Orbital angular momentum eigenfunctions

แชร์
ฝัง
  • เผยแพร่เมื่อ 5 ก.ย. 2024

ความคิดเห็น • 29

  • @abaykoshkimbay6380
    @abaykoshkimbay6380 2 ปีที่แล้ว +5

    Do not stop, you make really good and understandable videos, it would be better if you will solve some problems as examples for more understanding. Thanks for your work

    • @ProfessorMdoesScience
      @ProfessorMdoesScience  2 ปีที่แล้ว +3

      Glad you like the videos, and thanks for the suggestion! We are preparing some extra resources to go with the videos, including problems+solutions, we'll hopefully make an announcement soon!

  • @dineshhomedrawing2678
    @dineshhomedrawing2678 2 ปีที่แล้ว +4

    Best channel for quantum mechanics....🙏

  • @dannyx6095
    @dannyx6095 ปีที่แล้ว +2

    I am self learning QM and your videos are incredibly helpful. Thank you and please continue the good work!

    • @ProfessorMdoesScience
      @ProfessorMdoesScience  ปีที่แล้ว

      Great to hear they are useful, and good luck in your learning journey!

  • @lowerlowerhk
    @lowerlowerhk ปีที่แล้ว +2

    This is some intense brain gym. Thanks!

  • @mahmoudjbely1464
    @mahmoudjbely1464 3 ปีที่แล้ว +6

    Thank you for this elegant work!!!

  • @barrybai5123
    @barrybai5123 2 ปีที่แล้ว +2

    Fantastic explanation! When I subscribed the channel, there were only a few hundred subs. Now there are 6.56k ! You deserve more applauses than this. Hope it keeps growing.

  • @Lilliana1
    @Lilliana1 2 ปีที่แล้ว +2

    Fantastic Derivation
    I really like the using m=l and using that to find a basic relation. However it adds some work to calculate for general m. Having to recalculate normalization constant and then putting l=m wherever required will give general expression.
    But hey its a lot better than three confusing methods from Zettili

    • @ProfessorMdoesScience
      @ProfessorMdoesScience  2 ปีที่แล้ว

      There are indeed quite a few ways to get to the final expressions, and glad to hear our approach was clear! :)

  • @harrisonbennett7122
    @harrisonbennett7122 2 ปีที่แล้ว +2

    Some of the best content out there! I am so grateful for your videos!
    - An Undergraduate Mathematics student.

    • @ProfessorMdoesScience
      @ProfessorMdoesScience  2 ปีที่แล้ว

      Glad you like the videos! May we ask where you study?

    • @harrisonbennett7122
      @harrisonbennett7122 2 ปีที่แล้ว +1

      @@ProfessorMdoesScience Hi I sent you an email about referencing your videos :)

    • @ProfessorMdoesScience
      @ProfessorMdoesScience  2 ปีที่แล้ว +1

      @@harrisonbennett7122 Ups, somehow missed that; we have responded now! :)

  • @arpitadebnath4995
    @arpitadebnath4995 ปีที่แล้ว +1

    Hi! Your videos really helped me clear my concepts of orbital angular momentum. Thank you so much for posting such videos. At the same time, it is an earnest request to post similar videos on spin angular momentum as well. Thanks!

    • @ProfessorMdoesScience
      @ProfessorMdoesScience  ปีที่แล้ว +1

      Glad it helped! And we are about to start a new series looking at spin angular momentum for spin-1/2 particles, so stay tuned! :)

  • @richardthomas3577
    @richardthomas3577 ปีที่แล้ว +1

    Again, thank you!! Excellent!

  • @snjy1619
    @snjy1619 9 หลายเดือนก่อน +1

    😊😊😊😊 Please give the spin angular momentum playlist

  • @felipegiraldo1986
    @felipegiraldo1986 11 หลายเดือนก่อน

    Dear Prof M, where in the complete Spherical Harmonics Equation comes these factors from: Sq root ((l+m)!/(l-m)!) and the cosine derivative at the end respect to (sin(theta)^2l)

  • @karthikknarayanan776
    @karthikknarayanan776 ปีที่แล้ว +1

    Thanks for the excellent video! Can you explain why the raising and lowering operators preserve normalisation though? You mentioned this at 11:34 but it seems to me that since they are not unitary, they do not preserve normalisation.

    • @ProfessorMdoesScience
      @ProfessorMdoesScience  ปีที่แล้ว +1

      Thanks for the question, and let me clarify what we mean. The general properties of the angular momentum ladder operators are described in this video: th-cam.com/video/yGvfqRfw1BE/w-d-xo.html
      In that video, we show that for normalised states |l,m> and |l,m-1>, then the action of the lowering operator is:
      L- |l,m> = hbar*sqrt[l(l+1)-m(m-1)] |l,m-1>
      (note that in the video on ladder operators we actually use the general angular momentum J and the eigenvalues are labelled lambda and mu, but you can go to the especific form for orbital angular momentum using the replacements J --> L, lambda --> hbar^2*l(l+1) and mu --> hbar*m). In this step you are describing, we use the correct prefactor such that the states |l,m> and |l,m-1> are therefore each properly normalized. I hope this helps!

  • @rafeef1658
    @rafeef1658 ปีที่แล้ว +1

    Thanks a lot but sometimes I just find it a bit hard to understand your handwriting.

    • @ProfessorMdoesScience
      @ProfessorMdoesScience  ปีที่แล้ว

      Thanks for the feedback! What aspect do you find hard to understand? The text or the formulas? Something specific?