IQIS Lecture 2.3 - Single qubit interference (Hadamard-Phase-Hadamard)

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  • Опубліковано 9 вер 2024

КОМЕНТАРІ • 15

  • @alishirnia1221
    @alishirnia1221 3 роки тому +4

    Wonderful, clear lectures. Thank you.

  • @luisamariasandovalcalle5939
    @luisamariasandovalcalle5939 11 місяців тому

    You are an amazing teacher! I love your content!

  • @resh.official
    @resh.official Рік тому

    Hello, this must be told, I really appreciate your efforts! I am yet watching your very first lectures and Im not sure if you do explain in the following lectures. I just happened to think, It'd very helpful if you were giving some little tips or examples of fields of usage in quantum computations such QCs. Nevertheless, big fan of your lectures :)

  • @christinezuzart
    @christinezuzart 3 роки тому +1

    Can you please explain the math behind arriving at "cos phi/2 ket 0 - i sin phi/2 ket 1 "

    • @ArturEkert
      @ArturEkert  3 роки тому +2

      Please take a look at: thosgood.com/quantum-info/book/chapter2.html#single-qubit-interference

    • @christinezuzart
      @christinezuzart 3 роки тому

      Thanks for the resource link. Will check the said chapter.

    • @suvarnadhiraj
      @suvarnadhiraj 3 роки тому +1

      @@ArturEkert : I do not find explanation of how this particular equation is obtained. The notes just say to multiply the matrices, but the math is not shown. Can you provide a hint on how to arrive at the solution?

    • @peymanpahlevanzadeh4080
      @peymanpahlevanzadeh4080 3 роки тому

      @@ArturEkert I read the notes of the web page, but I could nut understand the point "cos phi/2 ket 0 - i sin phi/2 ket 1 "

    • @jameswinkleblack8846
      @jameswinkleblack8846 3 роки тому +13

      ​@@peymanpahlevanzadeh4080 There's just a bit of math omitted. Start at the state right after the phase shift is applied as in the video and apply the final Hadamard. As shown, you arrive at the state 1/2*( (1+exp(ix))*|0> + (1-exp(ix))*|1> ). (swapping phi for x for this explanation)
      Factor out an e(ix/2) from the overall state, and you have exp(ix/2)/2*( (exp(-ix/2)+exp(ix/2))*|0> + (exp(-ix/2)-exp(ix/2))*|1> ).
      Finally, use a form of Euler's formula (IE 2*cos(x/2) = exp(ix/2)+exp(-ix/2) and 2*i*sin(x/2) = exp(ix/2)-exp(-ix/2) ) to simplify both coefficients in front of the resultant |0> and |1> components. You end up with an additional exp(ix/2) relative to what is given by Ekert, but that is fine as these two states are equivalent in quantum mechanics :)

  • @Seiven2077
    @Seiven2077 Рік тому

    Timecodes:
    [00:05](ua-cam.com/video/eDR-OZZu99M/v-deo.html) Understanding single qubit interference in quantum information science
    [01:05](ua-cam.com/video/eDR-OZZu99M/v-deo.html) Characterizing the Hadamard gate
    [02:09](ua-cam.com/video/eDR-OZZu99M/v-deo.html) Understanding the phase gate in quantum circuits
    [03:05](ua-cam.com/video/eDR-OZZu99M/v-deo.html) The Hadamard gate is the most important gate in quantum interference.
    [04:02](ua-cam.com/video/eDR-OZZu99M/v-deo.html) Hadamard gate creates a superposition of 0 and 1 states
    [05:10](ua-cam.com/video/eDR-OZZu99M/v-deo.html) Hadamard gate creates superposition state
    [06:24](ua-cam.com/video/eDR-OZZu99M/v-deo.html) Phase gate determines final probability
    [07:30](ua-cam.com/video/eDR-OZZu99M/v-deo.html) Phase factors in quantum computing are crucial for quantum interference experiments.