Permutation operators in quantum mechanics

[@mayararamosdelima2914]

this is the best channel to learn QM, I don't even know how to begin to thank you Professor. Truly amazing work what y'all do. You def inspire me!

[@ProfessorMdoesScience]

Wow, thank you!

[@richardthomas3577]

Clear as a bell! Thank you! For suggestions, I bet you would do an awesome job of discussing Lie groups and Lie algebras!

[@ProfessorMdoesScience]

Glad you like it! And group theory is certainly a topic we'd like to cover, as it plays such a key role in physics :)

[@sagensoren55]

Very well explained sir

[@workerpowernow]

excellent explanation-all your videos outshine other explanations i've seen in standard physics texts

[@ProfessorMdoesScience]

Thanks for your kind words, and glad you enjoy them! :)

[@NilsSandell]

Don the three spaces V1, V2, V3 have to be the same? Other wise it doesn’t make sense to but a vector living in one of these spaces into another of them.

[@udvaschattopadhyay5389]

What happens for the equivalence at 2.22 if we have three balls in two boxes (3 particles in 2D single-particle Hilbert space) or three boxes with two balls? Can we say that one way of thinking is more general (I think switching the boxes for QM)?

[@ProfessorMdoesScience]

Good question! First I should say that I wouldn't take this analogy too far, it is simply to give an idea of what we are doing. This also means I haven't explored all its possible subtleties when relating to QM, so this may not be quite correct. In any case, I think that a simple way to think about the 3 balls and two boxes scenario is to actually still have three boxes, but two of them with the same "label" (so box 1, box 1, box 2). Does this help? Moving on from the analogy with boxes and balls, you may be interested in the "proper" way of doing this in quantum mechanics, which we explain in the playlist on second quantization: kzbin.info/aero/PL8W2boV7eVfnSqy1fs3CCNALSvnDDd-tb

[@udvaschattopadhyay5389]

@@ProfessorMdoesScience Yeah, great content .. Thanks for the reply.

[@rohitdeb1077]

In transposition operator when you prove the involuntary of the operator P21*P21 acting on u1u2 gives us first P21* u2u1, and again after that it should give us u2u1 as per your definition, how it gives us u1u2 state instead of u2u1?

[@rohitdeb1077]

Is it because of the fact that we can rearrange them?

[@ProfessorMdoesScience]

Yes, and in general it is easiest to see this if you keep both the state subindex label (inside the ket) and the vector space subindex label (outside the ket). Remember that the definition of P21 is that the first subindex "2" tells us that the particle associated with V1 (because it is the first entry) moves to V2 (because the subindex is "2"), and the particle associated with V2 (because it is the second entry) moves to V1 (because the subindex is "2"). This can be quite confusing in general, so the best strategy is to always re-order the states after the application of each permutation, so that they appear in the order V1,V2,... In your example, we get, for the first permutation: P21 |u_i>_1|u_j>_2 = |u_i>_2 |u_j>_1 = |u_j>_1|u_i>_2 We then apply the second permutation and get: P21 |u_j>_1|u_i>_2 = |u_j>_2 |u_i>_1 = |u_i>_1|u_j>_2, so that indeed: P21P21|u_i>_1|u_j>_2=|u_i>_1|u_j>_2. I hope this helps!

[@rohitdeb1077]

@@ProfessorMdoesScience yes , it is clear now , thank you for your response.

[@quantum4everyone]

Just a quick comment about the group theory. To show the product of two permutations is another permutation, just write in terms of transpositions and since the product of a string of transpositions is still a string of transpositions, you immediately have this rule holds. For the rearrangement theorem, the proof is by contradiction. Assume two permutations in the list are the same, so P_alpha Pi=P_alpha Pj for i not equal to j. Then by multiplying from the left by the adjoint of P_alpha, you immediately see that you must have Pi=Pj. But that is not true if i is not equal to j, so no two elements in the list can be the same.

[@ProfessorMdoesScience]

Thanks for the insights! We try to strike a fine balance between video length, content depth, and dependencies on other topics. Group theory is something that we've touched on in several videos without really going into it because it always feels like the videos will just become too long. However, we may one day start another series on mathematical methods and hopefully explore these ideas in more detail.

[@vaanff1942]

nicee

[@zoubirbecer6529]

Nicely explained, just please hold on when you speak.

[@ProfessorMdoesScience]

Thanks for the feedback!