Quantum phase estimation and the quantum Fourier transform | QuTech Academy

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  • Опубліковано 4 лют 2025

КОМЕНТАРІ • 9

  • @cyndicorinne
    @cyndicorinne Рік тому

    Kudos on another great explanation 💙💜

  • @charbeleid193
    @charbeleid193 3 місяці тому +2

    To be honest, this video just felt like someone reading an article to me. Thank you for making the videos, and I only mean this constructively, but as a person who comes to UA-cam to complement my readings, this was not very helpful.

  • @istvandentora3459
    @istvandentora3459 9 місяців тому

    how does it come out that the probability amplitude of the + state with coefficient 1/2 + e^(i*fi)/2 will be 1/2+cos(fi)/2? Or did you mean that the probability of 0 measurement output will be this much? But then it is a mistake to say that we measure in the X base.

  • @itanbar
    @itanbar 5 років тому +1

    One thing that i don't understand. When you do a controlled-U gate, why is the phase kicked back to the control qubit and not to the target qubit?

    • @wonka5004
      @wonka5004 4 роки тому +2

      Its because the target qubit(s) state is an eigenfunction of U.
      If |u> is the target qubit state, which is an eigenfunction of U with eigenvalue e^(i phi), then
      (|0> + |1>)|u> = |0 u> + |1 u>,
      which when acting on with controlled U gives
      |0 u> + U |1 u> = |0 u> + e^(i phi) |1 u> = (|0> + e^(i phi) |1>)|u>.

    • @MudahnyaFizik
      @MudahnyaFizik 3 роки тому +1

      You can refer to the previous video. He explained that.

    • @charbeleid193
      @charbeleid193 3 місяці тому

      Do the calculation it's not hard

  • @mohitchaudhari2650
    @mohitchaudhari2650 5 років тому

    great explanation man!

  • @RegistroDominio-l1b
    @RegistroDominio-l1b 2 місяці тому

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