Hello. at 17:55 the way you apply Lambda to the density matrix of the first qubit, I can't totally get it. how we can separate the density operator of one qubit within entangled state like this, like we sum over ho \tensor \sigma (separable state) . why we didnt apply lambda to the reduced density matrix? I think decomposition of the density matrix of the entire system in terms of classical states is vital here like exactely the representation stated earlier for the compund ZX system at 8:05. also i dont understand the way you apply lambda for each part like in the third term if we apply or multiply |0>
Hello at 16:35 isn't the trace of the density matrix equals 1? or ho here as you said earlier may be not in general a density matrix when the channel applied on it within a composed system? Thank you! at 16:41 cant get why Tr( ho) ensures linearity of the channel :(
Yes, the trace of a density matrix equals one - but the formula works for any 2-by-2 matrix rho, not just for density matrices. The channel is linear because the trace is linear: Lambda(aM + bN) = Tr(aM + bN) |0>
You're spending a lot of time on these. IBM quantum bubble is popping: check insider trading. It's common sense: people are still making better qubits while IBM is pressing numbers for investors. I do appreciate the learning material though, and that is why I inform someone who appears to be invested so much.
This guy is one of the better guys that can explain things!
Muchas Gracia Maestro. I WILL MAKE GOOD USE OF THE INFORMATION...
Thank you so much for Excellent Presentation.
Very insightful lesson (as always in these series)
astounding work professor
With all my respect for this lecture, please share the slides for lessons 09 & 10 with us.
First! Great to see another Qiskit video :D
Hope to know how channel, a communication concept, is introduced to quantum information. Any reference paper or book?
Hello. at 17:55 the way you apply Lambda to the density matrix of the first qubit, I can't totally get it. how we can separate the density operator of one qubit within entangled state like this, like we sum over
ho \tensor \sigma (separable state) . why we didnt apply lambda to the reduced density matrix?
I think decomposition of the density matrix of the entire system in terms of classical states is vital here like exactely the representation stated earlier for the compund ZX system at 8:05.
also i dont understand the way you apply lambda for each part like in the third term if we apply or multiply |0>
Hello at 16:35 isn't the trace of the density matrix equals 1? or
ho here as you said earlier may be not in general a density matrix when the channel applied on it within a composed system? Thank you!
at 16:41 cant get why Tr(
ho) ensures linearity of the channel :(
Yes, the trace of a density matrix equals one - but the formula works for any 2-by-2 matrix rho, not just for density matrices. The channel is linear because the trace is linear: Lambda(aM + bN) = Tr(aM + bN) |0>
@@John.Watrous Thank you for your patience with my many long questions, and for the excellent work you’ve done!
You're spending a lot of time on these. IBM quantum bubble is popping: check insider trading. It's common sense: people are still making better qubits while IBM is pressing numbers for investors. I do appreciate the learning material though, and that is why I inform someone who appears to be invested so much.