Quantum Spin (5) - Schrödinger's Equation & Matrix Exponentiation

[@tablettorrensabellan]

Thank you so much for this wonderfully and thorough explanation of spin, that completely demystify the otherwise magical explanations I've always found on the web and even in quantum mechanics books!

[@kellywshere]

This was just beautiful. I love how you even explain the mathematical concepts along the way with a clarity that is comparable to a 3b1b video

[@zhengrongqian1806]

That solved a lot of confusion I had, really great explanations! Thank you so much:)

[@yyc3491]

Just like watching a Hollywood movie! I really enjoy it. Thank you!

[@alikefayati1301]

Perfect teachinig!

[@harrisscook4773]

At 23:22 the negative one in the spin up state is referred to as an "unobservable" phase, but elsewhere in the series and even in the next chapter of this video the -1 scalar preceeding the spin up vector is cited as an eigenvalue which is an experimental observation under the Sigma sub-z observable. Clearly I am confused.

[@noahexplainsphysics]

Notice that |↑> and -|↑> both have the same eigenvalue under σ_z, which is +1 in both cases

[@harrisscook4773]

@@noahexplainsphysics Aha! Thanks.

[@ecdavek230]

Bravo Noah ! I was watching the vid and I paused at 23m34s to think about table just written out of : t vs |Psi(t) > At first, I was puzzled as no |y> states appeared in the table. And then it struck me - - IT’S PRECESSING ! It’s exactly as discussed in prior vid#4. I must say, that hit me with the same excitement I had when I was able to hear a tune for the first time from looking at sheet music. So I just had to take this moment to express my admiration and say ; Bravo ! That was kewl.

[@zhengrongqian1806]

What are the other cool things relevant to the Dirac belt trick?

[@AMADEOSAM]

Great explanation, many thanks!

[@randomdelhite937]

54:23 I did'nt get it. can anyone plz explain why it changed from 0 to 1.

[@MrNicePotato]

If you differentiate t^0 it becomes just 0 instead of t^-1, so the n=0 term doesn’t exist in the derivative.

[@SirTravelMuffin]

It looks like most people choose the negative of the Hamiltonian you provided, any reason for the choice of sign you made?

[@ernestpowell5012]

Thank you

[@ko-prometheus]

Привет, Очень познавательно!! Не хватает 3D анимации......