Recent content by Kenneth Adam Miller

  1. Kenneth Adam Miller

    I Exploring Entanglement: Is it Commutative?

    Fascinating. Thank you.
  2. Kenneth Adam Miller

    I Exploring Entanglement: Is it Commutative?

    Ok, so it's as though A now shares a total entanglement with all three, but BC sort of share a subspace determined by the metrics of their interaction, is that correct?
  3. Kenneth Adam Miller

    I What does the Hadamard correspond to in hardware?

    Nah, I don't know the reference off the top of my head. Yeah, it wasn't a hammer, it was firing one particle at another. I know there are more specifics to it, but this is just the way I remember it. I'll see if I can find it somewhere and post it here if I do.
  4. Kenneth Adam Miller

    I What does the Hadamard correspond to in hardware?

    OK, so the law of the preservation of energy demands that, as in the original example of the particle being split, that the total momentum before and after the interaction sum up to be the same. So, in the case that a particle is split, each particle of it is now in superposition because the...
  5. Kenneth Adam Miller

    I What does the Hadamard correspond to in hardware?

    Ok, maybe not experiments exactly, but perhaps learning examples. Hold on, I'm answering your other question.
  6. Kenneth Adam Miller

    I What does the Hadamard correspond to in hardware?

    Yeah, but there's not a fundamental discrepancy between the law of preservation of momentum and the observation of either position or momentum that has clearly been created here in order that entanglement arise. Entanglement is the eventual preservation of that law in the face of an implied...
  7. Kenneth Adam Miller

    I Exploring Entanglement: Is it Commutative?

    I've seen diagrams of quantum computer components at a high level that discusses multiplexing laser reflections over many qubits, and I have to believe that entanglement as a hardware operation has to be scaled to the many qubits by means of some operation that is applied to each of them...
  8. Kenneth Adam Miller

    I What does the Hadamard correspond to in hardware?

    Sure. https://en.wikipedia.org/wiki/Hadamard_transform That's the matrix definition.
  9. Kenneth Adam Miller

    I What does the Hadamard correspond to in hardware?

    No, I don't mean the Hamiltonian. That is something different; the hadamard is the quantum computing operator responsible for suspending qubits into entanglement.
  10. Kenneth Adam Miller

    I Why can't all bits of an entanglement be read at once?

    Suppose we have an entanglement state between many qubits. When we measure this state anywhere, the entanglement is destroyed. In quantum computing, we have two bits of parity for each qubit, and we have to re-setup the entanglement state for each qubit to read each piece of information...
  11. Kenneth Adam Miller

    I What does the Hadamard correspond to in hardware?

    As per the title, I'm wondering what exactly is going on at the quantum level that makes entanglement arise - what is the precursor? The hadamard, as I know it, is just a particular set of matrix values that make up an operator. From what I understand, in physics, entanglement is a naturally...
  12. Kenneth Adam Miller

    I Expectation value with imaginary component?

    AH! ##|z|^2 = zz*## Didn't know this!
  13. Kenneth Adam Miller

    I Expectation value with imaginary component?

    Yeah, I struggled with this problem because of the issue with understanding ##|a + ib| ^2## has only two terms. I thought there were three.
  14. Kenneth Adam Miller

    I Expectation value with imaginary component?

    The real part is unaffected. And that the complex conjugate of a real number is itself unaffected. Yeah, so this is the first time that I have started exercising this knowledge, so although I've read it several times, I haven't had to apply it.
  15. Kenneth Adam Miller

    I Expectation value with imaginary component?

    Ok, so if I did the calculations right using what you've said all the way through, I should get: ## <x | phi> = (1/sqrt(2)*<z+| + 1/sqrt(2)*<-z|)(1/2*|z+> + i*sqrt(3)/2*|-z>) = (1/(2*sqrt(2)) + i*sqrt(3))## And so ## |<x | phi>|^2 = (1/8 + 3/8) = 1/2 ##
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