Symmetries and quantum mechanics: Linear representations

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  • Опубликовано: 23 дек 2024

Комментарии • 14

  • @garywpearson1955
    @garywpearson1955 Год назад +5

    Great! Now I have a chance of understanding at least up to chapter 2 in Steven Weinberg's first volume on QFT. This is really helpful. Thank you!

  • @danfelev
    @danfelev 2 месяца назад

    Thank you for the great lecture! By any chance do you have the previous (first?) lecture recorded?

    • @tobiasjosborne
      @tobiasjosborne  2 месяца назад +1

      Here is the full playlist
      ruclips.net/p/PLDfPUNusx1ErdQhrdAzincNJKgTQahsX_&si=8k0blo5wqcJgb2J3

  • @Quiablo
    @Quiablo 2 месяца назад

    Very nice lecture. I would just like to make a suggestion. I think the example you used of Z2 to explain why every group can be interpreted as a permutation (i. e. as a subgroup of the permutatiom group Sn) is far too trivial... a more complex example would be much more interesting and iluminating (for starters, you used the only example where Sn is the same size as the group you want to represent, which is bascially never the case). For those struggling with this point, the key idea to understand why every group element can be represented as a permutation of the group elements is this. Consider an ordered list of all group elements. Now act on all elements of this list with one of the group elements (say g) from the left. You will get all the group elements again, and each one will appear only once, but the list will be in a different order (the fact that you will get all elements of the group and everyone of them only once follows from the group axioms; its not an immediate obvious fact but its not hard to understand why it has to be so). This means that acting on this list with the group element g is equivalent to performing a permutation of the group elements. This is valid for all group elements, so thats why you can always identify a group of n elements with a subgroup of Sn.

  • @rodrigo_ramoss
    @rodrigo_ramoss Год назад +1

    Hi, professor Osborne. Does the canonical commutation relation between the position and momentum operators arise from the quantization procedure mentioned in 19:35? If so, how does that happen? I guess that one would have first to establish a coordinate system in a small region of the manifold in order to make sense of such operators.
    Sorry if that was a bit of a trivial or nonsensical question, it was just something that came to mind while watching the lecture and that I would've asked if I were there. Anyway, thanks for making these lectures available :)

    • @tobiasjosborne
      @tobiasjosborne  Год назад +3

      This is quite a deep question. The way the CCRs arise in this language is via representations of the Heisenberg group. There is a theorem -- the Stone von Neumann theorem -- which ensures that CCRs are the only unitary representations of this group, which is the group of (projective) phase space translations. I hope this helps

    • @rodrigo_ramoss
      @rodrigo_ramoss Год назад

      @@tobiasjosborne Thanks for your reply

  • @kevinullmann583
    @kevinullmann583 Год назад +1

    Are all permutation reps just symmetry transformations of the regular rep?

    • @tobiasjosborne
      @tobiasjosborne  Год назад +1

      many thanks for your comment. Actually, not all permutation reps are symmetry transformations of the regular rep: in particular, if the set X is smaller than the group itself then the reps may be different. I hope this helps

  • @nadiryavuzkan.
    @nadiryavuzkan. Год назад +2

    Thanks a lot for the nice lecture.

  • @robin1826
    @robin1826 2 месяца назад

    Thank you!

  • @jacquessmeets4427
    @jacquessmeets4427 8 месяцев назад

    At 29.30, at the equation r(jk} ≡ < e{j)| ρ(s) | e(k)>, it is stated that, if you don't have a Hilbert space (so no inner product), you can still define matrices. But how then?? (without Hilbert space, you don't even have symmetry transformations). This really confuses me. Can you give some further explanation/clarification? Many thanks in advance.

    • @thomasbastos3869
      @thomasbastos3869 8 месяцев назад

      Every finite dimensional vector space V has a basis {e_i}. The effect of a linear operator T: V -> V on a basis vector is again another vector which can be written in terms of the same basis Te_i =a_{ij} e_j

    • @jacquessmeets4427
      @jacquessmeets4427 8 месяцев назад

      @@thomasbastos3869 Ah, Of course. You don't need an inner product to write a matrix rep. for an operator. You only need a basis. Thanks for your help.