8 votes
Accepted

How many bits do Alice and Bob needs to compare to make sure the channel is secure in BB84?

Your analysis of Eve's cheating doesn't seem quite right (although the final answer is correct). What you need to say is: Assume Alice prepares a particular state in one of the bases. You could assume ...
DaftWullie's user avatar
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6 votes
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Can quantum state tomography break bb84?

No, weak measurement and quantum tomography don't break BB84. I recommend that you create an explicit quantum circuit that implements the weak measurement or the quantum tomography, and check for ...
Craig Gidney's user avatar
6 votes
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Is the BB84 protocol an example of "quantum supremacy"?

This is an interesting question that reflects a conflation of some concepts in quantum information sciences. TL/DR - there is no task in BB84 that corresponds to what we when we speak of quantum ...
Mark Spinelli's user avatar
5 votes

How are eavesdroppers detected when using BB84 in the presence of noise?

The standard noisy approach is not to try to determine the presence of an eavesdropper as such, but to create a final key where, even if there is an eavesdropper, you can still be confident that the ...
GotCarter's user avatar
  • 401
5 votes
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Does the Lieb-Robinson bound constrain the speed of entanglement information transmission?

Entanglement does not transmit information, as follows from the No-communication theorem. Lieb-Robinson bound is a limit on speed at which perturbation propagates using short-range interactions, for ...
kludg's user avatar
  • 3,204
4 votes

What is the quantum state transmitted to Bob in BB84 protocol?

Application of Hadamard gates changes states $|0\rangle$ and $|1\rangle$ followingly: $\mathrm{H}|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ $\mathrm{H}|1\rangle = \frac{1}{\sqrt{2}}(|0\...
Martin Vesely's user avatar
4 votes

Is it a good idea to publish a research paper about the simulation of BB84 protocol?

First of all, it depends on the level of detail of your simulation. SimulaQron does not take into account noise, so I presume your simulation is only functional. Many BB84 functional simulations have ...
Michele Amoretti's user avatar
4 votes

Eavesdropping in case of the BB84 Protocol

Both are valid concerns. Regarding number 2, it is an assumption that Alice and Bob are reliably identified to each other and cannot be spoofed. At this point it is usually said "there are good ...
DaftWullie's user avatar
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4 votes

Can quantum state tomography break bb84?

Tomography generally speaking uses a collection of measurements to reproduce an underlying state. So you experimentally reproduce the same situation over and over, collect statistics and find the most ...
Steven Sagona's user avatar
3 votes
Accepted

What is the probability of detecting Eve's tampering, in BB84?

Firstly we have $$ \mathbb{P}[\text{Eve detected in round $i$}] = \mathbb{P}[\text{Eve chooses wrong basis and Bob chooses correct basis}]. $$ As we assume Alice, Bob and Eve choose their basis ...
Rammus's user avatar
  • 5,773
3 votes

How are eavesdroppers detected when using BB84 in the presence of noise?

If there is an evasdroping, an error rate is higher than some long-term average of a quantum communication channel used for a quantum key distribution. So, when the error rate is high, the key is ...
Martin Vesely's user avatar
3 votes

Show That an Intercept and Resend Attack on all qubits reduces mutual information

You've started the analysis along the right lines. In the case of no eavesdropper, Alice and Bob get perfectly correlated outcomes (note that we're post-selecting on the cases where Alice and Bob ...
DaftWullie's user avatar
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3 votes
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Doubt in CSS error correction step

We can think of an $n$-bit string as a "mask" that specifies a set of positions where the bit string is $1$. For an $n$-bit string $x$ denote with $s(x)\subset\{0, 1, \dots, n-1\}$ the set ...
Adam Zalcman's user avatar
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3 votes
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Hadamard gate with two inputs in the circuit for the BB84 protocol?

This is an example of a controlled gate. It means that the gate is applied to the target qubit (the one with the gate on) when the control qubit (the one with the small circle on) is 1. Another ...
met927's user avatar
  • 3,251
3 votes

Can Eve randomly guess the correct bases in the BB84 exchange of a one-time pad?

I like Norbert's comment about the chance of Eve guessing the message without any eavesdropping (it points out that the probability of Eve succeeding can never be made 0), but thought I should also ...
DaftWullie's user avatar
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3 votes
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Misunderstanding of QKD BB84 protocol

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, ...
Mariia Mykhailova's user avatar
3 votes
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How to correctly describe state preparation?

For applications such as BB84, it really doesn't matter which of the two descriptions you use, because from everybody's perspective apart from Alice, they do not have access to the first system, and ...
DaftWullie's user avatar
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3 votes
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Security of BB84 QKD protocol

You are correct she could guess the bases correctly however each time she has a probability of $1/2$ of selecting the correct basis. So if Alice sends $n$ qubits then the probability she selects all ...
Rammus's user avatar
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3 votes
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Doubt in understanding the use of Uhlmann's Theorem in BB84 security proof

The relation in 1. is essentially a statement of Uhlmann's theorem. Ulmann's theorem states: Given two density matrices $\rho$ and $\sigma$, we have that the fidelity $F(\rho, \sigma)$ is given by $$F(...
xzkxyz's user avatar
  • 416
2 votes
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How to deal with noise in BB84 protocol?

It is important to note that Step 5 is a classical step. Different protocols exist to correct keys, for instance the Cascade- and Winnow-protocols. With this you reveal bits of information in ...
nippon's user avatar
  • 1,517
2 votes

Show that if $b'$ and $b$ are uncorrelated, then $a'$ and $a$ are uncorrelated

So first, let's define a bit your notations. I guess (correct me if I'm wrong) that you consider Bob honest, and that what you denote by $\Psi_{a_k,b_k}$ is the BB84 qubit in basis $\{0,1\}$ if $b_k = ...
Léo Colisson's user avatar
2 votes
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Doubt in Simple proof of Security of the BB84 QKD

I'm not sure that I would claim the two situations are entirely equivalent - if Alice and Bob share Bell pairs, there are extra things they could do (e.g. testing Bell inequalities), but in terms of ...
DaftWullie's user avatar
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1 vote
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Derivations of Equations (12.205 - 12.207) in Nielsen Chuang

I take it you mean $$\begin{eqnarray} \rho_{v_k,x} & = \frac{1}{2^n} \sum\limits_{z} \left( \frac{1}{\sqrt{|C_2|}} \sum\limits_{w_1 \in C_2} (-1)^{z \cdot w_1} | v_k + w_1 + x \rangle \right) \...
ChrisD's user avatar
  • 316
1 vote
Accepted

Why is applying a Hadamard transform useful in the entanglement-based version of BB84

Eve can wait until Alice makes the announcement, but the announcement only happens once Bob has received the qubit, so in the simplest attack (where Eve just measures the qubit and then sends it on to ...
GotCarter's user avatar
  • 401
1 vote

Formal measurement of the probability of an outcome of a Qubit in the hadamard base

You got the bit right with the standard basis, so that's a good start. The problem that you're having with the Hadamard basis is that you're doing something weird with $M_+^\dagger M_+$. Since $M_+$ ...
DaftWullie's user avatar
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1 vote

Where can I find a detailled exposition of the Information theoretic part of BB84?

The actual secure protocol is a little bit different than the original protocol suggested in 84 and referenced to Micael Ben-Or. A good starting point might be this paper - https://link.springer.com/...
Dudu Ponar's user avatar
1 vote
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Heisenberg Uncertainty Principle for BB84 Protocol using Paulis Spin Matrices

First of all, Heisenberg principle is usually invoked to say that two operators don't commute and therefore if your qubit is in $|0\rangle$ which is an eigenstate of $\sigma_z$, you will have ...
Lorenzo's user avatar
  • 66
1 vote

BB84 Protocol doubt

Let us work this out for your particular example, and see why it works in the general case. We have: $$ \begin{array}{c|c|c|c|c|c|c|c|c} \text{index}&1&2&3&4&5&6&7&8\\\...
Tristan Nemoz's user avatar
  • 6,162
1 vote

Show when $a_k$ and $b_k$ are correlated when measuring in different bases, in the BB84 protocol

To start with, remember that there are only 4 possible states you could be talking about: $$ |\psi_{00}\rangle=|0\rangle,\qquad |\psi_{10}\rangle=|1\rangle,\qquad |\psi_{01}\rangle=|+\rangle,\qquad |\...
DaftWullie's user avatar
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1 vote
Accepted

Any good resources for learning about quantum networks and how to do a simulation of BB84 protocol?

A very good book about quantum networks (with a communication engineering flavour) is "Quantum Networking" by Rodney Van Meter. Regarding SimulaQron, look at this simple tutorial for getting ...
Michele Amoretti's user avatar

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