I started today with a quantum intro course starting here. I got freaked out by this griffiths chapter. Some friend of mine recommended this to me but you solved essentially most questions that I had and in hindsight you made me understand previous parts better. Thanks man, it's very clear. Even clearer than griffiths himself and he is very clear
@PankajBhambhani644 жыл бұрын
These videos are really Legendre-ary! Thank you for sharing these videos.
@dbf72829 Жыл бұрын
😭🤣
@santosfuentes151910 жыл бұрын
question: at 15:03 after you have divided out (- hbar squared RY over 2 m r squared) shouldn't r squared be canceled out? should you not have (1/R)dr(r^2dR) ? thanks!
@navneetmishra32083 жыл бұрын
Yup
@kgfcccvvhbbbc75653 жыл бұрын
@@navneetmishra3208 yes. No r squared. Teacher forget r squared. I forgive you.
@MiguelGarcia-zx1qj3 жыл бұрын
38:25 an easier to read depiction of spherical harmonics can be obtained by using saturation instead of value of light for the magnitude of the functions. So, the highest magnitudes would be represented as fully saturated colours, and low magnitudes would appear as mostly white (instead of black). I devised such a scheme, around 23 years ago, in order to represent said harmonics upon a sphere.
@Gamma_Digamma4 жыл бұрын
m and l are actually the quantum numbers where m is the magnetic quantum number giving the number of degenerate orbitals and l is the azimuthal quantum number giving which subshell were talking about (s, p, d or f)
@MarioRiveraS968 жыл бұрын
Thanks! You might have just saved my quantum mechanics course with this video!
@jacobvandijk6525 Жыл бұрын
And what have you done with it?
@cosmicnomad85752 жыл бұрын
That was excellent! Well done! Such a beautiful concept in my opinion!
@clopensets61044 жыл бұрын
I have been having trouble grasping the concept of spherical harmonics for days!!! My textbook didn't help in clarifying the mathematical details, as it mainly focused on the physical and experimental implications of mathematically predicted 'orbital configurations' (which were derived via spherical harmonics). This video helped clarify a lot of the mathematical components, which I was too lazy to research myself. THANKYOU FOR YOUR INSIGHTS!!
@fornasm10 күн бұрын
it might be worth to remind that: The problem with spherical co-ordinates (or any system of non-Cartesian co-ordinates) is that the direction from a point P along which one co-ordinate changes and the other two remain constant depends on the location of P . In Cartesian co-ordinates the directions and lengths of the basis vectors are independent of location. But if you are try to generalize this property of Cartesian co-ordinates to other co-ordinate systems it just does not hold.
@KingCrocoduck9 жыл бұрын
Mistake at 14:17 the r^2 term should have cancelled out
@KingCrocoduck9 жыл бұрын
nevertheless this video kicked ass
@MisterTutor20109 жыл бұрын
King Crocoduck perhaps you can do a video giving a foul mouth explanation of the rigid rotator as you did with the harmonic oscillator :)
@matthewlemke53105 жыл бұрын
@@MisterTutor2010 4 years ago you were triggered.
@MisterTutor20105 жыл бұрын
@@matthewlemke5310 What are you talking about? That video was hilarious.
@ifrazali30524 ай бұрын
@@matthewlemke53109 years ago lol How are you guys holding up?
@SlySkills6 жыл бұрын
Answers to your questions either in your video description or as a comment is the only thing these videos are missing to really check our understanding. Thank you so much for everything though. You've done more than enough and it's greatly appreciated.
@SlySkills6 жыл бұрын
Attempted Answers: Find P(1,1) (cos(theta)) = sin(theta) P(7) = 0 at 7 different points P(4,7) = 0 at 3 different points - This one is pretty much a guess, not sure how to compute. m = -3 then l = 3, only allowed value
@BPHSadayappanAlagappan2 жыл бұрын
Awesome 😎
@BPHSadayappanAlagappan2 жыл бұрын
I think if M=-3, L can be greater than 3.
@iqranasir92242 жыл бұрын
thank you for saving my semister 😍
@briannagopaul34937 ай бұрын
Thank you for these lectures, these make so much more sense!!
@paladin11477 ай бұрын
Amazing, just check the comments for certain errors or keep the Griffiths book next to you to verify. Thank you for the video
@fornasm10 күн бұрын
unfortunately, this lecture does not even seem a lecture on physics ... I hope it comes in next lectures
@bakirev4 жыл бұрын
I didn't quite get the reasoning of spherical coordinates not being independent of each other which causes the laplacian to be complicated. It was that vector r changes if you change phi for example. But r is just the distance right not the vector?
@dm32484 жыл бұрын
Thanks for such an amazing lecture..!!
@ameerbux786664 жыл бұрын
wow, only thing sad here is that ive only dicoverered this channel now. thanks for the vid, you explain great
@erikstephens63702 жыл бұрын
14:17 that term should be 1/R, not 1/Rr^2. The r's from the Y/r^2 and -h^2/2mr^2 cancel each other. Still, a great video.
@mojamamaduda8 жыл бұрын
thanks for this video,you are a legend
@scitwi91647 жыл бұрын
09:00 Are you sure about that? If the X part is a constant, then they Y+Z part is equal to the same constant with a negative sign. But X and Y can still change in various ways, just independent of x, right?
@clopensets61044 жыл бұрын
What do you mean? The 2nd derivative of X multiplied by the inverse of X, is only equal to the negative of the sum of the 2nd derivatives of Y and Z respectively, each multiplied by their respective inverses, when the potential energy equals the energy eigenvalue.
@clopensets61044 жыл бұрын
In 36:10 . Just out of curiosity, how do you get the normalisation factor from evaluating the double-integral underneath???
@movintarget3004 жыл бұрын
It's the same as the normalisation integral for more familiar Cartesian coordinates, except we are now in spherical polar coordinates, so dxdydz -> r^2sin(theta)d(theta)d(phi)d(r). I think you can just grind through this by plugging in (Y*Y) . This may be done in more detail in Griffiths Intro or try googling Jain QM pdf and it's done in there.
@BLVGamingY9 ай бұрын
why isn't at 22:30 the phi function an arbitrary sum of opposite exponentials as opposed to just one exponential nonetheless m ends up a whole number
@rajupdl64 жыл бұрын
Awesome lecture 👍
@takinyele9 жыл бұрын
Oh my god I love you so much
@seacaptain726 жыл бұрын
Is there somewhere I can find the solutions to the check your understanding questions at the end? I've done these on all the vids but haven't been able to check them
@rebekahshtayfman19677 жыл бұрын
I think you meant r^2*sin(theta)*dr*d(theta)*d(phi) at the 38 minute mark (approximately) By the way, thank you for the amazing videos. You're great!
@wonka50048 жыл бұрын
Great explanation 10/10
@hasnatahmad61644 жыл бұрын
one thing I didn't understand what are "pheeta" and "phi" here
@movintarget3004 жыл бұрын
Solutions of the wavefunction in spherical polar coordinates split into the dependence on pheeeta and phii. Edit: you can also solve for the r dependence but is less important in angular momentum which is... well, angular.
@إحساسفيزيائية4 жыл бұрын
Thanks 💜
@learningwithsaba61315 жыл бұрын
superb
@أزهارالحوامدة9 жыл бұрын
thank you
@ZeroXAlAttas9 жыл бұрын
Wow, this really helps~~ thanks!! :))
@navneetmishra32083 жыл бұрын
THANK YOU SO SO MUCH
@inperatieloos8 жыл бұрын
37:58 isn't it r^2*dr*sin([T])d[T]*d[F]?
@rap82098 жыл бұрын
+inperatieloos I think d(Omega)=Sin(T)d(T)d(Phi) so here he integrated over omega, not considering radial part
@fcmilsweeper99 жыл бұрын
Wait why do they have to be constant?
@Prometheus40969 жыл бұрын
+fcmilsweeper9 Without context, I think in the case you think about both describe the energy of the same system.
@Salmanul_4 жыл бұрын
Why spherical coordinates?
@HankGussman4 жыл бұрын
Orbitals can be described in terms of spherical coordinates.
@jacobvandijk6525 Жыл бұрын
Why do you describe your room with Cartesian coordinates? Because it is rectangular. Why do we describe the "room" of an electron with spherical coordinate? Because it is spherical (most of the time), they think.
@alexkumarsingh28054 жыл бұрын
29:11 0
@movintarget3004 жыл бұрын
Plot out cos(x) for 0 cos(pi)=-1.
@alexkumarsingh28054 жыл бұрын
Thank you
@herman97374 жыл бұрын
why cant weee just stick to cartesian.... jk spherical reveals nice intrinsic proterties of QM
@knowledgebankforfpscppscnt54865 жыл бұрын
22:18 how
@craigfowler70984 жыл бұрын
My brain hurts
@Salmanul_4 жыл бұрын
28:31, fractional derivatives exist so idk about that.
@shayanmoosavi91393 жыл бұрын
If m is not an integer the phi dependence of the wave function won't be single valued and our boundary condition which is Φ(0)=Φ(2π) won't be satisfied. Also fractional derivatives are extremely advanced stuff. This is an *introduction* to quantum mechanics which is for undergraduates. I've found this article, I hope it answers your question : en.m.wikipedia.org/wiki/Fractional_Schr%C3%B6dinger_equation Although I haven't understood a single word😂😂
@fornasm8 күн бұрын
I am sorry Prof. Carlson, I have appreciated your lectures so far because their math's treatment is clear and easy to follow. But now they are starting to luck with the important connection of math's to physics. After all this is a physics course, isn't it?! So why would one suddenly introduce spherical coordinates? At least it should have mentioned that major topic of original quantum mechanics was atomic physics and in atomics physics the "force" is central and 1/2 etc ... this is why it may be useful to introduce spherical harmonics: because the future "problems" you are going to treat are "atomic physics", correct?! Atomics physics has the peculiarity of introducing "symmetry" in the problems and this a rather important general problem in "physics" ... This is at least my opinion. If your share such opinion, don't you think it have been educative to remind that?! Otherwise, this course seems now turn into a math's course of solving partial derivative equations ... Many thanks
@MinuteMayne6 жыл бұрын
Bruh the way you write partials is wildly distracting.