I've read papers that consider the problem of sharing entanglement between A and C with an intermediate B. A and B share an entangled state and so do B and C. Then, entanglement swapping is used to create entanglement between A and C.

Now, in the optical fiber case, if A generates a Bell state and sends one half to B, B could simply transmit the received photon to C. This would create the required entanglement between A and C.

I understand that the motivation is that transmission may be lossy and hence one has to consider alternate techniques. But in principle, if the channel was perfect, is there any problem in doing what I suggested?

  • $\begingroup$ No problem at all! Of course, most of the interest in entanglement is because of the absence of such perfect channels, so if you give yourself a perfect quantum channel, you should possibly ask yourself why you want an entangled state.... $\endgroup$
    – DaftWullie
    Commented Jul 20, 2023 at 15:04
  • $\begingroup$ Thanks for your response. One could use the entanglement for superdense coding, which is what I had in mind. $\endgroup$
    – user401163
    Commented Jul 20, 2023 at 20:05

1 Answer 1


With a supply of pre-entangled pairs between multiple locations, the state may be shared using daisy-chaining and just a single measurement at the signal origin:

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$$\alpha(|00000\rangle + |11111\rangle) + \beta(|10000\rangle + 01111\rangle)$$

Measurement at $A$ collapses the entire daisy chain into:

Case $q_A = 0$

$$\alpha|0000\rangle + \beta|1111\rangle$$

Case $q_A = 1$

$$\alpha|1111\rangle + \beta|0000\rangle$$

Broadcasting the measurement result at $A$ now allows any other station $B,C,D,E$ to deduce $\alpha, \beta$ appropriately.


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