最近、Subhash Kakによる、より少ない古典的な通信コスト(より多くの量子リソース)を必要とするテレポーテーションプロトコルを紹介する論文を見つけました。別の答えを書いたほうがいいと思いました。
Kakは3つのプロトコルについて説明します。それらの2つは1 cbitを使用し、最後の1つは1.5 cbitを必要とします。しかし、最初の2つのプロトコルは異なる設定になっています。つまり、もつれた粒子は最初にアリスの研究室にあり(そしていくつかのローカル操作が実行されます)、その後、もつれた粒子の1つがボブの研究室に転送されます。これは、プロトコルが開始される前に、もつれた粒子がアリスとボブの間で事前共有される標準設定とは異なります。興味のある人は、1 cbitのみを使用するプロトコルを使用できます。1.5 cbit(fractional cbit)のみを使用する最後のプロトコルを説明しようとします。
X,Y,ZUXX,YZUXα|0⟩+β|1⟩|α|2+|β|2=1Y,ZU|000⟩+|111⟩
α|0000⟩+β|1000⟩+α|0111⟩+β|1111⟩
X,YZXYYZ
XOR
XOR=⎡⎣⎢⎢⎢1000010000010010⎤⎦⎥⎥⎥.
|00⟩→|00⟩|01⟩→|01⟩|10⟩→|11⟩|11⟩→|10⟩
α|0000⟩+β|1110⟩+α|0101⟩+β|1011⟩
X
α(|0000⟩+|1000⟩)+β(|0110⟩−|1110⟩)+α(|0101⟩+|1101⟩)+β(|0011⟩−|1011⟩)
XY
|00⟩(α|00⟩+β|11⟩)+|01⟩(α|01⟩+β|10⟩)+|10⟩(α|00⟩−β|11⟩)+|11⟩(α|01⟩−β|10⟩).
ZU
|00⟩, then both Alice and Bob do nothing.
(b) If Alice gets |10⟩, then Alice applies [100−1] and Bob does nothing.
(c) If Alice gets |01⟩, then Alice does nothing and Bob applies [0110].
(d) If Alice gets |11⟩, then Alice applies [100−1] and Bob applies [0110].
Basically, [1001], [100−1], [0110] and [0−110] can be appropiately used to alter the combined state of Z and U so that it becomes α|00⟩+β|11⟩. Note that if Alice gets |01⟩ or |11⟩, then Bob has to apply some unitary so that the combined state of Z and U is α|00⟩+β|11⟩.
Step 5: Apply Hadamard transform on the state of Z.
After applying the unitaries, the combined state of Z and U is α|00⟩+β|11⟩ (as mentioned above). So, after Step 5, the combined state of Z and U is,
α|00⟩+α|10⟩+β|01⟩−β|11⟩=|0⟩(α|0⟩+β|1⟩)+|1⟩(α|0⟩−β|1⟩).
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⟩ or |00⟩), Bob need not apply any unitary. Bob has to apply some unitary (specified prior to the protocool; say [0110]) 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 t1, send one cbit or at t2, 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.