Recent content by antonantal

Very good resources @gentzen. Thanks for sharing!

It is not clear yet which is the correct justification why entropy increases after measurement. Is it this: or this: The point of my previous post was to show that Nielsen / Chuang seem to have followed the same logic as I did.

Nielsen / Chuang: Quantum Computation and Quantum Information chapter 11.3.3 Measurements and entropy Then it goes on to calculate the entropy of the state after measurement: $$S(\rho^{'})=-tr(\rho^{'} \log\rho^{'})$$ I will now show that this is equivalent to the entropy that I have calculated...

My understanding is that error detection is not unitary (involves measurement) but error correction is. Of course one could put both detection and correction under the same umbrella of "error correction".

From Wikipedia: " Entropy measures the expected (i.e., average) amount of information conveyed by identifying the outcome of a random trial. " This is equivalent to what I have said before:

Sure, after measurement we can see which of the two states occurred. But to calculate the entropy we need to consider the probability of each state occurring. Similarly, after you toss a coin you can see if it is heads or tails, but to calculate the entropy you still consider the probability of...

Let's say we start with a qubit in a pure state (0 entropy), and we are on a certain branch of the many-worlds. Then an error occurs due to decoherence. That is because decoherence causes our branch to split into multiple branches, and each new branch lost some information which is now on the...

Related to my initial question, now I am trying to explain why at measurement we lose information and entropy increases: Consider the system to be a qubit in state ##\Psi=\alpha|0\rangle+\beta|1\rangle##, with ##\alpha,\beta \in (0,1)## Even though it is a superposition, it is still a single...

Quantum gates must be reversible. The usual justification for this is that in QM the time evolution of a system is a unitary operator which, by linear algebra, is reversible (invertible). But I am trying to get a better intuition of this, so I came up with the following explanation: In order to...

##\left(x_2x_3\right)\in\{0,1\}^2## just means that the string ##x_2x_3## is from the set of all 2-tuples of numbers belonging to ##\{0,1\}##. The notation is similar to how you would use ##R^n## for a coordinate space, just replace ##R## with ##\{0,1\}## and n with 2. Then, as you presumed, the...

These video lectures by Vazirani (as in Bernstein–Vazirani algorithm) are very good: https://www.youtube.com/channel/UCq9B8tT3oXl8BSyaoBPQXQw/playlists

The acceleration which increases the speed is provided by an electric force ##F_E=qE##. Since ##q## of a particle is small, in order to have a strong ##F_E## you need strong electric field ##E##. But there are limits to how strong an electric field can be generated, and this limits the...

I just want to expand a bit on this statement because not long ago I was struggling to understand how this is different from measuring a single qubit in state ##\frac{|0\rangle+|1\rangle}{\sqrt2}## which also gives with 50% probability 0 and with 50% probability 1. The difference is that in the...

If we represent the two particles as qubits, an entangled state would be for example:$$|\psi\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}$$At measurement, you randomly and equally likely get one of only two outcomes: both qubits 0 or both 1. Mathematically, it is because the state vector doesn't...

@sophiecentaur , you are correct, what I meant by "emitted at the same time" is just that "they are emitted in such a way that they could interfere at the screen".