "multiplication" of the state $|\sigma\rangle_s$ with $|\beta_{00}\rangle_{A,B}$, and using the combination rule $\langle u |_{\!\!X}\, | v \rangle_{\!\!X} = \langle u | v\rangle_{\!\!X} = \delta_{u,v}$ for symbols with a common subscript (this is mechanical calculation, skip to the end if you prefer): $$\begin{align*} \langle \beta_{xz} |_{s,A} &\Bigl[ |\sigma\rangle_s | \beta_{00}\rangle_{A,B} \Bigr] \\\\\\=&\; \tfrac{1}{\sqrt 2}\langle \beta_{xz} |_{s,A}\Bigl[\; \alpha_0 |0\rangle_s |0\rangle_s|0\rangle_A + \alpha_0 |0\rangle_s |1\rangle_s|1\rangle_A + \alpha_1 |1\rangle_s |0\rangle_s|0\rangle_A + \alpha_1 |1\rangle_s |1\rangle_s|1\rangle_A \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \langle 0|_s\langle x|_A + (-1)^z \langle 1|_s \langle 1 \oplus x|_A \Bigr] \times \\&\qquad\qquad \Bigl[\; \alpha_0 |0\rangle_s |0\rangle_A|0\rangle_B + \alpha_0 |0\rangle_s |1\rangle_A|1\rangle_B + \alpha_1 |1\rangle_s |0\rangle_A|0\rangle_B + \alpha_1 |1\rangle_s |1\rangle_A|1\rangle_B \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 \langle 0|0\rangle_s \langle x|0\rangle_A|0\rangle_B + \alpha_0 \langle 0|0\rangle_s \langle x|1\rangle_A|1\rangle_B \\&\qquad + \alpha_1 \langle 0|1\rangle_s \langle x|0\rangle_A|0\rangle_B + \alpha_1 \langle 0|1\rangle_s \langle x|1\rangle_A|1\rangle_B \\&\qquad + (-1)^z \alpha_0 \langle 1|0\rangle_s \langle 1 \oplus x |0\rangle_A|0\rangle_B + (-1)^z \alpha_0 \langle 1|0\rangle_s \langle 1 \oplus x |1\rangle_A|1\rangle_B \\&\qquad + (-1)^z \alpha_1 \langle 1|1\rangle_s \langle 1 \oplus x |0\rangle_A|0\rangle_B + (-1)^z \alpha_1 \langle 1|1\rangle_s \langle 1 \oplus x |1\rangle_A|1\rangle_B\Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 \langle x|0\rangle_A|0\rangle_B + \alpha_0 \langle x|1\rangle_A|1\rangle_B \\&\qquad + (-1)^z \alpha_1 \langle 1 \oplus x |0\rangle_A|0\rangle_B + (-1)^z \alpha_1 \langle 1 \oplus x |1\rangle_A|1\rangle_B\Bigr] \\\\\\=&\; \tfrac{1}{2} \delta_{x,0} \Bigl[\; \alpha_0 |0\rangle_B + (-1)^z \alpha_1 |1\rangle_B \Bigr] + \tfrac{1}{2} \delta_{x,1} \Bigl[\; \alpha_0 |1\rangle_B + (-1)^z \alpha_1 |0\rangle_B \Bigr] \\\\\\=&\; \tfrac{1}{2} \Bigl[\; \alpha_0 |x\rangle_B + (-1)^z \alpha_1 |1 \oplus x\rangle_B \Bigr]. \end{align*}$$ This calculation is doing nothing more than computing the marginal state on B, conditioned on obtaining the state $|\beta_{zx}\rangle_{\!s,A}$ on s and A. People still argue about the interpretation of this concept, but the mathematics itself is identical to what you would do for computing classical probability distributions on B conditioned on some property of s and A holding (such as the parity of s and A being some bit x ∈ {0,1}).

The scalar factor of $\frac12$ just represents the fact that the outcome $|\beta_{zx}\rangle$ of the measurement arises, independently of the values of x and z, with probability $(\frac12)^2 = \frac14$; the final state is some state on B which depends on the values of x and z. Without knowing what x and z are, Bob doesn't know the result of the measurement collapse — any more than he could guess Alice's message in the one-time pad protocol by staring at his own random bit. In fact, we can show that without knowledge of x and z, Bob's state looks perfectly random to him, just as it would have if he had measured B before Alice did anything, and just as the random bit in a one-time pad looks. But if Alice were to send x and z to him, he could perform operations on his state to obtain the state $|\sigma\rangle_B = \alpha_0 |0\rangle_B + \alpha_1 |1\rangle_B$. Which operations he performs depends on the values of x and z, just as whether Bob flips his bit in the one-time pad protocol depends on the message Alice sends him.

Completing the analogy to the one-time pad

The measurement above, in the basis of $|\beta_{zx}\rangle$ states, corresponds exactly to a parity measurement in the classical one-time pad scenario: the two bits x and z represent measured correlations between the spins s and A, in the standard basis and in the basis $|+\rangle \propto |0\rangle + |1\rangle$ and $|-\rangle \propto |0\rangle - |1\rangle$. In particular, you'll notice that whenever we obtain an outcome of $|\beta_{zx}\rangle$ where x = 0, Bob's final state is incorrect only by the sign of the $|1\rangle$ component; and when x = 1, then mapping each component of Bob's state from $|0\rangle$ to $|1\rangle$ (and vice-vera) gives him a state which is again correct up to a sign. The x bit corresponds exactly to the bit-parity measurement in a one-time pad classically; and without that bit, you could never hope to communicate classical information by teleportation. But on the plus side, although you have to send a bit of communication, whichever state s was in to begin with has been transmitted to Bob with perfect security!