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8

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 that's $|0\rangle$, but you can make the argument more generally. With 50% probability, Eve measures in the same basis that Alice prepared in (the 0/1 basis ...


5

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 yourself that it actually fails. The basic problem comes down to the fact that there is a trade-off between how much information you get and how likely you are to ...


4

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 example in spin lattice. I doubt it means something for protocols such as BB84; you can transmit quantum information by sending polarized photons, and photons ...


4

Application of Hadamard gates changes states $|0\rangle$ and $|1\rangle$ followingly: $\mathrm{H}|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ $\mathrm{H}|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$ Identity operator does not change the state in any way, i.e. $\mathrm{I}|0\rangle = |0\rangle$ and $\mathrm{I}|1\rangle = |1\rangle$. Hence ...


4

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 computation, so BB84 is not evidence of what researchers mean when they speak of "quantum supremacy". But historians will likely still consider the initial ...


4

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 been developed so far, and no one was so interesting to deserve publication in a research paper. Notice also that BB84 is the reference example in the ...


4

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 eavesdropper has negligible information about the key. So you aren't trying to distinguish between noise and eavesdropping, but pessimistically assuming that ...


3

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 of positions where $x$ has a $1$. The key observation is that the dot product $m\cdot z$ of $m$ with another $n$-bit string $z$ is zero when $z$ has an even ...


3

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 example of a gate like this you might have seen is the CNOT gate, that applies an x gate to the target qubit when the control is 1. This gate works in exactly the ...


3

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 point out a very different perspective - that of the context in which you would use quantum key distribution. In the context of today's communication, 1000 bits ...


3

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 independently and uniformly at random we have $$ \mathbb{P}[\text{Eve detected in round $i$}] = \frac12 \times \frac12 = \frac14. $$ So we also have $\mathbb{P}[\text{...


3

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 likely estimate for that state. In QKD, information is sent once and doesn't repeat. So for each unit of information, you're never ever able to collect enough ...


2

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 specific ways, due to which you learn if there are errors and where these are located.


2

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 the actual protocol applied, they're equivalent. One way to see this is to think about the protocol (more or less) as described: Alice first sends halves of ...


2

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 deleted and new one is distribuited again until noise level is at acceptable level. A natural noise can be reduced with classical error correction, something like ...


2

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 choose the same measurement bases). You could think of this as a probability table $$ \begin{array}{ccc} \text{Alice bit} & \text{Bob bit} & \text{...


1

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 uncertainty in the measurement of $\sigma_x$. However, let's check that Heisenberg principle is still holding even in this situation! Let's see Heisenberg's principle ...


1

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\\\hline x&0&1&1&1&0&1&0&0\\\hline y&1&1&0&1&0&0&0&1\\ \hline y'&0&1&1&1&0&...


1

To start with, remember that there are only 4 possible states you could be talking about: $$ |\psi_{00}\rangle=|0\rangle,\qquad |\psi_{10}\rangle=|1\rangle,\qquad |\psi_{01}\rangle=|+\rangle,\qquad |\psi_{11}\rangle=|-\rangle $$ It should be obvious that when $b'=b$, $a'=a$. This is because you're measuring the qubit in the same basis that it's prepared in (...


1

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 started: https://softwarequtech.github.io/SimulaQron/html/GettingStarted.html I strongly suggest to read the main paper on SimulaQron (https://arxiv.org/abs/...


1

There absolutely is. In fact, even in classical, there is the notion of computational security against polynomial time quantum adversaries. This is the whole point of post-quantum cryptography. This would let us keep using existing, classical, technology, but hopefully be secure against quantum-powered eavesdropping.


1

As far as I know, the error threshold depends on the security proof. Meaning, BB84 has different security proofs, each with different assumptions, resulting in different security threshold. The 'optimal' bound is the one that has the best 'error-rate'. Meaning, what is the highest noise(error) that can be tolerated, and still, the protocol can be proved ...


1

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 $b_k = 0$, and in basis $\{+,-\}$ if $b_k = 1$, whose "value" bit is $a_k$, i.e.: $$\Psi_{a_k,b_k} = H^{b_k}X^{a_k}|0\rangle$$ Then, Bob will measure in basis $b'_k$...


1

I think your misunderstanding is in the wording of the secret key rate; The PLOB-Bound offers an asymptotic, ultimate-upper bound on the secret key rate per use of a lossy bosonic channel. This bound on the secret key rate is computed as a regularisation, where one considers the infinite limit of $ n \rightarrow \infty$ transmissions across the channel in ...


1

How does Alice know the direction of the photon? You're talking about the photons as if we've just magically plucked them out of the air with no idea what polarisation they're in. In practice, we're producing the photons. A source of individual photons that can be produced 'on demand' is still an experimental challenge (not that I'm completely up to date ...


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