Using a fractional number of classical bits within quantum teleportation

Question

Recently, I heard that there can be transfer of rational classical bits (for example 1.5 cbits) from one party to another via quantum teleportation. In the Standard Teleportation Protocol, 2 classical bits and 1 maximally entangled shared resource state is required for perfect teleportation of the unknown state. But I do not understand how $1.x$ bits can be sent over in the classical channel.

1. Is that possible? If yes, could you give a brief explanation?

2. It'd be helpful if you could point me to some papers in which perfect teleportation is possible using fractional bits (and possibly extra quantum resources).

Some people might be wondering as to how this may be relevant to quantum computing. D. Gottesman and I.L. Chuang suggested that quantum teleportation will play an important role as a primitive subroutine in quantum computation. G. Brassard, S.L. Braunstein and R. Cleve showed that quantum teleportation can be understood as quantum computation.

Answer 1

I don't know for sure how you would achieve fewer than two bits of classical communication for a teleportation, but here's one way that you could have a non-integer number: if you teleport a qudit with dimension $d$ that is not a power of two. For each teleportation protocol, you'd have to send two dits of information, which you could represent in bits using $\lceil 2\log_2(d)\rceil$ bits. If you then repeat the protocol many times, you could combine the classical messages that you're sending and reduce it to $2\log_2(d)$ per teleportation protocol on average.

One possible route towards fewer than two bits of classical communication (if that's what you're after) is to use a combination of imperfect teleportation and non-universal teleportation (where we have some prior knowledge of what the state to be teleported could be). If your resource state is $\alpha|00\rangle+\sqrt{1-\alpha^2}|11\rangle$, then the probability of getting each measurement result in the teleportation protocol depends on the value of $\alpha$: teleporting a state $(\cos\frac{\theta}{2}|0\rangle+\sin\frac{\theta}{2}e^{i\phi}|1\rangle)$ gives the probailities of the four different Bell measurements, $$ |B_{xy}\rangle=\frac{1}{\sqrt{2}}\left(|0x\rangle+(-1)^y|1\bar x\rangle\right) $$ as $$ p_{xy}=\frac{1}{4}(1+(-1)^x(2\alpha^2-1)\cos\theta), $$ where $x$ and $y$ are single bits. Using the input distribution for the unknown quantum state, we can calculate the average value of $\sin\theta$.

For universal teleportation (where the input state could be any state), one has $\int_0^{\pi}\cos\theta\sin\theta d\theta=0$. In this case, the probabilities are all equal, and the best we can do is just to send the measurement result as two bits, $xy$.

Now imagine the case where $(2\alpha^2-1)\langle\cos\theta\rangle=\frac12$. Then, the probabilities are $(\frac38,\frac38,\frac18,\frac18)$. One can compress this information using, for example, Huffman encoding: $\{00,01,10,11\}\mapsto\{0,10,110,111\}$. This has an expected message length $\frac{15}{8}$. Thus, if you repeat this protocol many times, on average you send 1.875 bits per teleportation. This, of course, is just an example. Any value $(2\alpha^2-1)\langle\cos\theta\rangle>\frac13$ gives compression.

The trade-off is that unless $|\alpha|^2=|\beta|^2=\frac12$ (where you don't get any compression), the teleportation is imperfect.

Answer 2

I recently found a paper by Subhash Kak that introduces teleportation protocols that require lesser classical communication cost (with more quantum resource). I thought it'd be better to write a separate answer.

Kak discusses three protocols; two of them use 1 cbit and the last one requires 1.5 cbits. But the first two protocols are in a different setting, i.e the entangled particles are initially in Alice's lab (and a few local operations are performed), then one of the entangled particle is transferred to Bob's lab; this is unlike the Standard setting where the entangled particles are pre-shared between Alice and Bob before the protocol is even started. Interested people can go through those protocols that use only 1 cbit. I'll try to explain the last protocol that uses only 1.5 cbits (fractional cbits).

There are four particles, namely, $X, Y, Z$ and $U$. $X$ is the unkown particle (or state) that has to be teleported from Alice's lab to Bob's lab. $X, Y$ and $Z$ are with Alice, and $U$ is with Bob. Let $X$ be represented as $\alpha|0\rangle + \beta|1\rangle$, such that $|\alpha|^2+|\beta|^2=1$. The three particles $Y, Z$ and $U$ are in the pure entangled state $|000\rangle+|111\rangle$ (leaving the normalization constants for now).

So, the initial state of the whole system is: $$ \alpha|0000\rangle + \beta|1000\rangle + \alpha|0111\rangle + \beta|1111\rangle $$

Step 1: Apply chained XOR transformations on $X, Y$ and $Z$ (i) XOR the states of $X$ and $Y$ (ii) XOR the states of $Y$ and $Z$.

The $XOR$ unitary is given by: $$ XOR = \left[{\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}}\right]. $$

In other words, the state transformations are the following: $$ |00\rangle \rightarrow |00\rangle \\ |01\rangle \rightarrow |01\rangle \\ |10\rangle \rightarrow |11\rangle \\ |11\rangle \rightarrow |10\rangle \\ $$

After Step 1, the state of the whole system is: $$ \alpha|0000\rangle + \beta|1110\rangle + \alpha|0101\rangle + \beta|1011\rangle $$

Step 2: Apply Hadamard tranform on the state of $X$. $$ \alpha(|0000\rangle + |1000\rangle) + \beta(|0110\rangle - |1110\rangle) + \alpha(|0101\rangle + |1101\rangle) + \beta(|0011\rangle - |1011\rangle) $$

Step 3: Alice measures the state of $X$ and $Y$.

On simplifying the above representation, we get $$ |00\rangle(\alpha|00\rangle + \beta|11\rangle) + |01\rangle(\alpha|01\rangle + \beta|10\rangle) + |10\rangle(\alpha|00\rangle - \beta|11\rangle) + |11\rangle(\alpha|01\rangle - \beta|10\rangle). $$

Step 4: Depending on Alice's measurement outcome, appropiate unitaries are applied on $Z$ (by Alice) and $U$ (by Bob).

(a) If Alice gets $|00\rangle$, then both Alice and Bob do nothing.

(b) If Alice gets $|10\rangle$, then Alice applies $\left[{\begin{array}{cc}1 & 0 \\0 & -1 \end{array}}\right]$ and Bob does nothing.

(c) If Alice gets $|01\rangle$, then Alice does nothing and Bob applies $\left[{\begin{array}{cc}0 & 1 \\1 & 0 \end{array}}\right]$.

(d) If Alice gets $|11\rangle$, then Alice applies $\left[{\begin{array}{cc}1 & 0 \\0 & -1 \end{array}}\right]$ and Bob applies $\left[{\begin{array}{cc}0 & 1 \\1 & 0 \end{array}}\right]$.

Basically, $\left[{\begin{array}{cc}1 & 0 \\0 & 1 \end{array}}\right]$, $\left[{\begin{array}{cc}1 & 0 \\0 & -1 \end{array}}\right]$, $\left[{\begin{array}{cc}0 & 1 \\1 & 0 \end{array}}\right]$ and $\left[{\begin{array}{cc}0 & 1 \\-1 & 0 \end{array}}\right]$ can be appropiately used to alter the combined state of $Z$ and $U$ so that it becomes $\alpha|00\rangle + \beta|11\rangle$. Note that if Alice gets $|01\rangle$ or $|11\rangle$, then Bob has to apply some unitary so that the combined state of $Z$ and $U$ is $\alpha|00\rangle + \beta|11\rangle$.

Step 5: Apply Hadamard transform on the state of $Z$.

After applying the unitaries, the combined state of $Z$ and $U$ is $\alpha|00\rangle + \beta|11\rangle$ (as mentioned above). So, after Step 5, the combined state of $Z$ and $U$ is,

$$ \alpha|00\rangle + \alpha|10\rangle + \beta|01\rangle - \beta|11\rangle \\ = |0\rangle(\alpha|0\rangle + \beta|1\rangle) + |1\rangle(\alpha|0\rangle - \beta|1\rangle). $$

Step 6: Alice measures the state of $Z$.

Based on her measurement, she transmits one classical bit of information to Bob so that he can use an appropriate unitary to obtain the unkown state!

Discussion: So, how does the protocol require $1.5$ bits of clasiical communication? Cleary, Step 6 uses 1 cbit, and in Step 4, it is easy notice that for two outcomes (namely, $|10\rangle$ or $|00\rangle$), Bob need not apply any unitary. Bob has to apply some unitary (specified prior to the protocool; say $\left[{\begin{array}{cc}0 & 1 \\1 & 0 \end{array}}\right]$) if Alice gets the other two outcomes, and in those scenarios, Alice sends one cbit indicating that the unitary is to be used by Bob. So, it is mentioned that this has a computational burden of 0.5 cbits (because 50% of the time, Bob need not apply any unitary). Hence, the whole protocol requires only 1.5 cbits.

But, Alice must send that 1 cbit whether or not she gets those outcomes, right? Alice and Bob cannot agree on a particular time (after the protocol) when Alice sends that 1 cbit, and if Bob doesn't get that classical bit by that time, then he knows that he need not apply any unitary. These time dependent protcols are, in general, not allowed due to relativistic consequences (otherwise, you can even make the Standard protocol to use time for indicating information and reduce the classical communication cost to 1 cbit; for example, at $t_1$, send one cbit or at $t_2$, send one cbit). So, Alice must send that cbit everytime, right? In that case, the protcol requires 2 cbits (one in Step 4 and another in Step 6). I thought it'd be good if there was a discussion on this particular part.