I watched all your videos in two nights, your videos were great and I hope you can continue that process. I am waiting for your next videos, I am especially eager to see a video of you, with the subject of electromagnetic field quantization.Thank you very much
Thanks for watching! :) We are hoping to extend our series on second quantization to look at its application in a range of areas, thanks for the suggestion!
First off, thanks again for a great video. Second, thank you for no ads that we're forced to watch on our cellphones. Third, I believe that, in the physical world, commutation is not "perfect" because that's what Hamilton had to give up when he was working on creating quaternions. Also, it's from his work on quaternions that the dot product and cross product appear. If I have this third point correct, please confirm. Also, please create a video covering this, if you haven't already (I haven't seen all of your vides - yet.) Thank you.
I am no expert in quaternions. Having said that, I am aware of a few applications in quantum mechanics, but it remains unclear whether they provide any advantages. A good starting point to read about this may be the book "Quaternionic Quantum Mechanics and Quantum Fields" by Adler.
You will be able to re-write the full expression in various alternative forms, but you may have to write your entire expression to check it is consistent with the result in the slide. In general, we may not assume that CA = AC. I hope this helps!
While exercising and expanding the commutator [AB,CD] in the last relation at 7:01, I've written everything correct apart from the fact I've written AC[B,D] while you wrote CA[B,D]. Can you explain to me where have I gone wrong? To me, applying the mnemonic rule, A acts before B so it is to put before the commutator, and C acts before D so it is to put before the commutator as well, but why inverting their order? Thanks so much for everything.
Very good point! It looks like the mnemonic rule cannot be applied directly to an expression like this where we have products of operators in both entries in the commutator. To clarify, here is the expanded expression for the commutator, obtained by first "expanding" CD: [AB,CD]=C[AB,D]+[AB,C]D At this step we have moved C "through" AB. We can then expand AB in both terms to get: [AB,CD]=C(A[B,D]+[A,D]B) + (A[B,C]+[A,C]B)D = CA[B,D]+C[A,D]B+A[B,C]D+[A,C]BD For completeness, note that if we first "expand" AB, we get: [AB,CD]=A[B,CD]+[A,CD]B =...= AC[B,D]+A[B,C]D+C[A,D]B+[A,C]DB An expression that looks different to the one in the video, but that is in fact equivalent. I hope this helps!
Interesting question. It is possible to take it as a given fundamental relation, in which case you can then derive quantities such as the position representation of the momentum operator, p = -i*hbar*d/dx from it. But you can also take the opposite view, where the fundamental starting point is p = -i*hbar*d/dx and then derive the commutation relation from it. Overall, there is some fundamental assumption one needs to make, but whether that assumption is the canonical commutation relation itself or a related quantity is not fixed. I hope this helps!
@@ProfessorMdoesScience Thanks, I got it. But this sparks yet another question in me : how was this assumption arrived upon and how can one be sure that this gives the correct explanations of physical phenomena?
@@mikalgd4399 I would say the answer is the same as that for any other physical theory: experiment is the ultimate discriminator between competing theories based on different postulates.
Thank you very much, this was really easy to understand. But there was one problem I was facing...when we say operator in mathematics..we consider a function that has the same set as domain and co-domain( map from a set to itself). Is it different in here? I was confused what we mean by multiplying operators...I mean if I consider sine operator and cosine operator and multiply them ..what happens? I might have misunderstood something and this might be a very dumb question though..
Very good point! What we mean by "product" of two operator AB is the successive application of the two operators. Put another way, AB|psi> means that you first apply B on |psi> and then apply A to the result. I hope this helps!
@@abhishekkumbhar6118 I typically use Cohen-Tannoudji, Sakurai, and Shankar. Wikipedia is usually a good resource for these things too. I hope this helps!
Practice this formula useful to solve for tedious Commutation bracket problems for any power of Position & Momentum, analysed by Om Sir, Science Vision Institute, New Delhi.
Consider: AB-BA=(-1)(-1)AB+(-1)BA We can do this step because (-1)(-1)=1, so we are essentially multiplying by 1. Then (-1)(-1)AB+(-1)BA=(-1)[(-1)AB+BA] Here we take (-1) as a common factor. Finally, (-1)[(-1)AB+BA]=(-1)[BA+(-1)AB]=-(BA-AB) I hope this helps!
@@ProfessorMdoesScience Thank you so much for posting a response. This helped me to understand and now i do i feel silly for not realising it. Like a magic eye poster once you see the solution you can't stop seeing it!
Thanks for the feedback! They were just meant to be a quick refresher about operators and the definition of a commutator. I recommend that you follow the full series to build up the concepts from scratch: ru-vid.com/group/PL8W2boV7eVfnb10T_COKPozxEYzEKDwns We'll add an extra video in the next few weeks on "functions of operators" to complete the study of operators in quantum mechanics.
