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Thank you. If you want the slides you will hve to send me an email.

@@iitenergymaterialsgroup4816 can you send your email so i can messaging you

@@iitenergymaterialsgroup4816 can you send your email so i can message you 🙏🏼

Always start with the simplest problem, a spherical horse. Then you can move to something more complex, like the fine structure and Zeeman effect.

@@iitenergymaterialsgroup4816 But if your simplest model is far from reality (= too abstract, so there is no recognition). then it might not feel so simple anymore ;-)

@@jacobvandijk6525 Zeeman effect is not simple nor is it unphysical. Start simple end up with a real system and the student can understand all the way through. Believe me a 20 year old student will not get the idea if the spin-orbit couple is the first example. YYMV

@@iitenergymaterialsgroup4816 Haha, I agree with you on that last statement. P.S. I know the YMCA, but what is YYMV?

@@jacobvandijk6525 I mean YMMV (Your Mileage May Vary)

It is assumed to be 2 and thus cancels out a 2 in the denominator.

You are correct, it is a typographical error but the line below is correct.

Time stamp?

4:36, but I see now the operator got squared so it isn't replacing phi with -hbar*e^(ig)grad(Psi').

@@willemesterhuyse2547 That is correct.

Glad you find them useful.

Glad you find them useful.

al, is the partial wave coefficient. To have the solution to a problem of scattering, we need to find those coefficients. I believe, not sure tho, we find those coefficients by aplying the proper bound condiditions of our problem to the general solution, that's derived here.

You are very welcome

Agreed, very useful for exam prep. Wish there were more example problems available online though.

Not having the problems in the videos is deliberate.

It is not an iota. It is the "i" for a complex number and it can be removed as it is a global phase factor that cancels when computing the probability.

oh okay, i was wondering where my calculations went wrong. thank you.@@iitenergymaterialsgroup4816

I'm glad it was useful!

I'm glad you found it useful.

Thanks!

You very welcome.

You are very welcome

Normalization is performed in the usual way but integrating the probability density and setting it equal to 1. WHile the Griffiths book does not do the integral explicitly, I am sure you can find a book which has it done.

Thank you!

These details should be in our publication: doi.org/10.1016/j.jpowsour.2023.232852 and doi.org/10.3390/batteries9020115

I think you’re misunderstanding something here

@@gi99hf60 Could be. But it seems to me you can't explain it either.

@@jacobvandijk6525there was a global phase common factor taken out initially, now imbedded in the w_0. This is just an artifact of that. Due to normalization, it doesn’t matter what the absolute magnitude of c_a or c_b is, but the ratio of their squares. That’s why the initial global phase factor was taken out initially, to keep one coefficient at unity and just keep all the evolution info in the other. Then we only solve one equation, it was intentionally set up that way to simplify the calculation

@@jacobvandijk6525 physically, the probability of being in the lower state is still decreasing just by the coefficient of the other changing. You’ll have to normalize at the end anyways

You can always set the potential of the bottom of the well to whatever you like. If you set it at zero, as conventionally is done for the infinite well, all the energies are positive. Things are different with the finite well. There we generally set the bottom of the well to -V and all the energies are negative. However, you could just as well set the bottom of the well to zero and the "outside" to V. The solutions are thus shifted by V.

Thanks, I'll take a look

Happy to help!

Saying “potential” can mean both “electric potential” and “potential energy”

@@gi99hf60 When you are teaching, you must be clear about concepts.

You are very welcome!

Pleae tell me the time or slide number and I can better respond

You are welcome

Thank you, glad you found it helpful

You are welcome!

Couldn't agree more. It's a wonderful source of knowledge. Hope your son does well.

You are very welcome, I hope your son does well.

Reference for what in particular? Give me a time stamp and slide number.

If you give me a time stamp I can answer your question.

ft is the topmost angular momentum eigenfunction. If you apply L+ to raise it it just gives zero because you cannot go above the top function.

@@iitenergymaterialsgroup4816 Yep... absolutely... later realized myself.

If comes directly from the normalization of the entire 3D wavefunction. Recall that in the radial normalization integral you have |R|^2r^2 in the integrand. This is exactly |u|^2.

It helps if you tell me at what time in the video you are raising the question. I presume that you mean the part on two identical particles. In this case r_1 and r_2 are the positions of the two identical particles with respect to a common origin.

@peterchindove7146 Sorry about the delay, thanks!

You are correct

In order to get c_j in terms of c_0, you need to apply the recursion j time. Each time you get a power of 2 in the numerator and one higher value in the denominator, starting with 1 and ending with j. This is simply j!.

​@@iitenergymaterialsgroup4816ok 👍🏻

You are welcome!