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Пікірлер
@ostwald4107 13 күн бұрын
Sir will you determine the energy levels using the odd and even types
@iitenergymaterialsgroup4816 12 күн бұрын
I only do one of the two in the video, the other was a problem for the students to solve.
@juniorcyans2988 20 күн бұрын
After studying quantum physics for almost a semester, I still hadn't truly grasped the fundamental principles I should have understood. Every class, our teacher comes up with all sorts of confusing stuff, and I can't make sense of the logic behind it at all. As a busy junior student, I simply don't have the time to sit down and calmly read the textbook. Now, starting from scratch and watching your videos, I feel like I've suddenly gained clarity. I don't even know how to thank you enough!
@iitenergymaterialsgroup4816 18 күн бұрын
Thanks for the compliment, good luck!
@robestark2554 21 күн бұрын
Why does the good quantum number changes
@iitenergymaterialsgroup4816 21 күн бұрын
This is because the magnetic field is stronger than the spin-orbit coupling. Spin-orbit coupling has the total angular momentum as a good quantum number as well as m_j while the application of a strong magnetic field makes it more convenient to use the orbital momentum magnetic quantum number m_l and the spin magnetic quantum number m_s.
@EccentricTuber 22 күн бұрын
Sweet! This is the clearest explanation of the AB effect that I've encountered!
@iitenergymaterialsgroup4816 21 күн бұрын
Thank you!
@Kyoz 24 күн бұрын
🤍
@buginhogames 24 күн бұрын
why the psi a* psi b integral is zero in <x1>² but isnt in <x1 x2> ?
@iitenergymaterialsgroup4816 24 күн бұрын
Please give me a timestamp as to where I can find what you are referring to.
@buginhogames 24 күн бұрын
@@iitenergymaterialsgroup4816 at 10:36 the product is zero and at 12:54 it is not, but isnt it the same inner product?
@iitenergymaterialsgroup4816 21 күн бұрын
@@buginhogames The two inner products at those locations are not the same. ONe id <x1x2> and the other is <x1^2>
@jmdawlat 26 күн бұрын
Very well done! In fact this and the previous video were better than many others on the same topic. I appreciate it!
@iitenergymaterialsgroup4816 26 күн бұрын
Thanks
@Da_Jiraffe Ай бұрын
Personal timestamps for note taking 1:39 5:00
@warrior9673 Ай бұрын
Excuse me, at 1:58, shouldn't it be phi_n (0) at the very end of Phi(t)???
@juniorcyans2988 Ай бұрын
Very cool! I don't think my quantum class will show me these shapes. They just reminded me what I learned in my chemistry class.
@iitenergymaterialsgroup4816 Ай бұрын
Thanks!
@juniorcyans2988 Ай бұрын
I found a gold mine!!!
@iitenergymaterialsgroup4816 Ай бұрын
Glad you like it
@juniorcyans2988 Ай бұрын
This is the best video on this topic I've found so far. I feel lucky to find this channel! Thank you so much!
@iitenergymaterialsgroup4816 Ай бұрын
Glad you find it useful
@premkumarofficial007 2 ай бұрын
Thank you Brother
@AnitraWolinski-j1q 2 ай бұрын
Aliya Meadow
@BachVirgil-o7d 2 ай бұрын
99006 Bernhard Highway
@BUGSBunny-rz7jy 3 ай бұрын
perfect
@FluidRotor 3 ай бұрын
99464 Beier Underpass
@omar-to6pl 4 ай бұрын
shouldn't we have g_e in H'_hf (around 1:52)?
@iitenergymaterialsgroup4816 4 ай бұрын
It is assumed to be 2 and thus cancels out a 2 in the denominator.
@thecuriousphoton 4 ай бұрын
At 11:25 Its (-i)^l+1 instead of (-1)^l+1 in the second line
@iitenergymaterialsgroup4816 4 ай бұрын
You are correct, it is a typographical error but the line below is correct.
@willemesterhuyse2547 4 ай бұрын
Psi = e^(ig)Psi' not= -ihbar*e^(ig)grad(Psi'), so how can you apply the operator again to -ihbar*e^(ig)grad(Psi')?
@peterchindove7146 4 ай бұрын
Time stamp?
@willemesterhuyse2547 4 ай бұрын
4:36, but I see now the operator got squared so it isn't replacing phi with -hbar*e^(ig)grad(Psi').
@iitenergymaterialsgroup4816 4 ай бұрын
@@willemesterhuyse2547 That is correct.
@Al-Qaisi_Iraqi 4 ай бұрын
thank you so much
@Al-Qaisi_Iraqi 4 ай бұрын
Thank you so much
@iitenergymaterialsgroup4816 4 ай бұрын
Glad you find them useful.
@Al-Qaisi_Iraqi 4 ай бұрын
Thank you
@iitenergymaterialsgroup4816 4 ай бұрын
Glad you find them useful.
@sayanjitb 6 ай бұрын
Do you have any video on Bell's theorem? TIA
@manuelalmagro9837 6 ай бұрын
this playlist is amazing! thank you
@bigmonster_RB 7 ай бұрын
nice
@ashishbhargavkalita2013 7 ай бұрын
sir what is (al) that you used to expand cl
@carlosmurias3648 6 ай бұрын
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.
@SAMAHSAQUAQUE 7 ай бұрын
thaks a lot for this
@iitenergymaterialsgroup4816 7 ай бұрын
You are very welcome
@WeAreChecking 7 ай бұрын
Agreed, very useful for exam prep. Wish there were more example problems available online though.
@iitenergymaterialsgroup4816 7 ай бұрын
Not having the problems in the videos is deliberate.
@jackmortem4557 7 ай бұрын
where does the iota go in the equation for cb from 7:23 to 7:50 ?
@iitenergymaterialsgroup4816 7 ай бұрын
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.
@jackmortem4557 7 ай бұрын
oh okay, i was wondering where my calculations went wrong. thank you.@@iitenergymaterialsgroup4816
@shubhamanurag1156 7 ай бұрын
Thank You Sir for teaching this in a very simple way. 🙏
@iitenergymaterialsgroup4816 7 ай бұрын
I'm glad it was useful!
@DIGITALVHS 8 ай бұрын
Lmao my prof put "solve for the normalized molecular wave function of a hydrogen molecule" homework when we had a single lecture on molecular hydrogen ion and it didn't provide half of the concepts I just learned watching this video.
@iitenergymaterialsgroup4816 8 ай бұрын
I'm glad you found it useful.
@JoseRivas-ev2bn 8 ай бұрын
Hace pocos días conocí este experimento. Solo conozco algo de la teoría clásica del electromagnetismo. Pero noto algo extraño en el fenómeno que se trata de explicar con operaciones matemáticas muy complejas, para mi modesta formación matemática. No logro ver en qué parte del estudio se considera el propio campo magnético de la partícula en movimiento. Si se considera el propio campo magnético de la partícula, se entiende mejor que es dicho campo el que llega hasta el espacio donde está confinado el otro campo. De esta manera, se ve claramente que sí hay un área de interacción de ambos campos, y por supuesto una fuerza que cambia la trayectoria de la partícula.
@MUHAMMADARSLAN-m8y 9 ай бұрын
Brilliant competition
@ryounes487 9 ай бұрын
I always like ur videos
@iitenergymaterialsgroup4816 9 ай бұрын
Thanks!
@cowboyspaceranger 9 ай бұрын
Very hepful thank you! Great presentation.
@iitenergymaterialsgroup4816 9 ай бұрын
You very welcome.
@ryounes487 10 ай бұрын
Thank you 🙏
@iitenergymaterialsgroup4816 10 ай бұрын
You are very welcome
@lukschs1 10 ай бұрын
11:42 , Podría explicar como obtener la constante de normalización?
@iitenergymaterialsgroup4816 10 ай бұрын
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.
@hiteshdevalapelli2424 10 ай бұрын
clear cut stuff!!! Hatsoff!!!
@iitenergymaterialsgroup4816 10 ай бұрын
Thank you!
@TheWigglyman 10 ай бұрын
What x quantity of NMP did you use for y quantity of total powder ? I want to achieve same slurry viscosity as you ! Thanks
@iitenergymaterialsgroup4816 10 ай бұрын
These details should be in our publication: doi.org/10.1016/j.jpowsour.2023.232852 and doi.org/10.3390/batteries9020115
@santoshkumarsahu7814 10 ай бұрын
What will be the energy for bound state of a particle in a square well? Is E<0 or E>0?
@iitenergymaterialsgroup4816 10 ай бұрын
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.
@SampleroftheMultiverse 11 ай бұрын
Thanks I have been trying to describe the “U” shape wave that is produced in my model that is produced as the loading increases/ just before the wave function shifts to the next higher energy level. Your viewers might be interested in watching the test video of my model see the load verse deflection graph. kzbin.info/www/bejne/raOlpKSfepWpfZYsi=waT8lY2iX-wJdjO3
@iitenergymaterialsgroup4816 11 ай бұрын
Thanks, I'll take a look
@eliedemartelaer3117 Жыл бұрын
you saved my life thanks
@iitenergymaterialsgroup4816 Жыл бұрын
Happy to help!
@dcvlogs2822 Жыл бұрын
Thank you sir...
@iitenergymaterialsgroup4816 Жыл бұрын
You are very welcome!
@Amit1994-g9i Жыл бұрын
I dont understand how we take the expectation value of 1/J^2 mathematically and also have a <J> outside as if it was distributive under multiplication
@iitenergymaterialsgroup4816 Жыл бұрын
Pleae tell me the time or slide number and I can better respond
@triwenirana763 Жыл бұрын
Thank you sir 😊
@iitenergymaterialsgroup4816 Жыл бұрын
You are welcome
@agrajyadav2951 Жыл бұрын
absolute fire 🔥
@iitenergymaterialsgroup4816 Жыл бұрын
Thank you, glad you found it helpful
@alanknguyen Жыл бұрын
Very helpful, thank you very much!
@iitenergymaterialsgroup4816 Жыл бұрын
You are welcome!
@family-accountemail9111 Жыл бұрын
Amazing what the internet can find for me! I got my chem degree 40 yrs ago i only had 3 or 4 books now i can find multiple sources excellently presented explaing what I want. My son got stuck with his qm module in his astrophysics degree and I found this for him and he is much happier!
@iitenergymaterialsgroup4816 Жыл бұрын
You are very welcome, I hope your son does well.
@brainindependent6254 Жыл бұрын
reference please?
@iitenergymaterialsgroup4816 Жыл бұрын
Reference for what in particular? Give me a time stamp and slide number.
@hmp7181 Жыл бұрын
can you pls tell me why in l equal to 0 the first term is "null"
@iitenergymaterialsgroup4816 Жыл бұрын
If you give me a time stamp I can answer your question.