This is a topic that's not very popular (because maths) I can't say thanks enough for all of these videos which are so helpful
@ProfessorMdoesScience3 жыл бұрын
Glad you find them helpful! :)
@noelwass4738 Жыл бұрын
I do enjoy your content so much. I note that you have another video of the isotopic quantum harmonic oscillator where you mention about spherical harmonics. Everything is taught thoroughly and done in the right order at a leisurely pace. I will again be looking at ladder operations and the 1D harmonic oscillator. There is so much fascinating content here and I do enjoy learning quantum mechanics from this series.
@ProfessorMdoesScience Жыл бұрын
Glad you enjoy our videos! :)
@sandippaul4683 жыл бұрын
It's surprising we're getting the best internet course for free.
@ProfessorMdoesScience3 жыл бұрын
Thanks for watching! :)
@namanarora11383 жыл бұрын
Thank You for these Lectures:)
@ProfessorMdoesScience3 жыл бұрын
Thanks for watching! :)
@whilewecan2 жыл бұрын
Impressive. Thank you.
@ProfessorMdoesScience2 жыл бұрын
Glad you like it!
@sayanjitb2 жыл бұрын
Dear prof. Is it always possible to break a general state of 3D QHO into tensor products of three equivalent one-dimensional QHO states? thank you
@ProfessorMdoesScience2 жыл бұрын
In principle yes. But note that the story becomes more interesting when we have an isotropic 3D harmonic oscilator, as in that case the different energy eigenvalues have a large degree of degeneracy. Under these circumstances, the eigenfunctions are not uniquely defined, and different choices may be more convenient in different circumstances. For example, rather than using the three Cartesian 1D QHOs to build the 3D eigenfunctions, you can instead choose the eigenfunctions to also be eigenstates of orbital angular momentum, and then you can write the eigenfunctions in terms of spherical harmonics. We cover this alterantive view in this video: kzbin.info/www/bejne/oIC0hZt_eM9_otk and our next video (hopefully released soon) will also expand on this topic. I hope this helps!
@richardthomas35772 жыл бұрын
So helpful! You make this all seem so natural! (You take the "tense" out of tensor . . . sorry!)
@ProfessorMdoesScience2 жыл бұрын
Good one! :)
@drdca82633 жыл бұрын
If we have w_x=w_y=w_z=w (where I suppose those w should be \omega but w is easier to type), then the system should be rotationally symmetric, right? And, the eigenspaces for different eigenvalues should be degenerate, where, whenever n_x + n_y + n_z = n , the energy should be E_n = \hbar \omega (n + (3/2)), And where the dimension of the eigenspace should be, ah, how many ways can non-negative integers n_x , n_y , n_z add up to n? Placing 2 separators amongst a line of n things, gives (n+1)^2 / 2? Wait, no that can’t be right because that isn’t always an integer. Ah ok, no it should just be ((n+1) choose 2) - wait no, that’s not quite right because it excludes the case where the two separators have the same location. So, ((n+1)n/2) + (n+1) = ((n+1)(n+2)/2) , so ((n+2) choose 2). Ok. Ok, so, the dimension of the n-th eigenspace is ((n+2) choose 2), and I imagine that different superpositions of these basis states would give the corresponding basis states if we described the system in terms of a different basis-for-physical-space . Like, if we described the whole system as a tensor product of harmonic oscillators in the x’ , y’ , and z’ directions instead of in the x, y, and z directions. Hmm, Would that mean that with respect to one coordinate system, a system could be in a pure tensor state |n_x , n_y, n_z> , while in the x’ , y’ , z’ coordinate system, the components for the different coordinates would be entangled? (Though, that’s probably not as weird an idea when instead of talking about different physical systems, we are talking about a single particle.) I guess I need to go watch the video on angular momentum, which I imagine probably breaks down the energy eigenspaces by total angular momentum along with angular momentum in a given direction. (At lowest energy, the eigenspace has a single dimension, and iirc the only smooth rotationally symmetric functions which can be expressed as a product of an x component, a y component, and a z component, are gaussians, so that fits with n=0. For n=1 that gives (3 choose 2)=3 , so, I think that has 1 possible value of total angular momentum, and then the specific angular momentum in a given direction has 3 potential eigenvalues. 4 choose 2 is 6, uh... Wait, no, I think I’ve messed up and remembered things wrong. With n=0 there should be 1 possible value for total angular momentum, So, for n=1 I guess there should be 2? So the 3 is 1+2 ? Hm, I’m confused. Definitely not remembering. Definitely need to review. Good video.
@ProfessorMdoesScience3 жыл бұрын
Well well, you are anticipating our next few videos :) We are next going to explore the degeneracies for the isotropic (wx=wy=wz) oscillator, and after that we'll re-evaluate the whole problem in terms of central potentials. We'll keep you posted :)
@drdca82633 жыл бұрын
@@ProfessorMdoesScience I’ll look forward to it :)
@KOMHATHbIU3 жыл бұрын
Next video should be on linear chain model. But, I think that after the Bogoliubov transformation you will lose your audience somewhere in the Fock space...
@ProfessorMdoesScience3 жыл бұрын
We do hope to get to condensed matter physics in the future, but in the meantime we have already covered second quantization (kzbin.info/aero/PL8W2boV7eVfnSqy1fs3CCNALSvnDDd-tb) so we can already explore the Fock space ;)