I am studying the BB84 protocol and I think I have a slight misunderstanding. As far as I understand, if Bob measures the photon in the same basis as it was prepared by Alice, he is going to receive the exact bit Alice intended to send him without any randomness. If Alice and Bob have a basis mismatch, then bits received by Bob are going to be random.

Now consider this image I got from an article on the same subject at



Lets take the third bit 0. Here A prepares the photon in diagonal basis and Bob measures it in the same basis. Yet the bit Bob records is different from the one Alice sends. I do not understand how. I would really appreciate someone telling me where exactly I am misunderstanding the protocol.


1 Answer 1


This looks like a mistake in the image to me (or a reflection of Alice and Bob using faulty procedures to convert bits to qubits and vice versa, which seems much less likely).

Under ideal conditions, without the noise in the channel used to transmit the qubits and without the eavesdropper, if Alice and Bob use the same basis to encode the message in the qubit and to measure it, respectively, Bob will get the same bit as Alice encoded and sent. The third column even has the same letters for the polarization of the photon Alice sent and Bob measured, both are D - but Bob decodes D as 1, and in the last column Bob decodes D as 0 which he should do in the third column as well. (Similarly, in column 7 Alice encodes 1 as H in + polarization, while in the columns before and after she encoded 0 as H in the same polarization.)

  • $\begingroup$ Thank you. Seems the same to me. $\endgroup$
    – abirbhav
    Jul 7, 2023 at 7:56

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