??

"L as E is a conserved quantity in quantum and classical mechanics so it is interesting..." wow, conserved quantity are key to solve the dynamics of physics ... they are not just interesting they play major meaning in physics ... we use them to "label/identify" physical systems ...

Prof Carlson always good in concerning about maths, but I would have preferred more attention to Physics and avoid sentences as: " it would be nice to know how our L acts on this general psy(r, theta...) ..." Nice!? ... It seems we are doing all these calculations because it is "nice" ... isn't this Physics?! I thought we do all this to gain better understanding of Physics (geometry, what we experience physically ...). One of the major physical problem is considered here is atomic physics that has spherical symmetry ... this is major reason we are considering L and spherical coordinates etc

great. thanks.

... sorry, but has Prof. Carlson explained what are quantum numbers yet?? It seems this is not clarified, I have the feeling has not made any connection at all with classical mechanics here. As in classical mechanics to solve the dynamics of a particle in 3D one need more equations that are actually coming from other conservation law if they exist. Angular convervation law etc. There is not mention here yet Those are then what he quotes as "quantum numbers", but it seems he has not explaned them yet?! ???

How did you get that recurrence relation at 42:56 ?

😮😊🎉❤👍🙏👏👏👏

Finally, Prof Carlson clarifies the objective of its math's, and this is another good lecture. Thanks!

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

is this a course in mathematics of physics???????

unfortunately, this lecture does not even seem a lecture on physics ... I hope it comes in next lectures

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.

WOW, I had already done this, but here it is better done and clearer. 1/2 QFT is in this lesson. well done thanks!!

be careful: d2 psi/dx2 = -k2 psi would be equation of a harmonic oscillator if dx2 would be dt2 ....

How did you reduce f - g = a (1+ 2iq) - b (1 - 2iq) ???

Position uncertainty d < a < c < b Momentum urcentainty b < c < a < d (the opposite of above) I really hope that the actual answer is provided

I have a question out of the context. Which device are you using to deliver these lectures?

How to find the normalization constant B: psi is normalizable implies that the integral over all space (-infty to infty) of psi* psi = 1 Here psi = B*e^(kx) if x < 0 and B*e^(-kx) if x > 0. Note that psi* = psi in both cases. The integral over all space of psi* psi can be split as the sum integral1 + integral2, where integral1 goes from -infty to 0 of psi* psi and integral2 from 0 to infty of psi* psi. integral1 psi* psi = B^2 e^(2kx) = B^2 e^(2kx)/2k evaluated at lower bound -infty and upper bound 0: B^2/2k - 0 = B^2/2k integral2 psi* psi = B^2 e^(-2kx) = B^2 e^(-2kx)/-2k evaluated at lower bound 0 and upper bound infty = 0 - (B^2/-2k) = B^2/2k integral1 + integral2 = B^2(1/2k + 1/2k) = B^2/k which must equal one. Hence B^2 = k and we get B = sqrt(k) = sqrt(ma^2/h_bar) Alternatively one can use psi = B*e^(-k|x|) and that psi is an even function. The integral over all space then becomes 2 times the integral from 0 to infty, which is 2*integral2 in the above calculation. This leads to the same result 2*(B^2/2k) = 1 -> B = sqrt(k).

21:52 n should only take odd integer values no?

Thank you, professor, for this course.

The wave function collapses and becomes narrow, right? Could that possibly have any links with the idea that particles are different when unobserved and when observed, or that you can only know the velocity or the position but not both, or quantum superposition? Or maybe it's just a coincidence.

saw this on tiktok as a 1h vid and compressed to 20 pixels per frame, decided to watch it here after enduring the 16 minutes spent there lol

For the 18:30 Check your understanding problem I got: ρ(x, t) = c₁²•|X₁(x)|² + c₂²•|X₂(x)|², so, I don't find the time-dependence.

Thank you

Thank you

Thank you

I guess at 6:55 the inequality for a and b has to be reversed

Adieu

And that's a really ugly Psi Must fix.

Ill-defined?

did anyone understand the spin =wtf thing

I be watching this on my pz

my guess a) no istnt smooth, curves wrongon Left side b) smooth curves wrong on left side. looks good on right sidde c) All correct d) Has to curve When E<V so also bad

PSi looks very communist

I dont understand the check questions answer/ What is correct?

Anybody got the answer to the Understaning Check?

14:48 There is a star missing on the last blue psi for those who are confused :)

So is the continous distribution is not a wave funktion. Youd would have to square the wave funktion, and then integrate to get the expctation values for that right?

Check your understanding: . . . 1. Eigenvalue of Sy : hbar/2 and -hbar/2 ( just like Sx and Sz ) 2. Eigenvectors of Sy: y_plus = 1/sqrt(2) * ( 1, i ) # with eigenvalues hbar/2 y_minux = 1/sqrt(2) * ( 1, -i ) # with eigenvalues -hbar/2 3. x_plus state can be expressed in Sy eigenstates, as x_plus = ( 1/2 - i/2) * y_plus + ( 1/2 + i/2 ) * y_minus. So the possibilities is both 0.5 to attain one of the states of corresponding eigenvalue. 4. Same as before, 0.5 possibility to get both states and corresponding eigenvalues.

check your understanding: . . . 1. L2|f> = 2hbar^2|f>, L = 1, Lz could be -1, 0 , 1 ( all * hbar ) . A total of 3 states. 2. L2|f> = 15/4 hbar^2|f>, L = 3/2, Lz could be -3/2, -1/2, 1/2, 3/2 ( all * hbar ). A total of 4 states. 3. Lz|f> = 0, L2|f> could be any (hbar)^2 * l * ( l+1 ), l = 0, 1/2, 1, 3/2, .... 4. Lz|f> = 3/2 * hbar |f> could be any (hbar)^2 * l * ( l+1 ), l = 3/2, 2, 5/2, 3, ... Shouldn't there be a |f> at the end for the 4th question?

Check your understanding: . . . . 1. [x, Py] = 0 2. So we could measure x and Py as precise as we want. 3. [Px, Py] = 0 also. So we could measure Px and Py as precise as we want.

I kind of liked that proof of the roots of unity. that x^3=1 bit. You make y=0, the unity thing kicks in anyway.

Check your understanding: . . . . 1: H applies on the right state |ψm>, and return Em|ψm>. Em is just and number and get outside of the braket. So Hnm = Em<ψn|ψm> 2: Hnm is a diagonal matrix with values E1,E2...En on its diagonal line. 3: a+|ψm> and returns 1*|ψm+1>. a+nm = <ψn|ψm+1>. 4: a+nm is a off-diagonal matrix with values 1,1,1,1,1.... on the line just below the diagonal line.

Check Your Understanding at 19:30: . . . Part 1: a. Got (QR+RQ-2(µQ)(µR))/2, So it's not a commutator I guess. b. -(µQ)(µR) is an extra term, what does it mean? Part 2: Because µR is just a number, it could be moved outside of the braket to the front, and the rest part is just expectation of Q.

It is now 2024... I already forgot how much difference you can feel between 480p and 720p. This video is much sharper than the previous one.

Check Your understanding: Part 1: C2=0, C3=1, C4=0 Part 2: -a/pi

How do we know that psi-0 is equal to B? Isn't it that we can not use psi-1 or psi-2 equations to solve for psi-0 because they do not include x=0? Then why are we using them to find psi-0?

kzbin.info/www/bejne/gp6chI2cZ715ecUsi=5h5AElm81LIXkFdT took your videos

is this is the part of jee syllabus

Ah yes, my go to data set when I'm trying to understand discrete probability distributions. ~sexual partners of the general public